Uncategorized

You can Download Chapter 2 Solutions Questions and Answers, Notes, 2nd PUC Chemistry Question Bank with Answers Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Chemistry Question Bank Chapter 2 Solutions

Question 21.
Two elements A and B form compounds having formulas AB₂ and AB₄. When dissolved in 20 g of benzene (C₆H₆), 1 g of AB₂ lowers the freezing point by 2.3 K whereas 1.0 g of AB₄ lowers it by 1.3 K. Tin molar depression – constant for benzene is 5.1 K kg mol^-1. Calculate atomic masses of A and B.
Answer:


Suppose atomic masses of A and B are ‘a’ and ‘b’ respectively. Then
Molar mass of AB₂ = a + 2b = 110.87 g mol^-1
Molar mass of AB₄ = a + 4b =196.15 g mol^-1
Equation (ii) – Equation (i) gives
2b = 85.28 orb = 42.64
substituting in equation (i) we get
a + 2 × 42.64 = 110.87 or a = 25.59
Thus atomic mass A = 25.59 u
Atomic mass of B = 4.64 u.

Question 22.
At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of the solution is 1.52 bars at the same temperature, what would be its concentration?
Answer:
π = CRT
∴ In the first case,
4.98 = \(\frac{36}{180}\) × R × 300 = 60R
In the second case, 1.52 = C × R × 300
Dividing (ii) by (i), we get C = 0.061 M.

Question 23.
Suggest the most important type of intermolecular attractive interaction in the following pairs.

1. n-hexane and n-octane
2. I₂ and CCl₄
3. NaClO₄ and water
4. methanol and acetone
5. acetonitrile (CH₃CN) and acetone (C₃H₆O).

Answer:

1. London’s forces
2. London’s forces
3. Ion-dipole interactions
4. Intermolecular hydrogen bonding
5. Dipole-dipole interactions.

Question 24.
Based on solute-solvent interactions, arrange the following in order of increasing solubility in n*octane and explain. Cyclohexane, KCl, CH₃OH, CH₃CN.
Answer:
(a) Cyclohexane and n-octane both are non-polar. They mix completely in all proportions.
(b) KCl is an ionic compound, KCl will not dissolve in n-octane.
(c) CH₃OH is polar. CH₃OH will dissolve in n-octane.
(d) CH₃CN is polar but lesser than CH₃OH. Therefore, it will dissolve in n-octane but to a greater extent as compared to CH₃OH. Hence, the order is KCl < CH₃OH < CH₃CN < Cyclohexane.

Question 25.
Amongst the following compounds, identify which are insoluble, partially soluble and highly soluble in water?
(i) phenol
(ii) toluene
(iii) formic acid
(iv) ethylene glycol
(v) chloroform
(vi) pentanol.
Answer:
(i) Partially soluble (because phenol has a polar -OH group but aromatic phenyl, C₆H₅ – group)
(ii) Insoluble because toluene is nonpolar while water is polar
(iii) Highly soluble because formic acid can form hydrogen bonds with water.
(iv) Highly soluble because ethylene glycol can form hydrogen bonds with water
(v) Insoluble chloroform is an organic liquid
(vi) Partially soluble because-OH the group is polar but the large hydrocarbon part (C₅H₁₁) is nonpolar.

Question 26.
If the density of some lake water is 1.25g mL^-1 and contains 92 g of Na⁺ ions per kg of water, calculate the molality of Na+ ions in the lake.
Answer:
Number of moles in 92 g of Na⁺ ions

As these are present in 1 kg of water, by definition, molality = 4m.

Question 27
If the solubility product of CuS is 6 × 10^-16, calculate the maximum molarity of CuS in an aqueous solution.
Answer:
Maximum molarity of CuS in aqueous solution = solubility of CuS in mol L^-1
If S in the solubility of CuS in mol L^-1 , then

Question 28.
Calculate the mass percentage of aspirin (C₉H₈O₄) in acetonitrile (CH₃CN) when 6.5 g of C₉H₈O₄ is dissolved in 450 g of CH₃CN.
Answer:

Question 29.
Nalorphene (C₁₉H₂₁NO₃), similar to morphine, is used to combat withdrawal symptoms in narcotic users. The dose of nalorphine generally given is 1.5 mg. Calculate the mass of 1.5 x 10^-3 m aqueous solution required for the above dose.
Answer:
1.5 × 10^-3 m solution means that 1.5 × 10^-3 mole of nalorphine is dissolved in I kg of water.
Molar mass of C₁₉H₂₁NO₃ = 19 × 12 + 21 + 14 + 48 = 3119 mol^-1
∴ 1.5 × 10^-3 mole of C₁₉H₂₁NO₃;
= 1.5 × 10^-3 × 3119 = 0.467 g = 467 mg
∴ Mass of solution = 1000 g + 0.467 g = 1000.467g
Thus, for 467 mg of nalorphene, solution required =1000.467 g for 1.5 mg of nalorphene, solution required =\(\frac{1000.467}{467} \times 1.5 = 3.219\)

Question 30.
Calculate the amount of benzoic acid (C₆H₅COOH) required for preparing 250 mL of 0.15 M solution in methanol.
Answer:
0.15 M solution means that 0.15 mole of benzoic acid is present in 1 L
i.e. 1000 mL of the solution.
Molar mass of benzoic acid (C₆H₅COOH) = 72 + 5 + 12 + 32 + 1 =122 g mol^-1
∴ 0.15 mole of benzoic acid = 0.15 × 122g= 18.39
Thus, 1000 mL of the solution contain benzoic acid = 18.39
∴ 250 ml of the solution will contain benzoic acid = \(\frac{18.3}{1000}\) × 250= 4.575g

Question 31.
The depression in the freezing point of water observed for the same amount of acetic acid, trichloroacetic acid, and trifluoroacetic acid increases in the order given above. Explain briefly.
Answer:
The depression in freezing points are in the order:
acetic acid < trichloroacetic acid

Fluorine, being most electronegative, has the highest electron-withdrawing inductive effect. Consequently, trifluoroacetic acid is the Strongest acid while acetic acid is the weakest acid. Hence, trifluoroacetic acid ionizes to the largest extent while acetic acid ionizes to minimum extent to give ions in their solutions in water. Greater the ions produced, greater is the depression in freezing point. Hence the depression in freezing point is maximum for trifluoroacetic acid and minimum for acetic acid.

Question 32.
Calculate the depression in the freezing point of water when 10 g of CH₃CH₂ CHCICOOH is added to 250 g of water. K_a = 1.4 × 10^-3, K_f= 1.86 K kg mol^-1.
Answer:
Molar mass of CH₃CH₂CHCICOOH = 15 + 14 + 13 + 35.5 + 45 = 122.5 g mol^-1

If α is the degree of dissociation of CH₃CH₂CHCICOOH, then CH₃CH₂CHCLCOOH □ CH3CHXHCICOO^– + H +

To calculate vant’s Hoff factor:
CH₃CH₂CHCICOOH □ CH3CH₂HCICOO^– + H⁺

Question 33.
19.5 g of CH₂FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of fluoro acetic acid. K_f of water is 1.86 K kg mol^-1.
Answer:
Here, w₂ = 19.5 g, w₁, = 500g, K_f = 1.86 K kg mol^-1, (AT_f) obs = 1.0°

M₂ (Calculated) for CH₂ FCOOH = 14+19+45 = 78 g mol^-1
Vant Hoff factor (i) =

calculation of dissociation constant. Suppose the degree of dissociation at the given concentration is a.
Then

Taking volume of the solutions as 500ml,

Question 34.
Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 293 K when 25 g of glucose is dissolved in 450 g of water.
Answer:
Here, P° = 17.535 mm, w₂ = 25g,
w₁ = 450g
For solute (glucose, C₆H₁₂O₆, M₂ = 180 g mol^-1
For solvent (H₂O), M₁ = 18g mol^-1
Applying Raoult’s law,

substituting the given values, we get

substituting the given values, we get

Question 35.
Henry’s law constant for the molality of methane in benzene at 298 K is 4.27 x 10s mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.
Answer:
Here, K_H = 4.27 × 10⁵ mm
p = 760 mm
Applying Henry’s law
P = K_H x

i.e, mole fraction of methane in benzene = 1.78 x 10^-3

Question 36.
100 g of liquid A (molar mass 140 g mol^-1) was dissolved in 1000 g of liquid B (molar mass 180 g mol^-1). The vapour pressure of pure liquid B was found to be 500 torrs. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 Torr.
Answer:
Number of moles of a liquid

Number of moles of a liquid B (Solvent)

Mole fraction of B in the solution (x_A)

Mole fraction of B in the solution (x_B)
= 1-0.114 = 0.886
Also, given p⁰_B = 500 Torr
Applying Raoult’s law,
P_A = x_AP°_A = 0.114 × P°_A
P_B= x_B p°_B = 0.886 × 500 = 443 Torr
p Total = p_A + p_B
475 = 0.114 P°_A +443 or

Substituting this value in equation (i), we get p_A = 0.114x 280.7 Torr = 32 Torr

Question 37.
Benzene and toluene form ideal solution over the entire range of composition. The vapour pressures of pure benzene and toluene at 300 K are 50. 71mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in the vapour phase if 80 g of benzene is mixed with 100 g of toluene.
Answer:
Molar mass of benzene (C₆H₆ = 78 g mol^-1
Molar mass of toluene (C₆H₅CH₃) = 92 g mol^-1
∴ Number of moles in 80 g of benzene

Number of moles in 100 g of toluene

In the solution, mole fraction of benzene

=0.486
mole fraction of toluene = 1-0,486 = 0.514
p° Benzene = 50.71 mm, p° Toluene = 32.06mm
Applying Raoult’s law,
p Benzene= x_Tolucno × p°_Tolucne =0.514 × 32.06 mm = 16.48mm
Mole fraction of benzene in vapour phase

Question 38.
The air is a mixture of a number of gases. The major components are oxygen and, nitrogen with the approximate proportion of 20% is to 79% by volume at 298 K. The water is in equilibrium with air at a pressure of 10 atm. At 298 K if Henry’s law constants for oxygen and nitrogen at 298 K are 3.30 × 10⁷ mm and 6.51 × 10⁷ mm respectively, calculate the composition of these gases in water.
Answer:
Total pressure of air in equilibrium with water = 10 atm
As air contains 20% oxygen and 79% nitrogen by volumes
partial pressure of oxygen (P ) = \(\frac{20}{100}\) x 10 atm = 2atm
= 2 x 760 mm
= 1520 mm
partial pressure of nitrogen \(P_{N_{2}}\)
= \(\frac{79}{100}\) × 10 atm = 7.9 atm = 7.9 x 760 mm
= 6004 mm

K_H(O₂) = 3.30 x 10⁷mm,K_H (N₂) = 6.51 × 10⁷ mm
Applying Henry’s law

Question 39.
Determine the amount of CaCl (i= 2.47) dissolved in 2.51itre of water such that its osmotic pressure is 0.75 atm at 27° C.
Answer:

Molar mass of CaCl₂= 40 + 2x 35.5 = 111 gmol^-1
Amount dissolved = 0.0308 xlllg = 3.42g

Question 40.
Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K₂SO₄in 2 litre of water at 25° C, assuming that it is completely dissociated.
Answer:
K₂SO₄ dissolved = 0.025g
volume of solution = 2L
T = 250°C = 298 K
Molar mass of K₂SO₄ = 2 x 39 + 32 + 4 × 16= 174gmol^-1
As, K₂SO₄ dissociates completely as K₂SO₄ → 2K+ SO₄^2-
i.e., ions produced = 3 ∴ i =3

2nd PUC Chemistry Solutions Additional Questions and Answers

Question 1.
How does molarity of a solution change with temperature?
Answer:
Molarity decreases with increase in temperature because volume of solution increases with increase in temperature.

Question 2.
State Raoult’s law.
Ans:
It states that for a solution of volatile liquids its vapour pressure of each component is directly proportional to their mole fraction in the solution.
The relative lowering of vapour pressure is equal to mole fraction of solute in case of non-volatile solute.

Question 3.
Why is liquid ammonia bottle first cooled in ice before opening it?
Answer:
At room temperature, the vapour pressure of liquid ammonia is very high. On cooling, vapour pressure decreases. Hence the liquid ammonia will not splash out.

Question 4.
Sodium chloride and calcium chloride are used clear show from the roads. Why?
Answer:
Sodium chloride depresses the freezing point of water to such an extent that it cannot freeze to form ice. Therefore, it melts off easily at the prevailing temperature.

Question 5.
Two liquids A and B boil at 145°C and 190°C
respectively. Which of them has a higher vapour pressure of 80°C? (CBSE 2006)
Answer:
A being more volatile will have higher vapour pressure at 80°C.

Question 6.
Calculate the molality of sulphuric acid solution in which the mole fraction is 0.85.
Answer:

Question 7.
The molar freezing point depression constant for benzene is 4.90 kg mol’1. Selenium exists as a polymer. When 3.26 g of selenium is dissolved in 226 g of benzene, the observed freezing point is 0.112°C lower for pure benzene. Decide the molecular formula of selenium (At Wt. of selenium is 78.8 g mol^-1). (CBSE 2002)
Answer:

Question 8.
CCI₄ and water are immiscible whereas ethanol and water are miscible in all proportions. Correlate this behaviour with molecular structure of these compounds. (CBSE 2003,2001)
Answer:
CCI₄ is a non-polar covalent compound.
Water is a polar compound. CCI₄ can neither form H^– bonds with water molecules nor can it break H^– bonds between water molecules. Therefore, it is insoluble in water.

Ethanol is a polar compound and can form H’ bonds with water, which is a polar solvent, therefore it is miscible with water in all proportions.

Question 9.
The molarity of a solution of sulphuric acid is 1.35 M. Calculate its molarity (The density of the acid solution is 1.02 g cm³ )
Answer:
Let the solution be 1 litre or 1000 cm³
∴ Number of moles of H₂SO₄= 1.35
Wt. of solution = 1000 x 1.02 = 1020 g
Wt. of sulphuric acid = 1.35 x 98= 132.3g :
Wt. of water = 1020 – 132.3 = 887.79
Molality of H₂SO₄= ppp x 1000 = 1.52 m

From (i) M, = 110.82, from (ii) M2 = 196.15 AB4 – AB, = B2 196.15- 110.82 = B2 85.33 = B2 B = 42.665
Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B 110.82=Atomic mass of A+85.33 Atomic mass of A = 110.82 – 85.33 = 25.49 Atomic mass of A = 25.499 Atomic mass of B = 42.669

Question 10.
Two elements A and B form compounds having molecular formulas AB2 and AB4. When dissolved 20 g of C6H6, 1 g of AB2 lowers the freezing point by 2.3 K, whereas 1.0 g of AB„4 lower it by 1.3 K. The molar depression constant for benzene is 5.1 kg mol*1. Calculate the mass of A and B. (CBSE2004)
Answer:
Let the molar mass of AB₂ and AB₄ be M₁and M₂
Then, for AB₂,

From (I) M₁ = 110.82, from (ii) M₁ = 196.15
AB₄– AB₂= B₂
196.15 – 110.82 =B₂
85.33 = B₂
B = 42.665
Molar mass of AB2 = Atomic mass of A+ x 2 atomic mass of B
110.82 = Atomic mass of A+ 85.33
∴ Atomic mass of A = 110.82-85.33 = 25.49
Atomic mass of A = 25 .499
Atomic mass of B = 42.669

Students can Download Karnataka SSLC Maths Model Question Paper 6 with Answers (Old Pattern), Karnataka SSLC Maths Model Question Papers with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus SSLC Maths Model Question Paper 1 (Old Pattern)

Time: 3 Hours
Max Marks: 80

I. In the following questions, four choices are given for each question, choose and write the correct answer along with its alphabet: ( 1 × 8 = 8 )

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.

(8 × 1 = 8)
Question 1.
The distance between two points p(x₁, y₁) and q (x₂, y₂) is given by

Question 2.
The degree of polynomial p(x) = x² – 3x + 4x³ – 6 is
(A) 2
(B) 1
(C) 3
(D) 6

The Polynomial Root Calculator can find multiple roots to polynomial equations with just one click. To open the tool, click on it.

Question 3.
Which one of the following cannot be the probability of an event?
(A) \(\frac { 2 }{ 3 }\)
(B) -1.5
(C) 15%
(D) 0.7

Question 4.
The curved surface area of frustum of a cone is given by
(A) Π (r₁ + r₂) l
(B) Π (r₁ + r₂) h
(C) Π (r₁ – r₂) l
(D) Π (r₁ – r₂) h

Question 5.
The solutions for the equations x + y = 10 and x – y = 2 are
(A) x = 6, y = 4
(B) x = 4, y = 6
(C) x = 7, y = 3
(D) x = 8, y = 2

Question 6.
In the adjoining figure, TP and TQ are the tangents to the circle with centre O. The measure of ∠PTQ is

(A) 90°
(C) 70°
(B) 110°
(D) 40°

Question 7.
The coordinates of origin are
(A) (1, 1)
(B) (2, 2)
(C) (0, 0)
(D) (3, 3)

The Polynomial Roots Calculator will find the roots of any polynomial with just one click.

Question 8.
If the discriminant of quadratic equation b² – 4ac = 0 then the roots are
(A) Real and distinct
(B) Roots are equal
(C) No Real Roots
(D) Roots are unequal and irrational

(6 × 1 = 6)
Question 9.
State “Basic proportionality theorem”.

Question 10.
Identify the tangent to the circle in the adjoining figure and write its name.

Question 11.
State Euclid’s division lemma.

Question 12.
Find the number of zeroes of a polynomial p(x) from the graph given

Question 13.
Find the distance of the point p(3, 4) from the origin.

Question 14.
Express 140 as a product of prime factors.

(16 × 2 = 32)
Question 15.
How many two – digit numbers are divisible by 3?

Question 16.
∆ABC ~ ∆DEF , Area of ∆ABC = 64cm² and area of ∆DEF = 121cm². If EF = 15.4 cm, Find BC.

Question 17.
Solve for x and y : 2x + y = 6 and 2x – y = 2.

Question 18.
Five years ago, Gouri was thrice as old as Ganesh. Ten years later Gouri will be twice as old as Ganesh. How old are Gouri and Ganesh?

Question 19.
Find the area of the shaded region in the figure, where ABCD is a square of side 14 cm.

Question 20.
Construct a pair of tangents to a circle of radius 5 cm Which are inclined to each other at an angle of 60°.

Question 21.
Find the value of k, if the points A (2, 3), B(4, k) and C (6, -3) are collinear.

Question 22.
Prove 3 + √5 is irrational.

Question 23.
Find the zeroes of polynomial p(x) = 6x² – 3 – 7x.

Question 24.
Find the quadratic polynomial whose sum and product of zeroes are \(\frac { 1 }{ 4 }\) and -1 respectively.

Question 25.
Solve the equation 3x² – 5x + 2 = 0 by using the formula.

Question 26.
Evaluate : 2 tan² 45° + cos² 30° = sin² 60°

Question 27.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower is 30°, Find the height of the tower.

Question 28.
As observed from the top of a 100 m high lighthouse from the sea level the angle of depression of two ships are 30° and 45°. If one of the ships is exactly behind the other on the same side of the lighthouse, find the distance between the two ships (√3 = 1.73)

Question 29.
A die is thrown once. Find the probability of getting a number between 2 and 6.

Question 30.
The volume of a cube is 64 cm2. Find the total surface area of the cube.

(6 × 3 = 18)
Question 31.
Prove that “The tangent at any point of a circle is perpendicular to the radius through the point of contact”.
OR
Prove that “The lengths of tangents drawn from an external point to a circle are equal”.

Question 32.
Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it, whose sides are \(\frac { 2 }{ 3 }\) of the corresponding sides of the first triangle.

Question 33.
A two digit number is four times the sum of the digits. It is also equal to 3 times the product of digits. Find the number.
OR
The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and the denominator, then the sum of the new fraction and the original fraction is \(\frac { 29 }{ 20 }\) Find the original fraction.

Question 34.
If 4tanθ = 3, Evaluate \(\frac { 4sin\theta -cos\theta +1 }{ 4sin\theta +cos\theta -1 }\)
OR
If tan 2A = cot (A – 18°) where 2A is an acute angle. Find the value of A.

Question 35.
Calculate the median for the following data.

Class interval	Frequency (F)
0-20	6
20-40	8
40-60	10
60-80	12
80-100	6
100-120	5
120-140	3
	n = 50

OR
Calculate the mode for the following frequency distribution table

Class Interval	Frequency (F)
5-15	6
15-25	11
25-35	21
35-45	23
45-55	14
55-65	5
	n = 80

Question 36.
Construct ‘ogive’ for the following construction.

C31	0-3	3-6	6-9	9-12	12-15
F	9	3	5	3	1

(4 × 4 = 16)
Question 37.
The sum of four consecutive terms which are in an arithmetic progression is 32 and the ratio of the product of the first and the last term in the product of two middle terms is 7.15. Find the number.
OR
In an arithmetic progression of 50 terms, the sum of the first ten terms is 210 and the sum of the last fifteen terms is 2565. Find the arithmetic progression.

Question 38.
Prove that “In a right-angled triangle the square on the hypotenuse is equal to the sum of the square on the other two-siders”.

Question 39.
Solve the equations graphically.
2x – y = 2; 4x – y = 4

Question 40.
A wooden article was made by scooping out a hemisphere from one end of a cylinder and a cone from other ends as shown in the figure. If the height of the cylinder is 40 cm, a radius is 7 cm and height of the cone is 24 cm, find the volume of the wooden article.

Solutions

Solution 1.
(B) or (D)

Solution 2.
(C) 3

Solution 3.
(B) -1, 5

Solution 4.
(A) Π (r₁ + r₂) l

Solution 5.
(A) x = 6, y = 4

Solution 6.
(C) 70°

Solution 7.
(C) (0, 0)

Solution 8.
(B) Roots are equal

Solution 9.
If a line is drawn parallel to one side of a triangle, it divides the other two sides proportionally.

Solution 10.
RS

Solution 11.
Let a and b be any two positive integers.
Then there exist two unique whole nos. q and r such that a = bq + r, (0 ≤ r < b)

Solution 12.
4

Solution 13.

Solution 14.

Solution 15.
Nos. divisible by 3 = 12,15,18, 21, …….., 99
a = 12, d = 3, T = 99, n = ?
T_n = a + (n – 1)d
⇒ 99 = 12 + (n – 1) 3
⇒ 99 – 12 = 3n – 3
⇒ 87 + 3 = 3n
⇒ 90 = 3n
⇒ n = 30
30 two digits are divisible by 3.

Solution 16.

Solution 17.

Substitute the value of x in Eqn 1
2x + y = 6
⇒ 2(2) + y = 6
⇒ 4 + y = 6
⇒ y = 6 – 4 = 2
∴ x = 2, y = 2

Solution 18.
Let the age of Gauri be ‘x’ years.
Let the age of Ganesh be ‘y’ years.
5 years ago
(x – 5) = 3(y – 5)
⇒ x – 5 = 3y – 15
⇒ x – 3y = -15 + 5
⇒ x – 3y = -10 …..(1)
10 years later
(x + 10) = 2(y + 10)
⇒ x + 10 = 2y + 20
⇒ x – 2y = 20 – 10
⇒ x – 2y = 10 ……(2)

Substitute the value of y in equation 1
x – 3y = -10
⇒ x – 3(20) = -10
⇒ x – 60 = -10
⇒ x = -10 + 60
⇒ x = 50
Gouri’s Age = 50 yrs, Ganesha’s Age = 20 yrs.

Solution 19.

Solution 20.

Solution 21.

Solution 22.
Let us assume that (3 + √5) is rational

But √5 is irrational
It contradicts the fact that √5 is irrational
our assumption is wrong.
3 + √5 is irrational.

Solution 23.
p(x) = 6x² – 7x – 3
By splitting the middle term we get
6x² – 9x + 2x – 3
3x(2x – 3) + 1 (2x – 3)
(2x – 3)(3x + 1)
for zero’s of p(x) put 2x – 3 = 0 & 3x + 1 = 0
1. 2x – 3=0
2x = 3
x = \(\frac { 3 }{ 2 }\)
2. 3x + 1 = 0
3x = -1
x = \(\frac { -1 }{ 3 }\)
Zeros of p(x) are \(\frac { 3 }{ 2 }\) and \(\frac { -1 }{ 3 }\)

Question 24.
The quadratic polynomial is given by

Question 25.
3x² – 5x + 2 = 0
Compare it with the standard form

Solution 26.

Solution 27.

Solution 28.

Solution 29.
No. of all possible outcome = {1, 2, 3, 4, 5, 6}
n(S) = 6
Let E be the event of getting a no. between 2 and 6.
No. of favourable outcomes n(E) = {3, 4, 5} = 3
P(E) = \(\frac { n(E) }{ n(S) }\) = \(\frac { 3 }{ 6 }\)

Solution 30.
Volume of the cube = l³
64 = l³
l = 4
T.S.A. of the cube = 6l² = 6 × 4² = 6 × 16 = 96 Sq.cm.

Solution 31.
Theorem: The tangent at any point of a circle is perpendicular to the radius through the point of contact.

Proof: We are given a circle with centre O a tangent XY to the circle at a point R We need to prove that OP is perpendicular to XY.
Take a point Q on XY other the P and join OQ see in below fig.
The point Q must lie outside the circle. (Why? Note that if Q lies inside the circle XY will become a secant and not a tangent to the circle). Therefore OQ is longer than the radius OP of the circle. That is,
OQ > OR
Since this happens for every point on the line XY except the point R OP is the shortest of all the distances of the point O to the points of XY. So OP is perpendicular to XY. (as shown in theorem).
OR
Theorem: The lengths of tangents drawn from an external point to a circle are equal.

Proof: We are given a circle with centre O, a point P lying outside the circle and two tangents PQ, PR on the circle from P (see in fig.). We are required to prove that PQ = PR,
For this, we join OR OQ and OR. Then ∠OQP and ∠ORP are right angles because these are angles between the radii and tangents, and according to the theorem, they are right angles.
Now in right triangles OQP and ORP
OQ = OR (Radii of the same circle)
OP = OP (Common)
Therefore, ΔOQP = ΔORP (RHS)
This gives PQ = PR

Solution 32.

Steps of construction:

1. Draw BC = 6 cm.
2. With B as the centre & radius 4 cms. draw an arc.
3. With C as centre and radius, 5 cm cut the first arc at A
4. Join AB and AC to get ΔABC
5. Draw any ray BX below BC making an acute angle with BC.
6. Locate 3 points B₁, B₂, B₃ as BX such that BB₁ = B₁B₂ = B₂B₃
7. Join B₃C
8. Draw a line through B₂ parallel to B₃C intersecting BC at C’
9. Draw a line through C’ parallel to AC to intersect BA at A’
10. Now A’BC’ is the required Δ each of whose sides is \(\frac { 2 }{ 3 }\) of the corresponding sides of ΔABC

Solution 33.
Set the no. be xy.
10x + y = 4(x + y)
10x + y = 4x + 4y
10x – 4x = 4y – y
2x = y
10x + y = 3xy
10x + 2x = 3x(2x)
12x = 6×2
12 = 6x
x = 2
y = 2 × 2 = 4
No. = 24
OR

Solution 34.

Solution 35.

Solution 36.

Solution 37.

Solution 38.

Solution 39.

Solution 40.

Karnataka State Syllabus SSLC Maths Model Question Paper 2 (Old Pattern)

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.

(8 × 1 = 8)
Question 1.
The co-ordinates of the origin are
(a) (0, 0)
b) (1, 0)
(c) (0, 1)
(d) (1, 1)

Question 2.
If p(x) = x³ – 1 then p(-1) = ……..
(a) 2
(b) -2
(c) 1
(d) 0

Question 3.
The total number of all possible events when a dice is thrown once is.
(a) 4
(b) 5
(c) 6
(d) 8

Question 4.
The area of a square is given by
(a) l × b
(b) l × b × h
(c) πr²
(d) l²

Question 5.
If y = 2x – 1 and x + y = 5, then the value of x is
(a) 2
(b) -2
(c) 1
(d) -1

Question 6.
In the adjoining fig AB and AC are tangents from A. If ∠BAC = 80° then BQC

(a) 110°
(b) 100°
(c) 80°
(d) 60°

Question 7.
The graph of an equation cuts x-axis at two points. Then the no. of solutions of that equation is
(a) 0
(b) 1
(c) 2
(d) 3

Question 8.
If the discriminant of a quadratic equation b² – 4ac > 0 then the roots are
(a) equal
(b) imaginary
(c) unequal & irrational
(d) Real & District

(1 × 6 = 6)
Question 9.
State the converse of “Basic Proportionality theorem”

Question 10.
How are the radius and diameter of a circle related?

Question 11.
Find the HCF of 24 and 36 by factorisation.

Question 12.
If p(x) = x² – x – 2 find p(o)

Question 13.
Find the distance between the point p (4, 5) and Q (1, 2)

Question 14.
If a no. is divided by 5. which are the remainders?

(2 × 16 = 32)
Question 15.
How many two digits no. are divisible by 4?

Question 16.
∆ABC ~ ∆DEF, Area of ∆DEF = 25 Sqcm. and Area of ∆ABC = 100Sqcm., If DE = 16 cms find AB.

Question 17.
Solve: 3x + 2y = 8, x + 3y = 5

Question 18.
Five years ago A was 7 times as old as B. 20 yrs. later A will be twice as old as B. What are their present ages?

Question 19.
Find the are of the shaded region in the figure given.

Question 20.
Construct a tangent to a circle of radius 4 cms. at a point on the circumference.

Question 21.
If the coordinates of the vertices of a Δ are (1, 2), (3, 4) and (5, 6). find the perimeter of the triangle.

Question 22.
Prove that 2 + √3 ls irrational.

Question 23.
Find the zeros of the polynomial p(x) = 2x² – 3x – 5.

Question 24.
Find the quadratic polynomial whose sum & product of zeros are \(\frac { 1 }{ 3 }\) and 2 respectively.

Question 25.
Solve by using formula 5x² – 9x + 3 = 0

Question 26.
Evaluate : tan² 45° + 2 cos 30° – sin 60°

Question 27.
A tower stands vertically on the ground. From a point on the ground. 15 mts. away from the foot of the tower, the angle of elevation of the top of the tower is 60°. Find the height of the tower.

Question 28.
An observer 1.5 mts tall is 28.5 mts. away from a chimney. The angle of elevation by the top of the chimney from her eyes is 45°, What is the height of the chimney?

Question 29.
A dice is thrown once. Find the probability of getting a number lying between 3 & 6.

(6 × 3 = 18)
Question 30.
The T.S.A. of a cube is 96 Sqcms. Find the volume of the cube.

Question 31.
Prove that the tangents drawn to a circle from an external point are equal.
OR
Prove that tangents drawn to a circle from an external point subtend equal angles at the centre.

Question 32.
Construct a Δ of sides 6 cms, 9 cms and 12 cm and then another Δ similar to it, whose sides are \(\frac { 2 }{ 3 }\) of the corresponding sides of the first triangle.

Question 33.
Solve the following: 8x + 5y = 9, 3x + 2y = 4
OR
A fraction becames \(\frac { 1 }{ 3 }\) when 1 is subtracted from the numerator & it becomes \(\frac { 1 }{ 4 }\) when 8 is added to its denominator. Find the fraction.

Question 34.
If 4 tanθ = 3, Evaluate \(\left[ \frac { 5sin\theta -cos\theta +2 }{ 5cos\theta +sin\theta -2 } \right]\)

Question 35.
Calculate the median for the following data.

C.I.	f
20-40	4
40-60	5
60-80	6
80-100	7
100-120	8
	N=30

OR
Calculate the mode for the following frequency distribution.

C.I.	f
15-25	1
25-35	6
35-45	15
45-55	18
55-65	9
	N=50

Question 36.
Constant OGIVE for the following distribution

C.I.	f
3-6	3
6-9	6
9-12	4
12-15	2
15-18	5
	N=20

(4 × 4 = 16)
Question 37.
How many two digit numbers are divisible by 3?
OR
Find the 31st term of an AR whose 11th term is 38 and the 16th term is 73.

Question 38.
The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides Prove this.

Question 39.
Solve graphically : 3x – y = 5, 2x + y = 5

Question 40.
A metallic sphere of radius 4.2 cms is melted and recast into the shape of a cylinder of radius 6 cm. Find the height of the cylinder.

Solutions

Solution 1.
(a) (0, 0)

Solution 2.
(b) -2

Solution 3.
(c) 6

Solution 4.
(d) l²

Solution 5.
(a) 2

Solution 6.
(b) 100°

Solution 7.
(c) 2

Solution 8.
(d) Real & distinct.

Solution 9.
It states that “If a line divides the two sides of a triangle in proportion then that line is parallel to the third side

Solution 10.
diameter = 2 × radius

Solution 11.

24 = 2 × 2 × 2 × 3
36 = 2 × 2 × 3 × 3
HCF = 2 × 2 × 3 = 12

Solution 12.
p(x) = x² – x – 2
⇒ p(0) = 0² – 0 – 2
⇒ p(0) = -2

Solution 13.
Distance

Solution 14.
The Remainders are {0, 1, 2, 3, 4}

Solution 15.
The nos. are 12,16, 20, 24, 28, ……. ,96
a = 12, d = 4, T_n = 96, n = ?
T_n = a + (n – 1) d
⇒ 96 = 12 + (n – 1) 4
⇒ 96 – 12 = 4n – 4
⇒ 84 + 4 = 4n
⇒ 4n = 88
⇒ n = 22
There are 22 two digit nos. divisible by 4.

Solution 16.

Solution 17.

Solution 18.
Let the present ages of A = x years, B = y years.


Age of A = 40 years
Age of B = 10 years

Solution 19.

Solution 20.

Solution 21.

Solution 22.

Solution 23.
2x² – 3x – 5 = 0
2x² – 5x + 2x – 5 = 0
x[2x – 5] + 1[2x – 5] = 0
(2x – 5)(x + 1) = 0
if 2x – 5 = 0 or if x + 1 = o
2x = 5 or x = 1
x = \(\frac { 5 }{ 2 }\)
The zeroes are 1 and \(\frac { 5 }{ 2 }\)

Solution 24.

Solution 25.
It is in the form ax² + bx + c = 0
a = 5, b = -a, c = 3

Solution 26.

Solution 27.

AB = Tower
CB = distance of the point C from the foot of the tower.
ACB is a Right Angle Triangle
tan60° = \(\frac { AB }{ CB }\)
√3 = \(\frac { AB }{ 15 }\)
AB = 15√3 mtrs
Height of the tower = 15√3 Mtrs

Solution 28.

AB = Chimney
CD = Observer
ADE = angle of elevation
ADE is the Right angled ∆
AB = AE + EB = AE + 1.5
DE = CB = 28.5
tan45° = \(\frac { AE }{ DE }\)
1 = \(\frac { AE }{ 28.5 }\)
AE = 28.5
Height of the Chimney = AB = AE + EB = 28.5 + 1.5 = 30 mts.

Solution 29.
No. of all Possible outcomes n(s) = 6 {1, 2, 3, 4, 5, 6}
Let E be the vent of getting a no. between 3 and 6.
n(E) = {4, 5} = 2
p(E) = \(\frac { n(E) }{ n(S) }\) = \(\frac { 2 }{ 6 }\)

Solution 30.
T.S.A. of a cube = 6l²
96 = 6l²
⇒ l² = 16
⇒ l = √16 = 4
Volume of the cube = l³ = 4³ = 64 cc.

Solution 31.
Data: O is the centre of the circle. AB and AC are tangents from A

To prove :
AB = AC
Construction:
Join AO, OB and OC.
Proof:
In Δles AOB and AOC
OA is common
OBA = OCA = 90° (angle between radius and tangent)
OB = OC (radii)
By RHS Postulate
ΔAOB = ΔAOC
AB = AC
OR
Data: O is the centre of the circle AB and AC are tangents from A. Join OA, OB & OC.

To Prove : BQA = QOA
Proof:
In Δles AOB and AOC
OA is common
∠OBA = ∠OCA = 90° (angle between radius & tangent)
OB = OC (radii)
By RHS postulate
ΔAOB = ΔAOC
BOA = COA

Solution 32.
AXY is the required Δ

Solution 33.

Solution 34.

Solution 35.

Solution 36.

Solution 37.
List of two-digit nos.
divisible by 3 = 12, 15, 18, ……. , 99.
This is an AP
a = 12, d = 3, T_n = 99, n = ?
T_n = a + (n – 1 ) d
⇒ 99 = 12 + (n – 1)3
⇒ 99 – 12 = 3n – 3
⇒ 87 + 3 = 3n
⇒ 90 = 3n
⇒ n = 30
30 two digit nos. are divisible by 3.
OR

Substitute the value of d in equation 1
a + 10d = 38
⇒ a + 10(7) = 38
⇒ a + 70 = 38
⇒ a = 38 – 70
⇒ a = -32
⇒ T₃₁ = a + 30d
⇒ T₃₁ = -32 + 30(7)
⇒ T₃₁ = -32 + 210
⇒ T₃₁ = 178

Solution 38.
data: ΔABC || ΔPQR
To Prove that:

Solution 39.



Q = (2, 1) intersecting of two lines.
The two lines intersecting at the point (2, 1), so x = 2, y = 1 is the required solution of the pair of linear equations.

Solution 40.

Karnataka State Syllabus SSLC Maths Model Question Paper 3 (Old Pattern)

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.

(8 x 1 = 8)
Question 1.
The coordinates of the mid-point P of the join of the points A(x₁, y₁) and B(x₂, y₂) is

Question 2.
The quadratic polynomial among the following is
(a) p(x) = x² + 4x + 3
(b) q(x) = ax + b
(c) g(x) = 4x – 3
(d) f(x) = x³ + 4x² – 5x + 2

Question 3.
When a dice is thrown twice n(s) = ………
(a) 6
(b) 12
(c) 24
(d) 36

Question 4.
The volume of a sphere is given by
(a) \(\frac { 2 }{ 3 }\) πr³
(b) \(\frac { 4 }{ 3 }\) πr³
(c) 4πr³
(d) 2πr³

Question 5.
The values of x and y which satisfy the given equations are 2x – 3y = 2, 3x – 2y = 8
(a) x = 2, y = 2
(b) x = 3, y = 1
(c) y = 2 , y = 2
(d) x = 2, y = 4

Question 6.
In the given fig AB and AC are tangents from A and BAC = 80° value of BOA = …….

(a) 50°
(b) 60°
(c) 70°
(d) 90°

Question 7.
If A(3, 4) and B(5, 6) are two points, then the coordinates of the mid-point of AB are
(a) (4, 5)
(b) (3, 4)
(c) (5, 6)
(d) (6, 7)

Question 8.
The discriminant of the equation ax² + bx + c = 0
(a) c² – 4ab
(b) a² – 4ab
(c) b² – 4ac
(d) a² – 2ac

(1 × 6 = 6)
Question 9.
State Pythagoras Theorem

Question 10.
What is the length of the longest chord in a circle of radius 5 cms?

Question 11.
Find the HCF of 135 and 225.

Question 12.
From the graph, find the number of zeroes of the polynomial p(x).

Question 13.
Find the distance between A and B if the coordinates of A and B are (2, 3) and (5, 6) respectively.

Question 14.
Express 226 as a product of Prime facotrs.

(2 x 16 = 32)
Question 15.
How many two digit nos. are divisible by 5?

Question 16.
If a pair of corresponding sides of two triangles are 5 cms and 7 cms then find the ratio of the areas of triangles

Question 17.
Solve:
5x + 4y = 14 → (1)
4x + 2y = 10 → (2)

Question 18.
5 Years ago A was thrice as old as his son 20 years, later A will be twice as old as his son. Find their present ages.

Question 19.
Find the area of the sector OAPB

Question 20.
Draw a circle of radius 4 cms. construct a tangent to the circle from a point 10 cms away from the centre.

Question 21.
Find the value of K. If the points A (4, 6), B(8, 2k) and C(12, -6) are collinear.

Question 22.
Prove that 5 + √7 is irrational

Question 23.
Find the zeros of the polynomial p(x) = 6x² + 7x – 3

Question 24.
Find the quadratic polynomial, whose sum & Product of the zeros are \(\frac { 1 }{ 2 }\) & 2 respectively.

Question 25.
Solve by using formula 2x² – 5x + 3 = 0

Question 26.
Evaluate : 2 sin² 60° + 3 tan² 45° – cos² 30°

Question 27.
The angle of elevation of the top of a tower from a point on the ground, which is 60 M away from the foot of the tower is 30°. Find the height of the tower.

Question 28.
As observed from the top of a 200M height tower from the sea level the angle of depression of two boats are 30° and 45°, if one of the boats is exactly behind the other on the same side of the tower, find the distance between two boats (√3 = 1.73).

Question 29.
A dice is thrown once. Find the probability of getting a perfect square number.

Question 30.
A copper rod of diameter 1 cm and length 8 cms. is drawn into a wire of length 18 m of uniform thickness. Find the thickness of the wire.

(16 x 2 = 32)
Question 31.
Prove that in two concentric circles the chord of the larger circle, which touches the smaller circle is bisected at the point of contact.
OR
Prove that the tangent at any point of a circle is perpendicular to the radius through the point of contact.

Question 32.
Construct a triangle of side 6 cm., 9 cm and 1 cm. and then a triangle similar to it whose side, are \(\frac { 2 }{ 3 }\) of the corresponding sides of the first triangle.

Question 33.
Solve the following equations by reducing them to a pair of linear equation.

OR
5 pencils and 7 books together cost Rs. 50. whereas 7 pencils & 5 books together cost Rs. 46. Find the cost of 1 pencil and 1 book.

Question 34.
If sin 3A = cos (A – 26°). Where 3A is an acute angle, find the value of A.
OR

Question 35.
Calculate the median for the following data.

C.I.	f
0 – 20	4
20 – 40	6
40 – 60	8
60 – 80	10
80 – 100	4
100 – 120	5
120 – 140	3
	n = 40

OR
Calculate the mode for the following frequency distribution

C.I.	f
5 – 15	3
15 – 25	8
25 – 35	17
35 – 45	20
45 – 55	11
55 – 65	1
	n = 60

Question 36.
Construct ‘OGIVE’ for the following:

C.I.	f
0 – 3	8
3 – 6	2
6 – 9	4
9 – 12	2
12 – 15	1
	n = 17

(4 x 4 = 16)
Question 37.
Find the 11th term from the last term (towards the first-term) of the AR 10, 7, 4,…….., -62.
OR
In a flower bed there are 23 rose plants in the first row, 21 in the second, 19 in the third & so on. There are 5 rose-plants in the last row. How many rows are there in the flower bed ?

Question 38.
In a triangle if square as one side is equal to the sum of the squares on the other two sides, then the angle opposite to the first side is a right angle, prove this.

Question 39.
Solve Graphically :
2x – y = 5
4x – y = 13

Question 40.
Ram made a bird bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cms. Find the T.S.A. of the bird bath.

Solutions

Solution 1.
(b)

Solution 2.
(a) p(x) = x² + 4x + 3

Solution 3.
(d) 36

Solution 4.
(b) \(\frac { 4 }{ 3 }\) πr3

Solution 5.
(c) x = 4, y = 2

Solution 6.
(a) 50°

Solution 7.
(a) (4, 5)

Solution 8.
(c) b² – 4ac

Solution 9.
“In a right-angled triangle the square on the hypotenuse is equal to the sum of the squares as the other two sides”.

Solution 10.
10 cm.

Solution 11.

By division lemma
225 = 135 x 1 + 90
135 = 90 x 1 + 45
90 = 45 x 2 + 0
The last divisor is 45
HCF = 45

Solution 12.
4

Solution 13.

Solution 14.

Solution 15.
Nos. divisible by 5 are 10, 15, 20, ………, 95
They are in AP
a = 10, d = 5, T_n = 95, n = ?
T_n = a + (n – 1)d
⇒ 95 = 10 + (n – 1)5
⇒ 95 – 10 = 5n – 5
⇒ 85 + 5 = 5n
⇒ 90 = 5n
⇒ n = 18
18 two digit nos. are divisible by 5.

Solution 16.
Let ABC and PQR be the triangles.

Solution 17.

Substitute the value of x in eqn. 1.
5x + 4y = 14
⇒ 5(2) + 4y = 14
⇒ 10 + 4y = 14
⇒ 4y = 14 – 10
⇒ 4y = 4
⇒ y = 1
∴ x = 2, y = 1

Solution 18.
Let the age of A = x years
Let the age of son = y years
5 years ago
(x – 5) = 3(y – 5)
x – 5 = 3y – 15
x – 3y = -15 + 5
x – 3y = -10 …..(1)
20 years later
(x + 20) = 2(y + 20)
x + 20 = 2y + 40
x – 2y = 40 – 20
x – 2y = 20 …… (2)
From Eqn (1) & (2)

Present age of the father = 80 years
Age of the son = 30 years.

Solution 19.

Solution 20.

Solution 21.

Solution 22.
Let us assume that 5 + √7 is rational.

but √7 is irrational.
If contradicts the fact that √7 is irrational
our assumption is wrong.
(5 + √7) is irrational.

Solution 23.
p(x) = 6x² + 7x – 3
By splitting the middle term
6x² + 9x – 2x – 3
⇒ 3x(2x + 3) – 1(2x + 3)
⇒ (2x + 3) (3x – 1)
for zeroes of p(x) put
2x + 3 = 0 and 3x – 1 = 0
x = \(\frac { -3 }{ 2 }\) and x = \(\frac { 1 }{ 3 }\)
zeroes of p(x) are \(\frac { -3 }{ 2 }\) and x = \(\frac { 1 }{ 3 }\)

Solution 24.

Solution 25.
Compare this with ax² + bx + c = 0
a = 2, b = -5, c = 3

Solution 26.

Solution 27.

Solution 28.

Solution 29.
No. of all possible outcomes n(S) = 6
Let ‘E’ be the event of getting a perfect square no.
n(E) = 2 {1, 4}
p(E) = \(\frac { n(E) }{ n(S) }\) = \(\frac { 2 }{ 6 }\)

Solution 30.

Solution 31.

Data: Two concentric circles C₁ and C₂ have the same centre O.
AB is the chord of the larger circle C₂.
It touches C₁ at P
To prove AP = PB
Construction: Join OP
Proof: AB is the tangent to C₁ at P and OP is the radius.
OP ⊥ AB
AB is the chord for C₂ & OP ⊥ AB
OP is the bisector of AB, as the perpendicular from the centre bisects the chord.
i.e., AP = BR
OR

Data: ‘O’ is the centre of the circle XY is a tangent at P.
To prove: QP ⊥ XY
Construction: Take a point Q as XY other than P and join OQ.
Proof: The point P lies outside the circle (if it lies inside XY becomes a secant).
OQ is longer than OP (radius)
OQ > OP
Since this happens for every point on XY except the point P
OP is the shortest of all the distances from the point O to the points on XY.
So, OP is the perpendicular to XY.

Solution 32.

ΔABC is required Δ
ΔA’BC’ is required Δ

Solution 33.

Solution 34.
sin3A = cos(A – 26°)
⇒ sin 3A = cos (90 – 3A)
⇒ cos (90 – 3A) = cos (A – 26)
⇒ 90 – 3A = A – 26
⇒ 90 + 26 = A + 3A
⇒ 116 = 4A
⇒ A = 29°
OR

Solution 35.

Solution 36.

Solution 37.
If we write the given AP in the reverse order, then a = -62 and d = 3.
Now find the 11th term with these a and d.
so, a₁₁ = -62 + (11 – 1) x 3
a₁₁ = -62 + 30
a_n = -32
OR
The number of rose plants in 1st, 2nd, 3rd, ……… rows are
23, 21, 19, ………. 5
It forma an AP
Let the number of rows = n
a = 23, d = -2, a_n = 5
a_n = a + (n – 1)d
⇒ 5 = 23 + (n – 1)(-2)
⇒ 5 – 23 = -2n + 2
⇒ -18 – 2 = -2n
⇒ -20 = -2n
⇒ n = 10
No. of rows = 10

Solution 38.

AC² = AB² + BC²
we need to prove ∠B = 90°
construct ΔPQR , right angled at Q such that
PQ = AB & QR = BC
Now from ΔPQR , we have
PR² = PQ² + QR² (by pythogora’s theorem)
PR² = AB² + BC² (by construction)
AC² = AB² + BC² (data)
AC = PR
In ΔABC and ΔPQR
AB = PQ (by construction)
BC = QR (by construction)
AC = PR (Proved)
ΔABC = ΔPQR
∠B = ∠Q (CPCT)
∠Q = 90° (by Construction)
∠B = 90°

Solution 39.

Solution 40.

Let h be the height of the cylinder and r the common radius of cylinder
hemi-sphere T.S.A. of bird-bath = CSA of cylinder + CSA of the hemisphere.
= 2πrh + 2πr²
= 2πr (h + r)
= 2 x \(\frac { 22 }{ 7}\) x 30 (145 + 30)
= 2 x \(\frac { 22 }{ 7}\) x 30 x 175
= 33000 cm²

Karnataka State Syllabus SSLC Maths Model Question Paper 4 (Old Pattern)

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.

(1 × 8 = 8)
Question 1.
The distance between two points A (1, 7) and B (4, 2) is
(a) √34
(b) √43
(c) √68
(d) √86

Question 2.
If p(x) = x² + x + 1 then the value of p(-1) is
(a) -1
(b) 1
(c) 2
(d) -2

Question 3.
Which is correct for the event E?
(a) P(\(\bar { E }\)) = 1 + P(E)
(b) P(\(\bar { E }\)) = P(E)
(c) P(\(\bar { E }\)) = 1 – P(E)
(d) P(E) = 1 + P(\(\bar { E }\))

Question 4.
The volume of a cylinder is 154 cc. and the radius is 7 cms. then the height is
(a) 2 cms
(b) 3 cms
(c) 4 cms
(d) 1 cm

Question 5.
The value of x and y for the following equations are x + y = 15, x – 7 = 1
(a) (8, 7)
(b) (7, 8)
(c) (10, 5)
(d) (5, 10)

Question 6.
ABCD is a cyclic quadrilateral. A = 70°, then C = ?
(a) 100°
(b) 110°
(c) 120°
(d) 70°

Question 7.
OA is a radius in the circle. If coordinates of A are (2, 3) then OA =…

a) √9
b) √4
c) √13
d) √5

Question 8.
If the discriminant of a quadratic equation b² – 4ac < 0 then the roots are
(a) Real and distinct
(b) equal
(c) not real
(d) unequal&rational

(1 × 6 = 6)
Question 9.
In ΔABC, XY || BC, If AX = 2 cm, AB = 5 cms and AY = 4 cms. then AC = ………

Question 10.
Identify the largest chord in the given figure.

Question 11.
Find the HCF of 25 and 15

Question 12.
From the graph find the number of zeros of the polynomial p(x)

Question 13.
Which are the coordinates of the origin?

Question 14.
Express 200 as a product of prime factors.

(16 × 2 = 32)
Question 15.
How many 2 digit numbers are divisible by 9.

Question 16.
Diagonals of a trapezium ABCD with AB || DC intersect at O. If AB = 2CD. Find the ratio of the areas of triangles AOB and COD.

Question 17.
Solve for x and y
2x + y = 4
3x + 4y = 6

Question 18.
Half the perimeter of a rectangle whose length is 4 m. more than its width is 36 m. Find the dimensions of the garden.

Question 19.
Find the area of the shaded region in the fig given. If each side of the square is 14 cm.

Question 20.
Draw a circle of radius 4 cms. Draw two radii in it such that the angle between them is 130°. Construct two tangents at the ends of radii.

Question 21.
The coordinates of the vertices of a triangle are (2, 3), (4, 5) and (6, 9). Find its area.

Question 22.
Prove that (5 + √3) is irrational.

Question 23.
Find the zeros of the polynomial p(x) = 5x² – 2x – 3

Question 25.
Solve the given equation using formula x² – 6x – 4 = 0

Question 26.
If A = √2 – 1 show that \(\frac { tanA }{ 1+{ tan }^{ 2 }A } =\frac { \surd 2 }{ 4 }\)

Question 27.
From a point an the ground, the angles of elevation of the bottom and top of a tower fixed at the top of 20 m high building is 45° & 60° respectively. Find the height of the tower.

Question 28.
The angle of elevation of an aeroplane from a point on the ground is 60°. After 15 seconds flight the elevation changes to 30°. If the aeroplane is flying at a height of 1500√3 m, find the speed of the plane in km/hr.

Question 29.
12 defective pens are mixed with 132 good ones one pen is taken out at random from this lot. Find the probability that the pen is taken out is a good one.

(6 × 3 = 18)
Question 30.
Two cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

Question 31.
Prove that in two concentric circles the chord of the longer circle, which touches the smaller circle is bisected at the point of contact.
OR
Prove that the tangents drawn to a circle from an external point are equal.

Question 32.
Construct a Δ of sides, 8 cms., 10 cms & 12 cms. and then A similar to it, whose sides are \(\frac { 3 }{ 4 }\) the corresponding sides of the first Δ

Question 33.
90% and 97% acid solutions are mixed to get 21 Its. of 95% pure acid solution. Find the amount of each type of acid to be mixed to form the mixture.
OR
One says “give me a hundred friend I shall then become twice as rich as you”. The other says “If you give me 10,1 shall be 6 times as rich as you”. Find the amount of their capital.

Question 34.
If A, B, C are interior angles of a triangle ABC show that

Question 35.
Calculate the median for the following data

C.I.	f
65 – 85	4
85 – 105	5
105 – 125	13
125 – 145	20
145 – 165	14
165 – 185	8
185 – 205	4

OR
Calculate the mode for the following data.

C.I.	f
1000 – 1500	24
1500 – 2000	40
2000 – 2500	33
2500 – 3000	28
3000 – 3500	30
3500 – 4000	22
4000 – 4500	16
4500 – 5000	07

Question 36.
Draw the ‘less than type OGIVE for the following data.

C.I.	f
50 – 60	6
60 – 70	5
70 – 80	9
80 – 90	12
90 – 100	6

(4 × 4 = 16)
Question 37.
Which term of the A.P 45, 41, 37, 33, ………… is the first negative term?
OR

Question 38.
Prove that in a Δ, if square as one side is equal to the sum of the squares of the other two sides, then the angle opposite the first side is a right angle.

Question 39.
Solve Graphically
x + 3y = 6
2x – 3y = 12

Question 40.
The base radius and height of a right circular solid cone are 2 cms and 8 cms respectively. It is melted & recast into spheres of diameter 2 cms. each. Find the number of spheres formed.

Solutions

Solution 1.
(a) √34

Solution 2.
(b) 1

Solution 3.
(c) P(\(\bar { E }\)) = 1 – P(E)

Solution 4.
(d) 1 cm

Solution 5.
(a) (8, 7)

Solution 6.
(b) 110°

Solution 7.
(c) √13

Solution 8.
(c) not real

Solution 9.

Solution 10.
XY

Solution 11.
25 = 5 × 5
15 = 5 × 3

Solution 12.
2

Solution 13.
(0, 0)

Solution 14.

200 = 2 × 2 × 2 × 5 × 5 = 2³ × 5²

Solution 15.
Two digit Nos. are 18, 27, 36, ……. 99
It is in AP
a = 18, d = 9, a_n = 99, n = ?
a_n = a + (n – 1) d
⇒ 99 = 18 + (n – 1)9
⇒ 99 – 18 = 9n – 9
⇒ 81 + 9 = 9n
⇒ 90 = 9n
⇒ n = 10
There are 10 two digit nos.

Solution 16.
Area of ∆AOB and ∆COD
AÔB = COD (V.O.A.)
COO = OA (alternate angles)
By AA Similarity criterion
∆AOB ~ ∆DOC

Solution 17.

Substitute the value of x in equation 1
2x + y = 4
⇒ 2(2) + y = 4
⇒ 4 + y = 4
⇒ y = 4 – 4 = 0
x = 2, y = 0

Solution 18.

l + b = 36
⇒ l = 36 – b
l = b + 4
b + 4 = 36 – b
⇒ b + b = 36 – 4
⇒ 2b = 32
⇒ b = 16
l = b + 4
⇒ l = 16 + 4
⇒ l = 20
Length = 20 m
Width = 16 m

Solution 19.

Solution 20.
PA and PB are tangents to the circle at the ends of radii

Solution 21.

Solution 22.
Let us assume that (5 + √3)

RHS is rational
⇒ √3 is rational
But √3 is irrational
It is contradictory to the fact that √3 is irrational
Hence our assumption is wrong.
(5 + √3) is irrational.

Solution 23.
To get the zeros of the, polynomial
put p(x) = 0
⇒ 5x² – 2x – 3 = 0
⇒ 5x² – 5x + 3x – 3 = 0
⇒ 5x(x – 1) + 3(x – 1) = 0
⇒ (x – 1)(5x + 3) = 0
x – 1 = 0 or 5x + 3 = 0
x = 1 or x = \(\frac { -3 }{ 5 }\)
Zeros of the polynomial are 1 and \(\frac { -3 }{ 5 }\)

Solution 24.
Here a = a, b = -5, c = c
Let the zeros be α and β then

Solution 25.

Solution 26.

Solution 27.

Solution 28.

Solution 29.
No. of defective pens = 12
No. of good pens = 132
Total no. of pens = 144
No. of all possible outcomes = 144
Let E be the event that the pen taken out is a good one
No. of outcomes favourable to E = 132

Solution 30.

Let each edge of the cube be a
Volume of each cube = a³
64 = a³
⇒ a = 4
For the resulting cuboid l = 4 + 4 = 8 cm
breadth = 4 cm
Height = 4 cm.
Surface area of the resulting cuboid = 2(lb + bh + hl)
= 2[(8 × 4) + (4 × 4) + (4 × 8)]
= 2[32 + 16 + 32]
= 2 [80]
= 160 cm²

Solution 31.
C₁ and C₂ are the concentric circle with centre O A chord AB of the larger circle C touches the smaller circle C₂ at P

To prove: AP = BP
Construction: Join OP
Proof: AB is a tangent to the circle C₂ at P and OP is the radius.
OP ⊥ AB
AB is the chord of circle Q & OP ⊥ AB
OP is the bisector of chord AB
AP = PB
OR
Data: O is the centre of the circle. AB and AC are tangents from A

To prove AB = AC
Construction: Join OA, OB and OC
Proof: In ΔAOB and ΔAOC
ABO = ACO = 90° [Angle between radius & tangent]
OB = OC (Radii of the circle)
OA is common
By RHS criterian
ΔAOB = ΔAOC
AB = AC

Solution 32.
A’CC’ is the required triangle

Solution 33.
Let x Lts of 90% pure acid solution & Y Lts of 97% pure acide solution be mixed,
total volume of the mixture = (x + y) Lts
x + y = 21 ……… (1)
90% of x + 97% of y = 95% of 21

Substituting the value of y in eqn 1
x + y = 21
⇒ x + 15 = 21
⇒ x = 21 – 15 = 6
⇒ x = 6
Quantity of 90% pure acid solution = 6 Lts.
Quantity of 97% pure acid solution = 15 Lts.
OR
Let the amounts with them be respectively.
x and y
x + 100 = 2(y – 100)
⇒ x + 100 = 2y – 200
⇒ x – 2y = -200 – 100
⇒ x – 2y = -300 ……. (1)
y + 10 = 6(x – 10)
⇒ y + 10 = 6x – 60
⇒ y – 6x = -60 – 10
⇒ -6x + y = -70
⇒ 6x – y = 70 ……… (2)
From Equation 1
x = 2y – 300
Substituting the value of x in equation 2
6x – y = 70
⇒ 6(2y – 300) – y = 70
⇒ 12y – 1800 – y = 70
⇒ 11y = 70 + 1800
⇒ 11y = 1870
⇒ y = 170
Substitute the value Y in (1)
x = 2y – 300
⇒ x = 2(170) – 300
⇒ x = 340 – 300
⇒ x = 40
Their capitals are 40 Rs. & 170 Rs.

Solution 34.
Because A, B, C are the angles of a ∆ABC
A + B + C = 180°
⇒ B + C = (180 – A)

Solution 35.

Solution 36.

Solution 37.
a = 45
d = a₂ – a₁ = 41 – 45 = -4
Let nth term of the AP be the first negative term then a_n < 0
a + (n – 1)d < 0
⇒ 45 + (n – 1) (-2) < 0
⇒ 45 – 4n + 4 < 0
⇒ 49 – 4n < 0
⇒ 49 < 4n
⇒ n > \(\frac { 49 }{ 4 }\)
⇒ n > 12\(\frac { 1 }{ 4 }\)
least positive integral value of n = 13
13th term is the first negative term.
OR
Let ‘a’ be the first term and d the common difference
a₄ = a + (4 – 1) d = a + 3d
a₆ = a + 5d
a₇ = a + 6d
a₈ = a + 7d

Solution 38.

data : ABC is a ∆ in which AC² = AB² + BC²
To prove: ∠B = 90°
Construction: Construct ∆PQR right angled at Q such that
PQ = AB & RQ = CB
Proof: PR² = PQ² + RQ² (by Pythagoras theorem)
PR² = AB² + BC² (by construction)
But AC² = AB² + BC² (data)
AC = PR
In ∆ABC and ∆PQR
AB = PQ
BC = QR (by construction)
AC = PR (proved)
∆ABC = ∆PQR
∠B = ∠Q
but ∠Q = 90° (by construction)
∠B = 90°

Solution 39.

Solution 40.

Karnataka State Syllabus SSLC Maths Model Question Paper 5 (Old Pattern)

Four alternatives are given to each question. Choose an appropriate answer. Write it along with its alphabet.

(1 × 8 = 8)
Question 1.
The distance between the points (2, 3) and (4, 1) is
(a) 3√2
(b) 2√3
(c) √2
(d) 2√2

Question 2.
The cubic polynomial among the following is
(a) p(x) = x² – x – 2
(b) q(x) = 2x² + x – 7
(c) r(x) = x³ + x² – 1
(d) s(x) = x⁴ – x³ + x² – 2

Question 3.
The probability of a certain event is
(a) 0
(b) 1
(c) \(\frac { 1 }{ 2 }\)
(d) \(\frac { 1 }{ 3 }\)

Question 4.
The volume of a sphere of radius 21 cm is
(a) 38808 cm³
(b) 80838 cm³
(c) 83808 cm³
(d) 88380 cm³

Question 5.
If y = x – 2 and x + y = 8, then the values of x and y are respectively,
(a) -5, 3
(b) -3, 5
(c) 3, 5
(d) 5, 3

Question 6.
In the adjoining figure, AB and AC are tangents from A. Then ABC =?

(a) 30°
(b) 40°
(c) 50°
(d) 60°

Question 7.
The distance between points (a, b) and (-a, -b) is
(a) \(2\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\)
(b) \(\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\)
(c) 2\(\sqrt { { a }+{ b } }\)
(d) \(\sqrt { { a }+{ b } }\)

Question 8.
The roots of the equation 2x² – 200 = 0 are
(a) ±20
(b) ±10
(c) 20
(d) 10

(6 × 1 = 6)
Question 9.
State the converse of Basic proportionality theorem.

Question 10.
AB is a chord in a circle of radius 5 cms. OP ⊥ AB. If OP = 4 cms then AB = ……..

(a) 8 cm
(b) 10 cm
(c) 12 cm
(d) 6 cm

Question 11.
The product of HCF and LCM of two numbers is 75. Then the product of the numbers is
(a) 125
(b) 25
(c) 75
(d) 57

Question 12.
The zeroes of the Polynomial p(x) = x² + 4x + 3 are
(a) -3, -1
(b) 1, 3
(c) -1, 3
(d) -3, 1

Question 13.
If a point P (x, y) divides the line segment joining, A (x₁, y₁) and B (x₂, y₂) in the ratio m₁ and m₂, then the coordinates of P are

Question 14.
Express 120 as a product of prime factors.

(16 × 2 = 32)
Question 15.
Write the first term and common difference for the following A.P.
\(\frac { 1 }{ 3 }\) , \(\frac { 5 }{ 3 }\) , \(\frac { 9 }{ 3 }\) , \(\frac { 13 }{ 3 }\) , ……..

Question 16.
In the figure of ∆ABE = ∆ACD show that ∆ADE ~ ∆ABC

Question 17.
Solve for x and y
3x + 2y = 3
2x + 3y = 2

Question 18.
If twice the son’s age is added to the age of his father the sum is 90. If twice the father’s age is added to the age of the son the sum is 120. Find their ages.

Question 19.
In the figure given O is the centre of the bigger circle and AC is its diameter. Another circle with BA as the diameter is drawn.
If AC = 54 cms & BC = 10 cms. Find the area of the shaded region.

Question 20.
Construct a tangent of length 4 cm. from an external point to a circle of radius 3 cms.

Question 21.
Find the relation between x and y, such that the point (x, y) is equidistant from the points (3, 6) and (-3, 4).

Question 22.
Prove that 7 + √5 is irrational.

Question 23.
Find the quadratic polynomial, whose zeroes are (2 + √3) and (2 – √3)

Question 24.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time & product of its zeroes as 2, -7, -14 respectively.

Question 25.
Solve the equation 2x² – 5x + 3 = 0 using formula.

Question 26.
Evaluate:

Question 27.
From the top of a 7 m high building, the angle of elevation of the top of a tower is 60° & the angle of depression of its foot is 45°. Find the height of the tower.

Question 28.
From the top of a lighthouse, the angles of depression of two ships on the opposite sides of it are observed to be 30° and 60°. If the height of the lighthouse is h mts & the line joining the ships passes through the foot of the lighthouse find the distance between the ships.

Question 29.
A lot of 20 bulbs contains 4 defective ones, one bulb is drawn at random from the lot. What is the probability that the bulb is defective?

Question 30.
The sum of the radius of base and height of the solid right circular cylinder is 37 cms. If the total surface area of the solid cylinder is 1628 cm2. Find the volume of the cylinder.

(6 × 3 = 18)
Question 31.
A line drawn through the end point of a radius and perpendicular to it is a tangent to the circle. Prove this.

Question 32.
Draw a line segment of length 4.5 cms and divide it is the ratio 3 : 6 & measure each part.

Question 33.
Pocket money of A and B are in the ratio 6 : 5 & the ratio of their expenditure are in the ratio 4 : 3. If each of them saves Rs. 50 at the end of the month find their pocket money.
OR
The sum of a two-digit number & the no. obtained by reversing the digits is 99. If the no. obtained by reversing the digits is 9 more than the original no. find the no.

Question 34.
If tan2A = cot (A – 18°) where 2A is an acute angle, find A.
OR
Show that tan 48°. tan 23°. tan 42 °. tan 67° = 1

Question 35.
Calculate the median for the following data.

C.I.	f
1500 – 2000	14
2000 – 2500	50
2500 – 3000	60
3000 – 3500	86
3500 – 4000	74
4000 – 4500	62
4500 – 5000	48

OR
Calculate the mode for the following data.

C.I.	f
150 – 155	12
155 – 160	13
160 – 165	10
165 – 170	8
170 – 175	5
175 – 180	2

Question 36.
Contant OGIVE for the following distribution.

C.I.	f
50 – 60	6
60 – 70	5
70 – 80	9
80 – 90	12
90 – 100	6

(4 × 4 = 16)
Question 37.
If the 3rd and 9th terms of an A.P. are 4 and -8 respectively, which term of this AP is zero?
OR
If (2n + 3) is the nth term of an A.P. find (i) first term, (ii) common difference, (iii) 15th term.

Question 38.
Prove that in a right-angled triangle, the square as the hypotenuse is equal to the sum of the squares on the other two sides?

Question 39.
Solve the following in graphically:
3x – y = 3
2x + y = 2

Question 40.
The cost of painting the total outer surface of a closed cylindrical tank at 60 per square cm. is Rs. 237.60. The height of the tank is 6 times the radius of the base. Find the height & radius of the tank.

Solutions

Solution 1.
(d) 2√2

Solution 2.
(c) r(x) = x³ + x² – 1

Solution 3.
(b) 1

Solution 4.
(a) 38808 cm³

Solution 5.
(d) (5, 3)

Solution 6.
(c) 50°

Solution 7.
(a) \(2\sqrt { { a }^{ 2 }+{ b }^{ 2 } }\)

Solution 8
(b) ±10

Solution 9.
If a line divides the two sides of a triangle proportionally then that line is parallel to the third side.

Solution 10.
(d) 6 cms

Solution 11.
(c) 75

Solution 12.
(a) (-3, -1)

Solution 13.

Solution 14.

Solution 15.

Solution 16.
ΔABE = ΔACD (data)
AB = AC …… (1) (CPCT)
AE = AD (CPCT)
⇒ AD = AE ….. (2) (CPCT)
Dividing (1) & (2)
\(\frac { AB }{ AD }\) = \(\frac { AC }{ AE }\) …..(3)
also DAE = BAC ….. (4) (Common angle)
From (3) & (4)
ΔADE ~ ΔABC (SAS Similarity criterian)

Solution 17.

Substitute the value of y in equation (1)
3x + 2y = 3
⇒ 3x + 2(0) = 3
⇒ 3x + 0 = 3
⇒ 3x = 3
⇒ x = 1
∴ x = 1, y = 0

Solution 18.
Let the age of father = x years
Let the age of the son = y years

Substitute the value of y in equation 1
x + 2y = 90
⇒ x + 2(20) = 90
⇒ x + 40 = 90
⇒ x = 90 – 40 = 50
Age of the father = 50 years
Age of the son = 20 years

Solution 19.
Area of the shaded region = Area of the circle with AC as diameter – Area of the circle with AB as the diameter
Area of the shaded region

Solution 20.
Draw the rough figure & calculate OA using Pythagora’s theorem. Then construct the tangent.

Solution 21.
Let P → (x, y)
A → (3, 6) & B → (-3, 4)
PA = PB (data)
PA² = PB²
(3 – x)² + (6 – y)² = (-3 – x)² + (4 – y)²
⇒ 9 + x² – 6x + 36 + y² – 12y = 9 + x² + 6x +16 + y² – 8y
⇒ -6x + 36 – 12y = 6x + 16 – 8y
⇒ -12x – 4y + 20 = 0 (÷4)
⇒ 3x + y – 5 = 0
This is required relation.

Solution 22.
Let us assume that 7 + √5 is rational

RHS is rational
⇒ LHS is also rational but √5 is irrational.
It is contradictory to the fact that √5 is irrational.
our assumption is wrong
7 + √5 is irrational

Solution 23.
S = Sum of the zeros = 2 + √3 + 2 – √3 = 4
P = Product of the zeroes = (2 + √3)(2 – √3) = 4 – 3 = 1
The required quadratic polynomial is
k[x² – Sx + P] where ≠ 0 is real
⇒ k[x² – 4x + 1]

Solution 24.
Let α, β & γ be the zeroes of the required cubic polynomial

Solution 25.
2x² – 5x + 3 = 0
It is in the form ax² + bx + c = 0
a = 2, b = -5, c = 3

Solution 26.

Solution 27.

In the Rt. angled ΔABD
tan 45° = \(\frac { AB }{ BD }\)
1 = \(\frac { AB }{ BD }\)
BD – AB = 7m
AE = 7m
In the Rt. angled ΔAEC
tan60° = \(\frac { CE }{ AE }\)
√3 = \(\frac { CE }{ 7 }\)
CE = 7√3
Height of the tower = CD
= DE + CE
= AB + CE
= 7 + 7√3
= 7(√3 + 1)

Solution 28.

Solution 29.
Total no. of bulbs = 20
No. of all possible outcomes = 20
Let E₁ be the event that the bulb-drawn at random from the lot is defective
No. of outcomes favourable to E₁ is 4 Since there are 4 defective bulbs.

Solution 30.
r + h = 37
T.S.A. of the cylinder = 2πrh + 2πr² = 2πr (h + r)
But 2πr (h + r) = 1628

Solution 31.

Data: A radius OP of a circle is drawn. Through the circle is drawn. Through the end point P of this radius a line AB is drawn the perpendicular to radius OR
To prove: AB is a tangent to the circle at P
Proof: Let Q be any point different from P on this line.
Now OP ⊥AB (data)
OP is the shortest of all the distances from the point O to the line APB.
OP < OQ ⇒ OQ > Radius OP
Q is an exterior point of the circle i.e. Q lies outside the circle, for all positions of Q different from P
Line AB meets the circle only at the point P Line AB is a tangent to the circle at P

Solution 32.
AC : CB = 3 : 6
AB = 4.5 cms.
AC = 1.5 cms
CB = 3 cms.

Solution 33.
Let the pocket money of A and B respectively = 6x and 5x
Let their expenditure be 4y and 3y respectively.
Monthly savings of A and B = (6x – 4y) & (5x – 3y)

The pocket money of A = 6x = 6 × 25 = Rs. 150
The pocket money of B = 6x = 5 × 25 = Rs. 125
OR
Let the two digit no. be yx = 10y + x
as reversing the digits it becomes 10x + y
10y + x + 10x + y = 99
⇒ 11x + 11y = 99
⇒ x + y = 9 …….. (1)
According to the second condition
10x + y = 10y + x + 9
⇒ 9x – 9y = 9
⇒ x – y = 1 ……… (2)

Substitute the value of x in equation (2)
x – y = 1
⇒ 5 – y = 1
⇒ y = 4
The required no. is yx = 45

Solution 34.
tan 2A = cot (A – 18°)
cot(90 – 2A) = cot( A – 18) [∴ tan θ = cot (90 – θ)]
90 – 2A = A – 18
90 + 18 = A + 2A
108 = 3A
A = 36°
OR
LHS = tan 45° tan 23° tan 42° tan 67°
= tan(90 – 42) tan 23° tan 42° tan(90 – 23°)
= cot42° tan 23° tan 42° cot 23°

Solution 35.

Solution 36.

Solution 37.
Let the first term and common difference of the AP be a and d respectively.
T₃ = 4 ⇒ a + 2d = 4 …… (1)
T₉ = -8 ⇒ a + 8d = -8 ……. (2)
Solving Equation 1 & 2
d = -2
Substitute the value of d in equation 1
a + 2d = 4
⇒ a + 2(-2) = 4
⇒ a – 4 = 4
⇒ a = 8
Let the nth term be zero
T_n = 0
a + (n – 1)d = 0
⇒ 8 + (n – 1)(-2) = 0
⇒ 8 – 2n + 2 = 0
⇒ 10 – 2n = 0
⇒ 10 = 2n
⇒ n = 5
5th term of the AP = zero
OR
a. a_n = 2n + 3
put n = 1
a₁ = 2(1) + 3
⇒ a₁ = 5
first term = 5
Put n = 2
b. a₂ = 2(2) + 3
⇒ a₂ =4 + 3
⇒ a₂ = 7
d = a₂ – a₁
⇒ d = 7 – 5
⇒ d = 2
Common difference = 2
c. T_n = (2n + 3)
Put n = 15
T₁₅ = 2(15) + 3 = 30 + 3 = 33
Fifteenth term = 33

Solution 38.
data: ABC is a ∆ in which B = 90°
To Prove: AC² = AB² + BC²
Construction: Draw BD ⊥ AC
Proof: In ∆ABC and ∆ABD
A is common

Solution 39.

Solution 40.

Karnataka SSLC Maths Model Question Papers

Students can Download English Poem 10 Photograph Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 9 English Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Board Class 9 English Poem Chapter 10 Photograph

Photograph Questions and Answers, Summary, Notes

Comprehension:

C1. Answer the following in a sentence or two each:

Question 1.
How many people are there in the photograph?
Answer:
There are three people in the photograph.

Question 2.
How is the poet related to the people in the photograph?
Answer:
The people in the photograph are the poet’s mother and her cousins, Betty and Dolly.

Question 3.
Who was taking the snapshot?
Answer:
An uncle was taking the snapshot.

Question 4.
Is the mother described in the photo alive?
Answer:
No, she is dead.

Question 5.
Which aspect of the mother does the poet like very much?
Answer:
The poet has sweet memories of the smiling and sweet face of the mother which she likes. This nature of the mother is captured in the photo.

C2. Answer the following:

Question 1.
Why does the writer say
“And of this circumstance
There is nothing to say at all.
Its silence silences”?
Answer:
The poet says that her mother had been dead and now she finds herself in a situation in which there is nothing to be said but only emptiness. The silence of this situation silences her. In other words, she is left speechless. Fate has killed all the feelings in her.

By the phrase ‘of this circumstance’, the poet means the circumstance of the death of her mother. The poet puns upon the. word ‘silence’. Since there are no details, what she is left with is silence and this silence silences her.

There are different possible meanings of the use of the word ‘silence’ as verb: she cannot say anything as she does not know anything about the circumstance of the death; or she does not have words to express her feelings; or the loss of her mother has stifled her voice. If we go one step further, we can even wonder whether the silence is owing to the unnatural nature of the death of the mother.

Question 2.
Does the poet notice any change in the mother after the poet was bom? What do you think could have made the change in the mother’s face?
Answer:
Yes, the poet notices the change in the mother’s face after she was born. This could have been the outcome of sorrowful incidents or hardships in life. Age and ill health also might have made the mother lose the sweetness of her face and smile.

The poet is looking at her mother’s photograph which is indeed an old one. With it she can see how her mother looked when she was a little girl of twelve. The photo shows her on a beach with her two girl cousins who are younger than her, holding her hand.

It might have been windy at that time as their hair was flying on their faces when the uncle took the photograph. All the three smile through their flying hair. Looking at the photograph, the poet says that her mother had a sweet face, but it was a time before the poet was born.

The sea was washing their feet. The poet says that the sea has changed only a little but change has come about the ones whose feet it was washing. After 30 or 40 years, the mother would take out the photograph and take a look at it.

By that time, she was married and had a daughter. She would laugh a little and say “Look at Betty and Dolly, see how they have dressed for the beach”. By now, she can only remember those days. A huge change has come about her and she is no longer that small, innocent girl of twelve. After some years, when the poet’s mother dies, for the poet, her mother’s laughter becomes a thing of the past.

That’s why she says “the sea holiday was her past and mine is her laughter”. In the same way as the mother remembers her old days, the poet can remember her mother. The poem also shows that in due course of time, the two of them learned to live with their losses though the loss had made a permanent impression on their wry faces.

Question 3.
Why are the feet described as ‘transient feet’?
Answer:
The word ‘transient’ means fleeting/ passing or temporary. It is the opposite of permanent. The ‘feet’ are described with an adjective ‘transient’ to drive home the truth that the impression they make on the sand is transitory. They get washed off by the waves.

Everything about life is transient. Single incidents, as well as whole life itself, are transient. Only nature remains as a permanent feature around us. Thus, the poet presents a contrast between human life and the sea.

The poem lends itself to multiple interpretations with regard to the transitory nature of life. If the passage of time is one aspect of the transitory nature of life, sudden unexpected occurrences can also indicate how ephemeral everything is.

We can have a different reading of the poem here. At one point the children are posing for the photograph holding their aunt’s hand. At another, the sea washes their terribly transient feet. The word ‘terribly’ makes us wonder whether the children are swept away by the waves of the sea.

C3. Answer the following questions on your own.

Question 1.
What is the mood of the poet?
Answer:
Melancholy marks the utterances of the poet. The poet has a deep sense of loss on losing her mother and the tone of sadness is all-pervasive.

Question 2.
Which line in the poem do you like the most? Why?
Answer:
I like the line ‘All three stood still to smile through their hair’. On the one hand, it creates a powerful mental picture of the three human figures against the breeze on the seashore posing for the photograph. The golden time of childhood when children are filled with mirth and are free from all worries and complexities of life is powerfully pictured here.

Smiling through the hair is typical of people who pose on the seashore. The description captures the innocuous and blithe spirit of the children which is contrasted against the care and concern of old age. It shows the poet’s power of observation and description.

The line gets connected to the title also. A photograph becomes a metaphor for all that life captures and also loses. Life is transitory and we are likely to lose many things which we will remember only through photographs.

Question 3.
Is there any change in the life of the poet’s mother over the years? What kind of a person, you think, she was? Describe the mother in the poem in your own words.
Answer:
The three stanzas of the poem depict the three stages in the life of the mother – as a child with her cousins, as a mother looking at the old photograph, and as a memory for the daughter after her death. She had a smiling and sweet face in the photograph when she posed for it, holding the hands of her cousins.

However, the poet cannot remember witnessing the same cheerfulness on the mother’s face in her recollections of her mother after she was born. What could have been the reason for the change? Apparently, the poet’s mother had posed for the photograph with her cousins when she was young and was not yet bogged down by the responsibilities and hardships of life.

As people age, along with inevitable physical changes, they also experience a change in their mental make-up because of the challenges in life. The line,’… how they dressed us for the beach,’ indicates that the mother herself was quite young and followed the directions given by others when she accompanied her cousins to the beach.

Additional Questions:

Question 1.
What does the word ‘cardboard’ denote in the poem? Why has this word been used?
Answer:
The word ‘cardboard’ means a very stiff and thick paper. Here the cardboard is a part of the frame that keeps the photograph intact. Its use in the poem is ironical. It keeps the photograph of that twelve-year-old girl safe who herself was ‘terribly transient’. She had died years ago.

Question 2.
What has not changed over the years? What does this suggest?
Answer:
The sea has not changed over the years. It brings out the ‘transient’ nature of man when compared to nature and its objects. Time spares none. The pretty faces and the feet of the three girls are ‘terribly transient’ or mortal when compared to the ageless and the unchangeable sea.

Question 3.
The poet’s mother laughed at the snapshot. What did this laugh indicate?
Answer:
The poet’s mother would laugh at the snapshot as it would revive her memories of the old happy days on the sea beach and the strange way in which they were dressed up for the beach. The laugh indicates her youthful spirit.

Question 4.
What is the meaning of the line “Both wry with the laboured ease of loss”?
Answer:
Both the poet’s mother and the poet suffer a sense of loss. The mother has lost her childlike innocence and joyful spirit that the photograph captured years ago. For the poet the smile of her mother has become a thing of the past. Ironically, both labour to bear this loss with ease.

Question 5.
What does ‘this circumstance’ refer to?
Answer:
‘This circumstance’ refers to the death of the poet’s mother. The photograph of the dead mother brings sad nostalgic feelings in the poet. But the poet has nothing to say at all about this circumstance. The silence of the poet makes the pall of silence prevailing there still deeper.

Question 6.
The three stanzas depict three different phases. What are they?
Answer:
In the first stanza, the poet’s mother is shown as a twelve-year-old girl with a pretty and smiling face. She went paddling with her two cousins. This phase is before the poet’s birth. The second phase describes the middle-aged mother laughing at her own snapshot. The third phase describes the chilling pall of silence that the death of the mother has left in the life of the poet.

Question 7.
“The sea, which appears to have changed less, washed their terribly transient feet.” – How does the poet contrast the girls’ ‘terribly transient feet’ with the sea?
Answer:
All the girls standing on the beach have ‘terribly transient’ existence. They are mortal and suffer physical changes with the passage of time. The mother’s sweet face and her smile has disappeared years later. But the vast sea remains unchanged of seems ‘to have changed less’ in their comparison.

Question 8.
‘Both wry with the laboured ease of loss’ – What is the loss? Describe the irony in the situation.
Answer:
Actually, both the poet and her mother suffer a sense of loss. The mother loses her carefree childhood. She can’t have those moments of enjoyment again that she once experienced at the beach. She can’t be a sweet smiling girl of twelve again.

This is also the poet’s loss. Perhaps she will never see that smiling face and experience her laughter again in life. The irony of the situation is that both of them struggle to bear the loss with tolerable ease.

Multiple Choice Questions:

Question 1.
The poet looks at the photograph of her mother taken when she was
A) ten years old
B) twelve years old
C) an old woman
D) a young lady
Answer:
B) twelve years old

Question 2.
The mother had gone on a sea-holiday
A) with her father
B) with her mother
C) with her cousins
D) with her daughter
Answer:
C) with her cousins

Question 3.
The photograph was taken by her
A) father
B) cousin
C) sister
D) uncle
Answer:
D) uncle

Question 4.
The Phrase ‘Their terribly transient feet’ suggests that
A) their feet were moving
B) they were padding
C) life is not permanent
D) life is everlasting
Answer:
C) life is not permanent

Question 5.
The poet’s mother is dead and the poet has nothing to say about that situation because
A) she does not want to say anything
B) the silence of death has silenced her
C) she cannot relive the past
D) she knows her mother would come back.
Answer:
B) the silence of death has silenced her

Question 6.
The mother saw the photograph after …………. years.
A) twenty or thirty
B) twelve or thirteen
C) four or five
D) one or two.
Answer:
A) twenty or thirty

Question 7.
The names of the cousins are
A) Bertha and Pinky
B) Betty and Dolly
C) Rosa and Mary
D) Laila and Pinky.
Answer:
B) Betty and Dolly

Question 8.
The cardboard shows the pictures of
A) two schoolgirls
B) two real sisters
C) two neighbours
D) narrator’s mother and her two cousins.
Answer:
D) narrator’s mother and her two cousins.

Question 9.
The ‘big girl’ here means
A) the eldest cousin of the mother
B) the tallest of the girls
C) narrator’s mother
D) the heaviest of the girls.
Answer:
C) narrator’s mother

Question 10.
The sweet face the photograph showed was that of the
A) narrator’s cousin
B) narrator’s father
C) narrator’s mother
D) narrator’s brother.
Answer:
C) narrator’s mother

Question 11.
The photograph was taken
A) when the narrator was about twelve
B) about twelve years ago
C) when the narrator was a child
D) when the narrator was not even born.
Answer:
D) when the narrator was not even born.

Question 12.
Who were ‘Betty’ and ‘Dolly’?
A) narrator’s cousins
B) her sisters
C) her mother’s cousins
D) her neighbours.
Answer:
C) her mother’s cousins

Question 13.
‘Its silence silences’ means
A) death’s silence
B) silence only brings out deeper silence
C) poet’s silence
D) silence caused by the mother’s death gives birth to a pall of silence.
Answer:
D) silence caused by the mother’s death gives birth to a pall of silence.

Photograph by Shirley Toulson About The Poet:

Shirley Toulson was born in 1924 in Henley-on-Thames, England. She had a huge passion for writing and was greatly influenced by her father who was a writer too. She secured a B.A. in Literature from Brockenhurst College in London in the year 1953. Shortly, she took writing as a career but meantime also served as editor for many magazines.

Celtic Christianity, influenced her so greatly that most of her major works such as ‘Celtic Alternative’ in 1987 and ‘Celtic Year’ in 1993 were on that topic. But a whole lot of other works, including collections of poetry, essays on education and other publications on diverse topics brought her even more fame.

Photograph Summary in English

The poem ‘Photograph’ is an interesting piece describing how a photograph can set in motion recollections of different kinds. The photograph is also an indicator to the fact that everything in life is transitory. What is captured years ago on the photograph is a contrast to what people turn out to be years later.

The photograph can also be illusory in nature as it might not show exactly how the people being featured in the photograph felt at that particular point of time. The photograph can also refer to our desperate attempts to defy the passage of time and the inevitable changes that take place along with the passage of time.

The speaker begins the poem by saying, ‘A photograph: The cardboard shows me how it was.’ The photograph is of the speaker’s mother with her cousins. The photograph features the two girls holding the hands of the speaker’s mother and the speaker knows that the photograph was taken by an uncle. The speaker is struck by the realisation that her mother was sweet in her looks at the time the photo was taken.

This was even before the speaker was 4}orn. While referring to the sea washing the terribly transient feet of the people who posed for the photograph, the speaker introduces a contrast between the sea and the people.

People change with the passage of time, but the sea seems to have changed very little. This is an indicator to the permanence of nature and transience of man. Unfortunately, the change in human beings, both physically and emotionally, is for the worse.

The smiling, sweet face of the young girl is replaced by the melancholic face of the mother. Just as physical changes are inescapable, even mental deterioration also seems to be inevitable.

Twenty or thirty years passed since that photograph had been taken. Whenever mother looked at the photograph, she would laugh. She would point out to Dolly and Betty how they were dressed up (by their parents) for the sea-holiday.

That sea-holiday had become a thing of the past for the mother. And that sweet laughter of the mother which the photograph had captured had become the poet’s own past. Both of them suffered from a sense of loss. Ironically, both of them were labouring to ease the loss.

The poet says that her mother has been dead for years. She has been dead nearly as many years as she had lived. She (the poet) has nothing to say about the circumstances of her death or about the situation captured in the photo. Silence only brings out deeper silence and makes things mysterious.

The photograph describes three stages. In the first stage, the photograph shows the poet’s mother standing on the beach enjoying her holiday with her two girl cousins. She was 12 or so at that time.

The second stage takes us twenty or thirty years later. The mother would laugh at the way she and her cousins Betty and Dolly were dressed up for the sea-holidays. In the third stage, the poet remembers the mother with a heavy heart as the mother is already dead. The photograph revives a nostalgic feeling in the poet.

Thus, through an everyday, commonplace incident of a photo being clicked on the seashore, the poet is able to drive home philosophical reflections on the very nature of life.

Glossary:

paddling: moving like rowing.
transient: momentary; not lasting for long
wry: distorted.

Students can Download English Lesson 8 The Dinner Party Questions and Answers, Summary, Notes Pdf, KSEEB Solutions for Class 5 English Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.

Karnataka State Board Class 5 English Prose Chapter 8 The Dinner Party

The Dinner Party Questions and Answers, Summary, Notes

Comprehension:

I. Answer the following questions

Question 1.
Who are the hosts and guests in the story?
Answer:
A colonial official and his wife, army officers and government official and their wives.

Question 2.
Describe the place where the dinner is hosted
Answer:
Dining room, which has a bare marble floor, open rafters and wide glass doors opening onto a veranda.

Question 3.
What is the young girl discussing with the colonel?
Answer:
A woman’s unfailing reaction in any crisis and a man’s nerve control.

Question 4.
According to the colonel Who has greater self-control, the man or the woman?
Answer:
Man.

Question 5.
Why do you think that the American did not join the discussion?
Answer:
The American is a naturalist, who just observes others and studies their reactions. So he did not join the discussion.

Question 6.
What did the hostess want her servants to do? Why?
Answer:
The hostess wants her servant to place a bowl of milk on the Veranda Just outside the open doors. Because a cobra had just crawled over her feet.

Question 7.
Do you think this action was a routine one?
Answer:
No.

Question 8.
What made the naturalist think that there must be a cobra in the room?
Answer:
In India milk in a bowl means only one thing bait for a snake. He realizes there must be cobra in the room.

Question 9.
Where did the American search for the cobra?
Answer:
He looked up at the rafters – the likeliest place, but they were bare. The three of the room were empty, and in the fourth, the servants were waiting to serve the next course. So he deduced that the cobra might be under the table.

Question 10.
What did the American do to test the self-control of the people at the dinner party?
Answer:
When the American Naturalist concluded that the cobra was under the table his first impulse was to jump back and warn others. But he did not want to frighten others and the cobra into striking one of them. In a sobering tone, he announced that he just wanted to know what self – control everyone at the table had.

He challenged them that he will start counting to three hundred – that’s five minutes and asked everyone to not move a muscle, and those who moved will lose fifty rupees. All the twenty guests sit like stone images while he counted. When he was counting two hundred and eight – the cobra emerged from under the table and went to the veranda where the bowl of milk was placed. When the guests saw the cobra, they were shocked and started, screaming. But the American quickly jumped up and slammed the veranda doors safely shut.

Question 11.
Who shows greater self-control in the stray The American guest or the hostess?
Answer:
The Hostess.

Question 12.
How does the American figure out that there is a cobra under the table?
Answer:
The American naturalist, while watching the hostess observes a strange expression come over the face of the hostess. She was staring straight ahead, he muscles contracting slightly.

She gestured to the servant standing behind her chain and whispered something to him. The servant’s eyes widened and he quickly left the room. The American then noticed that the servant placed a bowl of milk on the veranda just outside the open doors. In India, Milk in a bowl means – bait for a snake.

He looked up at the rafters – the likeliest place, but they were bare. The three of the room were empty, and in the fourth, the servants were waiting to serve the next course. So he deduced that the cobra might be under the table.

Question 13.
The colonel believes that women behave very differently from men in a crisis. Do you agree with the colonet’s belief.
Answer:
Yes, I agree that women behave very differently from men in a crisis. Women have more patience though the cobra was crawling across Mrs. Wynnes’s feet she kept her self – control.

Question 14.
How do you describe the followings:
a) The American Naturalist:
He is the one who kept control of the difficult situation in the lesson.

b) Mrs. Wynnes:
She realised that a cobra was in the room. Because it was crawling across her foot. But she kept her self – control.

c) The colonel:
The colonel is a male-centric person. He believed that men have more nerve control than woman.

d) The servants:
The servants were obedient and discrete.

Language Exercise

I. Use the following words in sentences of your own.

• Forfeit: The student cheated in the exams, so he had to for diet an year of schooling.
• Sobar: The children were playing daily in the garden, but they all turned sober when they saw a cobra, crawling about.
• Gesture: The teacher gestured at Nagaraj and called him to solve the Maths problem on the blackboard.
• Stare: It is not right to stare at physically handicapped people.

Grammar:

In one of the earlier lessons you have learnt the functions of adjectives. In this lesson lets learn how the three different forms of degrees of adjective are formed.

The three different forms/degree of adjectives are

1. The Positive degree
2. The Comparative degree
3. The Superlative degree

Usually, the comparative degree is formed by the addition of ‘er’ to the positive degree form ex: tall-taller short – shorter
The superlative degree is usually formed by the addition of ‘est’.

ex: strong stronger strongest broad broader broadest near nearer nearest Now, form the comparative and superlative degrees of the adjectives given below

Positive Comparative Superlative

1. fine – finer – finest
2. sweet – sweeter – sweetest
3. dear – dearer – dearest
4. high – higher – highest
5. fair – fairer – fairest
6. long – longer – longest
7. weak – weaker – weakest
8. fair fairer fairest
9. cool – coller – coolest
10. low – lower – lowest
11. bold – bolder – boldest
12. few – fewer – fewest

II. Match the adjectives given in column ‘A’ with the words given in B’

A	B
1. Brilliant	i) sea
2. Sharp	ii) light
3. loud	iii) shower
4. juicy	iv) knife
5. heavy	v) fruit
6. busy	vi) noise
7. rough	vii) cut
	viii) street

Answer:

A	B
1. Brilliant	ii) light
2. Sharp	iv) knife
3. loud	vi) noise
4. juicy	v) fruit
5. heavy	iii) shower
6. busy	viii) street
7. rough	i) sea

III. Complete the phrases below using appropriate adjectives

1. a delicious meal
2. a rough sea
3. a cute child
4. a gold ring
5. a terrible accident
6. a barking dog
7. an interesting story
8. a tough task
9. a good friend
10. a beautiful hat

IV. Write five sentences of your own using any of the completed phrases above.

1. Nazir read a thrilling story about pirates.
2. My uncle has a harking dog.
3. Francis is a good friend.
4. Arabian sea is a rough sea.
5. My mother prepared a delicious meal.

The Dinner Party Summary In English

The lesson ‘The Dinner Party” written by Mona Gardner explains about the importance of self-control. A colonial official and his wife are giving a large dinner party for the army officers and government officials and their wives, and a visiting American naturalist. A spirited discussion springs up between a young girl and a woman.

The American notices the servant place a bowl of milk on the veranda just outside the open doors. He realizes there must be cobra in the room. The American told all that he would count three hundred, that was five minutes and not one of you is to move a muscle.

Those who move will forfeit fifty rupees. The twenty people sat like stone images by the ti pie the cobra emerged and went to the varanda where the bowl of milk kept. The colonel appreciated the American who has just shown us an example of perfect control.

The Dinner Party Summary In Kannada

Are you a fan of slot games? Do you enjoy the thrill of spinning the reels and hoping for a big win? Well, then you’re in luck because we have something special for you – 82 Slots. This online platform is your ultimate destination for all your slot game needs. With 82 Slots, you can embark on a journey of endless exploration and excitement as you discover a wide variety of slot games.

But what exactly is 82 Slots? How does it stand out from other online casinos? And why should you choose it as your go-to destination for slot games? In this blog post, we will answer all these questions and more as we dive into the world of 82 Slots.

Unveiling the Thrill: Exploring the Diverse World of 82 Slots

At its core, 82 Slots is an online casino that offers a wide range of slot games. But it’s much more than that. It’s a virtual paradise for anyone who loves the adrenaline rush of spinning the reels and watching the symbols align to form winning combinations.

What sets 82 Slots apart from other online casinos is its diverse selection of slot games. From classic fruit machines to modern video slots, they have it all. In fact, the name “82 Slots” itself is derived from the number of slot games offered on the platform. Yes, you read that right – there are 82 different slot games for you to choose from.

But it’s not just about quantity, it’s also about quality. 82 Slots partners with some of the top game developers in the industry to bring you the best possible gaming experience. You can find titles from popular providers such as NetEnt, Microgaming, and Play’n GO, among others.

One of the things that make 82 Slots stand out is its easy-to-use interface. Whether you are a seasoned player or a newbie, you will have no trouble navigating through the website and finding your favorite games. You can easily search for specific titles or browse through different categories to discover new ones.

See more: 82lottery
82 Slots: A Comprehensive Guide to Slot Game Variety

As mentioned earlier, 82 Slots offers a diverse range of slot games, from classic to modern. Let’s take a closer look at the different types of slot games you can find on this platform:

Classic Slots

For those who enjoy the old-school charm of slot machines, 82 Slots has a selection of classic slots to choose from. These games typically have three reels and feature traditional symbols like fruits and lucky sevens. They also have a nostalgic feel to them, making them a popular choice among players.

Some popular classic slot titles on 82 Slots include “Mystery Joker” by Play’n GO and “Fruit Spin” by NetEnt. These games offer simple gameplay and are perfect for beginners who want to get a taste of the slot game world.

Video Slots

Video slots are the most common type of slot games found in online casinos, and 82 Slots is no exception. These games have five or more reels and often come with advanced features like bonus rounds, free spins, and multipliers.

What sets video slots apart is their immersive graphics and sound effects. They come in various themes, from ancient civilizations to movie-inspired games, providing a unique experience with each title.

82 Slots has a vast collection of video slots, including popular titles like “Gonzo’s Quest” by NetEnt and “Book of Dead” by Play’n GO. These games offer exciting gameplay with the potential for big wins.

Progressive Jackpot Slots

If you’re feeling lucky and want to go for the big bucks, then progressive jackpot slots are the way to go. These games have a constantly increasing prize pool that can reach millions of dollars, making them a favorite among high rollers.

On 82 Slots, you can find progressive jackpot slots from top providers like Microgaming and NetEnt. Some popular titles include “Mega Moolah” and “Mega Fortune,” which have both made millionaires out of lucky players.

Branded Slots

For fans of movies, TV shows, or bands, 82 Slots has a selection of branded slots that feature popular themes and characters. These games are designed to give players an immersive experience with their favorite franchises.

Some of the branded slots available on 82 Slots include “Game of Thrones” by Microgaming and “Jimi Hendrix” by NetEnt. These games not only offer great graphics and sound effects but also have special features that are unique to the brand, making the gameplay even more exciting.

Embark on an Adventure: Discovering Hidden Gems in 82 Slots

With 82 different slot games to choose from, you might be wondering where to start. Well, let’s take a look at some hidden gems that you can find on 82 Slots:

“Vikings Go Berzerk” by Yggdrasil

This video slot takes you on a journey with fearless vikings as they battle sea monsters and search for treasure. With stunning graphics and an epic soundtrack, this game will keep you on the edge of your seat. And if you’re lucky, you might trigger the “Berzerk Mode” and get some extra wilds and free spins.

“Esqueleto Explosivo” by Thunderkick

If you’re looking for something unique, then this is the game for you. Set in a Mexican party atmosphere, this slot features singing skeleton heads that explode and make way for new symbols on the reels. With a catchy soundtrack and innovative gameplay, it’s no wonder that this game has become a fan favorite.

“Lucha Legends” by Microgaming

This slot game is all about Mexican lucha libre wrestling. With colorful characters and an upbeat soundtrack, it’s a fun and entertaining game to play. And with bonus features like wild symbols, multipliers, and free spins, you could walk away with some big wins.

82 Slots: A Treasure Trove of Slot Game Experiences

One of the things that make slot games so appealing is the variety of experiences they offer. From exciting themes and graphics to unique bonus features and big wins, each game has its own charm. And at 82 Slots, you can find all of these experiences in one place.

Thrills and Excitement

The thrill of spinning the reels and waiting for the outcome is what makes slot games so addictive. And 82 Slots offers plenty of opportunities for this kind of excitement. Whether you’re playing a classic slot or a modern video slot, every spin brings new possibilities and the potential for big wins.

In addition to that, many of the slot games on 82 Slots come with special features like free spins, multipliers, and bonus rounds, which add an extra layer of excitement to the gameplay. These features not only make the game more engaging but also increase your chances of winning.

Immersive Gameplay

With advanced technology and graphics, slot games have evolved from simple machines to immersive experiences. And 82 Slots is no exception. Each game on the platform has been carefully designed to provide players with a unique and engaging experience.

Whether it’s the graphics, sound effects, or storyline, every aspect of a slot game on 82 Slots is crafted to provide a fully immersive experience. So, get ready to be transported to different worlds and embark on exciting adventures with each spin.

Opportunity for Big Wins

Let’s be honest, we all play slot games for the chance to win big. And at 82 Slots, that chance is always present. With a wide variety of slot games to choose from, each with its own payout percentage and volatility, you can find the perfect game that suits your playing style and offers the potential for big wins.

Moreover, many of the slots on 82 Slots come with progressive jackpots, which means you have the chance to win life-changing sums of money with just one spin. And who knows, you could be the next lucky player to hit the jackpot.Spin, Win, and Explore: The Excitement of 82 Slots

Now that we’ve given you a glimpse into the world of 82 Slots, it’s time for you to experience it for yourself. So, grab your device, log in to the website, and get ready to spin, win, and explore the diverse world of slot games.

But before you do that, here are some things to keep in mind for an optimal gaming experience:

See more: 82lottery
Get Familiar with the Rules

Before you start playing a slot game, make sure you understand the rules and paytable. This will give you a better understanding of what to expect and how to increase your chances of winning.

To access the rules and paytable, simply click on the “i” button usually located in the game’s main menu or on the bottom left corner of the screen.

Set a Budget

It’s important to set a budget before you start playing. This will help you manage your funds and avoid overspending. Decide on the amount you’re willing to spend and stick to it. Remember, slot games are meant to be fun, so don’t chase losses and know when to stop.

Take Advantage of Bonuses and Promotions

One of the advantages of playing on 82 Slots is the various bonuses and promotions they offer. These can include free spins, cashback offers, and deposit bonuses. Keep an eye out for these promotions as they can give you extra playing time and increase your chances of winning.

Just make sure to read the terms and conditions of each bonus before accepting it. This will ensure that there are no surprises when it comes to wagering requirements or withdrawal limits.

From Classic to Modern: The Evolution of Slot Games in 82 Slots

Slot games have come a long way since their inception in the late 19th century. From mechanical machines with just a few symbols to advanced video slots with intricate graphics and special features, the evolution of slot games is truly fascinating.

And at 82 Slots, you can experience this evolution firsthand. With a wide selection of both classic and modern slot games, you can see how far the industry has come and appreciate the different styles and gameplay of each era.

Unlocking the Secrets: Tips and Strategies for 82 Slots

While slot games are mostly based on luck, there are a few tips and strategies that can help increase your chances of winning. Here are some things to keep in mind when playing on 82 Slots:

Bet Max (When Applicable)

Some slot games have a “Bet Max” button, which allows you to bet the maximum number of coins per spin. If you’re aiming to hit the jackpot, then this is a good option to consider. However, keep in mind that this strategy is only applicable if you have enough funds to support it.

Play Free Demo Versions

Most slot games on 82 Slots offer a free demo version that you can play without placing any bets. This is a great way to get familiar with the game’s mechanics and rules without risking any money. Plus, it’s also a good opportunity to test out different betting strategies and see which one works best for you.

Know When to Stop

As mentioned earlier, setting a budget is essential when playing slot games. But it’s equally important to know when to stop. It can be tempting to chase losses or continue playing after a big win, but it’s crucial to stick to your budget and not let emotions dictate your decisions.

82 Slots: Where Every Spin is an Opportunity

One of the things that make slot games so appealing is the fact that every spin brings new possibilities. And at 82 Slots, every spin is an opportunity to win big, explore new worlds, and have an exhilarating experience.

So, whether you’re a seasoned player looking for new and exciting titles or a newbie wanting to try out different slot game styles, 82 Slots is the perfect destination for all your slot gaming needs.

The Ultimate Slot Game Destination: 82 Slots

In conclusion, we can confidently say that 82 Slots is your ultimate destination for slot games. With its diverse selection of titles, immersive gameplay, and potential for big wins, it offers everything a slot game enthusiast could ask for.

So, what are you waiting for? Head over to 82 Slots now and embark on a thrilling journey of slot game exploration. Who knows, you might just hit the jackpot!

Slot games have long been a favorite pastime for many casino enthusiasts. With the rise of online casinos, players now have access to a vast array of slot games right at their fingertips. Among the plethora of online platforms, 82 Slots stands out as a premier destination for slot game exploration. Let’s delve into what makes 82 Slots a top choice for both seasoned players and newcomers alike.

Unveiling the Thrill: Exploring the Diverse World of 82 Slots

At 82 Slots, players are greeted with a diverse selection of slot games that cater to every preference. Whether you enjoy classic fruit machines, adventure-themed slots, or progressive jackpot games, you’ll find it all in one convenient location. The platform boasts a user-friendly interface that allows for easy navigation, making it simple to browse through the extensive collection of games on offer.

With themes ranging from ancient civilizations to futuristic worlds, each slot game tells a unique story and offers an immersive experience that keeps players coming back for more. Additionally, 82 Slots regularly updates its game library, ensuring that players always have access to the latest titles and innovations in the world of online slots.

Featured Slot Categories on 82 Slots:

• Classic Slots
• Video Slots
• Progressive Jackpot Slots
• Megaways Slots
• Branded Slots

Benefits of Playing on 82 Slots:

1. Wide variety of slot games to choose from.
2. User-friendly interface for seamless navigation.
3. Regular updates with new and exciting titles.
4. Diverse themes and gameplay styles to suit every player.

82 Slots: A Comprehensive Guide to Slot Game Variety

One of the standout features of 82 Slots is its comprehensive guide to slot game variety. Whether you’re a novice player looking to learn the ropes or a seasoned pro seeking new challenges, this platform has something for everyone. The site provides detailed information on each game, including its theme, features, and payout potential, allowing players to make informed decisions before diving into the action.

Moreover, 82 Slots offers valuable insights into the different types of slot games available, such as classic three-reel slots, multi-payline video slots, and high volatility games. By understanding the nuances of each game category, players can tailor their gaming experience to suit their preferences and maximize their chances of winning big.

Slot Game Variety Explained:

• Classic Slots: Simple gameplay with traditional symbols.
• Video Slots: Enhanced graphics, animations, and bonus features.
• Progressive Jackpot Slots: Accumulative prize pools for massive wins.
• High Volatility Slots: Higher risk but potential for larger payouts.
• Megaways Slots: Dynamic reels and thousands of ways to win.

Advantages of Exploring Slot Game Variety:

1. Discovering new gameplay styles and features.
2. Tailoring your gaming experience to suit your preferences.
3. Maximizing winning potential by choosing the right type of slot game.
4. Expanding your knowledge and skills in the world of online slots.

Embark on an Adventure: Discovering Hidden Gems in 82 Slots

While popular titles often take the spotlight, 82 Slots also offers players the opportunity to discover hidden gems and lesser-known slot games that deserve attention. These hidden treasures may not have the same level of recognition as blockbuster titles, but they often come with unique gameplay mechanics, innovative features, and hidden surprises waiting to be uncovered.

Exploring these hidden gems on 82 Slots can be a rewarding experience for players who enjoy stepping off the beaten path and trying something new. Whether it’s a quirky theme, a novel bonus round, or a distinctive reel layout, these hidden gems add an element of excitement and discovery to the slot game exploration journey.

Tips for Discovering Hidden Gems:

• Browse through the “New Releases” section for fresh titles.
• Explore categories beyond your usual preferences.
• Read player reviews and recommendations for underrated games.
• Take advantage of demo versions to test out unfamiliar titles.

Benefits of Finding Hidden Gems:

1. Uncovering unique gameplay experiences.
2. Enjoying innovative features and mechanics.
3. Adding variety to your slot game rotation.
4. Potentially discovering new favorite games.

See more: 82 lottery
82 Slots: A Treasure Trove of Slot Game Experiences

As players navigate the virtual halls of 82 Slots, they are met with a treasure trove of slot game experiences waiting to be explored. From the adrenaline rush of spinning the reels to the thrill of triggering bonus rounds and unlocking special features, each game offers a unique adventure that keeps players engaged and entertained for hours on end.

Whether you prefer the nostalgia of classic slots or the cutting-edge technology of modern video slots, 82 Slots caters to all tastes and preferences. The platform’s commitment to providing a diverse range of gaming experiences ensures that every player can find their perfect match and embark on a memorable journey filled with excitement and anticipation.

Notable Features of Slot Game Experiences on 82 Slots:

• Engaging gameplay mechanics that keep players entertained.
• Immersive themes and storylines that enhance the gaming experience.
• Interactive bonus rounds and special features for added excitement.
• Varied volatility levels to suit different playing styles.

Rewards of Exploring Slot Game Experiences:

1. Immersive and entertaining gameplay sessions.
2. Opportunities to trigger big wins and lucrative bonus rounds.
3. Engagement with diverse themes and storylines.
4. Flexibility to choose games that align with personal preferences.

Conclusion

In conclusion, 82 Slots emerges as a premier destination for slot game enthusiasts seeking a comprehensive and thrilling gaming experience. With its vast selection of games, informative guides, and hidden gems waiting to be discovered, the platform offers something for every player, whether they’re looking for familiar favorites or new adventures.

By exploring the diverse world of slot games on 82 Slots, players can unlock a treasure trove of experiences, from classic reels to modern marvels, and everything in between. So, why wait? Dive into the excitement of 82 Slots today and embark on a journey filled with spins, wins, and endless possibilities.

Betting can be a thrilling and rewarding experience, but it’s crucial to approach it with a strategic mindset. Whether you’re a seasoned gambler or just starting out, this handbook will equip you with the knowledge and tools you need to make informed and confident wagers. By understanding the fundamentals of betting, analyzing data, and managing your bankroll, you can increase your chances of success and enjoy the excitement of the game at 82lottery.

Understanding the Basics of Betting

The Power of Probability

• Probability is the foundation of betting. Understanding the likelihood of an outcome occurring is essential for making informed decisions.
• Learn how to calculate probability and use it to evaluate the risk and potential rewards of a bet.
• Explore the concept of expected value and how it can guide your betting strategy.

Deciphering Odds and Payouts

• Odds are the backbone of the betting industry, and understanding them is crucial for maximizing your winnings.
• Familiarize yourself with different types of odds, such as fractional, decimal, and American odds, and how to convert between them.
• Understand the relationship between odds and payouts, and learn how to identify value bets.

The House Edge and Its Implications

• The house edge is the advantage the bookmaker or casino has over the bettor, and it’s essential to understand its impact on your betting.
• Explore the different house edges across various betting markets and learn how to minimize the impact of the house edge on your overall strategy.
• Discover strategies for identifying and taking advantage of favorable house edge scenarios.

Gathering and Analyzing Data

Importance of Research and Data Collection

• Successful betting requires a deep understanding of the events, teams, or individuals you’re betting on.
• Learn how to gather and organize relevant data, such as performance statistics, historical trends, and expert analysis.
• Develop a systematic approach to data collection and storage to aid your decision-making process.

Advanced Data Analysis Techniques

• Go beyond basic statistics and delve into more sophisticated data analysis methods.
• Explore the use of regression analysis, machine learning, and other data-driven tools to identify patterns and trends.
• Understand how to interpret complex data visualizations and use them to inform your betting decisions.

Leveraging Online Resources and Tools

• Utilize online platforms, forums, and specialized betting websites to access a wealth of information and analytical tools.
• Discover how to effectively use statistical models, betting calculators, and expert predictions to enhance your decision-making.
• Stay up-to-date with the latest news, injuries, and other factors that can impact the outcome of events.

See more: 82 lottery

Developing a Winning Mindset

Bankroll Management Strategies

• Effective bankroll management is crucial for long-term success in betting.
• Learn how to set realistic betting limits, manage your risk, and optimize your stake sizes.
• Explore strategies for building and protecting your bankroll, including the use of diversification and hedging.

Emotional Control and Discipline

• Betting can be an emotional rollercoaster, and it’s essential to maintain a cool, rational approach.
• Understand the importance of controlling your emotions, such as fear, greed, and overconfidence, and develop techniques to manage them.
• Cultivate discipline in your betting habits, including setting clear goals, sticking to your strategy, and avoiding impulsive decisions.

Embracing a Growth Mindset

• Successful bettors are constantly learning and adapting to the ever-changing landscape of the betting industry.
• Adopt a growth mindset, where you are open to feedback, willing to experiment, and committed to continuous improvement.
• Explore ways to enhance your knowledge, such as seeking out mentors, attending industry events, and staying informed about the latest trends and developments.

Navigating Specialized Betting Markets

Sports Betting Strategies

• Dive into the world of sports betting, exploring the nuances of different sports, leagues, and events.
• Discover strategies for analyzing team and player performance, identifying value bets, and managing risk in sports betting.
• Learn how to leverage advanced metrics, injury reports, and expert analysis to gain an edge in sports betting.

Wagering on Esports and Virtual Sports

• The rise of esports and virtual sports has created new and exciting betting opportunities.
• Understand the unique characteristics of these markets, including the importance of player and team dynamics, game mechanics, and industry trends.
• Develop strategies for researching and analyzing esports and virtual sports data to make informed wagers.

Exploring Alternative Betting Markets

• Beyond traditional sports and esports, there are many other betting markets to explore, such as politics, entertainment, and current events.
• Learn how to identify and analyze the factors that influence these alternative markets, and develop strategies for capitalizing on them.
• Understand the potential risks and rewards associated with diversifying your betting portfolio across different markets.

Navigating the Regulatory Landscape

Understanding Licensing and Regulations

• Betting is a highly regulated industry, and it’s essential to familiarize yourself with the relevant laws and regulations in your jurisdiction.
• Explore the licensing requirements for bookmakers and online betting platforms, and understand the importance of operating within the legal framework.
• Stay informed about changes in regulations and be prepared to adapt your betting strategy accordingly.

Responsible Gambling Practices

• Responsible gambling is crucial for maintaining a healthy and sustainable betting lifestyle.
• Learn about the potential risks associated with problem gambling and develop strategies for identifying and addressing them.
• Discover resources and support services available to help you manage your gambling activities responsibly.

Ethical Considerations in Betting

• Betting should be approached with integrity and ethical principles in mind.
• Understand the importance of maintaining honesty, transparency, and fairness in your betting activities.
• Explore the ethical implications of issues such as match-fixing, insider information, and the use of technology in betting.

FAQs

What is the most important factor in successful betting?

The most important factor in successful betting is a deep understanding of the market you’re betting on. This involves thoroughly researching the relevant data, analyzing trends, and developing a well-informed strategy.

How can I manage my bankroll effectively?

Effective bankroll management is crucial for long-term success in betting. Some key strategies include setting realistic betting limits, diversifying your portfolio, and adjusting your stake sizes based on the risk and potential rewards of each bet.

What are some common mistakes that bettors should avoid?

Some common mistakes that bettors should avoid include chasing losses, making impulsive decisions, overestimating their knowledge or abilities, and failing to manage their emotions. It’s important to maintain discipline, stick to a well-tested strategy, and avoid letting emotions cloud your judgment.

How can I stay up-to-date with the latest trends and developments in the betting industry?

To stay informed about the latest trends and developments in the betting industry, it’s important to regularly consume content from reputable sources, such as industry publications, expert blogs, and specialized betting websites. Attending industry events and networking with other bettors can also be valuable for staying up-to-date.

What are the benefits of responsible gambling practices?

Responsible gambling practices not only help you maintain a healthy and sustainable betting lifestyle but also protect the integrity of the industry as a whole. By setting limits, avoiding problematic behavior, and seeking support when necessary, you can ensure that your betting activities remain enjoyable and within your control.

Conclusion

Betting with confidence is a skill that can be developed through a combination of knowledge, discipline, and a strategic mindset. By understanding the fundamentals of betting, gathering and analyzing data, and cultivating a winning mentality, you can increase your chances of success and enjoy the thrills of the betting world. Remember to always gamble responsibly, stay informed about the latest trends and developments, and approach your betting activities with integrity and ethical principles. With the right tools and mindset, you can confidently navigate the world of betting and achieve your goals.

Tuy nhiên, anh em đã biết về nguồn gốc trò chơi tài xỉu siêu hot hit này chưa? Nếu chưa thì hãy cùng 8xbet khám phá ngay trong bài viết sau đây nhé!

Nguồn gốc của tựa game tài xỉu ở các nước

Nguồn góc của trò chơi tài xỉu ở các nước

Tại mỗi quốc gia, trò chơi tài xỉu này lại có những điểm đặc biệt vì sự khác nhau giữa văn hóa và con người ở từng khu vực. Và sau đây, 8XBet sẽ đưa đến những thông tin thú vị về sự xuất hiện của tựa game này ở hai quốc gia có số lượng người chơi tài xỉu đông nhất là Trung Quốc và Mỹ nhé!

Nguồn gốc trò chơi tài xỉu ở Trung Quốc

Tính đến nay, các giới chuyên gia vẫn chưa nghiên cứu và tìm được câu trả lời chính xác cho nguồn gốc của trò chơi tài xỉu cực kỳ nổi tiếng này.

Họ chỉ biết rằng trò chơi đã có tuổi đời vô cùng lâu, áng chừng xuất hiện từ thời Trung Hoa cổ đại nhưng cụ thể về thời gian thì chưa xác định được.

Ban đầu, trò chơi được tham gia bởi những người lính sau khi kết thúc mỗi trận chiến. Lúc này thì trò chơi chỉ mang tính giải trí thông thường và không có bất cứ sự đặt cược tiền hay vật chất nào. Tại thời điểm đó thì người chơi sẽ thường sử dụng các vật dụng như đá, gỗ, xương động vật để tiến hành khắc dấu.

Điều đặc biệt chính là trò chơi tài xỉu không chỉ đơn thuần để giải trí mà nó còn như một hình thức tập luyện cho những người lính và võ sư thời đó bởi quy tắc trong trò chơi đòi hỏi người tham gia phải có sự tập trung cao độ và sự nghiêm túc để có thể dự đoán chính xác được kết quả.

Trước khi mang cái tên tài xỉu thì trò chơi này còn được gọi với nhiều tên khác như bát xí ngầu hoặc cặp xí ngầu vì nó chỉ sử dụng đến 2 viên xúc xắc.

Vượt qua thời gian thì trò chơi tài xỉu đã trở thành một tựa game quốc dân được ưa chuộng bởi giới thượng lưu của Trung Quốc và cả những người dân thường.

Nguồn gốc trò chơi tài xỉu ở Hoa Kỳ

Tại Hoa Kỳ thì trò chơi tài xỉu đã trở thành một biểu tượng hàng đầu tại các sòng bạc trực tuyến và truyền thống. Điều đáng ngạc nhiên ở đây chính là khu vực châu Âu nổi tiếng với những tựa game bài sử dụng 52 lá như Blackjack và Poker lại có chỗ đứng cho game bài tài xỉu. Điều này chính là một minh chứng rõ ràng nhất cho sức hấp dẫn của tựa game hot hit này.

Cụ thể tài xỉu ở Mỹ bắt đầu xuất hiện khi có những người dân từ Trung Quốc nhập cư vào. Mặc dù nó được chơi từ những năm đầu của thế kỷ 20 nhưng nó chỉ thực sự được công nhận và cung cấp tại những hệ thống trò chơi casino lớn nhỏ tại đất nước này từ đầu năm 2003.

Trò chơi tài xỉu hiện nay

Tính đến ngày nay thì tài xỉu đã phổ biến ở tất cả các quốc gia và khu vực trên thế giới, đặc biệt là châu Á. Với công nghệ ngày càng phát triển hiện đại thì hệ thống chơi game được tạo ra mạnh hơn và nhanh hơn cho phép người chơi trải nghiệm tại nhà vô cùng tiện lợi. 

Có thể nói công nghệ đã khiến trò chơi này được mở rộng lãnh thổ của mình và trở thành tựa game cá cược đổi thưởng số 1 trên thị trường.

Lý do khiến trò chơi tài xỉu được yêu thích?

Sau khi đã biết về nguồn gốc trò chơi tài xỉu thì chắc hẳn anh em cũng phần nào đoán ra được lý do tại sao trò chơi lại trở nên hot đến như vậy. Cụ thể như sau:

Cách chơi đơn giản

Điểm thu hút đầu tiên chính là cách chơi cực kỳ đơn giản, dù anh em có là tân thủ hay cao thủ thì vẫn có thể dễ dàng tham gia được. Cụ thể khi chơi thì bet thủ sẽ tiến hành đặt cược cho tổng điểm của 3 mặt xúc xắc, có thể lựa chọn cửa tài hoặc cửa xỉu tùy theo phán đoán.

Cửa cược không phức tạp

Đúng như tên gọi của trò chơi thì anh em sẽ có hai lựa chọn đặt cược là tài hoặc xỉu. Điều này có thể coi như một ưu điểm đối với các tân thủ bởi họ sẽ dễ dàng đặt cược và không cần phải phân tâm như những trò chơi có nhiều cửa cược khác nhau.

Có thể kiếm nhiều tiền từ tài xỉu

Trong trò chơi tài xỉu truyền thống thì tỷ lệ trả thưởng sẽ không quá cao, nhưng đối với hình thức chơi trực tuyến tại các nhà cái ngày nay thì khác hoàn toàn.

Tại sân chơi Sicbo games, anh em sẽ được hưởng mức thưởng hấp dẫn và có thể làm giàu nhanh chóng chỉ sau một vài ván.

Đặt cược mọi lúc mọi nơi

Bên cạnh lý do dễ chơi, phần thưởng lớn thì một ưu điểm khác khiến nhiều người yêu thích trò chơi tài xỉu chính là tính tiện lợi khi có thể tham gia ở bất cứ đâu, bất cứ khi nào và với bất cứ ai. Nếu anh em là tân binh thì chỉ cần mời bạn bè của mình cùng chơi với mức cược nhỏ.

Tổng hợp FAQ về trò chơi tài xỉu đổi thưởng

Câu hỏi thường gặp về trò chơi tài xỉu

Trò chơi tài xỉu thường được tổ chức ở đâu?

Người chơi có thể tham gia tại những sòng bạc truyền thống hoặc các sân chơi cá cược tài xỉu trực tuyến với tỷ lệ trả thưởng hấp dẫn cùng với nhiều chương trình khuyến mãi như nhà cái Sicbo games.

Có nên tham gia chơi tài xỉu không?

Chắc chắn là có, bởi tài xỉu không chỉ đem đến sự giải trí tuyệt vời mà nó còn là sàn đầu tư sinh lời vô cùng hiệu quả dành cho những anh em mong muốn làm giàu nhanh chóng.

Tỷ lệ trả thưởng khi chơi game tài xỉu là bao nhiêu?

Cụ thể đối với hai cửa cược tổng điểm tài xỉu thì tỷ lệ trả thưởng sẽ là 1:1, tiền thưởng sẽ tương đương với số tiền đặt cược ban đầu.

Trong bài viết trên, 8XBet đã cùng anh em tìm hiểu về nguồn gốc trò chơi tài xỉu cùng những lý do tại sao tựa game này lại được yêu thích đến như vậy.

Nếu như anh em đang phân vân không biết đặt cược vào trò chơi nào thì tài xỉu chính là một ứng cử viên sáng giá đưa anh em đến đỉnh cao thành công.

Mua kèo bóng đá đang dần trở thành xu hướng giải trí hấp dẫn hiện nay. Bên cạnh tính giải trí, bộ môn còn giúp anh em kiếm được nguồn lợi nhuận khủng. Vậy sảnh cược này mang lại những lợi ích gì và được chơi như thế nào? Hãy cùng tìm hiểu lời giải đáp qua bài viết dưới đây của 8xbet nhé!

Tìm hiểu sơ lược mua kèo bóng đá tại nhà cái là gì

Tìm hiểu sơ lược mua kèo bóng đá tại nhà cái là gì?

Mua kèo bóng đá là hình thức giải trí thú vị cho những ai yêu thích đến bộ môn thể thao vua. Qua việc sử dụng tiền của mình để đặt cược kết quả của một trận bóng. Bao gồm các trận đang hoặc sắp được diễn ra.

Và tất nhiên khi trận đấu kết thúc, nếu kết quả đặt cược trùng với tỷ số của trận đấu. Điều này có nghĩa là bạn đã cá độ thành công và dành chiến thắng. Người tham gia may mắn sẽ được nhận thưởng dựa vào tỷ lệ mà nhà cái đã đề ra từ trước.

Tuy nhiên khi chọn tham gia loại hình này, chúng ta cần coi đây chỉ là trò chơi giải trí. Anh em tuyệt đối không nên dùng toàn bộ tài sản của mình có để tham gia cá cược nhé.

Lợi ích nổi bật khi mua kèo nhà cái tại 8xbet

Sau khi tìm hiểu khái niệm được nêu ở nội dung trên, chắc hẳn mọi người đã nắm được phần nào đó về mua kèo bóng đá. Tuy nhiên, tại sao nhà cái 8xbet lại khiến đông đảo cược thủ tham gia loại hình giải trí này mỗi ngày? Chắc hẳn là vì quý hội viên tại đây đã được tận hưởng vô vàn lợi ích như sau:

Mua kèo dễ dàng, không gặp phải rủi ro nào

Quy luật và cách thực hiện mua kèo online vô cùng đơn giản. Cược thủ chỉ cần truy cập, đăng ký tài khoản game tại nhà cái. Sau đó, nạp tiền theo khả năng nguồn vốn của mình và dự đoán kết quả trận đấu. Chỉ với vài bước cơ bản như thế là bạn đã cá cược thành công, không lo gặp bất kỳ rủi ro nào như chơi theo kiểu truyền thống.

Xem thêm: Cách đọc kèo bóng đá tài xỉu chi tiết cho tân thủ

Xem thêm: Thể thao 8xbet – Nơi tận hưởng những trận cầu kịch tính nhất

Xem thêm: Kèo Man City – Top 3 kinh nghiệm soi kèo cực chuẩn

Xem thêm: Soi kèo Việt Nam – Kinh nghiệm soi kèo hiệu quả cho tân thủ

Nguồn lợi nhuận hấp dẫn

Đặc biệt, tỷ lệ nhận thưởng của bộ môn cá cược thể thao có mức lợi nhuận vô cùng hấp dẫn. Thậm chí chúng có thể lên đến hàng trăm triệu đồng chỉ sau 1 trận đấu nhỏ. Vì thế, rất nhiều đại gia đã xuất hiện chỉ bằng cách mua kèo bóng đá thôi đấy.

Nguồn lợi nhuận hấp dẫn

Tiền thắng thưởng rút về ngân hàng nhanh chóng

Sau khi may mắn thắng kèo, bạn hoàn toàn có thể rút tiền thưởng về tài khoản ngân hàng cá nhân. Khi tham gia tại 8xbet, quy trình này diễn ra vô cùng nhanh chóng và an toàn. Không những thế, sân chơi còn liên kết với các ngân hàng lớn Việt Nam, khiến giao dịch trở nên dễ dàng và hỗ trợ hội viên kịp thời khi gặp vấn đề.  

Hướng dẫn mua kèo bóng đá tại 8xbet

Mua kèo có thể tiến hành mọi lúc mọi nơi, chỉ cần thông qua điện thoại hoặc máy tính cá nhân. Người chơ tiến hành cá độ trên website hoặc bằng các app điện thoại của nhà cái 8xbet. Cụ thể như sau

• Bước 1: Truy cập vào website nhà cái 8xbet chính thống
• Bước 2: Bắt đầu tạo tài khoản cá cược bằng cách bấm vào mục “Đăng ký”. Sau đó, cung cấp tất tần tật thông tin theo yêu cầu của sân chơi trực tuyến.
• Bước 3: Nạp tiền để bắt đầu tham gia mua kèo, anh em cần chuyển tối thiểu 100,000 VND. Bằng các phương thức như Internet Banking, ví điện tử, thẻ cào điện thoại,…
• Bước 4: Sau khi đã hoàn thành các bước bên trên, cược thủ cần nhấn vào mục “Thể thao”. Tìm kiếm và click ô “Bóng đá” để chọn những trận đấu hợp lý và bắt đầu đặt kèo.

Hướng dẫn mua kèo bóng đá tại 8xbet

Một số kinh nghiệm khi mua kèo bóng đá

Một trong các yếu tố hàng đầu giúp mua kèo cá cược bóng đá chuẩn xác là kinh nghiệm. Sau đây, 8xbet sẽ chia sẻ cho các bạn một vài bí kíp đặt cược quan trọng. Đây đều là những điều được tổng hợp từ cao thủ chuyên nghiệp, chắc chắn sẽ giúp bạn có trận đấu thắng lợi. Cụ thể như sau:

• Liên tục theo dõi và cập nhật liên tục những thông tin liên quan đến các trận bóng đá. Đây là cơ sở quan trọng giúp anh em đánh giá sơ bộ về kết quả sắp tới. Đặc biệt khi tham gia giải trí bộ môn này, bạn không được đặt cược theo cảm tính.
• Phân tích, đánh giá từ các biến động của tỷ lệ kèo nhà cái. Từ đó, đưa ra những nhận định đúng đắn về trận đấu. Kết hợp thêm những kinh nghiệm từ các người chơi trên các diễn đàn, cộng đồng. 
• Nắm được cách quản lý khoa học số tiền vốn của mình. Cụ thể, biết dừng đúng lúc và không nên đặt dược theo đám đông. Bởi đây là những lý do quan trọng giúp anh em dễ rơi vào trường hợp trắng tay.

Một số kinh nghiệm khi mua kèo bóng đá

Kết luận

Bài viết trên đã hướng dẫn chi tiết cách mua kèo bóng đá tại 8xbet đơn giản nhất. Mặc dù đây chỉ là trò chơi giải trí, nhưng chúng vẫn mang lại cho bạn một nguồn thu nhập cực cao. Hy vọng anh em sẽ có thêm kinh nghiệm để sở hữu kết quả tốt nhất từ bộ môn này.

Nếu bạn tham gia vào cược bóng đá thì nên dành thời gian để nghiên cứu về các tỷ lệ kèo mà bạn sẽ đặt cược. Bạn chỉ cần hiểu rõ các thông tin này sẽ có cơ hội nhận được tiền thưởng siêu khủng. Nếu bạn chưa hiểu rõ về các thể loại kèo bóng đá, hãy cùng 8xbet tham khảo qua các thông tin bên dưới nhé.

Sơ lược cược bóng đá là gì?

Sơ lược cược bóng đá là gì?

Cược bóng đá là việc người chơi sẽ đặt cược vào 1 tỷ lệ kèo nhất định để dự đoán tỷ số cho các trận đấu sắp diễn ra. Phía nhà cái sẽ đưa ra các tỷ lệ cho người chơi chọn lựa với nhiều loại kèo đa dạng khác nhau. Người chơi nếu chiến thắng sẽ nhận được tiền thưởng, trường hợp thua cược thì bạn phải chịu thua hết tiền cược. 

Người chơi khi tham gia vào cá cược vào bóng đá thì 3 kèo đấu phổ biến nhất hiện nay, trong đó bao gồm kèo Châu Á, kèo Châu Âu và kèo tài xỉu. Đây là 3 loại kèo quen thuộc mà không ai là không biết khi tham gia chọn tỷ lệ kèo bóng đá. Để hiểu rõ hơn luật chơi và thể thức ăn cược của 3 loại kèo này thì người chơi cần phải nghiên cứu thật kỹ.

Các loại kèo cược bóng đá 8xbet phổ biến nhất hiện nay

Các loại kèo cược bóng đá 8xbet phổ biến nhất hiện nay

Trong cá cược kèo bóng đá thường sẽ có rất nhiều thể loại khác nhau để người chơi chọn lựa dễ dàng. Bạn nên tham khảo qua các loại kèo sau đây để có thể chọn lựa chính xác và dễ dàng nhận thưởng hơn.

Kèo Châu Á

Kèo châu Á là loại kèo chấp khá đa dạng với cách chơi đơn giản nên được nhiều người ưa chuộng hiện nay. Luật chơi của kèo châu Á được chia thành 2 đội là đội cửa trên và đội cửa dưới để người chơi chọn lựa đặt cược. 

Đội cửa trên là đội mạnh sở hữu khả năng thi đấu cao hơn so với đội cửa dưới. Họ sẽ chấp cửa dưới 1 bàn thắng hoặc nhiều hơn trong một số trường hợp nhất định. Trước khi đặt cược vào kèo châu Á thì người chơi cần có thời gian nghiên cứu thật kỹ về 2 đội bóng sẽ thi đấu.

Người chơi sẽ thực hiện đặt cược vào đội cửa trên hoặc đội cửa dưới để tìm tiền thưởng. Tùy thuộc vào số cược chấp ở kèo châu Á mà sau khi trận đấu kết thúc thì người chơi có thể nhận thưởng hoặc mất hết số tiền đã đặt cược. Tỷ lệ chiến thắng của kèo châu Á được nhà cái cập nhật cụ thể trước khi trận đấu bắt đầu. Người chơi sẽ nhận tiền thưởng tương ứng với tỷ lệ mà nhà cái đã đưa ra. Nếu thua cược thì bạn phải chịu thua hết số tiền đã đặt cược trước đó trên hệ thống.

Xem thêm: 8xbet đăng nhập

Kèo bóng đồng banh

Kèo đồng banh là loại kèo chấp trong các trận đấu mà 2 đội được nhận định ngang ngửa nhau về thực lực thi đấu. Điều này đồng nghĩa số bàn chấp bằng 0, người chơi có thể đặt cược vào bất kỳ đội bóng nào mà bạn nghĩ họ có khả năng chiến thắng đều được. Trong tình huống 2 đội bóng hòa nhau thì người chơi sẽ được hoàn trả lại số tiền đã đặt cược trước đó.

Kèo bóng chấp 0.25

Đây là loại kèo chấp mà đội mạnh hơn sẽ thực hiện cấp đổi yếu hơn 0.25 bàn thắng khi thi đấu. Trong trường hợp này thì người chơi sẽ đặt cược kèo trên nếu đội chấp giành được chiến thắng. Tình huống 2 đội hòa nhau thì bạn sẽ mất nửa tiền thưởng và người đặt cược đội yếu sẽ giành được tiền thưởng.

Kèo bóng đá chấp 0.5

Trong kèo chấp 0.5 thì tương ứng đội cửa trên sẽ chấp đội cửa dưới 0.5 bàn thắng. Người chơi nếu đặt cược dự đoán đội cửa trên giành được chiến thắng. Trường hợp kết quả giống bạn dự đoán thì người chơi sẽ giành được tiền thưởng. Trong trường hợp kết quả không giống với bạn dự đoán thì bạn sẽ thua hết tiền cược trước đó.

Kèo bóng đá chấp 0.75

Kèo bóng đá chấp 0.75

Loại kèo này thì đội cửa trên sẽ chấp cửa dưới 0.75 bàn thắng trước khi trận đấu bắt đầu. Người chơi nếu đặt cược cửa trên sẽ ăn toàn bộ tiền cược khi đội bóng này thắng khoảng 2 bàn thắng trở lên. Bạn sẽ thua 1 nửa tiền cửa khi kết quả của trận đấu là 1 bàn và thua hết tiền cược nếu không có bàn thắng nào.

Hướng dẫn người chơi tham gia cá cược bóng đá tại 8xbet

Hướng dẫn người chơi tham gia cá cược bóng đá tại 8xbet

Người chơi muốn tham gia đặt cược kèo bóng đá tại 8xbet thì bạn phải thực hiện chính xác 3 bước sau đây, cụ thể là:

• Bước 1: Người chơi sẽ đăng nhập vào tài khoản chơi game đã đăng ký trước đó tại 8xbet. Nếu chưa có tài khoản chơi game thì bạn phải thực hiện đăng ký lại từ đầu để có tài khoản đăng nhập.
• Bước 2: Người chơi vào hệ thống trò chơi và ấn chọn vào mục thể thao để tìm chọn tỷ kèo mà bạn muốn đặt cược.
• Bước 3: Bạn sẽ chọn kèo và đưa ra mức cược cụ thể, khi có kết quả hệ thống sẽ tiến hành trả thưởng cho người chơi.

Tóm lại, với các thông tin bên trên thì người chơi cũng đã có thêm kiến thức về cược bóng đá phổ biến hiện nay. Bạn nên xem qua cái tỷ lệ kèo bên trên để hiểu rõ trước khi tham gia đặt cược nhé.

Tien Len, also known as Thirteen or Vietnamese Poker, is a popular card game that originated in Vietnam. It has gained widespread popularity not just in Vietnam but also in other parts of the world, including the United States, where it is often referred to as “Chinese Poker.” Tien Len is a game of strategy, skill, and luck, making it both challenging and enjoyable to play.

If you are a fan of Tien Len and want to improve your game, you have come to the right place. In this article, we will reveal six good tips for playing Tien Len at 82lottery. These tips will help you understand the game better, develop your skills, and increase your chances of winning. So, let’s get started!

Tip #1: Familiarize Yourself with the Rules of Tien Len

Before diving into any game, it is essential to familiarize yourself with the rules. Tien Len follows a standard deck of 52 cards, with each player dealt 13 cards. The objective of the game is to be the first player to get rid of all your cards by creating sets and runs. Sets consist of three or four cards of the same rank, while runs consist of three or more consecutive cards of the same suit.

The game begins with the player holding the Three of Spades starting the round. The next player must then play a higher set or run of the same number of cards, or pass their turn if they do not have a higher combination. If all players pass, the player who played the last set or run can start a new round with any combination they choose. The game continues until one player gets rid of all their cards, and they are declared the winner.

Tips for Understanding the Rules

• Read through the rules carefully before playing.
• Start with a basic understanding of the game, and then move on to advanced strategies.
• Play a few practice rounds with friends or online before playing for money.

Tip #2: Observe Your Opponents’ Playing Style

Observing your opponents’ playing style is crucial in Tien Len. Understanding how they play their cards can give you an advantage by predicting their next move. Some players are aggressive and tend to play high-ranking cards, while others may be more conservative, holding onto their lower-ranked cards until the end.

Tips for Observing Your Opponents

• Pay attention to the cards your opponents play.
• Notice any patterns in their playing style.
• Look for any tells, such as facial expressions or gestures, that may indicate their next move.

Tip #3: Practice Strategic Card Management

In Tien Len, card management is crucial for success. You must know when to play your high-ranked cards and when to hold onto them. It is essential to keep track of which cards have been played and which are still in the game. This knowledge will help you make strategic decisions on which combinations to play and when.

Tips for Strategic Card Management

• Prioritize playing your sets and runs with higher-ranking cards first.
• Hold onto your low-ranked cards until the end of the round.
• Keep track of which cards have already been played to determine which ones are still in the game.

Tip #4: Learn How to Counter Your Opponents’ Moves

Being able to counter your opponents’ moves is a crucial skill in Tien Len. If you notice an opponent always playing high-ranked cards, you can use this to your advantage by saving your own higher-ranked cards to counter their moves. Similarly, if an opponent tends to hold onto their low-ranked cards until the end, you can play your own low-ranked cards early on to force them to use their higher-ranked cards.

Tips for Countering Your Opponents’ Moves

• Pay attention to your opponents’ playing style.
• Use this knowledge to predict their moves and counter them.
• Don’t be afraid to play higher-ranked cards early on to throw your opponents off.

Tip #5: Understand the Importance of Communication in Tien Len

Communication is an essential aspect of Tien Len, especially when playing with a partner. Strategically communicating with your partner can help you work together to defeat your opponents. It is also essential to pay attention to your opponents’ communication to understand their strategies and make informed decisions.

Tips for Effective Communication

• Communicate clearly and concisely with your partner.
• Develop a system of signals or code words to communicate your next move.
• Pay attention to your opponents’ communication to gain insight into their strategies.

Tip #6: Practice, Practice, Practice!

As with any game, practice makes perfect. The more you play Tien Len, the more you will understand the strategies and develop your skills. Don’t be discouraged by losses; instead, use them as learning opportunities to improve your game.

Tips for Practicing Tien Len

• Play regularly with friends or online.
• Analyze your games and identify areas for improvement.
• Learn from your mistakes and try different strategies.

FAQs

Q: Is it necessary to have a partner to play Tien Len?

A: No, you can also play Tien Len individually against other players.

Q: Can I play Tien Len with a deck of cards missing some cards?

A: No, Tien Len requires a full deck of 52 cards to play.

Q: Are there any variations of Tien Len worth trying?

A: Yes, there are variations such as Thirteen Cards or Big Two that add new elements to the game.

Q: Can I win consistently at Tien Len?

A: While luck plays a significant role in Tien Len, understanding and implementing effective strategies can increase your chances of winning.

Q: Is Tien Len only popular in Vietnam?

A: No, Tien Len has gained popularity in many other countries, including the United States.

See more: 82lottery login

Conclusion

Tien Len is a challenging and enjoyable game that requires strategy, skill, and a bit of luck. By following these six good tips for playing Tien Len at 82lottery, you can improve your game and increase your chances of winning. Remember to familiarize yourself with the rules, observe your opponents’ playing style, practice strategic card management, learn how to counter your opponents’ moves, understand the importance of communication, and most importantly, practice regularly. So gather your friends and get ready to play Tien Len like a pro!