1st PUC

Students can Download Basic Maths Chapter 16 Locus and its Equations Questions and Answers, Notes Pdf, 1st PUC Basic Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 1st PUC Basic Maths Question Bank Chapter 16 Locus and its Equations

Question 1.
Find the equation of the locus of a point which moves such that its distance from the point (-5,7) is twice its distance from (-3,1).
Answer:
Let A = (-5, 7) and B = (-3, 1) and P(y, b) be any point of the locus
By data PA = 2PB = PA² = 4PB²
⇒ (x + 5)⁵ + (y – 7)²
= 4[(x + 3)² + (y – 1)²]
⇒ x² + 10x + 25 + y² + 49 – 14y
= 4(x² + 9 + 6x + y² + 1 – 2y)
= 4x² + 36 + 24x + 4y² + 4 – 8y
= 3x² + 3y² + 14x + 6y – 34 = 0
which is the equation as to the locus.

Question 2.
Find the equation to the locus of the perpendicular bisector of the line joining A(3, 2) and B(4,1).
Answer:
The perpendicular bisector of the line joining A and B is the locus of the point which moves such that it is equidistant from A and B.
By data we have A = (3, -2), B = (4, 1)
Let P(x, y) be any point on the perpendicular bisector
Thus we have PA = PB = PA² = PB²
=> (x – 3)² + (y + 2)² = (x – 4)² + (y – 1 )²

⇒ -6x + 4y + 13
= -8x – 2y + 17
⇒ 2x + 6y – 4 = 0.
⇒ x + 3y – 2 = 0 is the equation of the locus of a point.

Question 3.
Find the equation of the locus of the point which moves such that the sum of its distances from (0, 5) and (0, -4) is 10 units.
Answer:
Let A = (0, 6) and B = (0, -6).
Let P(x, y) be the locus of the point; By data PA + PB = 10
⇒ \(\sqrt{x^{2}+(y-6)^{2}}+\sqrt{x^{2}+(y+6)^{2}}\) = 0
⇒ \(\sqrt{x^{2}+(y-6)^{2}}\) = 10 – \(\sqrt{x^{2}+(y+6)^{2}}\)
Squaring both sides


⇒ 10y – 12y – 100 = -20\(\sqrt{x^{2}+(y+6)^{2}}\)
⇒ -22y + – 100 = -20\(\sqrt{x^{2}+(y+6)^{2}}\) + by – 2
⇒ 11y + 50 = 10\(\sqrt{x^{2}+(y+6)^{2}}\)
squaring again we get
⇒ 121y² + 2500 + 110y
= 100(x² + y² + 36 + 12y)
= 100x² + 10y² + 3600 + 122y
⇒ 100x² – 21y² + 100y + 1100 = 0 is the locus of the equation.

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Karnataka 1st PUC Basic Maths Question Bank Chapter 15 Co-ordinate System in a plane

Question 1.
Find the reflection of (-4, 3) through x-axis, y-axis and through the centre.
Answer:
Reflection through x – axis → (-4, – 3)
Reflection through y – axis → (4,3)
Reflection through center → (4, -3)

Question 2.
Find ‘a’ if the distance between the points (a, 2) and (3, 4) is V8 units.
Answer:
Let P = (a, 2) and Q = (3, 4)
By data |PQ| = √8
⇒ √8 = \(\sqrt{(3-a)^{2}+(4-2)^{2}}\)
⇒ 8 = (3 – a)² + 2² = 8
⇒ 9 + a² – 6a + 4
⇒ a² – 6a + 5 = 0
⇒ (a – 5)(a – 1) = 0
⇒ a = 5 or a = 1

Question 3.
Show that the points (2, 2) (6, 3) and (4,11) form a right angled triangle.
Answer:
Let A = (2, 2), B = (6, 3), C = (4, 1)
Consider

By pythagorus theorem AC² = AB² + BC²

⇒ 85 = 17 + 68 = 85
∴ the points form a right angled triangle.

Question 4.
The points (x, 2) is equidistant from (8, -2) and (2, -2). Find the value of x.
Answer:
Let A = (2,2) B = (8,-2) C = (2,-2)
By data AB = AC = AB² = AC²
⇒ (8 – x)² + (-2 -2)² = (x – 2)² + (2 + 2)²

-16x + 4x = 4 -64 ⇒ – 12x = -60
∴ x = 5

Question 5.
If two vertices of an equilateral triangle are (3,4) and (-2,3). Find the co-ordinates of third vertex.
Answer:
Let A = (3, 4), B = (-2, 3) and C = (x, y)
bydata AB = BC = CA
⇒ AB² = BC² = CA²
consider BC² = CA²
⇒ (x + 2)² + (y – 3)² = (x – 3)² + (y – 4)²

4x + 6x – 6y + 8y = 25 – 13
∴ 10x + 2y =12
5x + y = 6 …… (1)
Again AB = BC
⇒ AB² = BC²
∴ 5(3 + 2)² + (4 – 3)² = (x + 2)² + (y – 3)² ………. (2)
from(1) y = 6 – 5x
∴ substitute (1) is (2)
(x + 2)² + (6 – 5x – 3)² = 26
x² + 4 + 4x + 9 + 25x² – 30x = 26
26x² – 26x – 13 = 0
⇒ 2x² – 2x – 1 = 0

Section Formula

Question 1.
Find the co-ordinates of the point which divide the line joining the points
(i) (1, -3) and (-3, 9) internally in the ratio 1 : 3 and the line joining the points
(ii) (2, -6) (4, 3) externally is the ratio 3 : 2.
Answer:
(i) Let A = (1, -3) B = (-3 9), Ratio is 1 : 3

(ii) Let A = (2, -6) B = (4, 3) ratio 3 : 2

Question 2.
Find the ratio in which (2, 7) divides the line joining the point (8, 9) and (-7, 4).
Answer:
Let
A = (x₁ y₁) = (8, 9)
B = (x₂ y₂) = (-7, 4)
p = (x₁ y₂) = (2, 7)
The ratio l : m in which p divides AB, is given by

⇒ The ratio is l : m = 2 : 3
∴ the points (2,7) divides internally

Question 3.
Determine the ratio in which the line 3x + y – 9 = 0 divide the segment joining the points (1, 3) and (2, 7).
Answer:
Let the of division k : 1 as and point then the coordinates of the points are \(\left(\frac{2 k+1}{k+1}, \frac{7 k+3}{k+1}\right)\)
But the point ‘C’ lies on 3x + y – 9 = 0

⇒ 4x – 3 = 0
⇒ k = \(\frac { 3 }{ 4 }\)
∴ The required ratio is 3 : 4 internally

Question 4.
Find the lengths of the medians of a triangle whose vertices are (3, 5) (5, 3) and (7, 7).
Answer:
Median is a tine joining the vertex of a triangle to the middle points of the opposite side.
Let A(3, 5) B(5, 3) C(7, 7) the vertices
Let D, E, F be the mid points of BC, CA, AB respectively

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Karnataka 1st PUC Basic Maths Question Bank Chapter 14 Standard Angles of Allied Angles

Question 1.
Find the value of
(i) cosec (-765°)
(ii) sec (-1305°)
(iii) sin 585°
Answer:
(i) -cosec 765°
= -cosec (360x² + 45°)
= -cosec 45°
= – √2

(ii) sec (1305°)
= sec (360.3 + 225)
= sec 225°
= sec (180 + 45°)
= -sec 45°
= – √2

(iii) sin (360 + 225°)
= sin 225°
= sin (180 + 45°)
= – sin 45°
= – √2

Question 2.
Evaluate: \(\frac { 4 }{ 3 }\)tan² 120° + 3 sin² 300° – 2cosec²240°- \(\frac { 3 }{ 4 }\)cot² 60°
Answer:
Now tan 120° = tan (180° – 60°)
= -tan 60°
= -√3
sin 300° = sin (360° – 60°)
= – sin 60° = \(\frac{-\sqrt{3}}{2}\)
cosec 240° = cosec (180° + 60°)
= – cosec 60° = \(\frac{-2}{\sqrt{3}}\)

Question 3.

Answer:

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Karnataka 1st PUC Basic Maths Question Bank Chapter 13 Angles and Trigonometric Ratios

Question 1.
Find the length of the arc and the area of the sector formed by an arc of a circle of radius 8 cm subtending an angle 25° as the centre.
Answer:

Question 2.
Find the angle between minute hand and the hourhand of a clock when the time is 5.40.
Answer:
Angle described by the hourhand is 12 hours = 360°
∴ per hour = \(\frac{360}{12}\) = 30°

Now the angle described by minute hand is 60° minutes = 360°
i.e., Angle in 40° minutes = 40 × 6° = 240°,

Question 3.
A circular wire of radius 6 cm is cut and bent so as to lie along the circumference of a loop whose radius in 66 cm. Find in degrees the angle which subtended at the centre of the loop.
Answer:
The length of the arc subtended is the circumference of the circle of radius 6cm.
∴ S = 2πr = 2π × 6 = 127 cm
radius of the loop = 66cm

Question 4.
The angles of a triangle are in A.P and the greatest is double the least. Express the angles in degrees and radians.
Answer:
Let the angle, be (a – d)° a° and (a + d)°, (a > d > 0°)
∴ (a – d) + a + (a + d) = 180° ⇒ 3a = 180° ⇒ a = 60°
i. e., (a – d) is the least and (a + d) is the greatest,
given (a + d) = 2 (a – d) = a + d = 2a – 2d
⇒ a – 3d = 0 ⇒ 60° – 3d = 0 ⇒ d = 20°

Question 5.
Express 25°. 30′ 10″ in radians.
Answer:

Trigonometric Ratios of Acute Angle

Question 1.
Prove that tan α\(\sqrt{1-\sin ^{2} \alpha}\) = sin α
Answer:
LHS : tan α . \(\sqrt{1-\sin ^{2} \alpha}\) = tan α . \(\sqrt{\cos ^{2} \alpha}\)
= \(\frac{\sin \alpha}{\cos \alpha}\) × cos α = sin α = RHS

Question 2.
Prove that \(\frac{1-\cos \theta}{\sin \theta}=\frac{\sin \theta}{1+\cos \theta}\)
Answer:

Question 3.
Prove that \(\sqrt{\frac{1-\sin A}{1+\sin A}}\) = sec A – tan A.
Answer:

Question 4.
Show that \(\frac{\tan A}{1-\cot A}+\frac{\cot A}{1-\tan A}\) = 1 + sec A cosec A.
Answer:

Question 5.
If x = a cos θ + sin θ. y = a sin θ – b cos θ. Show that x² + y² = a² + b²
Answer:
Consider x² = (a cos θ + b sin θ)²
⇒ x² = a²cos²θ + 2ab cos θ sin θ + b² sin²θ …….(1)
Again consider
y² = (a sin θ – b cos θ)² => y² = a² sin²θ – ab cos θ sin θ + b² cos² θ …….. (2)
(1) + (2) ⇒ x² + y² = a² + b²

Question 6.
If sin θ = \(\frac{3}{5}\) and θ is acute, find the value of
Answer:
By data sin θ = \(\frac{3}{5}\)
cos θ = \(\frac{3}{5}\) . tan θ = \(\frac{3}{4}\), cot = \(\frac{4}{3}\)

Trigonometric Ratios of some Standard Angles

Question 1.
Find the value of cos \(\frac{\pi}{3}\) – sin \(\frac{\pi}{6}\) – tan³ \(\frac{\pi}{4}\)
Answer:

Question 2.
Find value of 3 tan²30° + \(\frac{4}{3}\)sin²60° – \(\frac{1}{2}\)cosec²30° – \(\frac{1}{3}\)cos² 45°
Answer:

Question 3.
Find x, if \(\frac{x \csc ^{2} 30^{\circ} \cdot \sec ^{2} 45^{\circ}}{6 \cos ^{2} 45^{\circ} \sin 30^{\circ}}\) = tan² 45° – tan² 60°
Answer:

Question 4.
Show that cos² \(\frac{\pi}{4}\) – cos⁴ \(\frac{\pi}{6}\) + sin⁴ \(\frac{\pi}{6}\) + sin⁴ \(\frac{\pi}{3}\) = \(\frac{9}{16}\)
Answer:

Question 5.

Answer:

Heights and Distances

Question 1.
A person is at the top of tower 75 feet high from there he observes a vertical pole and finds the angles of depression of the top and the bottom of the pole which are 30° and 60° respectively. Find the height of the pole.
Answer:
Let AB be the tower = 75 feet,
CD be the pole = x
AX is ∥^lel to MC
∴ AĈM = XÂC = 30°
AX ∥^lel BD
Hence AD̂B = XÂD = 60°

Now from the right triangle ADB, we have
tan 60° = \(\frac{\mathrm{AB}}{\mathrm{BD}}\)
⇒ BD = \(\frac{\mathrm{AB}}{\tan 60^{\circ}}\) = BD = \(\frac{75}{\sqrt{3}}\)
tan 30° = \(\frac{\mathrm{AM}}{\mathrm{MC}}\)
⇒ AM = MC . tan 30°
⇒ AM = BD \(\frac{1}{\sqrt{3}}=\frac{75}{\sqrt{3}} \cdot \frac{1}{\sqrt{3}}\) = 25
Now x = CD = MB
⇒ x = AB – AM = 75 – 25 = 50 feet

Question 2.
A person standing on the bank of a river observe that the angle of elevation of the top of a tree on the opposite bank is 60°, when he returns 60 feet from the bank, he finds the angle to be 30°. Find the height of the tree and breadth of the river.
Answer:
PQ represent tree A and B be the points of observation. BQ is the breadth of the river. From the figure

tan 60° = \(\frac{P Q}{B Q}\)
⇒ PQ = BQ . √3
Again from the figure we have
tan 30° = \(\frac{P Q}{A Q}\)
⇒ AQ = \(\frac{P Q}{\tan 30^{\circ}}\) = AQ = √3PQ
⇒ AQ = √3 . (√3BQ) = 3BQ. Now BQ ⇒ AQ – AB
⇒ BQ = 3BQ – AB
⇒ 2BQ = 60
∴ BQ = 30° is the breadth of the river
∴ height of the tree = PQ = √3.BQ = √3.30 = 30√3

Question 3.
The angle of elevation of a stationary cloud from a point 3000 cm above the lake is 30° and the angle of depression of its reflection in the lake is 60°. What is the height of the cloud above the lake.
Answer:
D – the image of the cloud C.
AB the surface of the lake
Also CB = BD
Let P be the position of observation.
Thus by data AP = 3000 cm
By data, EP̂C = 30°, EP̂D = 60°


⇒ PE = √3 . CE ….. (1)

⇒ PE = \(\frac{\mathrm{ED}}{\sqrt{3}}\) ….. (2)
From (1) and (2) we get
⇒ √3CE = \(\frac{1}{\sqrt{3}}\)ED = 3CE = ED
⇒ 3(BC – BE) = EB + BD
⇒ 3(BC – 3000) = 3000 + BC (∵ BD = BC and BE PA = 3000)
⇒ 3BC – 9000 = 3000 + BC
⇒ 2BC = 12,000 = BC = 6000 cm
∴ height of the cloud = 600 cms

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Karnataka 1st PUC Basic Maths Question Bank Chapter 17 Straight Lines

Question 1.
Find the slope of the line joining the points (3, -4) and (-7,3).
Answer:

Question 2.
Show that the line joining points (2, -3) and (-5,1) is parallel to the line joining the points (7, -1) and (0,3) and perpendicular to the line joining the points (4,5) and (0, -2).
Answer:
Let m₁ be the slope of the lien l₁ joining points (2, -3) and (0, 3)

Let m₂ be the slope of the line l₂ going (7, -1) and (0,3)

⇒ m₁ = m₂
∴ L₁ and L₂ are parallel to each other.

Question 3.
L₁ and L₃ or l₂ l₃ are perpendicular to each other. The slope of a line is double the
slope of another line. If the tangent of the angle between the is \(\frac { 1 }{ 3 }\). Find the slope of the lines.
Answer:
If slope of one line is m, then the slope of the other is 2m.
Let the angle between them θ then tan θ = \(\frac { 1 }{ 3 }\)

⇒ \(\frac { 1 }{ 3 }\) = \(\frac{m}{1+2 m^{2}}\)
⇒ 1 + 2m² = 3m
⇒ 2m² – 3m + x = 0
⇒ (2m – 1)(m – 1 ) = 0
∴ 2m – 1 = 0 and m – 1 = 0
m = \(\frac { 1 }{ 2 }\) m = 1
∴ Slope are \(\frac { 1 }{ 2 }\), and 1.

On standard form of straight lines:

Question 1.
Find the equation of line passing through (-3, 5) with slope \(\frac { -1 }{ 5 }\)
Answer:
Let (x₁ y₁) = (-3, 5). slope m = \(\frac { -1 }{ 5 }\)
The equation of line is y – y₁ = m(x – x₁)
= y – 5 = \(\frac { -1 }{ 5 }\) (x + 3)
⇒ 5y – 25 = -x – 3
⇒ x + 5y – 22 = 0

Question 2.
Find the equation line passing through (-3, 2) and (11, -1).
Answer:
Let (x₁, y₁) = (-3,2), (x₂, y₂) = (11,-1)

⇒ \(\frac{y-2}{x+3}=\frac{-3}{14}\)
⇒ 14y – 28 = -3x – 9
⇒ 3x + 14y – 19 = 0 is the equation

Question 3.
Find the ratio in which the line join (1, 2) and (4, 3) is divided by the line joining the points (2,3) and (4, 1).
Answer:
Let A = (2,3) B = (4, 1)
∴ the equation of line is a given by \(\frac{y-3}{x-2}=\frac{1-3}{4-2}\)
⇒ \(\frac{y-3}{x-2}=\frac{-2}{2}\) = -1
⇒ y – 3 = -x + 2
∴ x + y – 5 = 0 is the equation of line
Let c = (1,2) and D(4,3) cut the line AB at P(x ,y) is the ratio 2 : 1 then


⇒ 4x + 1 + 3r + 2 – 5(r + 1) = 0
⇒ 2r – 2 = 0
⇒ r = 1
∴ the required ratio is 1 : 1

Question 4.
Express 9x – 4y – 13 = 0 in the intercept from
Answer:
Consider 9x – 4y = 13 = \(\frac{9}{13} x-\frac{4}{13} y\) = 1

∴ x – intercept = \(\frac{13}{9}\)
y – intercept = \(\frac{-13}{9}\)

Question 5.
If P ¡s the length of the perpendicular from the origin on a line, whose x and y intercepts are respectively a and b then show that = \(\frac{1}{p^{2}}=\frac{1}{a^{2}}+\frac{1}{b^{2}}\)
Answer:
Let the line any the axis at A and y – axis at B
Then by data OA = a and OB = b from the figure

Area of OAB = \(\frac { 1 }{ 2 }\) (O.A) (O.B) = \(\frac { 1 }{ 2 }\) ab
or = \(\frac { 1 }{ 2 }\) (AB).P
⇒ \(\frac { 1 }{ 2 }\) ab = \(\frac { 1 }{ 2 }\) (AB).P ⇒ a²b² = (AB)². p²
Also AB² = OA² + OB² ⇒ AB² = a² + b² (∴ because a²b² = (a² + b²) . p²

Intersection of two lines and Concurrency of Lines

Question 1.
Find the equation of line parallel to the y-axis and drawn through the point of intersection of x – 7y + 5 = 0 and 3x + y – 7 = 0.
Answer:
The equation of line through the point of intersection of the given lines is of the form x – 7y + 5 + k(3x + y – 7) = 0
∴ (1 + 3k)x + (k -7)y + 5 – 7k = 0 (1)
Since the lines are parallel to y-axis, coefficient of y = 0
i.e., k – 7 =0
∴ k = 7 substitute in (1) we get
22x – 44 = 0
∴ x – 2 = 0 is required equation

Question 2.
Find the equation of the line through the intersection of 5x – 3y = 1 and 2x + 3y – 23 = 0 and perpendicular to the line whose equation is 5x – 3y -1 = 0 and perpendicular to the line whose equation is 5x – 3y – 1 = 0.
Answer:
We have 5x – 3y – 1 = 0 ….. (1)
and 2x + 3y – 23 = 0 ….. (2) .
∴ the equation will be in the form
(5x – 3y – 1) + k(2x + 3y – 23) = 0
⇒ (5 + 2k)x + 3(k – 1)y – (23k + 1) = 0 …… (3)

∴ Slope of line = \(\frac{5+2 k}{3(k-1)}\)
Also 5x – 3y – 1 = 0
⇒ y = \(\frac{5}{3} x \frac{1}{3}\) …….. (1)
∴slope = \(\frac{5}{3}\)
Given (4) and (5) are perpendicular

⇒ k = -34. Substituting the value of k in (3), we get
(5x – 3y – 1) -34 (2x + 3y – 25) = 0
⇒ 5x – 68x – 3y – 102y – 1 + 782 = 0
⇒ 63x + 105y – 781 = 0

Question 3.
If lines whose equation are y = m₁ + c₁y = m₂x + c₂ and y = m₅x + c₃ meet is a point then prove that m₁( (c₂ – c₃) + m₂ (c₃ – c₁) + m₃(c₁ – c₂) = 0
Answer:
The equation of the given lines are
m₁x – y + c₁ = 0 ……. (1)
m₂x – y + c₂ = 0 …… (2)
m₃x – y + c₃ = 0 .. (3)

The three lines will be concurrent if the point of intersection of (1) and (2) lies on (3)

⇒ m₃(c₁ – c₂) – (m₂c₁ – m₁c₂) + c₃ (m₂ – m₁) =0, simplify,
⇒ m₁(c₂ – c₃) + m₂(c₃ – c₁) + m₃(c₁ – c₂) =0

Question 4.
For what values of k are the lines x – 2y + 1 = 0, 2x – 5y + 3 = 0 and 5x – 9y + k = 0 are concurrent.
Answer:
consider
x – 2y + 1 = 0 ……. (1)
2x – 5y + 3 = 0 ….. (2)
5x – 9y + k = 0 …….. (3)
Solve (1) and (2) \(\frac{x}{-6+5}=\frac{-y}{3-2}=1\)
⇒ \(\frac{x}{-1}=\frac{-y}{-1}=\frac{1}{-1}\)
⇒ x = 1, y = 1
point of intersection of (1) and (2) is (1,1),
⇒ (1, 1) lies on (3) = 5 – 9 + k = 0 = k = 4

Angle between the lines, Length of perpendicular

Question 1.
Find the angle between the lines 2x + 3y – 4 = 0 and 3x – 2y + 5 =0
Answer:
By data
2x + 3y – 4 = 0 … (1)
3x – 2y + 5 = 0 …… (2)
∴ Slope of (1) = \(\frac{-a}{b}=\frac{-2}{3}\) = m₁

∴ the angle between the lines is 90°

Question 2.
Determine the position of the points (2, 1) and (-1, 1) w.r.t the line 4x – ly + 1 = 0.
Answer:
4(2) -7(1) = -7 + 1 = 2 > 0
4( -1) – 7(1) = – 4 – 7 + 2 = -10 < 0
Since the two points are opposite in sign the two points lie an either sides of the given line 4x – 7y + 1 = 0

Question 3.
If points (9,8) and (-3,3) are equidistant from the line 5x + 2y + 7 = 0 find ‘a’.
Answer:
Distance from (a, 8) to the line 5x + 2y + 7 = 0

Given that the distance are equal

⇒ 5a + 23 = 2
or
5a + 23 = -2
⇒ a = \(\frac{-21}{5}\) and a = -5

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Karnataka 1st PUC Basic Maths Question Bank Chapter 8 Simple Interest, and Compound Interest

Question 1.
Find simple Interest on Rs 7,300 from 15 may 2009 to 8^th October 2009 at 10% p.a.
Answer:
Given P = Rs.7,300, R = 10%
Time = May + June + July + Aug + Sept. + Oct.
= 16 + 30 + 31 + 31 + 30 + 8 = l46days.

Question 2.
Determine the principal which will amount to Rs. 15,000 in 4 years at 8% p.a.
Answer:

Question 3.
In how many years will a sum of money double itself at 18.75% p.a simple interest.
Answer:
Let the principal be Rs. P, then amount = Rs. 2P
∴ SI = Amount – PrincipIe = Rs. 2P – P = P; R = 18.75% p.a

Question 4.
Kumar borrowed some many at the rate of 6% p.a for the first two years, at the rate of 9% p.a. for the next three years, and at the rate of 14% p.a. for the period beyond live years. If he pays a total interest of Rs. 11,400 at the end of nine years, how much money did he borrow.
Answer:
Let the sum borrowed be x. then

Hence sum borrowed = Rs. 12,000

Compound Interest

Question 1.
Find the CI amount and Cl on Rs. 12,000 for 3 years at 10% p.a. compounded annually.
Answer:
P = 12,000, n = 3,R = 10%p.a
We have amount A

∴ C.I = A.P = 15972 – 12000 = Rs. 3972

Question 2.
At what percent p.a. will a sum of Rs. 5,000 become Rs. 8,000 if the loan given for 4 years attract compound interest?
Answer:
Given A = Rs. 8000, P = Rs. 5000, n = 4, R = ?

Question 3.
Compute the CI. on Rs. 12,000 for 2 years at 20% p.a. when compounded half yearly.
Answer:
Given P = 12,000, R = 20% p.a. and n = 2

= 12000\(\left(\frac{11}{10}\right)^{4}\) = 12000 (1.4641) = 17569.20
A = Rs. 17,569.20
∴ Compound Interest = A – P = 17,569.20 – 12,000 = Rs. 5,569.20

Question 4.
A sum of money doubles itself at compound an Interest in 15 years. In how many years will it become eight times.
Answer:
Let the sum of the money be Rs. P is interested at the rate of R% p.a. It is given that the moñey double it self in 15 years

∴ the money will become 8 times in 45 years.

Question 5.
The population of a town in increasing at the rate of 5% p.a. What will be the population of the two as thus basic after two years if the present population is 20,0000.
Answer:
Given initial population = P = 20,000, R = 5%, n = 2
We have poplulation after 2 years = \(\mathrm{p}\left(1+\frac{\mathrm{R}}{100}\right)^{n}\)

Question 6.
In a factory the production of cars rose to 36,300 from 30,000 in 2 years. Find the rate of growth p.a.
Answer:
Let the rate of growth be R% p.a. present production A = 36500, previa production P = 30,000, n = 2

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Karnataka 1st PUC Basic Maths Question Bank Chapter 7 Linear Inequalities

Question 1.
Solve graphically 3x + 4y ≤ 60, x + 3y ≤ 30, x ≥ 0, y ≥ 0
Answer:
3x + 4y = 60

x + 3y + 30

(0, 0) satisfies 3x + 4y ≤ 60 and x + 3 ≤ 30

Question 2.
Solve 2 (2x + 3) – 10 £ 6(x – 2)
Answer:
Given 2 (2x + 3) – 10 £ 6(x – 2)
⇒ 4x + 6 – 10 ≤ 6x – 12 ⇒ 4x – 4 ≤ 6x – 12
⇒ 4x – 6x ≤ – 12 + 4 ⇒ -2x ≤ -8
⇒ x ≥ \(\frac{8}{2}\) ⇒ x ∈ [4, ∞] is the solution

Question 3.
Solve the equation \(\left|\frac{2}{x-4}\right|>1\),  x ≠ 4.
Answer:
we have \(\left|\frac{2}{x-4}\right|>1\),  x ≠ 4
⇒ \(\frac{2}{|x-4|}>1\) ⇒ 2 > |x – 4|
⇒ 4 – 2 < x < 4 + 2
⇒ 2 < x < 6. ∴ x e (2, 6) But x 4

Question 4.
Find all pairs of consecutive even positive integers both of which are larger than 8, such that their sum is less than 25.
Answer:
Let x be the smaller of the two consecutive even positive integers, then the other even integer is x + 2.
Given x > 8 and x + (x + 2) < 25.
⇒ x > 8, and 2x + 2 < 25.
⇒ x > 8, 2x < 23 ⇒ x > 8, x < \(\frac{23}{2}\)
⇒ 8 < x < ⇒ \(\frac{23}{2}\) x = 10,
∴ the required parity even integers is (10, 12)

Question 5.
In the first four papers each of 100 marks, Ravi got 95, 72, 73, 83 marks. If he wants an average of greater than or equal to 75 marks and less than 80 marks, find the range of marks he should score in the fifth paper.
Answer:
Let score be x in the fifth paper, then

Hence Ravi must score between 52 and 77 marks.

Solve and represent the following in equalities graphically 

Question 1.
x + y ≥ 4 : 2x – y > 0
Answer:

Question 2.
x + y ≤ 9, y > x, x ≥ 0
Answer:

Question 3.
2x – y > 1, x – 2y < – 1
Answer:

Question 4.
5x + 4y ≤ 20, x ≥ 1, y ≥ 2
Answer:

Question 5.
2x + y ≥ 4, x + y ≤ 3, 2x – y ≤ 6.
Answer:

Question 6.
x – 2y ≤ 3, 3x + 4y ≥ 12, x ≥ 0, y ≥ 1
Answer:

Question 7.
4x + 3y ≤ 60, y ≥ 2x, x ≥ 3, y ≥ 0
Answer:

Question 8.
x + 2y ≤ 10, x + y ≥ 1, x – y ≤ 0, x ≥ 0, y ≥ 0
Answer:

Question 9.
Solve the in equalities and represent the solutions graphically on number line.

Answer:
5(2x – 7) -3 (2x + 3) ≤ 0; 2x + 19 ≤ 6x + 47

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Karnataka 1st PUC Basic Maths Question Bank Chapter 6 Theory of Equations

Question 1.
Find the nature of roots of the equation 3x² – 4x + 9 = 0
Answer:
Nature of roots depends on b² – 4ac
⇒ a = 3, b = – 4, c = 9.
∴ b² – 4ac = (-4)² – 4 . (3) . 9 = 16 – 108 = – 92 < 0
∴ The roots are imaginary.

Question 2.
If a and B are the roots of 4x² + 3x – 5 = 0. Find the value of \(\frac{\alpha^{2}}{\beta}+\frac{\beta^{2}}{\alpha}\)
Answer:
a = 4, b = 3, c = – 5
Sum of roots = α + β = \(\frac{-b}{a}=\frac{-3}{4}\)
Product of roots = αβ = \(\frac{c}{a}=\frac{-5}{4}\)

Question 3.
Find three consecutive even natural numbers, such that the sum of their squares is 200.
Answer:
Let the required even numbers be x, x + 2, x + 4
By data x² + (x + 2)² + (x + 4)² = 200
⇒ x² + x² + 4x + 4 + x² + 8x + 16 – 200 = 0
3x² + 12x – 180 = 0 ⇒ x² + 4x – 60 = 0
⇒ (x + 10)(x – 6) =0
we require +ve the integers ∴ x = 6
⇒ 6,8, 10 be the numbers.

Question 4.
A piece of cloth costs Rs. 200. If the piece were 5mts longer and each meter of cloth costed Rs. 2 less, the cost of piece would have remained unchanged. How long is the piece and what is its original rate per meter.
Answer:
Let x be the length of the cloth, its cost = 200
∴ Rate of cloth per meter = \(\frac{200}{x}\)
Increased length of piece of cloth = (x + 5) mts
Now rate of cloth per meter = \(\frac{200}{21+5}\)
By date, old rate cloth – new rate of cloth = Rs 2.



⇒ x² + 5x – 500 = 0
⇒ x² + 25x – 20x – 500 = 0
⇒ x(x + 25) – 20(x + 25) = 0
⇒ x = 20 or -25 (neglected) [∴ x = 20mts]
length of cloth = 20 mts, rate per metre = Rs. 10.

Question 5.
Solve : x³ – 6x² + 11x – 6 = 0. since that ratio of two roots is 2 : 3
Answer:
By data α : β = 2 : 3
⇒ \(\frac{\alpha}{\beta}=\frac{2}{3}\) = α = 2β
from the given equation,

⇒ 5α³ – 12α² + 8 = 0
solving α = 2, ∴ α = 1
∴ The roots are 1,2,3

Question 6.
Solve by synthetic division that it has atleast one integral root between -3 and 3 for the equation x⁴ – 9x² + 4x + 12 = 0
Answer:
Let f(x) = x⁴ – 9X² + 4x + 12 = 0
Bu inspection x = ±1, ±2, ±3 are the roots or not
f(-3) = 0, f(-1) = 0, f(2) = 0
∴ -3, -1, 2 are the roots of the given equation
and
q
∴ x = – 3 is a root and x = – 2 is not a root
Try for x = – 1, x = -1 is a root
Now the quotient is x² – 4x + 4 = 0
⇒ (x – 2)² = 0 = x = 2, 2
Hence the form root are -3, -1,2, 2

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Karnataka 1st PUC Basic Maths Question Bank Chapter 11 Percentages of Profit and Loss Sales tax, Vat

Question 1.
Ravi purchased a pair of shoes costing ‘950. Calculate the total amount to be paid by him, if the rate of sales tax is 7%.
Answer:
Sales price = Rs. 950

Total amount paid by Ravi = ‘950 + ’57 = ‘1007.

Question 2.
Sneha purchased confectionery goods costing 165 on which the rate of sales tax is 6% and some tooth paste, cold cream, soap etc. costing ‘ 550 which the rate of sales tax is 10%. If she gives 1000 note to the shop keeper, What money will be return to Mrs Sneha?
Answer:
Price of confectionary goods with sales tax = ’65 + 6% + 165 = ‘174.90
Price of toothpaste, cold cream soap with tax = ‘550 + 10% of 550 = 550 + 55 = 605
∴ the average to be paid = 174.90 + 605 = 774.90
Amount to be returned by shop keeper = 1000 – 774.90 = ‘225.90 paise.

Question 3.
The price of an article inclusive of sales tax of 15% is 3450. Find its marked price if the sales tax is reduced to 6%. How much less does the customer pay for the article?
Answer:
Let the marked price = ‘x
∴ x + 15% of x = 3450
⇒ \(\frac{115 x}{100}\) = 3050 ⇒ x = 3000
∴ Market price of the article = ‘3000
Since new sales tax = 60%
Now the customer will pay = ‘3000 + 6% of ‘3000 = \(\frac{106}{100}\) × ‘3000 = 3180
∴ customer will pay for the article = 3450 – 3180 = 270 less

Problems on Overhead charges, Discount

Question 1.
A shop keeper buys an article for Rs. 1500 and spends20% of the cost on its packing transportation etc. Then he marked price on it. If he sells the article for ‘2,452.50 including 9% sales tax on the price marked. Find its profit as percent.
Answer:
Let marked price of the article be ‘x.
Sales tax price = 9% of x = \(\frac{9 x}{100}\)
According to given statement n + \(\frac{9 x}{100}\) = 2452.50
⇒ \(\frac{100 x+9 x}{100}\) = 2452.50 ⇒ x = 2452.50 × \(\frac{100}{109}\) = 2,250
∴ Marked price = ‘2,250 = selling price
Since the shopkeeper buys for ‘1500 and spends 20% of the cost as overhead.
∴ Total cost price of the article = ‘1500 + 20% ‘1500 = ‘1500 + ‘300= ‘1800
Profit = Selling price – total cost price = 2250 – 1800 = 450
∴ Profit % = \(\frac{450}{1800}\) × 100 = 22%

Question 2.
Prabha bought an article ‘374. Which included a discount of 15% on the market price and a sales tax of 10% on the reduced price. Find the market price of the article.
Answer:
Let the market price of the article be ‘x
∴ Discount 15% of x = \(\frac{15}{100} \times x=\frac{3 x}{20}\)
⇒ Remaining cost = x – \(\frac{3 x}{20}=\frac{17 x}{20}\)
Sales tax = 10%
∴ price paid by prabha = \(\frac{110}{100} \times \frac{17 x}{20}=\frac{187 x}{200}\)
Given \(\frac{187 x}{200}\) = 374 = x = 374 = \(\frac{200}{187}\) = 400
∴ Market price = ‘400

Question 3.
A Shopkeeper buys an article at a rebate of 20% on the printed price. He spends ” 40 on transportation of article. After charging a sales tax of 7% on the printed price he sells the article for ‘1,070. Find his gain on percent.
Answer:
Let printed price of the article be ‘ x
⇒ Sales tax on it = 7% of ‘x = \(\frac{7 x}{100}\)
By data x\(\frac{7 x}{100}\) = 1070 ⇒ x = 1000
∴ Printed price = ‘1000
Again the shop keeper buy the article at 20% rebate.
∴ The cost price to the shop keeper = ‘1000 – 20% of 1000 = ‘ 800
Since he spends ’40 on the transportation of the article
⇒ Total cost price = ‘800 + ’40 = ‘840
The selling price = printed price = ‘1000
⇒ Profit = ‘100 – ‘840 =’160
And profit = \(\frac{160}{840}\) × 100 = 19 \(\frac{1}{21} \%\)

Value-Added Tax (VAT)

Question 1.
A Shopkeeper sales an article at its marked price ‘7,500 and charges sales tax at the rate of 12% from the customer. If the shopkeeper pays a VAT of ‘180, calculate the price (inclusive of tax) paid by the shopkeeper.
Answer:
Since the shopkeeper sells the article for ‘7,500 and changes sales tax at the rate of 12%
∴ tax charged by the shopkeeper = 12% of ‘7500
= \(\frac{12}{100}\) x 7500 = 900
VAT = Tax charge – tax paid ,
⇒ ‘180 = ‘900 – Taxpaid
Tax paid by the shopkeeper = ‘ 900 – ‘180 =’ 720
If the shopkeeper buys the article for ‘ x,
Tax on it = 12% of ‘x = ‘720
⇒ x = ‘6,000 ∴ The price (inclusive of tax)
paid by the shopkeeper = ‘6000 + ‘720 = ‘6720.

Question 2.
During the financial year a shopkeeper purchased goods worth ‘4,15,000 and paid a total tax of ‘38,000. His sales during the period consisted of a taxable turnover of ‘50,000 for goods taxable at 5% and ‘3,20,000 for goods takable at 12%. He also sold tax exempted goods worth ‘ 45,000. Calculate his tax liability for the financial year.
Answer:
Goods taxable at 5% turover = ‘ 50,000
Tax = 50% of ‘50,000 = ‘ 2,500
of goods taxable at 125, turover = ‘3,20,000
tax = 12% of ‘3,20,000 = ‘38,400
Given tax exampled sales = ‘45,000
Total tax = ‘2,500 + ‘38,400 = ‘40,900
Tax paid = ‘38,000
‘ Net tax payable = Total tax changed – tax paid
= ‘40,900 – ‘38,000
=’2,900
∴ Tax liability (under VAT) = ‘2,900

Question 3.
A manufacturing company sold a commodity to its distributor for ‘22,000 including VAT. The distributor sold the commodity to a retailer for ‘ 22,000 excluding tax and the retailer sold it to the customer for ‘ 25,000 plan tax (under VAT). If the rate of tax is 10%. What was the:
(i) Sale price of the commodity for the manufacture?
(ii) An amount of tax received by the state government on the sale of the commodity?
Answer:
(i) Let the price of the commodity for the manufacture be ‘x, tax is 10%.
∴ tax charged by the manufactures = 10% of C.P

∴ The sale price of the commodity for the manufacture = 20,000

(ii) VAT collected by the manufacture = tax charged by the manufacture

since the distributor has sold the commodity tot he retailer for ‘22,000
∴ tax collected by the distributor = 10% of22,000
= \(\frac{10}{100}\) × 22000 = ‘2200 = paid by retailer
VAT to be deposited by the distributor = ‘2200 – ‘2000 = ‘200
Tax collected by retailer = 10% of 25000
= \(\frac{10}{100}\) × 25,000 = ‘2500
∴ VAT to be paid by the retailer = ‘2500 – ‘2200 = ‘300
∴ Amount of the (under VAT) received by the state government
= ‘2000 + ‘200 + ‘300 = ‘2500

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Karnataka 1st PUC Basic Maths Question Bank Chapter 10 Averages

Question 1.
A student asked to find the arithmetic mean of the numbers 4, 19, 17, 12, 34,16, 7, – 23,18, 21, 25, and x. Find the mean table 18. What should be the number in place of x.
Answer:

18 = \(\frac{196+x}{2}\) = 216 = 196 + x
∴ x = 216 – 196 = 20

Question 2.
The average of 30 results is 20 and the average of other 20 results is 30. Find the average of results taken to gather.
Answer:
Given n₁ = 30, x̄₁ = 20, n₂ = 20 , x̄₂ = 30

Question 3.
If the average of daily wages of workers of two factories is Rs. 53 and average wages to factory ‘A’ with 250 employees is Rs. 50. Find the average wage of factoring ‘B’ with 200 employees.
Answer:
Given X̄ = 53, X̄_A = 50, n_A = 250, X̄_B = 7, n_B = 200

i.e., 200 X̄_B = 23850 – 12500 = 11,350 ∴ X̄_B = \(\frac{11,350}{200}\) = 56/75
∴ Average wage of factory B = Rs. 56.75

Question 4.
Ten years ago the average age of the family of 4 members was 24 years. Two children have been born. The average age of the family is same to day. Find the present age of two children assuming that the children’s age differ by 2 years.
Answer:
10 year ago x = 24, n = 4 ∴ x = 24 × 4 = 96 year
If 10 years ago, total age of 4 member = 96 year
total age of member now = (96 + 10 × 4) year = 136 year
Total age of 6 member = 74 × 6 = 144 year
i.e., Sum of age of 2 children = 144 – 136 = 8 year
so x + x + 2 = 8 = x = 3 year
∴ The ages of children 3 and 4 year.