KSEEB SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources

KSEEB SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources.

Karnataka SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources

KSEEB SSLC Class 10 Science Chapter 16 Intext Questions

Text Book Part I Page No. 128

Question 1.

What changes can you make in your habits to become more environment-friendly?
We should switch off the electrical appliances when not in use. Water and food should not in wasted. Close the tap when not in use, Dump the objects made of plastic and glass in designated recycling boxes. Plastic, paper or glass must be recycled or reused and not dumped with other wastes. This is because objects made of plastic do not get decomposed easily. Besides soil fertility, they badly affect our environment. We should dispose the wastes safely and not disperse in public places. These are a few things that can be done to become more environment-friendly.

Question 2.
What would be the advantages of exploiting resources with short-term aims?
There should be a judicious use of natural resources as they are limited in nature. We should not exploit resources for our short term gains as this would only lead to depletion of natural resources for the present generation as well as generations to come. Hence, we say that there are hardly any advantages of exploiting natural resources for short term gains.

Question 3.
How would these advantages differ from the advantages of using a long-term perspective in managing our resources?
In the case of a long-time perspective in managing our resources, these resources will last for the generations to come. This management ensures uniform distribution among the people. It conserves the ‘ natural resources for many years and not just for a few years, as in the case of a short-term perspective in conserving natural resources.

Question 4.
Why do you think there should be equitable distribution of resources? What forces would be working against an equitable distribution of our resources?
Natural resources of the Earth must be distributed among the people uniformly so that each and every one gets his share of the resource. Human greed, corruption and the lobby of the rich and powerful are thq forces working against an equitable distribution of resources.

Text Book Part I Page No. 132

Question 1.
Why should we conserve forests and wildlife?

1. Destruction of forests not only affect on forest products but it affects the water resources and it is soil pollution.
2. Destruction of forest leads to shortage of fodder for animals, shortage of medicinal plants, shortage of fruits and nuts.
3. There is a shortage of valuable timbers such as sal, sandal wood etc.
4. Wild animals are also useful to us in many ways, hence we should conserve these by building sanctuaries and prohibited hunting.

Question 2.
Suggest some approaches towards the conservation of forests.

1. Cutting valuable trees should be prevented.
2. We should use forest products, such that there should not be damage to the environment.
3. All people should participate in the conservation of forest and wild animals.

Text Book Part I Page No. 135

Question 1.
One of the traditional systems of water harvesting used in our region is tanks. There may be other systems like ponds, water reservoirs etc.

Question 2.
Compare the above system with the probable systems in hilly/ mountainous areas or plains or plateau regions.
In plains, the water harvesting structures are crescent-shaped earthen embankments. These are low, straight and concrete. In hilly regions, the system of canal irrigation called Kulyths is used for water harvesting. This involves a collection of rain water in a stream, which is then diverted into manmade channels down the hill sides.

Question 3.
Find out the source of w ater in your region/locality. Is water from this source available to all people living in that area?
The source of water in our region is ground water. Water from the source is available to all the people living in that area.

KSEEB SSLC Class 10 Science Chapter 16 Textbook Exercises

Question 1.
What changes would you suggest in your home in order to be environment-friendly?

1. We must save water and electricity.
2. We should not waste food.
3. We should encourage reuse and recycling.
4. We must minimise the use of plastics.

Question 2.
Can you suggest some changes in your school which would make it environment friendly?

1. Enough plants and trees can be planted in the school.
2. Water should not be wasted but should be used judiciously.
3. Students should be taught to keep their classrooms and immediate surroundings neat and tidy.
4. Compost pits may be made in safe comers of the school, where biodegradable wastes may be dumped to prepare compost.
5. Minimising the usage of loudspeakers.
6. Organising seminars, quiz, essay competition, drawing competitions for spreading environmental awareness, celebrating Vanamahotsava… etc.

Question 3.
We saw in this chapter that there are four main stakeholders when it comes to forests and wildlife. Which among these should have the authority to decide the management of forest produce? Why do you think so?
The local people need large quantities of firewood, small timber and thatch. Bamboo is used to make slats for huts, and baskets for collecting and storing food materials. Implements for agriculture, fishing and hunting are largely made of wood, also forests are sites for fishing and hunting.

In addition to the people gathering fruits, nuts and medicines from the forests, there cattle also graze in forest areas or food on the fodder which is collected from forests.

Because of these reasons the people who live in or around forests have authority to decide the management of forest produce.

Question 4.
How can you as an individual contribute or make a difference to the management of
(a) forests and wildlife
(b) water resources
(c) coal and petroleum?
(a) Forest and wildlife:

1. We should protest against the cutting of trees (deforestation).
2. We should protest against the poaching of wild animals.
3. We should stop the annexation of forest land for our use.

(b) Water resources:

1. Turn the taps off while brushing or bathing and repair leaking taps.
2. We should practice rainwater harvesting.
3. We should avoid the discharge of sewage and other wastes into rivers and other water resources.

(c) Coal and Petroleum:

1. We should take a bus or practice car pooling to avoid excessi ve use of petroleum.
2. We should stop using coal.as a fuel (angithis).
3. We should use alternative source of energy such as hydro¬energy and solar energy instead of depending largely on coal and petroleum.

Question 5.
What can on as an individual do to reduce your consumption of the various natural resources?
Natural resources such as water, forests, coal and petroleum etc. are important for the survival of human beings. The ways in which we can reduce the

consumption of various natural resources are as follows:

1. We should stop the cutting of trees (deforestation).
2. We should use recycled paper to reduce the cutting down of trees
3. We should not waste water.
4. We should practice rainwater harvesting.
5. We should practice car pooling to avoid the excessive use of petroleum.
6. We should use alternative sources of energy such as hydro-energy and solar energy.

Question 6.
List five things you have done over the last one week to:
(a) conserve our natural resources.
(b) increase the pressure on our natural resources.
(a) To conserve our natural resources:

1. Travelled by a CNG bus for long distances and walk for short distances.
2. Used recycled paper.
4. Planted trees.
5. Harvested rainwater.

(b) To increase the pressure on our natural resources:

1. Used non-renewable resources of energy.
2. Wasted water.
3. Wasted electricity.
4. Used plastics and polythene bags for carrying goods.
5. Used escalators.

Question 7.
On the basis of the issues raised in this chapter, what changes would you incorporate in your life-style in a move towards a sustainable use of our resources?
One should incorporate the following changes in life-style in a move towards a sustainable use of our resources:

1. Stop cutting trees and practice plantation of trees.
2. Stop using plastic and polythene bags for carrying goods.
3. Use recycled paper.
5. Waste minimum amount of water while using and repair leaking taps.
6. Practice rainwater harvesting.
7. Avoid using vehicles for short distances. Instead, one can walk or cycle to cover short distances. To cover long distances, one should take a bus instead of using personal vehicles.
8. Switch off electrical appliances when not in use.
9. Use fluorescent tubes in place of bulbs to save electricity.
10. Take stairs and avoid using lifts.
11. During winters, wear an extra sweater to avoid using heaters.

1. Fill in the blanks

Question 1.
…… and …… are followed as means of protection of nature and natural resources.

Question 2.
Prticipation of the …… can indeed lead to the efficient management of forests.
local people

Question 3.
Irrigation methods like have been used in various parts of India since ancient times.
dams, tanks and canals.

Question 4.
…… and …… were formed from the degradation of biomass millions of years ago.
Coal, petroleum

Question 5.
Fossil fuels contain carbon along with …… also.
hydrogen, nitrogen and sulphur.

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KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4

Students can Download Class 10 Maths Chapter 9 Polynomials Ex 9.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka State Syllabus Class 10 Maths Chapter 9 Polynomials Ex 9.4

Question 1.
Verify that the numbers given alongside of the cubic polynomials below are their zeroes. Also verify the relationship between the zeroes and the coefficients in each case:
(i) 2x3 + x2 – 5x + 2; $$\frac{1}{2}$$, 1, – 2
(ii) x3 – 4x2 + 5x – 2; 2, 1, 1
Let P(x) = 2x3 + x2 – 5x + 2 on comparing with general polynomial P(x) = ax3 + bx2 + cx + d we get
a = 2, b = 1, c = – 5 and d = 2
Given zeroes $$\frac{1}{2}$$, 1, – 2

P(1) = 2 (1)3 + (1)2 – 5(1) + 2
= 2 × 1 + 1 – 5 + 2
= 2 + 1 – 5 + 2 = 5 – 5 = 0
P(1) = 0
P(- 2) = 2(- 2)3 + (-2)2 – 5(- 2) + 2
= 2 × – 8 + 4 + 10 + 2
= – 16 + 16
P(- 2) = 0
Hence 1/2, 1 & -2 are the zeroes of the given cubic polynomial
Again consider α = 1/2, β = 1 & γ = – 2

ii) Let P(x) = x3 – 4x2 + 5x – 2 on comparing with general polynomial
P(x) = ax3 + bx2 + cx + d
we get a = 1, b = – 4, c = 5 & d = 2.
Given zeroes 2, 1, 1
P(2) = (2)3 – 4(2)2 + 5(2) – 2
= 8 – 4 × 4 + 10 – 2
= 8 – 16 + 10 – 2
P(2) = 18 – 18 = 0
P(2) = 0
P(1) = (1)3 – 4(1)2 + 5(1) – 2
= 1 – 4 + 5 – 2
P(1) = 6 – 6 = 0
P(1) = 0
Hence, 2, 1 and 1 are the zeroes of the given cubic polynomial.
Again, consider α = 2, β = 1 & γ = 1
α + β + γ = 2 + 1 + 1 = 4
and α + β + γ = $$\frac{-b}{a}=\frac{-(-4)}{1}$$ = 4
αβ + βγ + γα = 2(1) + (1) (1) + (1) × (2)
= 2 + 1 + 2 = 5
αβ + βγ + γα = $$\frac{c}{a}=\frac{5}{l}$$ = 5
αβγ = 2 × 1 × 1 = 2
αβγ = $$\frac{-d}{a}=\frac{-(-2)}{1}=\frac{2}{1}$$ = 2

Question 2.
Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, – 7, – 14 respectively.
Let the cubic polynomial be P(x) = ax3 + bx2 + cx + d
Then sum of zeroes = $$\frac{-b}{a}$$ = 2
Sum of the product of zeroes taken two at a time = $$\frac{c}{a}$$ = – 7
Product of zeroes = $$\frac{-d}{a}$$ = – 14

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.
The given Polynomial is x3 – 3x2 + x + 1
comparing with Ax3 + Bx2 + Cx + D,
A = 1, B = – 3, C = 1 & D = 1
Let α = a – b, β = a & γ = a + b
α + β + γ = $$\frac{B}{A}=\frac{-(-3)}{1}$$ = 3
a – b + a + a + b = 3
3a = 3
a = $$\frac{3}{3}$$ = 1
a = 1
αβγ = $$\frac{-D}{A}$$ = – 1
(a – b) a (a + b) = – 1 Put a = 1
(1 – b) 1 (1 + b) = – 1
1 – b2 = – 1
1 + 1 = b2
b2 = 2
b = ±$$\sqrt{2}$$
∴ a = 1 & b = ±$$\sqrt{2}$$

Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ±$$\sqrt{3}$$. find other zeroes.
Two zeroes of the given polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± $$\sqrt{3}$$.
∴ x = 2 ± $$\sqrt{3}$$.
Therefore [(x – (2 + $$\sqrt{3}$$)] [x – (2 – $$\sqrt{3}$$)]
= (x – 2 – $$\sqrt{3}$$) (x – 2 + $$\sqrt{3}$$)
= [(x – 2) – $$\sqrt{3}$$][(x – 2) + $$\sqrt{3}$$)]
= (x – 2)2 – ($$\sqrt{3}$$)2
= x2 – 4x + 4 – 3 = x2 – 4x + 1
∴ It is factor of Polynomial apply division algorithm.

Other two zeroes are given by the Quotient
x2 – 2x – 35 = 0
x2 – 7x + 5x – 35 = 0
x(x – 7) + 5(x – 7) = 0
(x – 7)(x + 5) = 0
x – 7 = 0 (or) x + 5 = 0
x = 7 (or) x = – 5
∴ x = 7, – 5

Question 5.
If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.
Let us apply the division algorithm to the given Polynomial

But remainder is given to be x + a comparing coefficients of x in the remainder obtained and remainder given:
(2K – 9)x – K(8 – K)+ 10 = x + a
∴ 2K – 9 = 1
∴ 2K = 1 + 9
K = $$\frac{10}{2}$$
K = 5 and – K(8 – K) + 10 = a
Put K = 5
– 5(8 – 5) + 10 = a
– 5(3) + 10 = a
– 15 + 10 = a
a = – 5

KSEEB Solutions for Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

Students can Download Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4

Question 1.
Which term of the A.P: 121, 117, 113,……, is its first negative term?
[Hint: Find n for an < 0]
The given AP is 121, 117, 113,…….
a = 121, d = 117 – 121 = – 4
Let the nth term of the AP be the first negative term.
∴ an < 0
a + ( n – 1 )d < 0
121 + (n – 1) (- 4) < 0
121 – 4n + 4 < 0
125 < 4n
4n > 125

∴ n = 32.
Hence, 32nd term of the given AP is the first negative term.

Question 2.
The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP
Solution:

Question 3.
A ladder has rungs 25 cm apart. (see Fig.1.7 ). The rungs decrease uniformly ¡n length form 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2 $$\frac{1}{2}$$ m apart, what is the length of the wood required for the rungs?
[Hint: Number of rungs $$\frac{250}{25}$$ + 1].

Distance between top and bottom rung

Distance between every two rungs
= 25 cm [given]
Number of rungs (n) = + 1 for top most rung]
= 10 + 1 = 11
∴ n = 11
Because the rungs decrease uniformly in length from 45 cm at the bottom to 25cm at the top

Question 4.
The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the number of the houses following it. Find this value of x. [Hint: Sx-1 = S49 – Sx]
Solution:
We have the following consecutive numbers on the houses of a row; 1, 2, 3, 4, 5, ….. , 49.
These numbers are in AP, such that a = 1, d = 2 – 1 = 1, n = 49
Let one of the houses be numbered as x.
∴ Number of houses preceding it = x – 1
Number of houses following it = 49 – x
Now, the sum of the house-numbers preceding

Question 5.
A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $$\frac{1}{4}$$ m and a tread of $$\frac{1}{2}$$ m (see Fig 5.8). Calculate the total volume of concrete required to build the terrace.
Hint: Volume of concrete required to build the first step = $$\frac{1}{4} \times \frac{1}{2} \times 50 \mathrm{m}^{3}$$

Volume of concrete required to build the first step length × breadth × Height.

∴ Total volume of concrete required to build the terrace of 15 steps

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Students can Download Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

Karnataka State Syllabus Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7

Question 1.
The ages of two friends Ani and Biju differ by 3 years. Ani’s father Dharam is twice as old as Ani and Biju is tw ice as old as his sister Cathy. The ages of Cathy and Dharam differ by 30 years. Find the ages of Ani and Biju.
Let the ages of Ani and Biju be ‘x’ years and ‘y’ years respectively.
x – y = 3 (x > y) (or) y – x = 3 (y > x) → (1)
Age of Ani’s father Dharm = 2x years
Age of Biju’s sister = $$\frac{y}{2}$$ years.
According to the Question
2x – $$\frac{y}{2}$$ = 30
4x – y = 60 → (2)
case (i) when x – y = 3
x – y= 3 → (1)
4x – y = 60 → (2)
Subtract equation (1) and (2)

x = 19 years
Put x = 19 in equation (1)
x – y = 3
19 – y = 3
– y = 3 – 19
y = 16
Ani’s age = x = 19 years.
Biju’s age = y = 16 years.
case (ii) y – x = 3
– x + y = 3 → (1)
4x – y = 60 → (2)

x = 21
Put x = 21 in equation (1)
– 21 + y = 3
y = 3 + 21
y = 24
Ani’s age is 21 years and Biju age is 24 years.

Question 2.
One says, “Give me a hundred, friend! I shall then become twice as rich as you”. The other replies, “If you give me ten, I shall be six times as rich as you”. Tell me what is the amount of their (respective) capital? [From the Bijaganita of Bhaskara II]
[Hint: x + 100 = 2(y – 100), y + 10 = 6(x – 10)]
Solution:
Let the capital of 1st friend = ₹ x,
and the capital of 2nd friend = ₹ y
According to the condition,
x + 100 = 2(y -100)
⇒ x + 100 – 2y + 200 = 0
⇒ x – 2y + 300 = 0 … (1)
Also, 6(x – 10) = y + 10 ⇒ 6x – y – 70 = 0 …. (2)
From (1), x = -300 + 2y (3)
Substituting the value of x in equation (2), we get
6[-300 + 2y] – y – 70 = 0
⇒ -1870 + 11y = 0
$$\Rightarrow y=\frac{1870}{11}=170$$
Now, Substituting the value of y in equation (3), we get, x = – 300 + 2y
= – 300 + 2(170) = – 300 + 340 = 40
Thus, 1st friend has ₹ 40 and the 2nd friend has ₹ 170.

Question 3.
A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And, if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.
Let the actual speed of the train be x km/hr and the actual time taken be ‘y’ hours.
Distance = speed x time = x × y
= xy km
xy = (x + 10) (y – 2)
xy = xy – 2x + 10y – 20
2x- 10y = – 20
xy = (x – 10) (y + 3)
xy = xy + 3x – 10y – 30
3x – 10y = 30 → (2)
Subtract equation (1) and (2)

Put x = 50 in equation (1)
2(50) – 10y = – 20
– 10y = – 20 – 100
– 10y = – 120
y = $$\frac{-120}{-10}$$
y = 12
distance covered by train = xy
= 50 × 12 = 600 km.

Question 4.
The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.
Let the number of students in the class be x and the number of rows be y. Then the number of students in each row $$\frac{x}{y}$$ .
If 3 students are extra in row, then there would be 1 row less i,e when each row
has
Students then the number of rows is (y – 1)
∴ Total number of students
= number of rows x number of students in each row

and, if 3 students are less in a row, then there would be 2 rows more i.e, when each row has $$\left(\frac{x}{y}-3\right)$$ students, then the number of rows is (y + 2).
∴ Total number of students = Number of rows x number of students in each row.

Then equation (1) and (2) can be written as
a – 3y = – 3 → (3)
2a – 3y = 6 → (4)
Subtract equation (3) and (4)

a = 9
Put a = 9 in equation (3)
9 – 3y = – 3
– 3y = – 3 – 9
– 3 y = – 12
x = 9 × 4
x = 36
Hence number of students in class is 36 and number of rows is 4.

Question 5.
In a ∆ABC, ∠C = 3 ∠B = 2(∠A + ∠B). Find the three angles.
Solution:
Sum of angles of a triangle = 180°
∴ ∠A + ∠B + ∠C = 180° … (1)
∵ ∠C = 3∠B = 2(∠A + ∠B) … (2)
From (1) and (2), we have ∠A + ∠B + 2 (∠A + ∠B) = 180°
⇒ ∠A + ∠B + 2∠A + 2∠B = 180°
⇒ ∠A + ∠B = 60° …. (3)
Also, ∠A + ∠B + 3∠B = 180°
⇒ ∠A + 4∠B = 180° ….(4)
Subtracting (3) from (4), we get
∠A + 4∠B – ∠A – ∠B = 180°- 60°

Substituting ∠B = 40° in (4) we get,
∠A + 4(40°) = 180°
⇒ ∠A = 180° – 160° = 20°
∴ ∠C = 3∠B = 3 × 40° = 120°
Thus, ∠A = 20°, ∠B = 40° and ∠C = 120°.

Question 6.
Draw the graphs of the equations 5x – y = 5 and 3x – y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y-axis.
5x – y = 5 → (1)
3x – y = 3 → (2)
From equation (1)
y = 5x – 5

From equation (2)
y = 3x – 3

In the graph, we observe that the coordinates of the vertices’s of the triangle ABC formed by the two lines represented, by the A(0, -3) and B(0, -5).

Question 7.
Solve the following pair of linear equations:
(i) px + qy = p – q
qx – py = p + q
px + qy = p – q → (1)
qx-py = p + q → (2)
Multiply equation (1) p and equation (2) by q
p2x + pqy = p2 – pq → (3)
q2x – pqy = pq + q2 → (4)

Put x = 1 in equation (1).
p(1) + qy = p – q
qy = p – q – p
qy = – q
y = $$\frac{-q}{q}$$
y = – 1
Hence, the solution of the given pair of linear equation is x = 1, y = – 1

(ii) ax + by – c
bx + ay = 1 + c
ax + by = c
bx + ay = 1 + c
then ax + by – c = 0 → (1)
bx + ay – (1 + c) = 0 → (2)
Solve by cross multiplication method.

Hence, the solution of the given pair of linear equation is

ax + by = a2 + b2

Hence, the solution of the given pair of equations x = a, y = b.

(iv) (a – b)x + (a + b)y = a2 – 2ab – b2
(a + b) (x + y) = a2 + b2
(a – b)x + (a + b)y = a2 – 2ab – b2 → (1)
(a + b) (x + y) = a2 + b2
(a + b)x + (a + b)y = a2 + b2 → (2)
Subtract equation (1) and (2),

(a – b)x – (a + b)x = a2 – 2ab – b2 – a2 – b2 ax – bx – ax – bx = – 2ab – 2b2.

Put x = a + b in equation (1)
(a – b) (a + b) + (a + b)y = a2 – 2ab – b2
a2 – b2 + (a + b)y = a2 – 2ab – b2
(a + b)y = a2 – 2ab – b2 – a2 + b2
(a + b)y = – 2ab

Hence, solution of the pair of linear equation is x = a + b,

(v) 152x – 378y = – 74
– 378x + 152y = – 604
152x – 378y = – 74 → (1)
– 378x + 152y = – 604 → (2)

226x – 226y = – 678 divide by – 226
x + y = 3 → (3)
Subtract equation (1) and (2)

Put x = 2 in equation (3)
2 + y = 3
y = 3 – 2
y = 1
Hence, the solution of the given pair of linear equations is x = 2, y = 1

Question 8.
ABCD is a cyclic quadrilateral (see figure). Find the angles of the cyclic quadrilateral.

Solution:
∴ ∠A + ∠C = 180° and
∠B + ∠D = 180°
⇒ [4y + 20] + [- 4x] = 180°
⇒ 4y – 4x + 20° -180° = 0
⇒ 4y – 4x – 160° = 0
⇒ y – x – 40° = 0 … (1)

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Students can Download Class 10 Maths Chapter 2 Triangles Ex 2.6 Questions and Answers, Notes Pdf, KSEEB Solutions for Class 10 Maths helps you to revise the complete Karnataka State Board Syllabus and score more marks in your in examinations

Karnataka State Syllabus Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 1.
In Fig. 2.56, PS is the bisector of ∠QPR of ∆ PQR, Prove that $$\frac{\mathbf{Q S}}{\mathbf{S R}}=\frac{\mathbf{P Q}}{\mathbf{P R}}$$

Given: In the figure, PS is the bisector of

Question 2.
In Fig. 2.57, D is a point on hypotenuse AD of ∆ ABC, such that BD ⊥ AC, DM ⊥ BC and DN ⊥ AB. Prove that:
i) DM2 = DN.MC ii) DN2 = DM.AN

Given: D is a point on hypotenuse AC of ∆ ABC DM ⊥ BC, DN ⊥AB and BD ⊥ AC

ii) Consider ∆le ADN And DBN

Question 3.
In the figure, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2 – AB2 + BC2 + 2BC.BD

Solution:
∆ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced.
∵ In ∆ADB, ∠D = 90°
∴ Using Pythgoras Theorem, we have
AB2 = AD2 + DB2 ….. (1)
In right ∆ADC, ∠D = 90°
∴ Using Pythagoras Theorem, we have
= AD2 + [BD + BC]2
= AD2 + [BD2 + BC2 + 2BD.BC]
⇒ AC2 = [AD2 + DB2] + BC2 + 2BC – BD
⇒ AC2 = AB2 + BC2 + 2BC – BD [From (1)] Thus, AC2 = AB2 + BC2 + 2 BC.BD

Question 4.
In Fig. 2.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC. BD.

Given: In figure, ABC is a triangle in which ∠ABC < 90° and AD ⊥ BC
To prove: AC2 = AB2 + BC2 – 2BC.BD
Proof: In right triangle ABC ∠ D = 90°
AC2 = AD2 + DC2 [By pythagoras theorem]
= AD2 + (BC – BD)2 [CD = BC – BD]
= AD2 ± BC2 + BD2 – 2BC. BD
AC2 = AB2 + BC2 – 2BC.BD
∠D = 90° AB2 = AD2 + BD2 By Pythagoras theorem]

Question 5.
In the figure, AD is a median of triangle ABC and AM ⊥ BC. Prove that

Solution:

Question 6.
Prove that the sum of the squares of the diagonals of parallelogram is equal to the sum of the squares of its sides.

Given: In parallelogram ABCD
AB = CD & AD = BC
Construction: convert parallelogram into a rectangle and Draw AG ⊥ CD
To prove:
AC2 + BD2 = AB2 + BC22 + CD2 – AD2
Proof: consider ∆le BDF ∠D = 90°
BD2 = BF2 + FD2 = h2 + (x+d)2 — (1)
Consider ∆le AGC ∠G = 90°
AC2 = AG22 + GC2 = h2 + (x – d)2 — (2)
AC2+ BD2 = h2 + (x + d)2 + h2 + (x – d)2
2h2 + x2 + 2xd + d2 + x2 – 2xd + d2
2h2 + 2x2 + 2d2
= 2x2 + 2(h2 + d2)
= 2x2 + 2y2
= x2 + y2 + x2 + AC2 + BD2 =AB2 +BC2 i-CD2 + AD2.

Question 7.
In the figure, two chords AB and CD intersect each other at the point P. Prove that:
(i) ∆APC – ∆DPB
(ii) AP.PB = CP.DP

Solution:
We have two chords AB and CD of a circle. AB and CD intersect at P.
(i) In ∆APC and ∆DPB,
∴ ∠APC = ∠DPB ….. (1)
[Vertically opp. angles]
∠CAP = ∠BDP …… (2)
[Angles in the same segment]
From (1) and (2) and using AA similarity, we have
∆APC ~ ∆DPB

(ii) Since, ∆APC ~ ∆DPB [As proved above]
∴ Their corresponding sides are proportional,
$$\Rightarrow \frac{A P}{D P}=\frac{C P}{B P}$$
⇒ AP.BP = CP.DP, which is the required relation.

Question 8.
In Fig. 2.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that

Given: In figure, two chords AB and CD of a circle intersect each other at the point P. (when produced) outside the circle.
To prove: i) ∆ PAC ~ ∆ PDB
ii )PA. PB = PC . PD
Proof: i) we know that in a cyclic quadrilateral, the exterior angle is equal to the interior opposite angle.
consider ABCD ∆ PAC and ∆ PBD.
∠PAC = ∠PDB → (i)
∠PCA = ∠PBD → (2)
∆ PAC ~ ∆ PDB [A A similarity criterion]

ii) A PAC ~ ∆ PDB (Proved)
$$\frac{P A}{P D}=\frac{P C}{P B}$$ [Corresponding sides of the similar ∆le are proportional]
PA . PB = PC . PD

Question 9.
In the figure, D is a point on side BC of ∆ABC such that $$\frac{B D}{C D}=\frac{A B}{A C}$$. Prove that AD is the bisector of ∠BAC.

Solution:
Let us produce BA to E such that AE = AC
Join EC.

Question 10.
Nazima is fly fishing in a stream. The tip of her fishing rod ¡s 1.8 m above the surface of the water and the fly at the end of the string rests on the water 3.6 m away nad 2.4 m from a point directly under the tip of the rod. Assuming that her string (from the tip of her rod to the fly) is taut, how much string does she have out (see Fig. 2.64)?

If she pulls in the string at the rate of 5 cm per second, what will be the horizontal distance of the fly from her after 12 seconds?

AC = 3m
Hence, she has 3 m string out.
Length of the string pulled in 12 seconds
at the rate of5 cm/ sec 5 × 12cm = 60 cm = O.6 m.
∴ Length of remaining string left out
= AD = 3.0 – 0.6 = 2.4m

In right angled ∆ ABD ∠B = 90°
[pythagoras theorem]
= (2.4)2 – (1.8)2
= 5.76 – 3.24 = 2.52
BD = 12.52 = 1.59 m (approx)
Hence, the horizontal distance of the fly form Nazirna after 12 seconds = 1.2 + 1.59 = 2.79 m.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 2 Triangles Ex 2.6 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 2 Triangles Exercise 2.6, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.4.

Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.4

Question 1.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them :
(i) 2x2 – 3x + 5 = 0
(ii) 3x2 – $$4 \sqrt{3} x$$ + 4 = 0
(iii) 2x2 – 6x + 3 = 0
Solution:
i) 2x2 – 3x + 5 = 0
It is in the form of ax2 + bx + c = 0
a = 2, b = – 3 and c = 5
Nature of roots = b2 – 4ac
Discriminant = (- 3)2 – 4 × (2)(5)
= 9 – 40
= – 31
Discriminant = – 31 < 0
The given Quadratic equation has no real roots.

(ii) 3x2 – $$4 \sqrt{3} x$$ + 4 = 0
Here, a = 3, b = $$4 \sqrt{3}$$, c = 4

(iii) 2x2 – 6x + 3 = 0
Here, a = 2, b = -6, c = 3
b2 – 4ac = (-6)2 – 4(2)(3)
= 36 – 24 = 12
∴ b2 – 4ac = 12 > 0.
∴ It has two distinct roots.
2x2 – 6x + 3 = 0
Here, a = 2, b = -3, c = 3

Question 2.
Find the values of k for each of the following quadratic equations, so that they have two equal roots
(i) 2x2 + kx+3 = 0
(ii) kx(x – 2) + 6 = 0
Solution:
(i) Comparing the given quadratic equation with ax2+ bx + c = 0, we get a = 2,b = k,c = 3
∴ b2 – 4ac = (k)2 – 4(2)(3) = k2 – 24
∵ For a quadratic equation to have equal roots, b2 – 4ac = 0

(ii) kx(x – 2) + 6 = 0 ⇒ kx2 – 2kx + 6 = 0
Comparing kx2 – 2kx + 6 = 0 with ax2 + bx + c = 0, we get
a = k, b = -2k, c = 6
∴  b2 – 4ac = (-2k)2 – 4(k)(6) = 4k2 – 24k
Since, the roots are equal

But k cannot be 0, otherwise, the given equation is not quadratic. Thus, the required value of k is 6.

Question 3.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m2 ? If so, find its length and breadth.

Solution:
Let the breadth of rectangular mango grove is ‘x’ and length of rectangular mango grove is 2x
Area of rectangular mango grove = 800 m2
x (2x)= 800
2x2 = 800
x2 = $$\frac{800}{2}$$ = 400
x = ± $$\sqrt{400}$$
x = ± 20m
∴ breadth of mango grove x = 20 m
length of mango grove = 2x = 2 × 20 = 40 m
It is possible to construct a rectangular mango grove of length 40m and breadth 20m.

Question 4.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Solution:
Let the age of one friend = x years
∴ Age of other friend = (20 – x) years
Four years ago,
Age of one friend = (x – 4) years
Age of other friend = (20 – x – 4) years
= (16 -x) years
According to the condition,
(x – 4) × (16 – x) = 48
⇒ 16x – 64 – x2 + 4x = 48
⇒ – x2 + 20x – 64 – 48 = 0
⇒ -x2 + 20x – 112 = 0
⇒ x2 – 20x + 112 = 0 …(1)
Comparing equation (1) with ax2 + bx + c = 0, we get a = 1, b = – 20 and c = 112

Since, b2 – 4ac is less than 0.
∴ The quadratic equation (1) has no real roots.
Thus, the given situation is not possible.

Question 5.
Is it possible to design a rectangular park of perimeter 80 m. and area 400 m∴? If so, find its length and breadth.
Solution:

Let the length of rectangular park is ‘x’
Perimeter of the rectangular park = 80m
2 (length + breadth) = 80
length + breadth = $$\frac{80}{2}$$ = 40
area of rectangle = l × b
x (40 – x) = 400
40x – x2 = 400
x2 – 40x + 400 = 0
x2 – 20x – 20x + 400 = 0
x (x – 20) – 20 (x – 20) = 0
(x – 20) (x – 20) = 0
x – 20 (or) x – 20 = 0
x = 20 m
length of rectangular park = 20 m and breadth of rectangular park = (40 – x)
= 40 – 20 = 20m
∴ It is possible to design rectangular park of length 20 m and breadth is 20 m.
∴ The park is a square having 20m side.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.4, drop a comment below and we will get back to you at the earliest

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2

Question 1.
Express each number as a product of its prime factors :
(i) 140
(ii) 156
(iii) 3825
(iv) 5005
(v) 7429
Solution:

Question 2.
Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91
(ii) 510 and 92
(iii) 336 and 54
Solution:
(i) 26 and 91
a = 26, b = 91

26 = 2 × 13
91 = 7 × 13
LCM of (26, 91) = 2 × 13 × 7 = 182
HCF of (26, 91) = 13
Verification
LCM of (26, 91) × H C F (26, 91) = 182 × 13 = 2366
product of two numbers = 26 × 91 = 2366
Hence LCM × HCF = Product of two numbers.

(ii) 510 and 92
a = 510, b = 92

(iii) 336 and 54
a = 336, b = 54

Question 3.
Find the LCM and HCF of the following integers by applying the prime factorization method.
(i) 12, 15 and 21
(ii) 17, 23 and 29
(iii) 8, 9 and 25
Solution:
(i) 12, 15 and 21
12 = 2 × 2 × 3
15 = 3 × 5
21 = 3 × 7
∴ H.C.F. = 3
L.C.M. = 3 × 2 × 2 × 5 × 7 = 420
∴ HCF × LCM = Product of three numbers
3 × 420 = 12 × 15 × 21
1260 = 1260

(ii) 17, 23 and 29
17 = 17 × 1
23 = 23 × 1
29 = 29 × 1
∴ H.C.F. = 1
L.C.M. = 17 × 23 × 29 = 11339
∴HCF × LCM = Product of three numbers
1 × 11339 = 17 × 23 × 29
11339 = 11339

(iii) 8 = 2 × 2 × 2 = 23
9 = 3 × 3 = 32
25 = 5 × 5 = 52
L C M of (8, 9, 25) = 23 × 32 × 52
= 8 × 9 × 25 = 1800
LCM of (8, 9, 25)= 1800
HCF of(8, 9, 25)= 1
(∵ 8, 9 & 25 havg no common Factor)

Question 4.
Given that HCF(306,657) = 9, find LCM (306,657).
Solution:
Here, HCF of 306 and 657 = 9 and we have
LCM × HCF = Product of the numbers.
∴ LCM × 9 = 306 × 657

Thus, LCM of 306 and 657 is 22338.

Question 5.
Check whether 6n can end with the digit 0 for any natural number ‘n’.
Solution:
If the number ends with digit 0, then the prime factorization of the number must contain the factors of 10 as 2 × 5.
Now the prime factorization of 6n = (2 × 3)n = 2n × 3n
Since it does not contain the factors of 10 as 2 × 5, thus n cannot end with the digit 0, for any natural number n.

Question 6.
Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Solution:
We have
7 × 11 × 13 + 13 = 13[(7 × 11) + 1] = 13[78]
i. e., 13 × 78 cannot be a prime number because it has factors 13 and 78.
∴ 13 × 78 is a composite number.
Also 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = 5[7 × 6 × 4 × 3 × 2 × 1 + 1] which is also not a prime number because it has a factor 5
Thus, 7 × 11 × 13 + 13 and
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.

Question 7.
There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time and go in the same direction. After how many minutes will they meet again at the starting
Solution:
Sonia takes 18 minutes to drive one round of the field.
Ravi takes 12 minutes for the same.
Suppose they both start at the same point and at the same time.
Time to meet both again at the starting point,
For this, we have to find out HCF of 18 and 12.
18 = 2 × 3 × 3 = 2 × 32
12 = 2 × 2 × 3 = 22 × 3
∴ H.C.F. = 2 × 3 = 6
∴ After 6 minutes, they both meet at the same point.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Exercise 8.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2

KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2.

Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2

Question 1.
Fill in the blanks In the following table,
given that ‘a’ is the first term, ‘d’ is the common difference and an the nth term of the A.P.

Solution:
(i) Given
a = 7, an = a + (n – 1 )d
d = 3, an = 7 + (8 – 1)3
n = 8, an = 7 + 7 x 3
an = ?, an = 7 + 21
an = 28
∴ nth term of an A.P = 28.

(ii) Given
a = – 18 an = a + (n – 1)d d = ? 0 = – 18 + (10 – 1)d
n = 10
an = 0
18 = 9d
d = $$\frac { 18 }{ 9 }$$
d = 2
Common difference = 2

(iii) Given
a = ?, a$$n$$ = a + (n – 1)d
d = – 3, – 15 = a + (18 – 1) – 3
n = 18, – 15 = a + 17 x – 3
a$$n$$ = – 5, – 15 = a – 51 = a
36 = a
a = 36
First term = 36.

(iv) Given
a = – 18.9, an a = a +(n – 1)d
d = 2.5, 3.6 = – 18.9 + (n – 1)2.5
n = ? 3.6 + 18.9 = (n – 1)2.5
n – 1 = $$\frac { 225 }{ 25 }$$
n – 1 = 9
n = 9 + 1
n = 10
an = 3.6
3.6 + 18.9 = (n – 1)2.5

(v) Given
a = 3.5, an = a + (n – 1)d
d = 0, an = 3.5 + (105 – 1)0
n = 105, an = 3.5 + 104 x 0
an = ?, an = 3.5

Question 2.
Choose the correct choice in the following and justify:
(i) 30th term of the AP: 10,7,4, , is, ….,
(A) 97
(B) 77
(C) -77
(D) -87

(ii) 11th term of the AP: -3, $$-\frac{1}{2}$$, 2, …. ,is
(A) 28
(B) 22
(C) -38
(D) -48$$\frac{1}{2}$$
Solution:
(i) (C): Here, a = 10, n = 30
∵ T10 = a + (n – 1)d and d = 7 – 10 = -3
∴ T30 = 10 + (30 – 1) × (-3)
⇒ T30 = 10 + 29 × (-3)
⇒ T30 = 10 – 87 = -77

(ii) (B): Here, a = -3, n = 11 and

Question 3.
In the following APs find the missing terms in the boxes:

Solution:
(i)
a = 2, a + d =?, a + 2d = 26
a + 26 = 26
2 + 2d = 26
2d = 24
∴ d=12 . .
∴ a + d = 2 + 12 = 14
∴ Ans: 14

(ii) a2 = 13, a4 = 3, a? and a3 = ?
a2 = a + d
13 = a + d
a = 13 – d ➝ (1)
a4 = a + 3d
3 = a + 3d ➝ (2)
Putting eqn (1) in eqn (2)
3 = 13 – d + 3d
3 – 13 = 2d
2d = – 10
d = $$\frac{10}{2}$$
d= – 5
Put d = – 5 in eqn (1)
a = 13 – (- 5) = 13 + 5 = 18
a = 18
Third term a3 = a + 2d
a3 = 18 + 2 x (- 5)
= 18 – 10 = 8
a3 = 8
Hence missing terms are 18 and 8.

(iii) a = 5, a + d =?, a + 2d =?

(iv) a = – 4, a6 = 6
a6 = a + 5d
6 = – 4 + 5d
6 + 4 = 5d
d = $$\frac{10}{5}$$ = 2
d = 2
a2 = a + d = – 4 + 2 = – 2
a3 = a + 2d = – 4 + 2(2) = – 4 + 40
a4 = a + 3d = – 4 + 3(2) = – 4 + 6 = 2
a5 = a + 4d = – 4 + 4(2) = – 4 + 8 = 4
Hence missing terms are – 2, 0, 2 & 4

(v) a =? a + d = 38, a + 2d =?
a + 3d =?, a + 4d =?a + 5d = 22

4d = -60
$$\mathrm{d}=-\frac{-60}{4} \quad=-15$$
a + d = 38
a – 15=38
a = 38 + 15 = 53
a + 2d = 53 + 2(-15) = 53 – 30 = 23
a + 3d = 53 + 3(-15) = 53 – 45 = 8
a + 4d =53 + 4(-15) = 53 – 60 = 7
∴ 53, 23, 8, -7.

Question 4.
Which term of the AP: 3, 8, 13, 18, is 78?
Solution:
Let the nth term be 78
Here, a = 3 ⇒ T1 = 3 and T2 = 8
∴ d = T2 – T1 = 8 – 3 = 5
And, Tn = a + (n- 1 )d
⇒ 78 = 3 + (n – 1) × 5 ⇒ 78 – 3 = (n -1) × 5
⇒ 75 = (n – 1) × 5 ⇒ (n – 1) = 75 ÷ 5 = 15
⇒ n = 15 + 1 = 16
Thus, 78 is the 16th term of the given AP.

Question 5.
Find the number of terms In each of the following APs:
(i) 7, 13, 19, ………… 201
(ii) $$18,15 \frac{1}{2}, 13, \dots \dots 47$$
Solution:
i) Given a = 7
d = 13 – 7 = 6 = 205
an = a + (n – 1)d
205 = 7 + (n – 1)6
205 – 7 = 6n – 6
198 + 6 = 6n
6n = 198 + 6
6n = 204
n = $$\frac{204}{6}$$ = 34
n = 34
Hence, number of terms of a given AP is 34.

(ii) $$18,15 \frac{1}{2}, 13, \dots \dots 47$$
$$a=18, d=a_{2}-a_{1}=\frac{31-36}{2}=\frac{-5}{2}$$
an = -47, n = ?
an = a + (n – 1) d
$$-47=18+(n-1)\left(\frac{-5}{2}\right)$$
$$-47-18=(n-1)\left(\frac{-5}{2}\right)$$
$$(n-1)\left(\frac{-5}{2}\right)=-65$$
$$n-1=-65 \times \frac{-2}{5}$$
n – 1 = -13 × -2
n – 1 = + 26
∴ n = 26 + 1
∴ n = 27

Question 6.
Check whether -150 is a term of the AP:
11, 8, 5, 2…
Solution:
For the given AP,
we have a = 11, d = 8 -11 = -3
Let -150 be the nth term of the given AP
∴ Tn = a + (n – 1 )d
⇒ -150 = 11 + (n – 1) × (-3)
⇒ -150 – 11 = (n – 1) × (-3)
⇒ -161 = (n – 1) ⇒ (-3)

But n should be a positive integer.
Thus, -150 is not a term of the given AP

Question 7.
Find the 31st term of an AP whose 11th term Is 38 and the 16th term is 73.
Solution:
a = 38, a16 = 83 a31 =?
an = a + (n – 1) d
a16 = a + (16 – 1) d
a + 15d = 83
38 + 15d = 83
15d = 83 – 38
15d = 45
$$d=\frac{45}{15}$$
∴ d=3.
∴ an = a + (n – 1)d
a31 = 38 + (31 – 1) 3
= 38 + 30 × 3
= 38 + 90
∴ a = 128.

Question 8.
An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term.
Solution:
Here, n = 50, T3 = 12, Tn = 106
⇒ T50 = 106
Let the first term = a and the common difference = d
Tn = a + (n – 1 )d
T3 = a + 2d = 12 …(1)
T50 = a + 49d = 106 …(2)
Subtracting (1) from (2), we get

Question 9.
If the 3rd and the 9th terms of an AP are 4 and -8 respectively, which term of this AP is zero?
Solution:
a3 = 4, a9 = -8, an = 0, n =?
a3 = a + 2d = 4 ………….. (1)
a9 = a + 8d = -8 …………. (2)
From equation (1) – equation (2).

6d = 12
$${d}=\frac{-12}{6}$$
∴ d = -2
a + 2d = 4
a – 2(2) = 4
a – 4 =4
∴ a = 4 + 4
∴ a = 8
an = a + (n – 1) d
= 8 + (n – 1) (-2)
= 8 – 2n + 2
= 10 – 2n = 0 ∵ an = 0
$$n=\frac{10}{2}$$
∴ n = 5
∴ 5th term of this AP is Zero.

Question 10.
The 1 7th term of an AP exceeds its 17th term by 7. Find the common difference.
Solution:
a17 = a10 + 7
a + 16d = a + 9d + 7
a + 16d – a – 9d = 7
7d = 7
d = $$\frac{7}{7}$$ = 1
Common difference = d = 1

Question 11.
Which term of the AP: 3, 15, 27. 39, … will be 132 more than its 54th term?
Solution:
3, 15, 27, 39, ………… an =?, n =?
an = a54 + 132
a = 3, d = 15 – 3 = 12
an = a54 + 132
an = a + 53d + 132
3 + 53(12) + 132
= 3 + 636 + 132
∴ an = 771
an = a + (n – 1) d = 771
= 3 + (n – 1)12 = 771
3 + 12n — 12 = 771
12n – 9 = 771
12n = 771 + 9
12n = 780
$$n=\frac{780}{12}$$
∴ n = 65.
∴ 65th term is 132 more than its 54th term.

Question 12.
Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?
Solution:
Let for the 1st AP, the first term = a and common difference = d

And for the 2nd AP, the first term = a and common difference = d

According to the condition,

Question 13.
How many three-digit numbers are divisible by 7?
Solution:
First three-digit number divisible by 7
105 and the List number Is 994.
∴ AP is 105, 112. 119 994.
a = 105, d = 112 – 105 = 7. an = 994.
n =?
a + (n – 1)d = an
105 + (n – 1) 7 = 994
105 + 7n – 7 = 994
7n + 98 = 994
7n = 994 – 98
7n = 896
$$n=\frac{896}{7}$$
∴ n = 128.
∴ Numbers with 3 digits divisible by 7 are 128.

Question 14.
How many multiples of 4 lie between 10 and 250?
Solution:
Multiples of 4. after 10 are 12, 16, 20….
Multiples of 4 upto 250 is 248
∴ A.P. is 12, 16, 20, …….. 248
a = 12, d = 16 – 12 = 4
n =?
a = a + (n – 1) d = 248
12 + 4n – 4 = 248
4n + 8 = 248
4n = 248 – 8
4n = 240
$$n=\frac{240}{4}$$
∴ n = 60
∴ Multiples of 4 lie between 10 and 250 is 60.

Question 15.
For what value of n, are the nth terms of two APs: 63,65,67 … and 3,10,17, …. equal?
Solution:

Thus, the 13th terms of the two given AP’s are equal.

Question 16.
DetermIne the AP whose third term is 16 and 7th term exceeds the 5th term by 12.
Solution:
a3 = a + 2d
16 = a + 2d ➝ (1)
7th term = 5th term + 12
a + 6d = a + 4d + 12
a – a + 6d – 4d = 12
2d = 12
d = $$\frac{12}{2}$$ = 6
put d = 6 in eqn (1)
16 = a + 2 (6)
16 – 12 = a
a = 4
Hence required AP a, a + d, a + 2d
4, 4 + 6, 4 + 2(6),…
4, 10, 16, 22, …..

Question 17
Find the 20th term from the Last term of the AP: 3, 8, 13, …………., 253.
Solution:
3, 8, 13, …., 253.
Given a = 3
d = a2 – a = 8 – 3
d = 5
an = 253
an= a + (n – 1)d
253 = 3 + (n – 1)5
253 – 3 (n – 1)5
n – 1 = $$\frac{250}{5}$$ = 50
n = 50 + 1 = 51
20th term from last term is 51 – 19 = 32
a32 = a + 31 d
= 3 + 31 x 5 = 3 + 155
a32 = 158
20th term from the last term is 158

Question 18.
The sum of the 4th and 8th terms of an AP is 24 and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.
Solution:
Let the first term = a and the common difference = d

Question 19.
Subba Rao started work in 1995 at an annual salary of Rs. 5000 and received an Increment of Rs. 200 each year. in which year did his income reach Rs. 7000?
Solution:
Payment of Subba Rao in the year 1995 = Rs. 5000
increment = Rs. 200
∴ The payment he received in the year 1996 is Rs. 5.200

a = 5000, d = 5200 – 5000 = 200,
an = 7000, n=?
a + (n – 1)d = an
5000 + (n – 1) 200 = 7000
5000 + 200n – 200 = 7000
200n + 4800 = 7000
200n = 7000 – 4800
200n = 2200
$$n=\frac{2200}{200}$$
∴ n = 11
Value of ‘n’ is 11
∴ From 1995 to 10 years means 2005. his salary becomes Rs. 7,000.

Question 20.
Ramkall saved Rs. 5 In the first week of a year and then Increased her weekly savings by Rs. 1.75. If in the nüI week. her weekly savings become Rs. 20.75. find ‘n’.
Solution:
Let the saving in the first week = ₹ 5
The savings in the next week = ₹ 5 + ₹ 175 = 6.75
saving in the next week = ₹ 6.75 + ₹ 1.75 = Rs.8.5
∴ AP, 5, 6.75, 8.5, …………. 20.75
a = 5, d = 1.75 an = 20.75 & n = ?
an = a + (n – 1)d
20.75 = 5 + (n – 1)1.75
20.75 – 5 = (n – 1)1.75
15.75 = (n – 1) 1.75
n – 1 = $$\frac{15.75}{1.75}$$
n – 1 = 9 = 9 + 1 = 10
n = 9 + 1
n = 10

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Exercise 1.2, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3

KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.3.

Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.3

Question 1.
Find the roots of the following quadratic equations, if they exist, by the method of completing the square:
(i) 2x2 – 7x + 3 = 0.
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + $$4 \sqrt{3} x$$ + 3 = 0
(iv) 2x2 + x + 4 = 0
Solution:
(i) 2x2 – 7x + 3 = 0
Dividing all terms by 2,

(ii) 2x2 + x – 4 = 0
Dividing all terms by 2,

(iii) 4x2 + $$4 \sqrt{3} x$$ + 3 = 0
Dividing all terms by 4,

(iv) 2x2 + x + 4 = 0
Dividing all terms of the equation by 2,

Here ‘x’ has no fixed value because $$\sqrt{-31}$$ is not a square root.

Question 2.
Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.
(i) 2x2 – 7x + 3 = 0.
(ii) 2x2 + x – 4 = 0
(iii) 4x2 + $$4 \sqrt{3} x$$ + 3 = 0
(iv) 2x2 + x + 4 = 0
Solution:
(i) 2x2 – 7x + 3 = 0.
Here, a = 2, b = -7, c = 3

(ii) 2x2 + x – 4 = 0
Here, a = 2, b = 1, c = 4

(iii) 4x2 + $$4 \sqrt{3} x$$ + 3 = 0
Here, a = 4, b = $$4 \sqrt{3}$$, c = 3

(iv) 2x2 + x + 4 = 0
Here, a = 2, b = 1, c = 4

Question 3.
Find the roots of the following equations:

Solution:

Taking negative sign, x = $$\frac{3-1}{2}$$
Thus, the required roots of the given equation are 2 and 1.

Question 4.
The sum of the reciprocals of Rehman’s ages, (in years) 3 years ago and 5 years from now is $$\frac{1}{3}$$. Find his present age.
Solution:
Let the present age of Rehman be ‘x’ years.
After 5 years his age will be (x + 5)
3 years back his age was (x – 3)

(x – 7) (x + 3) = 0
If x – 7 = 0, then x = 7
If x + 3 = 0, then x = -3
∴ x = 7 OR x = -3
∴ Present age of Rehman is 7 years.

Question 5.
In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.
Solution:
Let Shefali’s marks in mathematics is Y and marks in English is (30 – x)
Shefali’s gets 2 more marks in mathematics then x + 2 and 3 marks less in English is (30 – x – 3)
Product of Shefali’s marks = 210
(x + 2) (30 – x – 3) = 210
(x + 2) (27 – x) = 210
27x + 54 – x2 – 2x = 210
– x2 + 25x + 54 = 210
x2 – 25x + 210 – 54 = 0
x2 – 25x + 156 = 0
x2 – 13x – 12x + 156 = 0
x(x – 13) – 12 (x – 13) = 0
(x – 13) (x – 12) = 0
x – 13 = 0 (or) x – 12 = 0
x = 13 (or) x = 12
Therefore, Shefali’s marks in Mathematics =13
Marks in English = 30 – 13 = 17 (or) Shefali’s marks in Mathematics = 12 then.
Marks in English = 30 – 12 = 18

Question 6.
The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field.
Solution:

Let shorter side of rectangle ABCD be x metre,
Then, longer side of rectangle ABCD,
AB = (x + 30) m.
Diagonal AC = (x + 60) m.
In ⊥∆ABC, ∠B = 90°.
As per Pythagoras theorem,
AB2 + BC2 = AC2
(x + 30)2 + (x)2 = (x + 60)2
x2 + 60x + 900 + x2 = x2 + 120x + 3600
2x2 + 60x + 900 = x2 + 120x + 3600
2x2 – x2 + 60x – 120x + 900 – 3600 = 0
x2 – 60x – 2700 = 0

x2 – 90x + 30x – 2700 = 0
x(x – 90) + 30 (x – 90) = 0
(x – 90) (x + 30) = 0
If x – 90 = 0, then x = 90
If x + 30 = 0, then x = -30
Shorter side of rectangle, BC = x = 90 m.
Longer side of rectangle, AB = x + 30 = 90 + 30= 120 m.
Diagonal of rectangle, AC = x + 60 = 90 + 60 = 150 m.

Question 7.
The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.
Solution:
Let the larger number be x.
Since, (smaller number)2 = 8(larger number)
⇒ (smaller number)2 = 8
⇒ smaller number = $$\sqrt{8 x}$$
According to the condition,

Thus, the smaller number = 12 or -12
Thus, the two numbers are 18 and 12 or 18 and -12.

Question 8.
A train travels 360 km at a uniform speed. If the speed had been 5 km/h more, it would have taken 1 hour less for the same journey. Find the speed of the train.
Solution:
Let initial speed of the train be x km/h,
Then, time taken to cover 360 km
If the speed had been 5 km/h more,
$$\frac{360}{x}$$ Hr.
then time taken to cover is $$\frac{360}{x+5}$$ Hr.

x(x + 365) = 360 (x + 5)
x2 + 365x = 360x + 1800
x2 + 365x – 360x – 1800 = 0
x2 + 5x – 1800 = 0
x2 + 45x – 40x – 1800 = 0
x(x + 45) – 40 (x + 45) = 0
(x + 45) (x – 40) = 0
If x + 45 = 0, then x = – 45
If x – 40 = 0, then x = 40
∴ Initial speed of a train is 40 km/hr.

Question 9.
Two water taps together can fill a tank in $$9 \frac{3}{8}$$ hours. The tap of a larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Solution:
Let the time required to fill up the tank for tap having a larger diameter be ‘x’ Hr. For ‘x’ hour part of the tank filling is 1.
For $$9 \frac{3}{8}$$ hr. part of the tank filling ………..?
i.e., $$\frac{75 \times 1}{8 x}$$ part
Time required for smaller diameter is = (x – 10)Hr.
For (x – 10)Hr part of the tank, filling is 1
For $$9 \frac{3}{8}$$ Hr. part of the tank is …………?

∴Time required for larger diameter, x = 25 Hr.
∴ Time required for smaller diameter = x – 10 = 15 Hr.

Question 10.
An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11km/h more than that of the passenger train, find the average speed of the two trains.
Solution:
Let Average speed of passenger train be = x km/h.
. But Average speed of Express train be = (x + 11) km/h.
Time required for Passenger:
For travelling x km time required is 1 Hr
For travelling 132 km time required …….?
$$\frac{132}{x}$$ Hr.
Time required for Express train :
For travelling (x + 11) km is 1 Hr.
For travelling 132 km …………?
i.e., $$\frac{132}{(x+11)}$$ Hr
Express train required 1 Hr. less comparing to Passenger train. .

x(x + 143) = 132 (x+ 11)
x2 + 143x = 132x + 1452
x2 + 143x – 132 x – 1452 = 0
x2 + 11x – 1452 = 0
x2 + 44x – 33x – 1452 = 0
x(x + 44) – 33(x + 44) = 0
(x + 44) (x – 33) = 0
If x + 44 = 0, then x = -44
If x – 33 = 0, then x = 33
∴ Average speed of passenger train is 33 km/hr.
∴ Average speed of Express train = x + 11 = 33 + 11 = 44 km/hr.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.3 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Exercise 10.3, drop a comment below and we will get back to you at the earliest.

KSEEB SSLC Class 10 Science Solutions Chapter 15 Our Environment

KSEEB SSLC Class 10 Science Solutions Chapter 15 Our Environment are part of KSEEB SSLC Class 10 Science Solutions. Here we have given Karnataka SSLC Class 10 Science Solutions Chapter 15 Our Environment.

Karnataka SSLC Class 10 Science Solutions Chapter 15 Our Environment

KSEEB SSLC Class 10 Science Chapter 15 Intext Questions

Text Book Part I Page No. 140

Question 1.
What are trophic levels? Give an example of a food chain and state the different trophic levels in it.
Several hierarchical level in an ecosystem are called trophic levels. They are based on following facts:

• Energy and its transportation in form of food from one end to other.
• Biomass balance in ecosystem.
• Food chain of pond is a simple food chain. Example:

Weeds → Small pond animals → Fish (Rohu) → Man → Big fishes

Trophic level of a Pond:

1. Producer: Weeds and plants of pond like ‘duck weed’, ‘lotus’ or ‘typha’ are at lowest level or bottom of a pond which produce food for pond animals and energy transfer.
2. Primary consumers: Algae, bacteria, Phytoplankton eat producer and conserve the energy in their body.
3. Secondary consumers: Fishes, crabs etc. are those which feeds on primary consumers.
4. Top consumers: Bigger fishes and humans are top consumers.

Question 2.
What is the role of decomposers in the ecosystem?
Decomposers has very important place in any ecosystem as every ecosystem contains dead and decaying materials which is of no use and most of the time harmful for the system too. Decomposers act over these materials and change or break them to the components easily invisible to the environment. Hence, decomposers act over dead and decayed materials and convert them to the components which can be easily mixed up with other environmental factors.

Text Book Part I Page No. 142

Question 1.
Substances that are broken down by biological processes are said to be biodegradable. Eg: paper and peel of a fruit. But plastic leather etc. are not broken down. These are called Non-biodegradable.

Question 2.
Give any two ways in which biodegradable substances would affect the environment.

1. Leaves of the plants decay and reduces soil fertility,
2. Bio-degradable substances have carbon. When this is burnt, CO2 and CO are produced and causes air pollution.

Question 3.
Give any two ways in which non-biodegradable substances would affect the environment.

1. As these are not decomposing, they cause air pollution and water pollution.
2. Plastic enters stomach of many animals and causes death of animals.

Text Book Part I Page No. 144

Question 1.
What is ozone and how does it affect any ecosystem?
Ozone is an atmospheric gas and its chemical formula is 03. It is formed by the union of three atoms of oxygen.

Formation of Ozone:
1. In the presence of UV rays, oxygen molecules break down to its component or forms to nascent oxygen atom.

2. This oxygen atom combines with another molecule of oxygen and forms ozone molecule.

Effect of Ozone:

Ozone is not good for health or to be consumed as it is poisonous in nature but it plays an important role in environment protection. The living organisms and factors which prevails life at earth, like water and atmospheric gases get distracted by UV rays coming from sun. Ozone forms a layer over our atmosphere and absorb harmful UV rays.

Question 2.
How can you help in reducing the problem of waste disposal? Give any two methods.
(a) Waste disposal and management is a big issue nowadays as new technologies produce many useful products but at the end :t turn to a garbage because all the things are non-biodegradable in nature. Waste management require 3R principle i.e., Reduce, reuse and recycle.

(b) Waste disposal can be managed by:

1. Categorizing the waste and disposing accordingly.
2. Collecting things which can be recycled after use and send them to recycling units.
3. Making manure by unrequired, spoiled and left over food products.

KSEEB SSLC Class 10 Science Chapter 15 Textbook Exercises

Question 1.
Which of the following groups contain only biodegradable items?
(a) Grass, flowers and leather.
(b) Grass, wood and plastic.
(c) Fruit-peels, cake and lime-juice.
(d) Cake, wood and grass.
(b) Grass, wood and plastic.

Question 2.
Which of the following constitute a food-chain?
(a) Grass, wheat and mango.
(b) Grass, goat and human.
(c) Goat, cow and elephant.
(d) Grass, fish and goat.
(b) Grass, goat and human.

Question 3.
Which of the following are environment-friendly practices?
(a) Carrying cloth-bags to put purchases in while shopping.
(b) Switching off unnecessary lights and fans.
(c) Walking to school instead of getting your mother to drop you on her scooter
(d) All of the above.
(d) All of the above.

Question 4.
What will happen if we kill all the organisms in one trophic level?
Ecosystem will get disturbed due to killing of organisms in one trophic level:
1. If producers are killed: No energy will be converted to food for primary consumers, so they will die, since primary consumers are not available, secondary and top consumers as well will collapse.

2. If primary consumers are killed: Secondary and top consumers will die immediately in absence of food (in form of primary consumer) and if all fauna will die, population will stop and humans will also be effected in absence of fauna: Similarly, CO2 balance will be reduced and cause depletion of flora,

Question 5.
Will the impact of removing all the organisms in a trophic level be different for different trophic levels? Can the organisms of any trophic level be removed without causing any damage to the ecosystem?

• No, same effect will be caused and that is disturbance in ecosystem.
• No, there is no way to remove a trophic level organism without harming ecosystem.

Question 6.
What is biological magnification? Will the levels of this magnification be different at different levels of the ecosystem?
Some non-biodegradable substances (like pesticides) somehow enters the food chain or web and since, they do not get decomposed at any level and get accumulated in the bodies of top level consumers, this phenomenon is called biological magnification.

Yes, at lower level it is lesser and at higher it is more.

Question 7.
What are the pro Problems caused by the non-biodegradable wastes that we generate?
Technological development leads to a better and comfortable life a way and garbage enhancement in other way. We always think of a product which last for long time or a durable product which do not get affected by atmospheric changes like waterproof heat resistance product etc. But after use due to its nature, it do not get degraded and remain garbage for very long time.

Problem generated by garbage:

1. They react with other materials and atmospheric gases and form many harmful by products and chemicals.
2. Harm the animals which mistakenly eat them.
3. Acquire huge land and engage manpower to manage them all the time.
4. Pollution rises.

Question 8.
If all the waste we generate is biodegradable, will this have no impact on the environment?
Yes, biodegradable waste will be decomposed with lesser time and its component will be mixed with environment. Hence, it will leave no impact on environment.

Question 9.
Why is damage to the ozone layer a cause for concern? What steps are being taken to limit this damage?
Ozone is an atmospheric gas and its chemical formula is O3.

Damage to the ozone layer will lead:

1. More UV rays to enter the atmosphere.
2. Damage to flora and fauna.
3. Diseases to human beings related to exposure to sun.

Damage control. Protection to ozone layer can be initiated by:

1. Main reason for ozone layer depletion is pollution. Hence, ceasing pollution is a must to stop the depletion.
2. Coolant like CFC must be banned which are considered to be most harmful to ozone layer.

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 15 Our Environment will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 15 Our Environment, drop a comment below and we will get back to you at the earliest.