KSEEB SSLC Class 10 Maths Solutions Karnataka State Syllabus

Expert Teachers at KSEEBSolutions.com has created Karnataka SSLC KSEEB Solutions for Class 10 Maths Pdf Free Download in English Medium and Kannada Medium of KTBS Karnataka State 10th Standard Maths Textbook Solutions Answers Guide, Textbook Questions and Answers, Notes Pdf, Model Question Papers with Answers, Study Material are part of KSEEB SSLC Class 10 Solutions.

Here we have given KSEEB Karnataka State Board Syllabus Class 10 Maths Textbook Solutions based on NCERT Syllabus. Students can also read Karnataka SSLC Maths Model Question Papers with Answers 2020-2021 Pdf.

Karnataka State Board Syllabus for Class 10 Maths Solutions

KSEEB Solutions for Class 10 Maths

10th Standard Maths Textbook Solutions Karnataka State Syllabus in English Medium

KSEEB Solutions For Class 10 Maths Chapter 1 Arithmetic Progressions

KSEEB 10th Maths Solutions Chapter 2 Triangles

SSLC Maths Solutions Class 10 KSEEB Chapter 3 Pair of Linear Equations in Two Variables

10th Standard Karnataka State Syllabus Maths Guide Chapter 4 Circles

Karnataka State Board 10th Maths Solutions Chapter 5 Areas Related to Circles

SSLC Maths Solutions KSEEB Chapter 6 Constructions

10th Maths In Kannada Solutions for Class 10 Chapter 7 Coordinate Geometry

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers

10th Maths Syllabus State Board Karnataka Solutions Chapter 9 Polynomials

SSLC Mathematics Solution Part 2 KSEEB Chapter 10 Quadratic Equations

Maths Guide For Class 10 State Syllabus Karnataka Chapter 11 Introduction to Trigonometry

Maths Solutions for Class 10 State Syllabus Karnataka Chapter 12 Some Applications of Trigonometry

Karnataka SSLC Maths Text Book Solutions Chapter 13 Statistics

Karnataka State Board Syllabus for Class 10 Maths textbook Solutions Chapter 14 Probability

Karnataka SSLC Maths Notes Pdf Chapter 15 Surface Areas and Volumes

Karnataka State Board Syllabus for Class 10 Maths Solutions in Kannada Medium

10th Standard Maths Textbook Solutions Karnataka State Syllabus in Kannada Medium

KSEEB Solutions For Class 10 Maths Chapter 1 Arithmetic Progressions

KSEEB 10th Maths Solutions Chapter 2 Triangles

SSLC Maths Solutions Class 10 KSEEB Chapter 3 Pair of Linear Equations in Two Variables

10th Standard Karnataka State Syllabus Maths Guide Chapter 4 Circles

Karnataka State Board 10th Maths Solutions Chapter 5 Areas Related to Circles

SSLC Maths Solutions KSEEB Chapter 6 Constructions

10th Maths In Kannada Solutions for Class 10 Chapter 7 Coordinate Geometry

KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers

10th Maths Syllabus State Board Karnataka Solutions Chapter 9 Polynomials

SSLC Mathematics Solution Part 2 KSEEB Chapter 10 Quadratic Equations

Maths Guide For Class 10 State Syllabus Karnataka Chapter 11 Introduction to Trigonometry

Maths Solutions for Class 10 State Syllabus Karnataka Chapter 12 Some Applications of Trigonometry

Karnataka SSLC Maths Text Book Solutions Chapter 13 Statistics

Karnataka State Board Syllabus for Class 10 Maths textbook Solutions Chapter 14 Probability

Karnataka SSLC Maths Notes Pdf Chapter 15 Surface Areas and Volumes

We hope the given Karnataka SSLC KSEEB Solutions for Class 10 Maths Pdf Free Download in English Medium and Kannada Medium of KTBS Karnataka State 10th Standard Maths Textbook Solutions Answers Guide, Textbook Questions and Answers, akub koyyur 10th maths Notes Pdf, Model Question Papers with Answers, Study Material will help you.

If you have any queries regarding KSEEB Karnataka State Board Syllabus Class 10th Std Maths Text Book Solutions based on NCERT Syllabus, drop a comment below and we will get back to you at the earliest.

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?
Answer:

  1.  We must refuse to buy products that harm us and the environment.
  2.  We should minimise the use of electricity and water.
  3.  We should encourage recycling of things.

Question 2.
What would be the advantages of exploiting resources with short-term aims?
Answer:
With the human population increasing at a tremendous rate due to improvement in health-care, thedem and for all resources is increasing at an exponential rate. The management of natural resources requires a long term perspective so that these will last for the generations to come and will not merely be exploited to the hilt for short term gains.

Question 3.
How would these advantages differ from the advantages of using a long term perspective in managing our resources?
Answer:
If resources are used in accordance with short term aims, present generation will be able to utilize the resources properly for overall development. But if we plan to use resources with long term aims, not only the present generation is benefited but also the future generations will also be able to utilize resources for fulfilling its necessities. Thus it would be better to use our natural resources with a long term perspective so that it could be used by the present generation as well as conserved for future use.

Question 4.
Why do you think that there should be equitable distribution of resources? What forces would be working against an equitable distribution of our resources?
Answer:
Nature shows no partiality. Natural resources belong to all and these resources should be used judiciously. Equitable distribution of resources will benefit both poor as well as rich people.
Human greed, corruption, and the lobby of the rich and powerful are the forces working against an equitable distribution of our resources.

Text Book Part I Page No. 132

Question 1.
Why should we conserve forests and wildlife?
Answer:

  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.
Answer:

  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.
Find out about the traditional systems of water harvesting/ management in your region.
Answer:
We must dug small pits and lakes, put in place simple water shed systems, built small earthen dams, constructed dykes, sand and limestone reservoirs, set up root top water collecting units. These are the traditional systems of water harvesting/management in our region.

Question 2.
Compare the above system with the probable systems in hilly/ mountainous areas or plains or plateau regions.
Answer:
In the above mentioned places check dams are built because here water harvesting is difficult.

Question 3.
Find out the source of water in your region/locality. Is water from this source available to all people living in that area?
Answer:
Tube wells and river water (Tungabhadra) are the water sources available to all people in our area. There are different sources in different places. In some places there is too much shortage of water because of failure of rain recently.

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?
Answer:

  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?
Answer:

  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?
Answer:
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 and
(c) coal and petroleum?
Answer:
a) Forests and wild animals.

  1. cutting valuable trees should be avoided by destroying forest affects the quality of soil and water resources.
  2. Hunting should be prohibited.
  3. There should be wild sanctuaries which gives protection for wild animals.

b) Water Resources:
Answer:

  1. Water resources should be free from pollution.
  2. Excess usage of water should be avoided.

c) Coal and Petroleum:
Answer:
We should minimise the use of coal and petroleum, because these are fossil fuels. By burning these there are ill effects such as air pollution and acid rainfall etc.

Question 5.
What can you as an individual do to reduce your consumption of the various natural resources?
Answer:

  1. We must have come across the five R’s to save the environment: Refuse, Reduce, Reuse, Repurpose and Recycle.
  2. We should encourage tree plantation programmes.
  3. We must reduce the burning of fossil fuels.
  4. Encouragement should be given for harvesting the water.

Question 6.
List five things you have done over the last one week to —
(a) conserve our natural resources.
Answer:
We should travel in bus instead of using own vehicles or we should practice walking, we must use LED bulbs or fluorescent tubes in our homes. We must use the lift or taking the stairs, wearing an extra sweater or using a heating device (heater or sign) on cold days.

(b) increase the pressure on our natural resources.
Answer:

  1. We should grow Number of trees around our house.
  2. Reducing own vehicles by using public transport system or by, walking.
  3. There should not be more factories.
  4. We must prevent soil erosion.
  5. We must reduce the usage of vehicles to avoid air pollution.

Question 7.
On the basis of the issues raised in this chapter, what changes would you incorporate in your lifestyle in a move towards sustainable use of our resources?
Answer:
We need to change our lifestyles so that we can use natural resources on a sustainable basis. The changes which can be brought about are as follows:

  • Stop cutting trees and start planting trees.
  • Use LED bulbs and fluorescent tubes.
  • Take the stairs and avoid using lifts.
  • During summers use bamboo made fans avoid air coolers and electricians.
  • Use more of public transport.
  • Let our conscience be always alert not to pollute the environment from any of our activities.

KSEEB SSLC Class 10 Science Chapter 16 Additional Questions and Answers

1. Fill in the blanks

Question 1.
…… and …… are followed as means of protection of nature and natural resources.
Answer:
traditions, customs and rituals.

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

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

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

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

We hope the given KSEEB SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources will help you. If you have any query regarding Karnataka SSLC Class 10 Science Solutions Chapter 16 Sustainable Management of Natural Resources, drop a comment below and we will get back to you at the earliest.

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
Answer:
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
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 1
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
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 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

KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4

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.
Answer:
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
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 3
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 4

Question 3.
If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.
Answer:
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 = ±\(\sqrt{2}\)

KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4

Question 4.
If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ±\(\sqrt{3}\). find other zeroes.
Answer:
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.
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 5
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.
Answer:
Let us apply the division algorithm to the given Polynomial
KSEEB Solutions for Class 10 Maths Chapter 9 Polynomials Ex 9.4 6
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 9 Polynomials Ex 9.4

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]
Answer:
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
Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 1
∴ n = 32.
Hence, 32nd term of the given AP is the first negative term.

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

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.
Answer:
Let the first term be ‘a’ and common difference is ‘d’
a3 + a7 = 6.
a + 2d + a + 6d = 6
2a + 8d = 6 divide by 2
a + 4d = 3
a = 3 – 4d → (1)
a3 × a7 = 8
(a + 2d) (a + 6d) = 8
(3 – 4d + 2d) (3 – 4d + 6d) = 8
(3 – 2d) (3 + 2d) = 8
Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 2
Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 3
Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 4

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

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].

Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 5
Answer:
Distance between top and bottom rung
Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 6
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
Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 7

Question 4.
The houses of a row are numbered consecutively from Ito 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint: Sx-1 = S49 – Sx]
Answer:
The numbers of houses are 1, 2, 3, 4, ……… 49
a = 1 & d = 1
sum of n terms of a AP
S = \(\frac{n}{2}\)[2a + (n – 1)d]
Sum of number of houses preceding xth house is equal to Sx-1 i.e.,

Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 8
Sum of numbers of houses following xth house is equal to S49 – Sx.
Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 9

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

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 = \(\frac{1}{4} \times \frac{1}{2} \times 50 \mathrm{m}^{3}\)
Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 10
Answer:
Volume of concrete required to build the first step length × breadth × Height.Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 11
∴ Total volume of concrete required to build the terrace of 15 steps
Karnataka State Syllabus Class 10 Maths Chapter 1 Arithmetic Progressions Ex 1.4 12

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

KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2

Students can Download Class 10 Maths Chapter 14 Probability Ex 14.2 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 14 Probability Ex 14.2

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Answer:
Number of all possible out comes
= 5 × 5 = 25
n(S) = 25
(i) Let E1 be a event favourable to visit same day for shop [Tue, Tue; Wed, Wed; Th, Thu; Fri, Fri; Sat, Sat]
n(E1) = 5
Probability of event both visit shop same day
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 1

(ii) Let E2 be a event favourable to visit shop both on consecutive days
E2 = {T, W; W, T; W, Th; Th, W; Th, F; F, Th; F, S; S, F} n(E2) = 8
Probability of a event favourable to visit both consecutive days
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 2

(iii) Probability of both will visit shop different days = 1 – Probability both visit shop same day
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 3

KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2

Question 2.
A die is numbered in such a way that its faces show the numbers 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws:
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 4
What is the probability that the total score is
Answer:
(i) even?
(ii) 6?
(iii) at least 6?
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 5
Total number of possible out comes
∴ n(S) = 36

(i) Let E1 be a event of sum is an even number.
∴ E1 = [2, 4, 4, 4, 4, 8, 4, 4, 8, 4, 6, 6, 4, 6, 6, 8, 8, 12] ∴ n(E1) = 18
Probability of sum is even number
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 6

(ii) Let E2 be a event of sum is equal to 6. ∴ n (E2) = 4
Probability of sum is equal to 6
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 7

(iii) Let E3 be a event of sum of out comes favourable to score atleast 6.
n (E3) = 15
Probability of sum atleast 6
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 8

Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is double that of a red ball, determine the number of blue balls in the bag.
Answer:
Let the number of blue balls = x.
∴ total number of balls = 5 + x.
P (drawing red ball) = \(\frac{5}{x+5}\)
P (drawing blue ball) = \(\frac{x}{x+5}\)
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 9
x = 2 × 5
x = 10
Hence the number of blue balls in a bag is 10.

KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2

Question 4.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Answer:
Total number of balls in box =12
∴ n(S) = 12, number of out comes favourable to event drawing black ball = n(E) = x
Probability of drawing black ball
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 10
If 6 more balls put in a box then total number of balls = n(S) = 6 + 12 = 18
∴ n(S)=18
Number of out comes favourable to event drawing black ball = n (E) = x + 6
Probability of drawing black ball
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 11
According to Question from (1) and (2)
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 12
6x + 36 = 18x
36 = 18x – 6x
12x = 36
x = \(\frac{36}{12}\) = 3
x = 3

Question 5.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is \(\frac{2}{3}\) Find the number of blue balls in the jar.
Answer:
Total number of marbles in the jar = 24
Let the number of blue marbles is x and the number of green marbles is 24 – x
Probability of marble drawn is green
KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2 13
24 – x = 16
x = 24 – 16
x = 8
∴ Number of blue marble is 8.

KSEEB Solutions for Class 10 Maths Chapter 14 Probability Ex 14.2

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.
Answer:
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)

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 1
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)
Adding equation (1) and (2)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 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.

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

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)].
Answer:
Let the amount of their respective capitals be ₹x and ₹y
x + 100 = 2(y – 100)
x+ 100 = 2y – 200
x – 2y = – 300 → (1)
and y + 10 = 6 (x – 10)
y + 10 = 6x – 60
– 6x + y = – 70 multiply by – 2
12x – 2y = 140 → (2)
Subtract equation (1) and (2)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 3
x = 40
Put x = 40 in equation (1)
40 – 2y = – 300
– 2y = – 300 – 40
– 2y = – 340
y = \(\frac{340}{2}\)
y = 170
∴ amounts of their respective capitals are ₹40 and ₹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.
Answer:
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)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 4
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.

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

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.
Answer:
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
hasKSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 5
Students then the number of rows is (y – 1)
∴ Total number of students
= number of rows x number of students in each row
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 6

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.
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 7
Then equation (1) and (2) can be written as
a – 3y = – 3 → (3)
2a – 3y = 6 → (4)
Subtract equation (3) and (4)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 8
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.
Answer:
Given
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 9

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.
Answer:
5x – y = 5 → (1)
3x – y = 3 → (2)
From equation (1)
y = 5x – 5
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 10
From equation (2)
y = 3x – 3
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 11
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).

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

Question 7.
Solve the following pair of linear equations:
(i) px + qy = p – q
qx – py = p + q
Answer:
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)
Adding equation (3) and (4)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 12
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
Answer:
ax + by = c
bx + ay = 1 + c
then ax + by – c = 0 → (1)
bx + ay – (1 + c) = 0 → (2)
Solve by cross multiplication method.
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 13
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 14
Hence, the solution of the given pair of linear equation is
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 15

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 16
ax + by = a2 + b2
Answer:
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 17
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
Answer:
(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),
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 18
(a – b)x – (a + b)x = a2 – 2ab – b2 – a2 – b2 ax – bx – ax – bx = – 2ab – 2b2.
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 19
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
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 20
Hence, solution of the pair of linear equation is x = a + b,
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 21
(v) 152x – 378y = – 74
– 378x + 152y = – 604
Answer:
152x – 378y = – 74 → (1)
– 378x + 152y = – 604 → (2)
Adding equation (1) and (2)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 22
226x – 226y = – 678 divide by – 226
x + y = 3 → (3)
Subtract equation (1) and (2)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 23
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

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

Question 8.
ABCD is a cyclic quadrilateral (see Fig.3.7). Find the angles of the cyclic quadrilateral.

KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 24

Answer:
We know that the opposite angles of a cyclic Quadrilateral are supplementary,
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 25
4y + 20 (4x) = 180°
4y – 4x = 180 – 20
4y – 4x = 160 divide by 4
y – x = 40 → (1)
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 26
3y – 5 + (- 7x + 5 ) = 180°
3y – 5 – 7x + 5 = 180°
3y – 7x = 180° → (2)
From equation (1)
y = 40 + x → (3)
Put equation (3) in equation (2)
3(40 + x) – 7x = 180°
120 + 3x – 7x = 180°
– 4x = 60°
x = \(\frac{60}{-4}\)
x = – 15
Put x = -15 in equation (1)
y – x = 40
y = 40 4 – x
y = 40 – 15
y = 25°
KSEEB Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.7 27

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

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 \(\frac{\mathbf{Q S}}{\mathbf{S R}}=\frac{\mathbf{P Q}}{\mathbf{P R}}\)
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 1
Answer:
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 2
Given: In the figure, PS is the bisector of

KSEEBKSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 3 Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 3

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

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
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 4

Answer:
Given: D is a point on hypotenuse AC of ∆ ABC DM ⊥ BC, DN ⊥AB and BD ⊥ AC
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 5

ii) Consider ∆le ADN And DBN
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 6

Question 3.
In Fig. 2.58, ABC is a triangle in which ∠ABC > 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC. BD
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 7
Answer:
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 + CD2 [Pythagoras theorem]
AC2 = AD2 + (BD + CB)2 [DC = BD + BC]
AC2 = AD2 + BD2+ BC2 +2.BD. BC
AC2 = AB2 + BC2 + 2BD.BC [In right angle ∆le ADB AB2 = AD2 + BD2]

Question 4.
In Fig. 2.59, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC. BD.
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 8
Answer:
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
[In right angle ADB with
∠D = 90° AB2 = AD2 + BD2 By Pythagoras theorem]

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 5.
In Fig. 2.60,AD is a median of a triangle ABC and AM I BC. Prove that:
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 9
Answer:
Given: In figure, AD is a median of a triangle ABC and AM ⊥ BC.
∴ D is mid-point of BC (∵ AD is median)
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 10
Proof:
i) In right triangle AMC ∠M = 90°
∴ AC2 = AM2 + MC2 [By Pythagoras theorem]
= AM2 +(MD + DC)2
= AM2 + MD2 + DC2 + 2MD. DC
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 11
[In right angle ∆le AMD with ∠M = 90°
AM2 + MD2 = AD2 (By pythagoras theorem)]
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 12
ii) In right triangle AMB ∠M = 90°
AB2 = AM2 + MB2 [Pythagoras
theorem]
= AM2 + (BD – MD)2
= AM2 + BD2 + MD2 – 2BD . MD
= AM2 + MD2 + BD2 – (2 BD)MD
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 13

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.
Answer:
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 14

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)
Adding (I) and (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 Fig. 2.61, two chords AB and CD intersect each other at the point P.
Prove that:
i) ∆ APC ~ ∆ DPB
ii) AP. PB = CP.DP
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 15
Answer:
Given: In figure, two chords AB and CD intersect each other at the point P.
To prove: i) ∆ APC ~ ∆ DPB
ii) AP. PB = CP . DP
Proof: i) In ∆ APC and ∆ DPB
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 16

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
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 17
Answer:
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.
Therefore, for cyclic quadrilateral ABCD.
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

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

Question 9.
In Fig. 2.63, 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.

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 18
Answer:
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 19
Given: In figure, D is a point on side BC. of ∆ ABC such that \(\frac{B D}{C D}=\frac{A B}{A C}\)
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 20

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)?
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 21
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?
Answer:
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 22
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
KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6 23

In right angled ∆ ABD ∠B = 90°
AD2 = AB2 + BD2
BD2 = AD2 – AB2
[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.

KSEEB Solutions for Class 10 Maths Chapter 2 Triangles Ex 2.6

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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 1

(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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 2

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

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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 3

(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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 4
But k cannot be 0, otherwise, the given equation is not quadratic. Thus, the required value of k is 6.

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

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.
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 5
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
length × breadth = 800
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.

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

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
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 7
Since, b2 – 4ac is less than 0.
∴ The quadratic equation (1) has no real roots.
Thus, the given situation is not possible.

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

Question 5.
Is it possible to design a rectangular park of perimeter 80 m. and area 400 m2? If so, find its length and breadth.
Solution:
KSEEB SSLC Class 10 Maths Solutions Chapter 10 Quadratic Equations Ex 10.4 6
Let the length of rectangular park is ‘x’
Perimeter of the rectangular park = 80m
2 (length + breadth) = 80
length + breadth = \(\frac{80}{2}\) = 40
breadth = (40 – x)
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:
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 1
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 2

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
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 3
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
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 4

(iii) 336 and 54
a = 336, b = 54
KSEEB SSLC Class 10 Maths Solutions Chapter 8 Real Numbers Ex 8.2 5

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, 23 and 29 are prime numbers
LCM of (17, 23, 29) = 17 × 23 × 29
= 391 × 29
= 11339
HCF of (17, 23, 29) = 1
(∵ 17, 23 and 29 have no common Factor)

(iii) 8, 9 and 25
8 = 2 × 2 × 2= 23
9 = 3 × 3 = 32
24 = 5 × 5 = 52
∴ H.C.F. = 1
L.C.M. = 23 × 32 × 52 = 8 × 9 × 25 = 1800
∴ HCF × LCM = Product of three numbers
1 × 1800= 8 × 9 × 25
1800 = 1800

Question 4.
Given that HCF (306, 657) = 9, find LCM (306, 657).
Solution:
H.C.F × L.C.M = a × b
(a, b) × (a, b) = a × b
9 × x = 306 × 657
x = \(\frac{306 \times 657}{9}\)
∴ x = 22338
∴ L.C.M of (306, 657) = 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:
i) 7 × 11 × 13 + 13 = 13(77 + 1)
= 13 × 78
∴ It is a composite number
[It is having more than two factors]

ii) 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
=5[7 × 6 × 4 × 3 × 2 × 1 + 1]
= 5 (1008 + 1)
= 5 × 1009 = 5045
∴ It is a composite number.

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.
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 7
Solution:
(i) a = 7, d = 3, n = 8, an =?
an = a + (n – 1) d
a8= 7 + (8 – 1) 3
= 7 + 7 × 3
= 7 + 21
∴ a8 = 28

(ii) a = -18, d =?, n = 10, an = 0
an = a + (n – 1) d
0 = -18 + (10 – 1) d
0 = -18 + 9d
18 = 9d
9d = 18
\(\mathrm{d}=\frac{18}{9}=2\)

(iii) a =?, d = -3, n = 18, an = -5
an = a + (n – 1) d
-5 = a + (18 – 1) (-3)
= a + 17(-3)
-6 = 1 – 51
∴ a = -5 + 51 = 46

(iv) a = -18.9, d = 2.5, n =? an = 3.6
an = a + (n – 1) d
3.6= -18.9 + (n – 1) (2.5)
3.6= -18.9 + 2.5n – 2.5
3.6 = 2.5n – 21.4
2.5n = 3.6 + 21.4
2.5n = 25
\(n=\frac{25}{2.5}=\frac{250}{25}\)
∴ n = 10.

(v) a = 3.5, d = 0, n = 105, an =?
an = a + (n – 1) d
= 3.5 + (105 – 1) (0)
= 3.5+ 104 × 0
= 3.5 +0
∴ 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
Solution:
a = 10. d = 7 – 10 = -3, n = 30, a30 =?
an = a + (n – 1) d
a30 = 10 + (30 – 1)(-3)
= 10 + 29(-3)
= 10 – 87
∴ a30 = 77
∴ Ans: (C) -77

(ii) 11th term of the A.P.
\(-3,-\frac{1}{2}, 2, \dots \ldots,\) is
A) 28
B) 22
C) -38
D) \(-48 \frac{1}{2}\)
Solution:
\(a=-3, \quad d=-\frac{1}{2}-(-3)=-\frac{1}{2}+3=2 \frac{1}{2}\)
n = 11, a11 =?
an = a + (n – 1) d
\(a_{11}=-3+(11-1)\left(2 \frac{1}{2}\right)\)
\(=-3+10\left(\frac{5}{2}\right)\)
= -3 +25
∴ a11 = 22
∴ Ans: (B) 22

Question 3.
In the following APs find the missing terms in the boxes:
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 1
Solution:
(i) KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 2
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) Here, a =?, a + d = 13, a + 2d=?,
a + 3d = 3
a + d + 2d = 3
13 + 2d = 3
2d = 3 – 13
2d = -10
∴ d = -5
a + d = 13
a + (-5) = 13
a – 5 = 13
∴ a – 13 + 5 = 18
∴ a = 18
a + 2d =?
= 18 + 2(-5)
= 18 – 10
a + 2d = 8
∴ Ans: 18, 8

(iii) a = 5, a + d =?, a + 2d =?
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 1

(iv) a = -4, a +d=? a + 2d =?
a+3d = ? a + 4d =? a + 5d = 6
a + 5d = 6
-4 + 5d = 6
5d = 6 + 4
5d = 10
\(\mathrm{d}=\frac{10}{5}\)
∴ d=2.
a + d = -4 + 2 = -2
a + 2d = -4 + 2 (2) = -4 + 4 = 0
a + 3d = -4 + 3 (2) = -4 + 6 = 2
a + 4d = -4 + 4 (2) = -4 + 8 = 4
∴ Ans: -2, 0, 2, 4

(v) a =? a + d = 38, a + 2d =?
a + 3d =?, a + 4d =?a + 5d = 22
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 2
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:
a = 3, d = 8 – 3 = 5, an = 78, n =?
an = a + (n – 1) d
78 = 3+(n—1)(5)
78 = 3 + 5n – 5
78 = 5n – 2
5n = 78 + 2
\(n=\frac{80}{5}\)
∴ n = 16

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) 7, 13, 19, ………. 201
a = 3, d = 13 – 7 = 6, an = 201. n =?
a + (n – 1) d = an
3 + (n – 1) 6 = 201
3 + 6n – 6 = 201
6n – 3 = 201
6n = 203
\(n=\frac{204}{6}\)
∴ n = 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 A.P: 11, 8, 5, 2, ………..
Solution:
11, 8, 5, 2, ……….. -150
a = 11, d = 8 11 = -3. an = -150.
a + (n – 1) d = an
11 + (n – 1) (-3) = -150
11 – 3n + 3 = -150
-3n + 14 = -150
-3n = -150 – 14
-3n = -164
3n = 164
\(n=\frac{164}{3}\)
Here value of ‘n’ is not perfect. Hence -150 is not a term of the A.P.

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:
n = 50, a3 = 12, an = 106, a29 =?
a11 = a + (n – 1) d
a50 = a + (50 – 1) d = 106
∴ a + 49d = 106 ……………… (1)
a3 = a + 2d = 12 ………………. (2)
Subtracting equation (2) in equation (1).
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 3
Substituting the value of d.
a + 2d = 12
a + 2(2) = 12
a + 4 = 12
∴ a = 12 – 4
a = 8.
∴ an = a + (n – 1) d
a29 = 8 + (29 – 1) 2
= 8 + 28 × 2
= 8 + 56
∴ a29 = 64

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).
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 4
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, d =?
a + 16d = a + 9d + 7
a +1 6d – a – 9d = 7
7d = 7
\(\mathrm{d}=\frac{7}{7}\)
∴ = 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 1 000th terms?
Solution:
Having common difference ‘d’, the APs
1st set a, a+d, a+2d
2nd set b, b + d, b + 2d
100th term of 1st set – 100th term of 2nd set = 100
∴ a + 99d – (b + 99d) = 100
a + 99d – b – 99d = 100
a – b= 100
Similarly.
1000th term of 1st set = 1 + 999d
1000 th term of 2nd set = b + 999d
Their difference
= a + 999d – (b + 999d)
= a + 999d – b – 999d
= a – b.

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;
63, 65, 67,……….
a = 63. d = 65 – 63 =2. an =?
nth term of this is
an = a + (n – 1)d
= 63 + (n – 1) 2
= 63 + 2n – 2
an= 2n + 61 …………….. (i)
3, 10, 17, ………….
a = 3, d = 10 – 3 = 7, an =?
an = a + (n – 1)d
= 3 + (n – 1) 7
= 3 + 7n – 7
an = 7n — 4 ………….(ii)
Here. nth terms of second AP are equal.
∴ equation (i) = equation (ii)
2n + 61 = 7n – 4
2n – 7n = -4 – 61
5n = 65
5n =65
\(n=\frac{65}{5}\)
∴ n = 13
∴13th terms of the two given APs are equal.

Question 16.
DetermIne the AP whose third term is 16 and 7th term exceeds the 5th term by 12.
Solution:
a = 16, a7 = a5 + 12, A.P =?
a7 = a5 + 12
a + 6d = a + 4d + 12
a + 6d – a – 4d = 12
2d = 12 ∴ d = 6.
a + 2d = 16
a + 2(6) = 16
a + 12 = 16
∴ a = 16 – 12 ∴ a = 4
a = 4, d = 6.
∴ A.P.:
a, a + d, a + 2d, …………………
4, 4 + 6, 4 + 12, …………….
4, 10, 16, ………….

Question 17
Find the 20th term from the Last term of theAP: 3, 8, 13, …………., 253.
Solution:
3, 8, 13, ………….., 253
a = 3. d = 8 – 3 = 5, an = 253
20th term from the last term of the AP starting from 253 =?
253, 258, 263, ………… a20 =?
a = 253, d = 258 – 253 = 5, n = 20
an = a + (n – 1) d
a20= 253 + (20 – 1) 5
= 253 + 19 × 5
= 253 + 95
∴ a20 = 348
∴ 20th term from the last term of the AP is 348.

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:
a4 + a8 = 24 …………. (1)
a6 + a10 = 44 ………… (2)
But A.P is a, a + d, a + 2d
from equation (1).
a4 + a8 = 24
a + 3d + a + 7d = 24
2a + 10d = 24 ………….. (3)
from equation (2).
a6 + a10 = 44
a + 5d + a + 9d = 44
2a + 14d = 44 ………… (4)
Subtracting eqn. (4) from equation (3)
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 5
4d = 20
\(d=20 / 4\)
∴ d = 5
Substituting the value of d in equation (3)
2a + 10d = 24
2a + 10(5) = 24
2a + 50 = 24
2a = 24 – 50
2a = -26
a = 26/2 ,
∴ a = -13 .
∴ AP: a, a + d, a + 2d, ………..
-13, -13 + 5, -13 + 2(5), ………
-14, -8, -3, …………

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
KSEEB SSLC Class 10 Maths Solutions Chapter 1 Arithmetic Progressions Ex 1.2 6
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:
Savings in the First week = Rs. 5.
Savings in the Second week 5 + 1.75 = Rs. 6.75
Savings in ‘nth week is Rs. 20.75
∴ A.P. 5, 6, 75 , …………. , 20.75
a = 5, d = 6.75 – 5 = 1.75, an = 20.75.
n =?
a + (n – 1) d = an
5 + (n – 1) (1.75) = 20.75
5 + 1.75n – 1.75 = 20.75
1 .75n + 3.25 = 20.75
1.75n = 20.75 – 3.25
1.75n = 17.5
\(n=\frac{17.5}{1.75}\)
\(n=\frac{1750}{175}\)
∴ n = 10
∴ In the 10th week, her savings becomes Rs. 20.75.

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.

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