{"id":55071,"date":"2020-11-16T13:11:48","date_gmt":"2020-11-16T13:11:48","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=55071"},"modified":"2023-09-19T02:23:21","modified_gmt":"2023-09-19T02:23:21","slug":"ncert-solutions-for-class-10th-maths-chapter-3-pair-of-linear-equations-in-two-variables","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/ncert-solutions-for-class-10th-maths-chapter-3-pair-of-linear-equations-in-two-variables\/","title":{"rendered":"NCERT Solutions for Class 10th Maths: Chapter 3 Pair of Linear Equations in Two Variables"},"content":{"rendered":"\n

Class 10: Maths Chapter 3 solutions. Complete Class 10 Maths Chapter 3 Notes.<\/p>\n\n\n\n

NCERT Solutions for Class 10th Maths: Chapter 3 Pair of Linear Equations in Two Variables<\/strong><\/h2>\n\n\n\n

NCERT 10th Maths Chapter 3, class 10 Maths Chapter 3 solutions<\/p>\n\n\n\n

Page No. 28<\/p>\n\n\n\n

Exercise 3.1<\/h4>\n\n\n\n

1. Aftab tells his daughter, “Seven years ago, I was seven times as old as you were then. Also, three years from now, I shall be three times as old as you will be.” (Isn’t this interesting?) Represent this situation algebraically and graphically.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let present age of Aftab be x<\/em>
And, present age of daughter is represented by y<\/em>
Then Seven years ago,
Age of Aftab = x <\/em>-7
Age of daughter = y<\/em>-7
According to the question,
(x <\/em>– 7)  = 7 (y <\/em>\u2013 7 )
x <\/em>\u2013 7 = 7 y <\/em>\u2013 49
x<\/em>– 7y<\/em> = – 49 + 7
x <\/em>\u2013 7y<\/em> = – 42 \u2026(i)<\/strong>
x<\/em> = 7y<\/em> \u2013 42
Putting y<\/em> = 5, 6 and 7, we get
x<\/em> = 7 \u00d7 5 – 42 = 35 – 42 = – 7
x<\/em> = 7 \u00d7 6 – 42 = 42 \u2013 42 = 0
x<\/em> = 7 \u00d7 7 \u2013 42 = 49 \u2013 42 = 7<\/p>\n\n\n\n

Three years from now ,
Age of Aftab = x <\/em>+3
Age of daughter = y <\/em>+3
According to the question,
(x <\/em>+ 3) = 3 (y <\/em>+ 3)
x <\/em>+ 3 = 3y <\/em>+ 9
x <\/em>-3y <\/em>= 9-3
x <\/em>-3y<\/em> = 6 \u2026(ii)<\/strong>
x<\/em> = 3y <\/em>+ 6
Putting, y <\/em>= -2,-1 and 0, we get
x<\/em> = 3 \u00d7 – 2 + 6 = -6 + 6 =0
x<\/em> = 3 \u00d7 – 1 + 6 = -3 + 6 = 3
x<\/em> = 3 \u00d7 0 + 6 = 0 + 6 = 6<\/p>\n\n\n\n

Algebraic representation
From equation (i)<\/strong> and (ii)<\/strong>
x <\/em>\u2013 7y<\/em> = \u2013 42 \u2026(i)<\/strong>
x <\/em>– 3y<\/em> = 6 \u2026(ii)<\/strong>
Graphical representation<\/p>\n\n\n\n

\n
\"Graphical
Graphical representation, NCERT Solutions for Class 10th Maths: Chapter 3 Pair of Linear Equations in Two Variables<\/figcaption><\/figure>\n<\/figure>\n\n\n\n

NCERT 10th Mathematics Chapter 3, class 10 Mathematics Chapter 3 solutions<\/p>\n\n\n\n

2. The coach of a cricket team buys 3 bats and 6 balls for Rs 3900. Later, she buys another bat and 3 more balls of the same kind for Rs 1300. Represent this situation algebraically and geometrically.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let cost of one bat = Rs x<\/em>
Cost of one ball = Rs y<\/em>
3 bats and 6 balls for Rs 3900 So that
3x <\/em>+ 6y<\/em> = 3900 \u2026 (i)<\/strong>
Dividing equation by 3, we get
x <\/em>+ 2y<\/em> = 1300
Subtracting 2y<\/em> both side we get
x<\/em> = 1300 \u2013 2y<\/em>
Putting y<\/em> = -1300, 0 and 1300 we get
x<\/em> = 1300 \u2013 2 (-1300) = 1300 + 2600 = 3900
x <\/em>= 1300 -2(0) = 1300 – 0 = 1300
x<\/em> = 1300 \u2013 2(1300) = 1300 \u2013 2600 = – 1300<\/p>\n\n\n\n

x<\/em><\/td>3900<\/td>1300<\/td>-1300<\/td><\/tr>
y<\/em><\/td>-1300<\/td>0<\/td>1300<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Given that she buys another bat and 2 more balls of the same kind for Rs 1300
So, we get
x<\/em> + 2y <\/em>= 1300 \u2026 (ii)<\/strong>
Subtracting 2y both side we get
x<\/em> = 1300 \u2013 2y<\/em>
Putting y<\/em> = – 1300, 0 and 1300 we get
x<\/em> = 1300 \u2013 2 (-1300) = 1300 + 2600 = 3900
x<\/em> = 1300 \u2013 2 (0) = 1300 – 0 = 1300
x<\/em> = 1300 \u2013 2(1300) = 1300 \u2013 2600 = -1300<\/p>\n\n\n\n

x<\/em><\/td>3900<\/td>1300<\/td>-1300<\/td><\/tr>
y<\/em><\/td>-1300<\/td>0<\/td>1300<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n

Algebraic representation
3x <\/em>+ 6y<\/em> = 3900 \u2026 (i)<\/strong>
x <\/em>+ 2y<\/em> = 1300 \u2026 (ii)<\/strong>
Graphical representation,<\/p>\n\n\n\n

\n
\"Graph<\/figure>\n<\/figure>\n\n\n\n

NCERT 10th Maths Chapter 3, class 10 Maths Chapter 3 solutions<\/p>\n\n\n\n

3. The cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160. After a month, the cost of 4 kg of apples and 2 kg of grapes is Rs 300. Represent the situation algebraically and geometrically.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let cost each kg of apples = Rs x<\/em>
Cost of each kg of grapes = Rs y<\/em>
Given that the cost of 2 kg of apples and 1kg of grapes on a day was found to be Rs 160
So that
2 x <\/em>+ y<\/em> = 160 \u2026 (i)<\/strong>
2x<\/em> = 160 – y<\/em>
x<\/em> = (160 \u2013 y<\/em>)\/2
Let y<\/em> = 0 , 80 and 160,  we get
x<\/em> = (160 \u2013 ( 0 )\/2 = 80
x<\/em> = (160- 80 )\/2 = 40
x<\/em> = (160 \u2013 2 \u00d7 80)\/2 = 0<\/p>\n\n\n\n

Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
So we get
4x<\/em> + 2y <\/em>= 300 \u2026 (ii)<\/strong>
Dividing by 2 we get
2x<\/em> + y<\/em> = 150
Subtracting 2x<\/em> both side, we get
y<\/em> = 150 \u2013 2x<\/em>
Putting x<\/em> = 0 , 50 , 100 we get
y<\/em> = 150 \u2013 2 \u00d7 0 = 150
y<\/em> = 150 \u2013 2 \u00d7  50 = 50
y<\/em> = 150 \u2013 2 \u00d7 (100) = -50<\/p>\n\n\n\n

Algebraic representation,
2x <\/em>+ y<\/em> = 160 \u2026 (i)<\/strong>
4x<\/em> + 2y<\/em> = 300 \u2026 (ii)<\/strong>
Graphical representation,<\/p>\n\n\n\n

\n
\"NCERT
NCERT Solutions for Class 10th Maths: Chapter 3 Pair of Linear Equations in Two Variables<\/figcaption><\/figure>\n<\/figure>\n\n\n\n

NCERT 10th Maths Chapter 3, class 10 Maths Chapter 3 solutions<\/p>\n\n\n\n

Exercise 3.2<\/h4>\n\n\n\n

1. Form the pair of linear equations in the following problems, and find their solutions graphically.<\/strong>
<\/p>\n\n\n\n

(i) 10 students of Class X took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let number of boys = x<\/em>
Number of girls = y<\/em>
Given that total number of student is 10 so that
x <\/em>+ y<\/em> = 10
Subtract y both side we get
x<\/em> = 10 \u2013 y<\/em>
Putting y <\/em>= 0 , 5, 10 we get
x<\/em> = 10 \u2013 0 = 10
x<\/em> = 10 \u2013 5 = 5
x<\/em> = 10 \u2013 10 = 0<\/p>\n\n\n\n

Given that If the number of girls is 4 more than the number of boys
So that
y<\/em> = x <\/em>+ 4
Putting x<\/em> = -4, 0, 4, and we get
y <\/em>= – 4 + 4 = 0
y <\/em>= 0 + 4 = 4
y <\/em>= 4 + 4 = 8<\/p>\n\n\n\n

Graphical representation<\/p>\n\n\n\n

\n
\"3.2-1-1,
3.2-1-1, NCERT Solutions for Class 10th Maths: Chapter 3 Pair of Linear Equations in Two Variables<\/figcaption><\/figure>\n<\/figure>\n\n\n\n

Therefore, number of boys = 3 and number of girls = 7.<\/p>\n\n\n\n

(ii) 5 pencils and 7 pens together cost Rs 50, whereas 7 pencils and 5 pens together cost Rs 46. Find the cost of one pencil and that of one pen.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

Let cost of pencil = Rs x<\/em>
Cost of pens = Rs y<\/em>
5 pencils and 7 pens together cost Rs 50,
So we get
5x<\/em> + 7y<\/em> = 50
Subtracting 7y<\/em> both sides we get
5x<\/em> = 50 \u2013 7y<\/em>
Dividing by 5 we get
x<\/em> = 10 – 7 y \/5
Putting value of y<\/em> = 5 , 10 and 15 we get
x<\/em> = 10 \u2013 7 \u00d7 5\/5 = 10 \u2013 7 = 3
x<\/em> = 10 \u2013 7 \u00d7 10\/5 = 10 \u2013 14 = – 4<\/p>\n\n\n\n

x<\/em> = 10 \u2013 7 \u00d7 15\/5 = 10 \u2013 21 = – 11<\/p>\n\n\n\n

Given that 7 pencils and 5 pens together cost Rs 46
7x<\/em> + 5y<\/em> = 46
Subtracting 7x<\/em> both side we get
5y<\/em> = 46 \u2013 7x<\/em>
Dividing by 5 we get
y<\/em> = 46\/5 – 7x<\/em>\/5
y = 9.2 \u2013 1.4x<\/em>
Putting x<\/em> = 0 , 2 and 4 we get
y<\/em> = 9.2 \u2013 1.4 \u00d7 0 = 9.2 \u2013 0 = 9.2
y<\/em> = 9.2 \u2013 1.4 (2) = 9.2 \u2013 2.8 = 6.4
y<\/em> = 9.2 \u2013 1.4 (4) = 9.2 \u2013 5.6 = 3.6<\/p>\n\n\n\n

Graphical representation<\/p>\n\n\n\n

\n
\"Graphical
Graphical representation, NCERT Solutions for Class 10th Maths: Chapter 3 Pair of Linear Equations in Two Variables<\/figcaption><\/figure>\n<\/figure>\n\n\n\n

Therefore, cost of one pencil = Rs 3 and cost of one pen = Rs 5.<\/p>\n\n\n\n

2. On comparing the ratios a<\/em>1<\/sub>\/a<\/em>2<\/sub> , b<\/em>1<\/sub>\/b<\/em>2<\/sub> and c<\/em>1<\/sub>\/c<\/em>2<\/sub>, find out whether the lines representing the following pairs of linear equations intersect at a point, are parallel or coincident.<\/strong><\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 5x<\/em> \u2013 4y<\/em> + 8 = 0
7x<\/em> + 6y<\/em> \u2013 9 = 0<\/p>\n\n\n\n

Comparing these equation with
a<\/em>1<\/sub>x<\/em> + b<\/em>1<\/sub>y<\/em> + c<\/em>1<\/sub> = 0
a<\/em>2<\/sub>x<\/em> + b<\/em>2<\/sub>y<\/em> + c<\/em>2<\/sub>= 0<\/p>\n\n\n\n

We get
a<\/em>1<\/sub> = 5, b<\/em>1<\/sub> = -4, and c<\/em>1<\/sub> = 8
a<\/em>2<\/sub> =7, b<\/em>2<\/sub> = 6 and c2<\/sub> = -9
a<\/em>1<\/sub>\/a2<\/sub> = 5\/7,
b1<\/sub>\/b<\/em>2<\/sub> = -4\/6 and
c<\/em>1<\/sub>\/c<\/em>2<\/sub> = 8\/-9
Hence, a<\/em>1<\/sub>\/a<\/em>2<\/sub> \u2260 b1<\/sub>\/b2<\/sub><\/p>\n\n\n\n

Therefore, both are intersecting lines at one point.<\/p>\n\n\n\n

(ii) 9x<\/em> + 3y<\/em> + 12 = 0
18x<\/em> + 6y<\/em> + 24 = 0
Comparing these equations with<\/p>\n\n\n\n

a<\/em>1<\/sub>x<\/em> + b<\/em>1<\/sub>y<\/em> + c<\/em>1<\/sub> = 0<\/p>\n\n\n\n

a<\/em>2<\/sub>x<\/em> + b<\/em>2<\/sub>y<\/em> + c<\/em>2<\/sub>= 0
We get
a<\/em>1<\/sub> = 9, b<\/em>1<\/sub> = 3, and c<\/em>1<\/sub> = 12
a<\/em>2<\/sub> = 18, b<\/em>2<\/sub> = 6 and c<\/em>2<\/sub> = 24
a<\/em>1<\/sub>\/a<\/em>2<\/sub> = 9\/18 = 1\/2
b<\/em>1<\/sub>\/b<\/em>2<\/sub> = 3\/6 = 1\/2 and
c<\/em>1<\/sub>\/c<\/em>2<\/sub> = 12\/24 = 1\/2
Hence, a<\/em>1<\/sub>\/a<\/em>2<\/sub> = b<\/em>1<\/sub>\/b<\/em>2 <\/sub>=c<\/em>1<\/sub>\/c<\/em>2<\/sub>
Therefore, both lines are coincident<\/p>\n\n\n\n

(iii) 6x<\/em> \u2013 3y<\/em> + 10 = 0
2x<\/em> \u2013 y<\/em> + 9 = 0
Comparing these equations with<\/p>\n\n\n\n

a<\/em>1<\/sub>x<\/em> + b<\/em>1<\/sub>y<\/em> + c<\/em>1<\/sub> = 0<\/p>\n\n\n\n

a<\/em>2<\/sub>x<\/em> + b<\/em>2<\/sub>y<\/em> + c<\/em>2<\/sub>= 0<\/p>\n\n\n\n

We get
a<\/em>1<\/sub> = 6, b<\/em>1<\/sub> = -3, and c<\/em>1<\/sub> = 10
a<\/em>2<\/sub> = 2, b<\/em>2<\/sub> = -1 and c2<\/sub> = 9
a<\/em>1<\/sub>\/a<\/em>2<\/sub> = 6\/2 = 3\/1
b<\/em>1<\/sub>\/b<\/em>2<\/sub> = -3\/-1 = 3\/1 and
c<\/em>1<\/sub>\/c<\/em>2<\/sub> = 12\/24 = 1\/2
Hence, a<\/em>1<\/sub>\/a2<\/sub> = b<\/em>1<\/sub>\/b<\/em>2 <\/sub>\u2260c<\/em>1<\/sub>\/c<\/em>2<\/sub>
Therefore, both lines are parallel<\/p>\n\n\n\n

3. On comparing the ratios a<\/em>1<\/sub>\/a<\/em>2<\/sub> , b<\/em>1<\/sub>\/b<\/em>2<\/sub> and c<\/em>1<\/sub>\/c<\/em>2<\/sub> find out whether the following pair of linear equations are consistent, or inconsistent.<\/strong><\/p>\n\n\n\n

(i) 3x<\/em> + 2y<\/em> = 5 ; 2x<\/em> \u2013 3y<\/em> = 7
(ii) 2x<\/em> \u2013 3y<\/em> = 8 ; 4x<\/em> \u2013 6y<\/em> = 9
(iii) 3\/2x<\/em> + 5\/3y<\/em> = 7 ; 9x <\/em>\u2013 10y<\/em> = 14
(iv) 5x<\/em> \u2013 3y<\/em> = 11 ; \u2013 10x<\/em> + 6y<\/em> = \u201322
(v) 4\/3x<\/em> + 2y<\/em> =8 ; 2x<\/em> + 3y<\/em> = 12<\/p>\n\n\n\n

Answer<\/strong><\/p>\n\n\n\n

(i) 3x<\/em> + 2y<\/em> = 5 ; 2x<\/em> \u2013 3y<\/em> = 7
a<\/em>1<\/sub>\/a<\/em>2<\/sub> = 3\/2
b<\/em>1<\/sub>\/b<\/em>2<\/sub> = -2\/3 and
c<\/em>1<\/sub>\/c<\/em>2<\/sub> = 5\/7
Hence, a<\/em>1<\/sub>\/a<\/em>2<\/sub> \u2260 b1<\/sub>\/b2<\/sub>
These linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.<\/p>\n\n\n\n

(ii) 2x<\/em> \u2013 3y<\/em> = 8 ; 4x<\/em> \u2013 6y<\/em> = 9
a<\/em>1<\/sub>\/a<\/em>2<\/sub> = 2\/4 = 1\/2
b<\/em>1<\/sub>\/b<\/em>2<\/sub> = -3\/-6 = 1\/2 and
c<\/em>1<\/sub>\/c<\/em>2<\/sub> = 8\/9
Hence, a<\/em>1<\/sub>\/a2<\/sub> = b<\/em>1<\/sub>\/b<\/em>2 <\/sub>\u2260c<\/em>1<\/sub>\/c<\/em>2<\/sub>
Therefore, these linear equations are parallel to each other and thus have no possible solution. Hence, the pair of linear equations is inconsistent.
(iii) 3\/2x<\/em> + 5\/3y<\/em> = 7 ; 9x <\/em>\u2013 10y<\/em> = 14
a<\/em>1<\/sub>\/a<\/em>2<\/sub> = 3\/2\/9 = 1\/6
b<\/em>1<\/sub>\/b<\/em>2<\/sub> = 5\/3\/-10 = -1\/6 and
c<\/em>1<\/sub>\/c<\/em>2<\/sub> = 7\/14 = 1\/2
Hence, a<\/em>1<\/sub>\/a2<\/sub> \u2260 b<\/em>1<\/sub>\/b<\/em>2<\/sub>
Therefore, these linear equations are intersecting each other at one point and thus have only one possible solution. Hence, the pair of linear equations is consistent.<\/p>\n\n\n\n

(iv) 5x<\/em> \u2013 3y<\/em> = 11 ; \u2013 10x<\/em> + 6y<\/em> = \u201322
a<\/em>1<\/sub>\/a<\/em>2<\/sub> = 5\/-10 = -1\/2
b<\/em>1<\/sub>\/b<\/em>2<\/sub> = -3\/6 = -1\/2 and
c<\/em>1<\/sub>\/c<\/em>2<\/sub> = 11\/-22 = -1\/2
Hence, a<\/em>1<\/sub>\/a2<\/sub> = b<\/em>1<\/sub>\/b<\/em>2 =<\/sub>c<\/em>1<\/sub>\/c<\/em>2<\/sub>
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.<\/p>\n\n\n\n

(v) 4\/3x<\/em> + 2y<\/em> =8 ; 2x<\/em> + 3y<\/em> = 12
a<\/em>1<\/sub>\/a<\/em>2<\/sub> = 4\/3\/2 = 2\/3
b<\/em>1<\/sub>\/b<\/em>2<\/sub> = \/3 and
c<\/em>1<\/sub>\/c<\/em>2<\/sub> = 8\/12 = 2\/3
Hence, a<\/em>1<\/sub>\/a2<\/sub> = b<\/em>1<\/sub>\/b<\/em>2 =<\/sub>c<\/em>1<\/sub>\/c<\/em>2<\/sub>
Therefore, these linear equations are coincident pair of lines and thus have infinite number of possible solutions. Hence, the pair of linear equations is consistent.<\/p>\n\n\n\n

NCERT 10th Maths Chapter 3, class 10 Maths Chapter 3 solutions<\/p>\n\n\n\n

4. Which of the following pairs of linear equations are consistent\/inconsistent? If consistent, obtain the solution graphically:<\/strong><\/p>\n\n\n\n