NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise – 3.1
The exercise solutions provided here are created with a vision to assist the students in their first term exam preparation. Exercise solution cover all the questions given in Exercise 3.1 of the NCERT textbook. These solutions are researched and prepared by our subject experts. Studying this study material will aid you in solving different questions on distance from a point.
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Q1. 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 you will be” Represent this situation algebraically and graphically.
Solution:-
A.T.Q
(x-7) = 7y – 49
X = 7y – 42
Putting y = 5 , 6 and 7 we get
X = 7 x 5 – 42 = 35 – 42 = -7
X = 7 x 6 – 42 = 42 – 42 = 0
X = 7 x 7 – 42 = 49 – 42 = 7
| x | -7 | 0 | 7 |
| y | 5 | 6 | 7 |
A.T.Q
(x + 3) = 3(y + 3)
X = 3y + 6
Putting , y = -2 ,-1 and 0 , we get
X = 6
| x | 0 | 3 | 6 |
| y | -2 | -1 | 0 |
Algebraic representation
From equation (I) and (ii)
X – 7y = -42
X – 3y = 6
Graphical representation.
NCERT Solutions For Class 10 Math Chapter – 3 Linear equation in two variables Exercise – 3.1
Q1. 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 you will be” Represent this situation algebraically and graphically.
Solution:-
A.T.Q
(x-7) = 7y – 49
X = 7y – 42
Putting y = 5 , 6 and 7 we get
X = 7 x 5 – 42 = 35 – 42 = -7
X = 7 x 6 – 42 = 42 – 42 = 0
X = 7 x 7 – 42 = 49 – 42 = 7
| x | -7 | 0 | 7 |
| y | 5 | 6 | 7 |
A.T.Q
(x + 3) = 3(y + 3)
X = 3y + 6
Putting , y = -2 ,-1 and 0 , we get
X = 6
| x | 0 | 3 | 6 |
| y | -2 | -1 | 0 |
Algebraic representation
From equation (I) and (ii)
X – 7y = -42
X – 3y = 6
Graphical representation.
Q2. 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.
Solution:-
A.T.Q
3x + 6y = 3900 (i)
Dividing equation by 3 , we get
X + 2y = 1300
Subtracting 2y both side we get
X = 1300 – 2y
Putting y = -1300 , 0 and 1300 we get
x = 1300 – 2(-1300)
= 1300 + 2600
= 3900
X = 1300 – 2(0)
= 1300 – 0
= 1300
| x | 3900 | 1300 | -1300 |
| y | -1300 | 0 | 1300 |
X + 2y = 1300
X = 1300 – 2y
X = 1300 – 2(-1300)
= 1300 + 2600
= 3900
X = 1300 – 2(0)
= 1300 – 2(1300)
= -1300
| x | 3900 | 1300 | -1300 |
| y | -1300 | 0 | 1300 |
Algebraic representation
3x + 6y = 3900
X + 2y = 1300
Graphical representation,
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Q3. The cost of 2 kg of apples and 1 kg 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.
Solution:-
2x + y = 160 (i)
2x = 160 – y
X = (160 – y)/2
X = 0
| x | 80 | 40 | 0 |
| y | 0 | 80 | 160 |
4x + 2y = 300 (ii)
Dividing by 2 we get
2x + y = 150
Y = 150 – 2x
Y = 150 – 2 x 0
Y = 150
Y = 150 – 2 x 50 = 50
Y = 150 – 2 x (100) = – 50
| x | 0 | 50 | 100 |
| y | 150 | 50 | -50 |
Algebraic representation,
2x + y = 160 (i)
4x + 2y = 300 (ii)
Graphical representation,
Q2. 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.
Solution:-
A.T.Q
3x + 6y = 3900 (i)
Dividing equation by 3 , we get
X + 2y = 1300
Subtracting 2y both side we get
X = 1300 – 2y
Putting y = -1300 , 0 and 1300 we get
x = 1300 – 2(-1300)
= 1300 + 2600
= 3900
X = 1300 – 2(0)
= 1300 – 0
= 1300
| x | 3900 | 1300 | -1300 |
| y | -1300 | 0 | 1300 |
X + 2y = 1300
X = 1300 – 2y
X = 1300 – 2(-1300)
= 1300 + 2600
= 3900
X = 1300 – 2(0)
= 1300 – 2(1300)
= -1300
| x | 3900 | 1300 | -1300 |
| y | -1300 | 0 | 1300 |
Algebraic representation
3x + 6y = 3900
X + 2y = 1300
Graphical representation,
Q3. The cost of 2 kg of apples and 1 kg 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.
Solution:-
2x + y = 160 (i)
2x = 160 – y
X = (160 – y)/2
X = 0
| x | 80 | 40 | 0 |
| y | 0 | 80 | 160 |
4x + 2y = 300 (ii)
Dividing by 2 we get
2x + y = 150
Y = 150 – 2x
Y = 150 – 2 x 0
Y = 150
Y = 150 – 2 x 50 = 50
Y = 150 – 2 x (100) = – 50
| x | 0 | 50 | 100 |
| y | 150 | 50 | -50 |
Algebraic representation,
2x + y = 160 (i)
4x + 2y = 300 (ii)
Graphical representation,