Access Answers for SAT Maths Chapter 3 – Pair of Linear Equations in Two
Variables
Exercise 3.1
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.
Solutions: Let the present age of Aftab be ‘x’.
And, the present age of his daughter be ‘y’.
Now, we can write, seven years ago,
Age of Aftab = x-7
Age of his daughter = y-7
According to the question,
x−7 = 7(y−7)
⇒x−7 = 7y−49
⇒x−7y = −42 ………………………(i)
Also, three years from now or after three years,
Age of Aftab will become = x+3.
Age of his daughter will become = y+3
According to the situation given,
x+3 = 3(y+3)
⇒x+3 = 3y+9
⇒x−3y = 6 …………..…………………(ii)
Subtracting equation (i) from equation (ii) we have
(x−3y)−(x−7y) = 6−(−42)
⇒−3y+7y = 6+42
⇒4y = 48
⇒y = 12
The algebraic equation is represented by
x−7y = −42
x−3y = 6
For, x−7y = −42 or x = −42+7y
The solution table is
,For, x−3y = 6 or x = 6+3y
The solution table is
The graphical representation is:
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.
Solutions: Let us assume that the cost of a bat be ‘Rs x’
And,the cost of a ball be ‘Rs y’
According to the question, the algebraic representation is
,3x+6y = 3900
And x+3y = 1300
For, 3x+6y = 3900
Or x = (3900-6y)/3
The solution table is
For, x+3y = 1300
Or x = 1300-3y
The solution table is
The graphical representation is as follows.
, 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.
Solutions:Let the cost of 1 kg of apples be ‘Rs. x’
And, cost of 1 kg of grapes be ‘Rs. y’
According to the question, the algebraic representation is
2x+y = 160
And 4x+2y = 300
For, 2x+y = 160 or y = 160−2x, the solution table is;
For 4x+2y = 300 or y = (300-4x)/2, the solution table is;
Variables
Exercise 3.1
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.
Solutions: Let the present age of Aftab be ‘x’.
And, the present age of his daughter be ‘y’.
Now, we can write, seven years ago,
Age of Aftab = x-7
Age of his daughter = y-7
According to the question,
x−7 = 7(y−7)
⇒x−7 = 7y−49
⇒x−7y = −42 ………………………(i)
Also, three years from now or after three years,
Age of Aftab will become = x+3.
Age of his daughter will become = y+3
According to the situation given,
x+3 = 3(y+3)
⇒x+3 = 3y+9
⇒x−3y = 6 …………..…………………(ii)
Subtracting equation (i) from equation (ii) we have
(x−3y)−(x−7y) = 6−(−42)
⇒−3y+7y = 6+42
⇒4y = 48
⇒y = 12
The algebraic equation is represented by
x−7y = −42
x−3y = 6
For, x−7y = −42 or x = −42+7y
The solution table is
,For, x−3y = 6 or x = 6+3y
The solution table is
The graphical representation is:
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.
Solutions: Let us assume that the cost of a bat be ‘Rs x’
And,the cost of a ball be ‘Rs y’
According to the question, the algebraic representation is
,3x+6y = 3900
And x+3y = 1300
For, 3x+6y = 3900
Or x = (3900-6y)/3
The solution table is
For, x+3y = 1300
Or x = 1300-3y
The solution table is
The graphical representation is as follows.
, 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.
Solutions:Let the cost of 1 kg of apples be ‘Rs. x’
And, cost of 1 kg of grapes be ‘Rs. y’
According to the question, the algebraic representation is
2x+y = 160
And 4x+2y = 300
For, 2x+y = 160 or y = 160−2x, the solution table is;
For 4x+2y = 300 or y = (300-4x)/2, the solution table is;
