Question 3 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.
Class 10 - Math - Pair of Linear Equations in Two Variables Page 44
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Let cost each kg of apples = Rs x
Cost of each kg of grapes = Rs y
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 + y = 160 … (i)
2x = 160 - y
x = (160 – y)/2
Let y = 0 , 80 and 160, we get
x = (160 – ( 0 )/2 = 80
x = (160- 80 )/2 = 40
x = (160 – 2 × 80)/2 = 0
x80400y080160
Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
So we get
4x + 2y = 300 … (ii)
Dividing by 2 we get
2x + y = 150
Subtracting 2x both side, we get
y = 150 – 2x
Putting x = 0 , 50 , 100 we get
y = 150 – 2 × 0 = 150
y = 150 – 2 × 50 = 50
y = 150 – 2 × (100) = -50
x050100y15050-50
Algebraic representation,
2x + y = 160 … (i)
4x + 2y = 300 … (ii)
Cost of each kg of grapes = Rs y
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 + y = 160 … (i)
2x = 160 - y
x = (160 – y)/2
Let y = 0 , 80 and 160, we get
x = (160 – ( 0 )/2 = 80
x = (160- 80 )/2 = 40
x = (160 – 2 × 80)/2 = 0
x80400y080160
Given that the cost of 4 kg of apples and 2 kg of grapes is Rs 300
So we get
4x + 2y = 300 … (ii)
Dividing by 2 we get
2x + y = 150
Subtracting 2x both side, we get
y = 150 – 2x
Putting x = 0 , 50 , 100 we get
y = 150 – 2 × 0 = 150
y = 150 – 2 × 50 = 50
y = 150 – 2 × (100) = -50
x050100y15050-50
Algebraic representation,
2x + y = 160 … (i)
4x + 2y = 300 … (ii)
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