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Question
Vijay had some bananas, and he divided them into two lots A and B. He sold the first lot at the rate of Rs 2 for 3 bananas and the second lot at the rate of Re 1 per banana, and got a total of Rs 400. If he had sold the first lot at the rate of Re 1 per banana, and the second lot at the rate of Rs 4 for 5 bananas, his total collection would have been Rs 460. Find the total number of bananas he had.
Solution
Let the number of bananas in lots A and B be x and y respectively
Case I: Cost of the first lot at the rate of ₹ 2 for 3 bananas + Cost of the second lot at the rate of ₹ 1 per banana = ₹ 400
⇒ `2/3x + y` = 400
⇒ `2x + 3y` = 1200 ......(i)
Case II: Cost of the first lot at the rate ₹ 1 per banana + Cost of the second lot at the rate of ₹ 4 for 5 bananas = Amount received
⇒ `x + 4/5y` = 460
⇒ 5x + 4y = 2300 ......(ii)
On multiplying in equation (i) by 4 and equation (ii) by 3 and then subtracting them, we get
(8x + 12y) – (15x + 12y) = 4800 – 6900
⇒ – 7x = – 2100
⇒ x = 300
Now, put the value of x in equation (i), we get
2 × 300 + 3y = 1200
⇒ 600 + 3y = 1200
⇒ 3y = 1200 – 600
⇒ 3y = 600
⇒ y = 200
∴ Total number of bananas = Number of bananas in lot A + Number of bananas in lot B
= x + y
= 300 + 200
= 500
Hence, he had 500 bananas.
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