Let the number of bananas in lot 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 = Amount received
Case II :
Cost of the first lot at the rate of ₹ 1 per banana + Cost of the second lot at the rate of ₹ 4 for 5 bananas = Amount received
On multiplying in Eq. (i) by 4 and Eq. (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 Eq. (i), we get
2(300) + 3y = 1200
3y = 600
y = 200
∴ Total number of bananas = Number of bananas in lot Number of bananas in lot B = x + y
x + y = 300 + 200 = 500
Hence, he had 500 bananas.