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 Rs. 2 for 3 bananas + Cost of the second lot at the rate
Of Rs.1 per banana = amount received
⇒ 23x+y=400
⇒ 2x+3y=1200 ...(i)
Case II: Cost of the first lot at the rate of Rs.1 per banana and the second lot at the rate of Rs. 4 for 5 bananas = amount received
⇒ x+45y=460
⇒ 5x+4y=2300 ...(ii)
On multiplying Eq. (i) by 4 and Eq (ii) by 3 and then subtracting them, we get
8x + 12y = 4800
15x + 12y = 6900
-7x = - 2100
x = 300
Now, put the value of x in Eq (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.