A shopkeeper has three types of apples (A, B, and C) with the following stock and individual weight.
Type A: 50 apples in stock, 40 gm each.
Type B: 40 apples in stock, 60 gm each.
Type C: 60 apples in stock, 25 gm each.
He selects a certain number of apples of each of the three types and places them into a crate K. The number of apples chosen for each type are consecutive multiples of 5. The number of C-type apples in K is the minimum among all.
If the total weight of all the apples in the crate is 2050 gm, then how many type B apples did the shopkeeper put in the crate?

Correct option is C

The number of apples chosen for each type are consecutive multiples of 5.
5x, 5(x+1) and 5 (x+2)
C-type apples in K is the minimum among all
C-type apples = 5x
There are two possible cases
Case I
A-type apples are maximum
A-type apples = 5 (x+2)
B-type apples = 5(x+1)
C-type apples = 5x
Casel II
B-type apples are maximum
B-type apples = 5 (x+2)
A-type apples = 5(x+1)
C-type apples = 5x
We take Case I
Total weight = 40×5(x+2)+60×5(x+1)+25×5x
= 200x + 400 + 300x + 300 + 125x
= 625x + 700
ATQ,
2050 = 625x + 700
1350 = 625x
x = 2.16
2.16 is not integer
B-type apples = 5 (2.16+1) = 15.8
x ≠ 2.16
We take Case II
Total weight = 60×5(x+2)+40×5(x+1)+25×5x
= 300x + 600 + 200x + 200 + 125x
= 625x + 800
ATQ,
2050 = 625x + 800
1250 = 625x
x = 2
Type B apples = 5(x+2)
= 5(2+2)
= 20

Exam Hall Method: