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Question:

A donation box can receive only cheques of ₹100, ₹250, and ₹500. On one good day, the donation box was found to contain exactly 100 cheques amounting to a total sum of ₹15250. Then, the maximum possible number of cheques of ₹500 that the donation box may have contained, is

Explanation:

Let the number of 100, 250 and 500 notes be a, b and c respectively.

⇒ a + b + c = 100 …(1)

⇒ 100a + 250b + 500c = 15250
⇒ 2a + 5b + 10c = 305 …(2)

(2) – 2 × (1)
⇒ 3b + 8c = 105
⇒ c = $\frac{105 - 3 b}{8}$ = $\frac{104 + 1 - 3 b}{8}$ = 13 + $\frac{1 - 3 b}{8}$

c will be maximum when b is minimum. The least value b can take such that c is integer is 3.

∴ c = 13 + $\frac{1 - 9}{8}$ = 12

Hence, 12.

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