IPMAT Indore 2026 · Quantitative Ability · Question 29

We need non-negative integers with

2a+5b+10c=50

Step 1 — note a parity constraint. 2a and 10c are both even and 50 is even, so 5b must be even — meaning b is even. Write b=2b':

2a+10b'+10c=50 \Rightarrow a+5b'+5c=25 \Rightarrow a=25-5\left( b'+c \right)

Step 2 — let s=b'+c. We need a\ge 0, so s\le 5.

Step 3 — count. For each value of s, the pair \left( b',c \right) can be chosen in s+1 ways:

1+2+3+4+5+6=21

Answer: Option 4 — 21

Shortcut. The parity observation — that the number of 5-rupee coins must be even — halves the problem instantly. After the substitution it reduces to counting pairs summing to at most 5, which is the triangular number \left( \genfrac{}{}{0pt}{}{7}{2}\right)=21.