IPMAT Indore 2023 · Quantitative Ability · Question 13

Step 1 — introduce two multipliers. From the ratios,

a=2k, b=5k, c=5m, d=2m

for positive integers k and m.

Step 2 — use the total.

2k+5k+5m+2m=2023 \Rightarrow 7\left( k+m \right)=2023 \Rightarrow k+m=289

Step 3 — write the target in one variable.

a+c=2k+5m=2k+5\left( 289-k \right)=1445-3k

Step 4 — maximise. The expression decreases as k grows, so take k as small as possible. Since a=2k must be a positive integer, the minimum is k=1:

a+c=1445-3=1442

Check: k=1,m=288 gives a=2, b=5, c=1440, d=576. Sum =2+5+1440+576=2023 ✔ and a+c=1442 ✔

Answer: 1442

Shortcut. Note that 2023=7\times 289 — the 7 is exactly 2+5 from each ratio, which is why the problem works at all. That gives k+m=289 in one line.

Then a+c=2k+5m. Since the m coefficient (5) is larger, you want m big and k small, so set k to its floor of 1:

a+c=2\left( 1 \right)+5\left( 288 \right)=2+1440=1442

A caution worth absorbing: “positive integers” means the minimum is k=1, not k=0. If zero were allowed the answer would be 1445. One word in the question moves the answer by 3.