IPMAT Indore 2024 · Quantitative Ability · Question 37

Step 1 — name the quantities. Let apples =a, so bananas =a+10, and oranges =r.

Step 2 — write the “55% overall” condition.

0.7a+0.6\left( a+10 \right)+0.5r=0.55\left( a+\left( a+10 \right)+r \right)

Step 3 — expand and simplify.

1.3a+6+0.5r=0.55\left( 2a+10+r \right)=1.1a+5.5+0.55r

0.2a+0.5=0.05r \Rightarrow \boxed{r=4a+10}

Step 4 — apply the divisibility condition. a+\left( a+10 \right)=2a+10 must be a multiple of 11:

2a+10\equiv 0 (\mod 11) \Rightarrow 2a\equiv 1 (\mod 11) \Rightarrow a\equiv 6 (\mod 11)

(since 2\times 6=12\equiv 1).

Step 5 — require whole fruits sold. 70\% of a must be a whole number, so 10∣a. (Then 60\% of a+10 and 50\% of r=4a+10 are automatically integers.)

Step 6 — combine. We need a\equiv 0 (\mod 10) and a\equiv 6 (\mod 11). Testing a=10,20,\ldots: a=50 gives 50=4\left( 11 \right)+6 ✔

r=4\left( 50 \right)+10=210

Check: apples 50, bananas 60 (sum 110=11\times 10 ✔), oranges 210; total 320. Sold =35+36+105=176=0.55\times 320 ✔

Answer: Option 3 — 210

Shortcut. Reduce to one linear relation, then use the options. The weighted-average condition collapses to the clean

r=4a+10

Now work backwards from the options rather than solving congruences: each option gives a=\left( r-10 \right)/4, so

Two are non-integers, leaving 190 and 210. Check the multiple-of-11 rule: a=45\Rightarrow 2a+10=100 ✘; a=50\Rightarrow 110=11\times 10 ✔.

Answer 210 in under a minute — this is the standard way to beat “minimum value with divisibility” questions.