IPMAT Indore 2024 (MCQ) - A fruit seller had a certain number of apples, bananas, and oranges at the start of the day. The number of bananas was 10 more than the number of apples, and the total number of bananas and apples was a multiple of 11. She was able to sell 70% of the apples, 60% of bananas, and 50% of oranges during the day. If she was able to sell 55% of the fruits she had at the start of the day, then the minimum number of oranges she had at the start of the day was | PYQs + Solutions

IPMAT Indore 2024

Algebra

Linear Equation

Hard

A fruit seller had a certain number of apples, bananas, and oranges at the start of the day. The number of bananas was 10 more than the number of apples, and the total number of bananas and apples was a multiple of 11. She was able to sell 70% of the apples, 60% of bananas, and 50% of oranges during the day. If she was able to sell 55% of the fruits she had at the start of the day, then the minimum number of oranges she had at the start of the day was

190

210

180

220

Correct Option: 2

Step 3 (assuming

K

to be a multiple of

10

) might confuse you:

Let us assume that

K=1

. Then

A+B=11

. But, this does not satisfy equation (1) as the value of

A

does not come out as a whole number. This is true for all positive odd numbers.

Let's put

K=2

. This gives:

A+B=22

From (1),

\newline

2A + 10 = 22

A=6, B =16

But then,

60\%

16

does not come out as a whole number. So is the case with

4, 6

and

8

Hence,

10

becomes the least value that suffices all the conditions.