Problem Solving

Rama went to the market and bought some apples, mangoes and bananas. He bought 42 fruits in all. The number of bananas is less than half the number of apples; the number of mangoes is more than one-third the number of apples and the number of mangoes is less than three-fourths the number of bananas.

1. How many apples did Rama buy?
(A) 20 (B) 23 (C) 26 (D) 28

2. How many bananas did Rama buy?
(A) 8 (B) 9 (C) 10 (D) 11

[spoiler]OA: 1st question : B & 2nd question : D[/spoiler]

So the first equation you get is a + b + m = 42.

Then with less than and greater than I tend to make them equal for the sake of making it easier.

So B = .5a; a = 1/3m; m = 3/4b

First I'm going to solve for bananas as when i know that I know apples is a little more than double that.

So b = .5a which equals a = 2b (as stated above);

So 2B + b + (3/4)B = 42 = (15/4)b = 42 = 15B = 168 which equals [spoiler]B = 11[/spoiler]

So you take that and plug it in for [spoiler]11 = .5A which is a 22 but remember bananas are less than half of so it's going to be greater than 22 and since fruits are whole it's 23.[/spoiler]

Hope that helps.

coolhabhi wrote:Rama went to the market and bought some apples, mangoes and bananas. He bought 42 fruits in all. The number of bananas is less than half the number of apples; the number of mangoes is more than one-third the number of apples and the number of mangoes is less than three-fourths the number of bananas.

1. How many apples did Rama buy?
(A) 20 (B) 23 (C) 26 (D) 28

2. How many bananas did Rama buy?
(A) 8 (B) 9 (C) 10 (D) 11

Best way: Process Of Elimination and calculated with assumption that "Banana<=apple/2 (max)"

1. If we carefully observe, both (C) & (D) ruled out because 26/2=12 and 28/2=14
2. Now we left with 20 0r 23 apples, so to satisfy the condition "the number of mangoes is more than one-third the number of apples and the number of mangoes is less than three-fourths the number of bananas." ONLY B& D will qualify for the answer.