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Re: In a get together, every 3 guests used a bowl of soup between them, [#permalink]
OA Explanation:

Easiest Approach: divide the options by 2, 3 and 4 separately and then see if they add up to 91 or not!

[We can eliminate A as it is not divisible by 3 and 4, and also option C as it is not divisible by 3]

A. \(\frac{70}{2} + \frac{70}{3} + \frac{70}{4} = 35 + 23.33 + 17.5 = 75.83\)

B. \(\frac{72}{2} + \frac{72}{3} + \frac{72}{4} = 36 + 24 + 18 = 78\)

C. \(\frac{80}{2} + \frac{80}{3} + \frac{80}{4} = 40 + 26.67 + 20 = 86.67\)

D. \(\frac{84}{2} + \frac{84}{3} + \frac{84}{4} = 42 + 28 + 21 = 91\)

E. Cannot be None of the Above, as we have a match as option D

Conventional Approach:

People sharing 1 bowl of Soup, Rice and Meat are in 3 : 2 : 4
We need to make number of people equal here, so we take LCM of 2, 3 and 4 = 12
i.e. (3)(4) : (2)(6) : (4)(3)

Now, 12 people require 4 + 6 + 3 bowls in total = 13 bowls
Since we have 91 bowls, we have a multiplication factor of 7 (13 x 7 = 91)

So Total number of people must be 12 x 7 = 84

Hence, option D
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