Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
Re: In a get together, every 3 guests used a bowl of soup between them,
[#permalink]
26 Feb 2021, 00:08
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. 702+703+704=35+23.33+17.5=75.83
B. 722+723+724=36+24+18=78
C. 802+803+804=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