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29 Jun 2021, 07:00
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E
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Difficulty:
95%
(hard)
Question Stats:
49% (02:58) correct
51%
(03:23)
wrong
based on 125
sessions
History
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Not Attempted Yet
Out of 200 food items, 80 are sweet, 60 are spicy, 60 are tangy. Among these, there are 40 sweet and spicy items, there are 50 spicy and tangy items, and there are 50 tangy and sweet items. What is the ratio of the maximum possible food items of all 3 types and the minimum possible food items of all 3 types?
(A) 5:3
(B) 4:3
(C) 3:2
(D) 2:3
(E) 1:1
(A) 5:3
(B) 4:3
(C) 3:2
(D) 2:3
(E) 1:1
DashingNightmare2
Current Student
Joined: 30 May 2021
Last visit: 19 Jun 2022
Posts: 101
Given Kudos: 77
Location: India
Concentration: Marketing, Technology
Schools: ISB '23 (A)
GMAT 1: 750 Q50 V41
GPA: 2.96
30 Jul 2021, 09:20
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Steps to find values,
1. Let there are x items that are all sweet spicy and tangy
2. Now we can take care of all the values for 2 at a time among (sweet, spicy and tangy) see attachment
for reference
3. Now remaining 1 at a time can also be found.
for maximum possible food items of all 3 types(we have to maximize x) ,
we have to choose x such that none of 40-x, 5-x and 50-x becomes negative
=> 40-x>=0 => x<=40
Hence maximum possible food items of all 3 types = 40
For minimum possible food items of all 3 types(we have to minimize x),
We have to chose minimum x such that none of x-10, x-30 , x-40 becomes negative
Here we have to chose x>=40 such that none of the values becomes negative.
Hence minimum possible food items of all 3 types = 40
Required Ratio =40:40 =1:1
Hence E.
Thank you for reading the long post , hope it helped
1. Let there are x items that are all sweet spicy and tangy
2. Now we can take care of all the values for 2 at a time among (sweet, spicy and tangy) see attachment
for reference
3. Now remaining 1 at a time can also be found.
for maximum possible food items of all 3 types(we have to maximize x) ,
we have to choose x such that none of 40-x, 5-x and 50-x becomes negative
=> 40-x>=0 => x<=40
Hence maximum possible food items of all 3 types = 40
For minimum possible food items of all 3 types(we have to minimize x),
We have to chose minimum x such that none of x-10, x-30 , x-40 becomes negative
Here we have to chose x>=40 such that none of the values becomes negative.
Hence minimum possible food items of all 3 types = 40
Required Ratio =40:40 =1:1
Hence E.
Thank you for reading the long post , hope it helped
Attachments
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24 Jan 2025, 11:12
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