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04 Dec 2018, 02:36
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
67% (02:30) correct
33%
(02:42)
wrong
based on 347
sessions
History
Date
Time
Result
Not Attempted Yet
One cup of nuts that contains exactly half peanuts and half cashews is added to a bowl of nuts that is exactly one third peanuts, one third cashews, and one third almonds. This results in a three-cup mixture of nuts. What fraction of the new nut mixture is peanuts?
A. 2/9
B. 5/18
C. 1/3
D. 7/18
E. 4/9
A. 2/9
B. 5/18
C. 1/3
D. 7/18
E. 4/9
04 Dec 2018, 04:24
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One cup of 1/2 p, 1/2 c is added to two cups of 1/3 p, 1/3 c, 1/3 a in order to obtain 3 cups of final mixture. Let's pick numbers for the number of nuts we have in the mixtures to find out the correct proportion.
If we have 1p and 1c in the first mixture (1 cup), then we need 4 nuts in the second mixture (2 cups). Not doable with integers, let's move on.
2p and 2 c: we need 8 nuts. Move on.
3p and 3c: we need 12 nuts. It works as it's a multiple of 3. We'd have 4p, 4c and 4a.
Now let's count the number of p and the total nuts:
p=7
total=18
Answer D: 7/18
If we have 1p and 1c in the first mixture (1 cup), then we need 4 nuts in the second mixture (2 cups). Not doable with integers, let's move on.
2p and 2 c: we need 8 nuts. Move on.
3p and 3c: we need 12 nuts. It works as it's a multiple of 3. We'd have 4p, 4c and 4a.
Now let's count the number of p and the total nuts:
p=7
total=18
Answer D: 7/18
General Discussion
ShankSouljaBoi
Joined: 21 Jun 2017
Last visit: 17 Apr 2024
Posts: 622
Given Kudos: 4,090
Location: India
Concentration: Finance, Economics
Schools: Rotman '23 (D) IIMA PGPX'22 (D) Desautels '23 (D) Schulich Sept '23 (D) WBS (D) IIML IPMX'25 (A)
GMAT 1: 660 Q49 V31
GMAT 2: 620 Q47 V30
GMAT 3: 650 Q48 V31
GPA: 3.1
WE:Corporate Finance (Non-Profit and Government)
Products:
Schools: Rotman '23 (D) IIMA PGPX'22 (D) Desautels '23 (D) Schulich Sept '23 (D) WBS (D) IIML IPMX'25 (A)
GMAT 3: 650 Q48 V31
Posts: 622
04 Dec 2018, 21:28
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Let C be cup capacity and B be bowl capacity.
Now after addition of C to B ----> B + C = 3C ----> B=2C
Total Almonds% = (B/3 + C/2)/B+C
----> Substituting B=2C -----> 7/18
Now after addition of C to B ----> B + C = 3C ----> B=2C
Total Almonds% = (B/3 + C/2)/B+C
----> Substituting B=2C -----> 7/18
