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alphonsa
Joined: 22 Jul 2014
Last visit: 25 Oct 2020
Posts: 106
Given Kudos: 196
Concentration: General Management, Finance
Schools: Cox '19 (D) Babson '19 (A$) Katz '19 (D) Eller"19 (A$) Northeastern '19 (A$) Krannert '19 (A$) ISB '18 (D) Rutgers '19 (A) Olin '19 (I)
GMAT 1: 670 Q48 V34
WE:Engineering (Energy)
Products:
Schools: Cox '19 (D) Babson '19 (A$) Katz '19 (D) Eller"19 (A$) Northeastern '19 (A$) Krannert '19 (A$) ISB '18 (D) Rutgers '19 (A) Olin '19 (I)
GMAT 1: 670 Q48 V34
Posts: 106
29 Aug 2014, 05:17
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
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Question Stats:
60% (02:08) correct
40%
(02:18)
wrong
based on 233
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In a garden, the ratio of the number of apples, bananas and cherries is 7:5:2. If the number of cherries is greater than or equal to 5, how many apples are there in the garden?
(1) The sum of the number of cherries and bananas is less than 29
(2) The sum of the number of apples and bananas is less than 37.
Source: 4Gmat
(1) The sum of the number of cherries and bananas is less than 29
(2) The sum of the number of apples and bananas is less than 37.
Source: 4Gmat
Pure algebraic questions like this one are no longer part of the GMAT Data Sufficiency syllabus.
For relevant Data Sufficiency questions, please check the Data Sufficiency forum:
Data Sufficiency Forum
Data Sufficiency Forum
29 Aug 2014, 05:36
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In a garden, the ratio of the number of apples, bananas and cherries is 7:5:2. If the number of cherries is greater than or equal to 5, how many apples are there in the garden?
Apples:Bananas:Cherries = 7:5:2. Since given that the number of cherries is greater than or equal to 5 and from the ratio we know that the number of cherries is a multiple of 2, then the least number of cherries is 6. So, A:B:C = 7x:5x:2x, where \(x\geq{3}\).
(1) The sum of the number of cherries and bananas is less than 29 --> 5x + 2x = 7x < 29 --> \(3\leq{x}\leq{4}\) --> the number of apples could be 7x = 7*3 = 21 or 7x = 7*4 = 28. Not sufficient.
(2) The sum of the number of apples and bananas is less than 37 --> 7x + 5x = 12x < 37 --> since \(x\geq{3}\), then x can only be 3 --> the number of apples is 7x = 7*3 = 21. Sufficient.
Answer: B.
Apples:Bananas:Cherries = 7:5:2. Since given that the number of cherries is greater than or equal to 5 and from the ratio we know that the number of cherries is a multiple of 2, then the least number of cherries is 6. So, A:B:C = 7x:5x:2x, where \(x\geq{3}\).
(1) The sum of the number of cherries and bananas is less than 29 --> 5x + 2x = 7x < 29 --> \(3\leq{x}\leq{4}\) --> the number of apples could be 7x = 7*3 = 21 or 7x = 7*4 = 28. Not sufficient.
(2) The sum of the number of apples and bananas is less than 37 --> 7x + 5x = 12x < 37 --> since \(x\geq{3}\), then x can only be 3 --> the number of apples is 7x = 7*3 = 21. Sufficient.
Answer: B.
