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Updated on: 24 Jul 2019, 05:17
Originally posted by bytatia on 16 Aug 2014, 18:56.
Last edited by Sajjad1994 on 24 Jul 2019, 05:17, edited 1 time in total.
Last edited by Sajjad1994 on 24 Jul 2019, 05:17, edited 1 time in total.
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
15%
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Question Stats:
81% (01:15) correct
19%
(01:41)
wrong
based on 561
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In what proportion must rice at $0.8 per pound be mixed with rice at $0.9 per pound so that the mixture costs $0.825 per pound?
(A) 1 : 3
(B) 1 : 2
(C) 1 : 1
(D) 2 : 1
(E) 3 : 1
(A) 1 : 3
(B) 1 : 2
(C) 1 : 1
(D) 2 : 1
(E) 3 : 1
Source: Nova GMAT
Difficulty Level: 600
Difficulty Level: 600
29 Jan 2018, 11:11
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bytatia
We can let x = the number of pounds of the $0.8 per pound flour and y = the number of pounds of the $0.9 per pound flour and create the equation:
0.8x + 0.9y = 0.825(x + y)
800x + 900y = 825x + 825y
75y = 25x
3y = x
3/1 = x/y
Answer: E
imsahlahno
Joined: 15 Sep 2013
Last visit: 14 Apr 2019
Posts: 27
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Concentration: Strategy, Entrepreneurship
Schools: Anderson '18 Darden '18 Johnson '18 ISB '17 Kellogg '18 NUS '18 Haas '18 McCombs '18 (WL) Ross '18 (II)
GMAT 1: 680 Q47 V36
GMAT 2: 740 Q50 V40
GPA: 3.65
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Schools: Anderson '18 Darden '18 Johnson '18 ISB '17 Kellogg '18 NUS '18 Haas '18 McCombs '18 (WL) Ross '18 (II)
GMAT 2: 740 Q50 V40
Posts: 27
06 Sep 2014, 03:10
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Using weighted average method:
Let x be the proportion in which the $0.8 per pound flour is mixed with $0.9 per pound flour.
Thus 0.8*x + 0.9*(1-x) = 0.825
0.9 – 0.1x = 0.825
0.1x = 0.075
x = 0.75
Thus ¾ of the mixture contains $0.8 per pound flour. Thus ratio of both flours is 3:1
If one has knowledge of weighted averages… the method by acegmat1 solves the questions in approx. 45 seconds…
Let x be the proportion in which the $0.8 per pound flour is mixed with $0.9 per pound flour.
Thus 0.8*x + 0.9*(1-x) = 0.825
0.9 – 0.1x = 0.825
0.1x = 0.075
x = 0.75
Thus ¾ of the mixture contains $0.8 per pound flour. Thus ratio of both flours is 3:1
If one has knowledge of weighted averages… the method by acegmat1 solves the questions in approx. 45 seconds…
