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02 Dec 2019, 09:05
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
78% (02:10) correct
22%
(02:30)
wrong
based on 288
sessions
History
Date
Time
Result
Not Attempted Yet
Airplane fuel worth $126 per pound and $135 per pound is mixed with a third variety of high grade fuel in the ratio 1:1:2. If the resultant mixture is worth $153 per pound, the price of the third variety of fuel per pound will be:
A. $169.50
B. $170
C. $175.50
D. $180
E. $205
A. $169.50
B. $170
C. $175.50
D. $180
E. $205
Ansh777
Joined: 03 Nov 2019
Last visit: 06 Jun 2023
Posts: 53
Given Kudos: 128
Location: India
Schools: Booth '24 Columbia CBS '24 Darden '24 Fuqua '24 Haas '24 Harvard '24 LBS '24 Stanford '24 Kellogg '24 Wharton '24
GMAT 1: 710 Q50 V36
Schools: Booth '24 Columbia CBS '24 Darden '24 Fuqua '24 Haas '24 Harvard '24 LBS '24 Stanford '24 Kellogg '24 Wharton '24
GMAT 1: 710 Q50 V36
Posts: 53
02 Dec 2019, 11:24
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Rate/Pound $126 ; $135 ; Let:X ; $153
Quantity 1K + 1K + 2K = 4K
Total Cost= 126*K+135*K+X*2K = 153*4K => X=175.5
Answer: C
Quantity 1K + 1K + 2K = 4K
Total Cost= 126*K+135*K+X*2K = 153*4K => X=175.5
Answer: C
02 Dec 2019, 12:22
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Let F= cost/pound of 3rd fuel type
Let x= amount of each fuel in the mixture
126x+135x+F*2x=153*4x=612x
612x-126x-135x=F*2x
351x=F*2x
175.5=F
Ans=C
Let x= amount of each fuel in the mixture
126x+135x+F*2x=153*4x=612x
612x-126x-135x=F*2x
351x=F*2x
175.5=F
Ans=C
