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10 Nov 2020, 10:45
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A
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Difficulty:
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Question Stats:
86% (01:32) correct
14%
(01:48)
wrong
based on 328
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A truck is filled to 1/4 of its maximum weight capacity. An additional y pounds are added such that the truck is now filled to 7/8 of its capacity. In terms of y, what is the maximum weight capacity of the truck, in pounds?
A. 8y/5
B. 7y/5
C. 5y/8
D. 5y/4
E. 5y/3
A. 8y/5
B. 7y/5
C. 5y/8
D. 5y/4
E. 5y/3
10 Nov 2020, 10:51
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let max capacity of truck be 32
1/4th of max weight is 8
later 8+y = 7/8 * 32 ; 8+y = 28 ; y=20
substitute y in given options
8y/5 ; 8*20 /5 ; 32
option A is correct
1/4th of max weight is 8
later 8+y = 7/8 * 32 ; 8+y = 28 ; y=20
substitute y in given options
8y/5 ; 8*20 /5 ; 32
option A is correct
Bunuel
10 Nov 2020, 20:40
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Let the maximum capacity= x
Filling it to 1/4th of its capacity = \(\frac{x}{4}\)
Adding y pounds = \(\frac{x}{4} + y\) and this is now 7/8th of the capacity = \(\frac{7x}{8}\)
Therefore \(\frac{x}{4} + y = \frac{7x}{8}\)
\(y = \frac{7x}{8} - \frac{x}{4} = \frac{5x}{8}\)
Therefore x = 8y/5
Option A
Arun Kumar
Filling it to 1/4th of its capacity = \(\frac{x}{4}\)
Adding y pounds = \(\frac{x}{4} + y\) and this is now 7/8th of the capacity = \(\frac{7x}{8}\)
Therefore \(\frac{x}{4} + y = \frac{7x}{8}\)
\(y = \frac{7x}{8} - \frac{x}{4} = \frac{5x}{8}\)
Therefore x = 8y/5
Option A
Arun Kumar
