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04 Mar 2025, 02:26
Kudos
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If you choose numbers x = 1, y = 1
Then B can complete the job in an hour
A and B together can complete the job in half an hour.
In half an hour, B can complete only half of the work.
So the work saved by B with help of A is 1 - 1/2 = 1/2
Fraction of job saved by B with help of A = 1/2 / 1 = 1/2
If plugin x = 1 and y = 1, you should get 1/2
Options A B and D are eliminated right off the bat as x - y = 0
C) (x + y)/(xy)
Option C (1+1)/1 = 2. Eliminated as it doesn't give 1/2
E) y/(x + y)
Option E 1/(1+1) = 1/2
Hence option E is the answer.
Algebraic method.
B complete the job in y hours
A and B together can complete the job in xy/(x+y) hours
A = RT. A = 1
1 = RT -> 1 = (1/x+1/y) T -> 1 = (x+y/xy) T -> T = xy/(x+y)
Amount of work B can finish in xy/(x+y) hours is A = 1/y * xy(x+y) -> A = x/(x+y)
Amount of work B can save in xy/(x+y) hours is 1 - x/(x+y) = (x + y - x ) / (x + y) = y/(x+y) -> Option E
Then B can complete the job in an hour
A and B together can complete the job in half an hour.
In half an hour, B can complete only half of the work.
So the work saved by B with help of A is 1 - 1/2 = 1/2
Fraction of job saved by B with help of A = 1/2 / 1 = 1/2
If plugin x = 1 and y = 1, you should get 1/2
Options A B and D are eliminated right off the bat as x - y = 0
C) (x + y)/(xy)
Option C (1+1)/1 = 2. Eliminated as it doesn't give 1/2
E) y/(x + y)
Option E 1/(1+1) = 1/2
Hence option E is the answer.
Algebraic method.
B complete the job in y hours
A and B together can complete the job in xy/(x+y) hours
A = RT. A = 1
1 = RT -> 1 = (1/x+1/y) T -> 1 = (x+y/xy) T -> T = xy/(x+y)
Amount of work B can finish in xy/(x+y) hours is A = 1/y * xy(x+y) -> A = x/(x+y)
Amount of work B can save in xy/(x+y) hours is 1 - x/(x+y) = (x + y - x ) / (x + y) = y/(x+y) -> Option E
enigma123
06 Mar 2026, 00:28
Kudos
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