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07 May 2021, 11:16
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A
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Difficulty:
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Question Stats:
76% (02:26) correct
24%
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wrong
based on 136
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Machine A, B, and C, working together at their respective constant rates, can do a certain job in 6 hours. The rate of machine A is twice that of machine B, and the rate of machine C is 50% less than the rate of machine B. How long would it take machines B and C, working at their respective constant rates, to compare the same job?
A. 14
B. 15
C. 16
D. 17
E. 18
A. 14
B. 15
C. 16
D. 17
E. 18
09 Oct 2024, 06:54
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AJSIPA
Machine A, B, and C, working together at their respective constant rates, can do a certain job in 6 hours. The rate of machine A is twice that of machine B, and the rate of machine C is 50% less than the rate of machine B. How long would it take machines B and C, working at their respective constant rates, to compare the same job?
A. 14
B. 15
C. 16
D. 17
E. 18
Assuming the rates of machines A, B, and C are denoted by a, b, and c, respectively, we want to find the value of x from the equation b + c = 1/x, given that:
a + b + c = 1/6
a = 2b
c = b/2
a = 2b
c = b/2
Substituting a = 2b and c = b/2 into the first equation gives:
2b + b + b/2 = 1/6
b = 1/21
b = 1/21
Thus, c = 1/42, and we have:
b + c = 1/x
1/21 + 1/42 = 1/x
x = 14.
1/21 + 1/42 = 1/x
x = 14.
Answer: A
General Discussion
12 May 2021, 10:02
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In this question on rates, the solution becomes easier if we take some simple variables to represent the respective rates of the machines A, B and C. This way, an equation can be formed using these variables which can then be solved to determine the values of the variables and hence the answers.
Let Machine A work at the rate of a units/per hour, machine B at b units/hour and machine C at c units/hour.
Let the total work representing the job be 6x units. Since this work is being completed in 6 hours by all three machines working together,
Combined rate = \(\frac{Total work }{ Total time}\) = \(\frac{6x }{ 6}\) = x units per hour.
The cardinal equation to solve questions on Rates (Time & Work) is Work = Time x Rate. Remember, if nothing else works, this equation will always work for you.
Back to the question now!
Since the combined rate of the three machines is a+b+c, we can say a+b+c = x. We also know,
a = 2b and c = (1/2) b.
Substituting these values in the equation, we have, 2b + b + \(\frac{b}{2}\) = x. On solving, we get, b = \(\frac{2x }{7}\) and therefore c = \(\frac{x}{7}\).
Since we want B and C to complete the job, combined rate of B and C = \(\frac{2x}{7}\) + \(\frac{x}{7}\) = \(\frac{3x}{7}\) units per hour.
Therefore, time taken to complete the job (of 6x units) = \(\frac{Work }{ Rate}\) = 6x / (3x/7) = 14 hours.
The correct answer option is A.
Notice how I took the total work as 6x units (instead of the uber popular 1 unit) so that I didn’t have to deal with a lot of fractions in my calculations.
Hope that helps!
Aravind BT
Let Machine A work at the rate of a units/per hour, machine B at b units/hour and machine C at c units/hour.
Let the total work representing the job be 6x units. Since this work is being completed in 6 hours by all three machines working together,
Combined rate = \(\frac{Total work }{ Total time}\) = \(\frac{6x }{ 6}\) = x units per hour.
The cardinal equation to solve questions on Rates (Time & Work) is Work = Time x Rate. Remember, if nothing else works, this equation will always work for you.
Back to the question now!
Since the combined rate of the three machines is a+b+c, we can say a+b+c = x. We also know,
a = 2b and c = (1/2) b.
Substituting these values in the equation, we have, 2b + b + \(\frac{b}{2}\) = x. On solving, we get, b = \(\frac{2x }{7}\) and therefore c = \(\frac{x}{7}\).
Since we want B and C to complete the job, combined rate of B and C = \(\frac{2x}{7}\) + \(\frac{x}{7}\) = \(\frac{3x}{7}\) units per hour.
Therefore, time taken to complete the job (of 6x units) = \(\frac{Work }{ Rate}\) = 6x / (3x/7) = 14 hours.
The correct answer option is A.
Notice how I took the total work as 6x units (instead of the uber popular 1 unit) so that I didn’t have to deal with a lot of fractions in my calculations.
Hope that helps!
Aravind BT
