Machines J, K, and M, working together at their respective constant

Question

Machines J, K, and M, working together at their respective constant rates, produce c parts in 4 hours. Machine J, working alone at its constant rate, produces c parts in 10 hours, and machine K, working alone at its constant rate, produces c parts in 12 hours. How many hours does it take machine M, working alone at its constant rate, to produce c parts?

A. 11
B. 14
C. 15
D. 18
E. 20

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2024-01-26_15-41-04.png [ 56.61 KiB | Viewed 6685 times ]

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Official Answer

C

Bunuel · Accepted Answer

Assume that the rates of machines J, K, and M, are j, k, and m, respectively. Then, we are given that:

j + k + m = 1/4.
j = 1/0.
k = 1/12.

Substituting the value of j and k in the first equation gives 1/10 + 1/12 + m = 1/4, which results in m = 1/15.

Therefore, since time is the reciprocal of the rate, it takes 15 hours for machine M to produce c parts.

Answer: C.

Abhishek009 · Answer

Let the total work 'C' be LCM  (4,10 & 12) = 60

So, Efficiency of  J + K + M = 15 ,Efficiency of  J = 6  & Efficiency of  K = 5

Thus, Efficiency of  M = 15 - ( 6 + 5) = 4

So, Time required by M will be  \frac{60}{4} = 15 , Answer must be (C)

unraveled · Answer

Machines J, K, and M, working together at their respective constant rates, produce c parts in 4 hours. Machine J, working alone at its constant rate, produces c parts in 10 hours, and machine K, working alone at its constant rate, produces c parts in 12 hours. How many hours does it take machine M, working alone at its constant rate, to produce c parts?

A. 11
B. 14
C. 15
D. 18
E. 20

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Attachment:

2024-01-26_15-41-04.png [ 56.61 KiB | Viewed 6685 times ]

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Official Answer

C

A. 11
B. 14
C. 15
D. 18
E. 20

Although we can solve this using the work/time method, trying some other logic. 
Weighted average formula might work in this but in an inverse manner.
We know that if A,B and C contribute to an average rate of all of them together then that average is inclined towards the rate that has maximum weight among the three. For example: If A is rate of apples bought per unit, B for orange and C for pineapples. In this case the average rate of per unit fruit would be inclined towards the rate that is maximum among the three. If A > B > C then average of the three would be nearer to A.

Similarly, for this problem if average is 4 hrs and J and K are 10 and 12 then M would be more than 12. 
This is because the relation in work and time is inverse, the more a machine works the less the time it takes to complete a task.
Here among the answer choices only 11 is less than 12 therefore this much of thinking is not worth as we are yet to find something more logical to reach an answer as we have four more choices.

Inverse relation logic can thus be used. 
1/4 = 0.25
1/10 = 0.1 
1/12 = 0.08..
Hence which among the four choices result in 0.25 - 0.18.. = 0.06..
Only C satisfies as 1/15 = 0.06..
Finally, the decimals are close so it is little tricky to choose the right answer.

Here we now go back to the work time formula where
1/j+1/k+1/m = 1/4
This gives m = 15.

Answer D.

Confide · Answer

Work of J + Work of K + Work of M = c parts
4c/10 + 4c/12 + 4c/t = c
4/10 + 4/12 + 4/t = c
6t + 5t + 60 = 15t
4t = 60
t = 15

Keep it simple

NEYR0N · Answer

let c = 1 , also - assuming you have memorized basic fractions -- > 1/4 - 1/10 - 1/12 = 0.25 - 0.1 - 0.083 = 0.149smthng = 0.15

Machines J, K, and M, working together at their respective constant

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Machines J, K, and M, working together at their respective constant

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