Machine M, working alone at its constant rate, processes one ton of

Question

Machine M, working alone at its constant rate, processes one ton of a chemical product in 3/4 of the time that Machine N, working alone at its constant rate, does. If the two machines, working simultaneously at their respective constant rates, process one ton of the chemical product in 6 hours, how many hours will it take for Machine N working alone to process this amount?

A. 8
B. 9
C. 10
D. 12
E. 14

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Official Answer

E

A. 8
B. 9
C. 10
D. 12
E. 14

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modiassumenda · Answer

Solution: 
Let the time taken by  Machine N  alone be  x hours .
According to the question,  Machine M  takes only  3/4  of N's time.
So,
Time taken by M = (3/4)x hours
 Step 1:  Write the work rates.
Rate of N = 1/x ton per hour
Rate of M = 1/((3/4)x)
= 4/(3x) ton per hour
 Step 2:  Use the combined work rate.
Together, they finish 1 ton in 6 hours.
So,
1/x + 4/(3x) = 1/6
 Step 3:  Simplify.
Take the LCM (3x):
3/(3x) + 4/(3x) = 1/6
7/(3x) = 1/6
42 = 3x
x = 14

Final Answer: E. 14 hours

Edskore · Answer

Work rate problem, and I actually like this one because the setup trips a lot of people even though the algebra is short.

1. Let N's time alone = n hours. The problem says M takes 3/4 of N's time, so M's time alone = (3/4)n hours. Notice M is faster than N here, since it takes less time.

2. Rates add when things work together. Rate of M = 1/((3/4)n) = 4/(3n). Rate of N = 1/n.

3. Combined they finish in 6 hours, so combined rate = 1/6. Set up: 4/(3n) + 1/n = 1/6.

4. Common denominator on the left: 4/(3n) + 3/(3n) = 7/(3n). So 7/(3n) = 1/6, which gives 3n = 42, so n = 14.

5. That's N's time alone, exactly what's being asked. Answer is E, 14 hours.

The trap here, at least the one I fell into the first time I saw this style question, is reading "3/4 of the time" and just assuming that means multiply 6 by something simple to land on 12. It's tempting because 12 is sitting right there as answer D and it "feels" like the round number a work problem should spit out. But GMAT rate problems basically never let you skip setting up the actual rate equation, there's no shortcut that avoids the fractions.

Quick check: plug n=14 back in, M's time is 10.5 hours, rate 2/21, N's rate is 1/14 which is 1.5/21, add them and you get 3.5/21 = 1/6. Checks out.

One-liner: whenever you see "X takes a fraction of the time Y takes," translate that into an actual rate before you touch the combined-rate equation, don't try to shortcut it in your head.

onlyPlanA · Answer

Let the time in which Machine N complete the work be t, then time in which Machine M completes the work = 3/4t

We are given that:
1/t + 1/(3/4t) = 1/6
1/t + 4/3t = 1/6
So, t = 14 hours.

IMO, answer is E.

icandoit4u · Answer

Let hours worked by N be N 
    
 Time 
 Rate 
 Work 
 
  M 
 3/4*N 
 1/(3N/4) 
 1 
 
  N 
 N 
 1/N 
 1 
 
  M+N 
 6 
 1/6 
 1

rate of M + rate of N= rate of M+N
4/3n+1/n=1/6
7/3n=1/6
n=14

Bunuel · Answer

Official Explanation:

https://www.youtube.com/watch?v=m_T_v5wqOS8

Machine M, working alone at its constant rate, processes one ton of

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Machine M, working alone at its constant rate, processes one ton of

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