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let the amount of work be 3000
rate of work for 10 students = 3000/60 = 50
rate of work per student = 50/10 =5

now let x be the number of student to complete the work in 50 days
According to the question 3000 = (5x)*50
or x = 12

hence 2 more student will be required

option A
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Bunuel
Ten theater students are to begin work on a set for a musical. Working at a constant rate, they will finish the set in 60 days, but the director of the musical decides they need to finish in 50 days. How many more theater students, working at the same constant rate, must work to finish the set on time?

A. 2
B. 3
C. 5
D. 10
E. 12

Hi

M1D1H1/W1=M2D2H2/W2

M=Men
D=Days
H=hours
W=work

It is given that MensDays=10*60

In order to finish the work in 50 days we need 10+M

So 10*60=(10+M)*50

M=2

A
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Bunuel
Ten theater students are to begin work on a set for a musical. Working at a constant rate, they will finish the set in 60 days, but the director of the musical decides they need to finish in 50 days. How many more theater students, working at the same constant rate, must work to finish the set on time?

A. 2
B. 3
C. 5
D. 10
E. 12

We are given that 10 theater students will finish a set in 60 days. Since rate = work/time, the rate of those students is 1/60 = 1/60. However, since the set needs to be completed in 50 days, we can create the following proportion to determine how many more students are needed to complete the job in 50 days:

10 students is to a rate of 1/60 as x students is to a rate of 1/50.

10/(1/60) = x/(1/50)

600 = 50x

12 = x

Thus, 2 more students will be needed to complete the set in 50 days.

Answer: A
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Bunuel
Ten theater students are to begin work on a set for a musical. Working at a constant rate, they will finish the set in 60 days, but the director of the musical decides they need to finish in 50 days. How many more theater students, working at the same constant rate, must work to finish the set on time?

A. 2
B. 3
C. 5
D. 10
E. 12

\(\frac{50-days}{60-days}\)= \(\frac{5}{6}\).
To finish the job in \(\frac{5}{6}\) of the normal time, the crew requires \(\frac{6}{5}\) of the normal number of students -- an increase of \(\frac{1}{5}\):
\((\frac{1}{5})(10) = 2\) more students

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The job requires 60*10 = 600 total days of work. So to do the job in 50 days, we need 12 people working simultaneously, or 2 extra people.
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Hi All,

We're told that 10 theater students are to begin work on a set for a musical. Working at a constant rate, they will finish the set in 60 days, but the director of the musical decides they need to finish in 50 days. We're asked for the number of ADDITIONAL theater students we would need, working at the same constant rate, to finish the set on time. This question can be approached in a number of different ways, including by focusing on the 'total work' needed to complete the job.

In these types of situations, it often helps to start with the total work|output needed to complete the given task. Here, we would need 10 students who each work for 60 days...

(10 students)(60 days each) = 600 total days of student-work needed to complete the job

Thus, 1 student would 600/1 = 600 days
2 students would take 600/2 = 300 days each
3 students would take 600/3 = 200 days each
Etc.

We can actually use this same logic based on the total number of days...
600/50 = 12 students needed

Since we already have 10 students, we would need 2 additional students...

Final Answer:

GMAT assassins aren't born, they're made,
Rich
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W=R*T
10 students do 10 units of work (1u/student) 10=R*60 and R=1/6
now, keep same 10 u of work and keep same constant rate but lower t (t=50):
10=(1/6)*50 --> 10=8 (how many units are missing in the RHS to reach 10? 2 units. (2 units is equivalent to 2 students (see first assumption above)). Thus, we need 2 extra students.
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