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Ten theater students are to begin work on a set for a musical. Working

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Ten theater students are to begin work on a set for a musical. Working  [#permalink]

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New post 07 Mar 2017, 11:30
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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

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Ten theater students are to begin work on a set for a musical. Working  [#permalink]

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New post Updated on: 08 Mar 2017, 11:07
1
Let the Rate of 1 student = R

As per given information

10 * R * 60 = W (Total Work)
R=W/600

To complete the work in 50 days, lets say we need 'N' students working at the same rate

N*R*50 = W
replacing value of R above,
N*(W/600)*50 =W
Solving for N we get N=12

Answer is A. 2

Originally posted by quantumliner on 07 Mar 2017, 12:05.
Last edited by quantumliner on 08 Mar 2017, 11:07, edited 1 time in total.
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Ten theater students are to begin work on a set for a musical. Working  [#permalink]

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New post Updated on: 08 Mar 2017, 04:59
2
the rate of each worker is 1/600
let the workers needed be N..

then N x (1/600) = 1/50

solving for N gives N= 12.

hence 12-10 = 2 more students required

ans A
thanks

Originally posted by mohshu on 07 Mar 2017, 22:23.
Last edited by mohshu on 08 Mar 2017, 04:59, edited 2 times in total.
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Re: Ten theater students are to begin work on a set for a musical. Working  [#permalink]

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New post 08 Mar 2017, 02:27
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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Re: Ten theater students are to begin work on a set for a musical. Working  [#permalink]

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New post 08 Mar 2017, 09:26
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Bunuel wrote:
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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Re: Ten theater students are to begin work on a set for a musical. Working  [#permalink]

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New post 10 Mar 2017, 10:42
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Bunuel wrote:
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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Re: Ten theater students are to begin work on a set for a musical. Working  [#permalink]

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New post 22 Feb 2019, 12:41
Bunuel wrote:
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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Re: Ten theater students are to begin work on a set for a musical. Working  [#permalink]

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New post 22 Feb 2019, 15:17
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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Re: Ten theater students are to begin work on a set for a musical. Working  [#permalink]

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New post 24 Feb 2019, 16:06
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...

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Re: Ten theater students are to begin work on a set for a musical. Working  [#permalink]

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New post 03 Mar 2019, 09:26
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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Re: Ten theater students are to begin work on a set for a musical. Working   [#permalink] 03 Mar 2019, 09:26
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