A piece of work can be done by 10 boys . . ..

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Vincen: Legendary Member; Posts: 2898; Joined: Thu Sep 07, 2017 2:49 pm; Thanked: 6 times; Followed by:5 members

A piece of work can be done by 10 boys . . ..

by Vincen » Tue Dec 19, 2017 7:17 am

A piece of work can be done by 10 boys in 6 days working 4 hours each day. How many minimum possible number of boys are needed to complete another work which is three times more if they work for same integer number of hour and day, say x hour and x days?
(1) 5
(2) 10
(3) 12
(4) 15
(5) 30

The OA is option D.

I got confused here. I'd like the explanation of an expert. Thanks in advanced.

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Source: Beat The GMAT — Problem Solving | Tue Dec 19, 2017 7:17 am

GMATWisdom: Master | Next Rank: 500 Posts; Posts: 100; Joined: Wed Nov 29, 2017 4:38 pm; Thanked: 14 times

by GMATWisdom » Fri Dec 22, 2017 11:39 am

Vincen wrote:A piece of work can be done by 10 boys in 6 days working 4 hours each day. How many minimum possible number of boys are needed to complete another work which is three times more if they work for same integer number of hour and day, say x hour and x days?
(1) 5
(2) 10
(3) 12
(4) 15
(5) 30

The OA is option D.

I got confused here. I'd like the explanation of an expert. Thanks in advanced.

If 10 boys can do a work in 6 days while working 4 hours a day
then one boy will do the same work in 10x6x4= 240 hours.

the new work is three times more
=> one boy will do three times more work in 240 + 240x3 =960 hours

further let Y be the minimum number of boys working x hours a day for x days to complete the new work
then Y = 960/x^2 or Y*x^2 = 960

now we have to find the maximum value of x which should be an integer so as to get the Y as an integer

that value we can find by making factors of 960 in the form Y*x^2
960 x 1^2
240 x 2^2
120 x 4^2
15 x 8^2
Thus the maximum value of x is 8 and the minimum value of Y is 15

Hence option (4) is correct.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Sun Sep 08, 2019 10:02 am

Vincen wrote:A piece of work can be done by 10 boys in 6 days working 4 hours each day. How many minimum possible number of boys are needed to complete another work which is three times more if they work for same integer number of hour and day, say x hour and x days?
(1) 5
(2) 10
(3) 12
(4) 15
(5) 30

The OA is option D.

I got confused here. I'd like the explanation of an expert. Thanks in advanced.

Since 10 boys work for 6 days and 4 hours each day, the job requires 10 x 6 x 4 = 240 man-hours. Since the other job requires three times more time, it requires four times as much time, or a total of 240 x 4 = 960 man-hours. Now, if we let n = the number of boys needed to complete this job for x hours per day and x days, we have:

nx^2 = 960

Since n and x are positive integers and we want the minimum value of n, we need to maximize the value of x. So let's factor 960:

960 = 16 x 60 = 4^2 x 4 x 15 = 4^2 x 2^2 x 15 = 8^2 x 15

We see that the largest perfect square that divides into 960 is 8^2, so x = 8, and hence, the minimum value of n must be 15.

Answer: D/4

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