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A piece of work can be done by 10 boys in 6 days working 4 hours each

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A piece of work can be done by 10 boys in 6 days working 4 hours each [#permalink] New post   04 May 2016, 07:35
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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
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D
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Kurtosis
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Re: A piece of work can be done by 10 boys in 6 days working 4 hours each [#permalink] New post   04 May 2016, 08:01
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Total work done = 10 * 6 * 4 = 240 units

New work = Three times more than current work

New work = Current work + 3(Current work)
New work = 4*240 = 960 units

Number of boys * x * x = 960

960 can be factorized as 2^6 * 3 * 5 --> 2^6 can be expressed as x^2 i.e. (2^3)^2

Minimum number of boys = 3 * 5 = 15

Answer: D
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Re: A piece of work can be done by 10 boys in 6 days working 4 hours each [#permalink] New post   06 May 2016, 22:47
Vyshak -- I understand everything up to: "# of boys * x * x = 960"

How are you able to replace (2^3)^2 for x^2 like that? I'd think each boy does (1)(6)(4) = 24 units of work, so 960 / 24 = 40 boys

Can you help me understand please? Thank you :D

Michael
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Re: A piece of work can be done by 10 boys in 6 days working 4 hours each [#permalink] New post   06 May 2016, 22:59
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mdacosta wrote:
Vyshak -- I understand everything up to: "# of boys * x * x = 960"

How are you able to replace (2^3)^2 for x^2 like that? I'd think each boy does (1)(6)(4) = 24 units of work, so 960 / 24 = 40 boys

Can you help me understand please? Thank you :D

Michael


Hi Michael,

The question states that the number of hours worked and the number of days taken to complete the work are same. When you take 6 hrs and 4 days or vice versa, it violates the given condition. When we factorize we should express one number as a square of an integer. We are able to express only 2^3 in this way.
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Re: A piece of work can be done by 10 boys in 6 days working 4 hours each [#permalink] New post   07 May 2016, 01:36
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Hello all! This is my first post, please excuse if I am not following protocol etc :)

I misunderstood this statement: another work which is three times more - is this phrased correctly?
I understood that the work was 3x times 240, ie 720 (which is 5 boys working 12 hours over 12 days)

I understand how "three times" is not exactly the same as "three times more", however, this is how people speak when they mean 3x times; they use both to mean the same thing.

English is not my first language and so it is quite possible that I misunderstand the subtlety of this. My concern is more if I should look out for something like this on the test, or if it will be more clear in the GMAT?
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Re: A piece of work can be done by 10 boys in 6 days working 4 hours each [#permalink] New post   07 May 2016, 01:38
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Vyshak I don't seem to understand why have you considered New work = Current work + 3(Current work)

In question stem it is mentioned that How many minimum possible number of boys are needed to complete another work which is three times more
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Re: A piece of work can be done by 10 boys in 6 days working 4 hours each [#permalink] New post   07 May 2016, 01:51
anurag16 wrote:
Vyshak I don't seem to understand why have you considered New work = Current work + 3(Current work)

In question stem it is mentioned that How many minimum possible number of boys are needed to complete another work which is three times more


Hi,

If new work(x') is three times current work then you can consider x' = 3x. If new work is 3 times more than current work then it means that the difference between new work and current work is three times.
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A piece of work can be done by 10 boys in 6 days working 4 hours each [#permalink] New post   07 May 2016, 01:53
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Vyshak Then my solution will be something like this. Correct me if I am wrong:

Rate=?; Time= 24hrs; Work=1
So Rate= 1/24 for 10 boys
So Rate= 1/240 for 1 boy

Now let there be n boys for new work
Rate= n/240; Time=t ; New Work= 1+3
So (n/240)*t=4
So nt=960
Now for n to be minimum t should be maximum.
960=2^6 * 3 * 5
Since days=hrs, we can only have days =2^3 and hrs=2^3
so 960/(2^3*2^3)=15
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Re: A piece of work can be done by 10 boys in 6 days working 4 hours each [#permalink] New post   07 May 2016, 02:35
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chetan2u 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


Hi,
the most important point here is to note that 3 times more than x means x+3x or 4x..

First find the boys-hour required for initial work \(= 10*6*4 = 240\)
Now the new work is 3 times more, so boys hour required \(= 240+3*240 = 960\)..
Now x*x*boys = \(960 = 64 *3*5 = 8^2 *3*5.\).
We are looking for minimum possible number of boys, equate two sides
so x= 8 and boys = 15

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Re: A piece of work can be done by 10 boys in 6 days working 4 hours each [#permalink] New post   23 Feb 2022, 11:25
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