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# Two workers A and B are engaged to do a work. A working alone takes 8

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Manager
Joined: 09 Jan 2016
Posts: 106
GPA: 3.4
WE: General Management (Human Resources)
Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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10 Mar 2017, 22:30
4
1
16
00:00

Difficulty:

55% (hard)

Question Stats:

71% (02:41) correct 29% (02:51) wrong based on 187 sessions

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Two workers A and B are engaged to do a work. A working alone takes 8 hours more to complete the job than if both worked together. If B worked alone, he would need 4.5 hours more to complete the job than they both working together. What time would they take to do the work together ?

A. 4 hours
B. 5 hours
C. 6 hours
D. 7 hours
E. 8 hours
Manager
Joined: 06 Dec 2016
Posts: 245
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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11 Mar 2017, 20:58
2
4
I did it this way and got the same result

1/x+8 + 1/x+4.5 = 1/x
1/x+4.5 = 1/x - 1/x+8
1/x+4.5 = x+8-x/(x(x+8)
1/x+4.5 = 8/(x(x+8)
x(x+8) = 8(x+4.5)
x² + 8x = 8(x+4.5)
x² = 40
x is equal to approximately 6.

##### General Discussion
Manager
Joined: 17 May 2015
Posts: 249
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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11 Mar 2017, 02:22
3
2
Chemerical71 wrote:
Two workers A and B are engaged to do a work. A working alone takes 8 hours more to complete the job than if both worked together. If B worked alone, he would need
4.5 hours more to complete the job than they both working together. What time would they take to do the work together ?

A. 4 hours
B. 5 hours
C. 6 hours
D. 7 hours
E. 8 hours

Let A works at the rate of 'a' units/hr and B works at the rate of 'b' units/hr. A and B complete the job in x hrs.

A working alone takes 8 hours more to complete the job than if both worked together.
=> a(x+8) = (a+b)x
=> ax + 8a = ax + bx
=> 8a = bx or a = bx/8 ---(1)
If B worked alone, he would need 4.5 hours more to complete the job than they both working together

=> b(x+4.5) = (a+b)x
=> bx +4.5b = ax + bx
=> 4.5b = ax or a = 4.5b/x ---(2)

From (1) and (2), we have
$$\frac{bx}{8} = {4.5b}{x} \Rightarrow x^{2} = 4.5*8 \Rightarrow x = 6$$

Manager
Joined: 09 Jan 2016
Posts: 106
GPA: 3.4
WE: General Management (Human Resources)
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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11 Mar 2017, 03:44
ganand wrote:
Chemerical71 wrote:
Two workers A and B are engaged to do a work. A working alone takes 8 hours more to complete the job than if both worked together. If B worked alone, he would need
4.5 hours more to complete the job than they both working together. What time would they take to do the work together ?

A. 4 hours
B. 5 hours
C. 6 hours
D. 7 hours
E. 8 hours

Let A works at the rate of 'a' units/hr and B works at the rate of 'b' units/hr. A and B complete the job in x hrs.

A working alone takes 8 hours more to complete the job than if both worked together.
=> a(x+8) = (a+b)x
=> ax + 8a = ax + bx
=> 8a = bx or a = bx/8 ---(1)
If B worked alone, he would need 4.5 hours more to complete the job than they both working together

=> b(x+4.5) = (a+b)x
=> bx +4.5b = ax + bx
=> 4.5b = ax or a = 4.5b/x ---(2)

From (1) and (2), we have
$$\frac{bx}{8} = {4.5b}{x} \Rightarrow x^{2} = 4.5*8 \Rightarrow x = 6$$

Thank you very much .
Manager
Joined: 17 May 2015
Posts: 249
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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12 Mar 2017, 03:15
matthewsmith_89 wrote:
I did it this way and got the same result

1/x+8 + 1/x+4.5 = 1/x
1/x+4.5 = 1/x - 1/x+8
1/x+4.5 = x+8-x/(x(x+8)
1/x+4.5 = 8/(x(x+8)
x(x+8) = 8(x+4.5)
x² + 8x = 8(x+4.5)
x² = 40
x is equal to approximately 6.

Hi,
This is perfectly fine. There is a typo in your solution.

x² = 40 It should be
$$x^{2}$$= 8*4.5 = 36 => x = 6

Thanks.
Manager
Joined: 09 Jan 2016
Posts: 106
GPA: 3.4
WE: General Management (Human Resources)
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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12 Mar 2017, 05:29
2
ganand wrote:
matthewsmith_89 wrote:
I did it this way and got the same result

1/x+8 + 1/x+4.5 = 1/x
1/x+4.5 = 1/x - 1/x+8
1/x+4.5 = x+8-x/(x(x+8)
1/x+4.5 = 8/(x(x+8)
x(x+8) = 8(x+4.5)
x² + 8x = 8(x+4.5)
x² = 40
x is equal to approximately 6.

Hi,
This is perfectly fine. There is a typo in your solution.

x² = 40 It should be
$$x^{2}$$= 8*4.5 = 36 => x = 6

Thanks.
Thank you very much for solving.
Intern
Joined: 10 Dec 2014
Posts: 45
Concentration: Marketing
GMAT 1: 750 Q50 V41
WE: Marketing (Manufacturing)
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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07 Jul 2017, 16:23
2
Let a be the number of hours A completes work and b be the number of hours B completes work. Then (1/a) is A's rate per hour and (1/b) be B's rate per hour.
Then the rate if A and B work together is (1/a)+(1/b)=(a+b)/(ab). Time that A and B complete work is 1/[(a+b)/(ab)] = (ab)/(a+b) (this is what we want to find out)

As A works alone takes 8 hours more than A and B work together, then: a - (ab)/(a+b) = (a^2)/(a+b) = 8 (*)
As B works alone takes 4.5 hours more than A and B work together, then: b - (ab)/(a+b) = (b^2)/(a+b) = 4.5 (**)

From (*) (**), we have [(a^2)/(a+b)]*[(b^2)/(a+b)] = [(ab)^2]/[(a+b)^2] = 8*4.5 = 8*9/2 = 36
then (ab)/(a+b) = sqrt (36) = 6 => Answer

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Senior Manager
Joined: 28 Jun 2015
Posts: 290
Concentration: Finance
GPA: 3.5
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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07 Jul 2017, 18:04
1
Chemerical71 wrote:
Two workers A and B are engaged to do a work. A working alone takes 8 hours more to complete the job than if both worked together. If B worked alone, he would need
4.5 hours more to complete the job than they both working together. What time would they take to do the work together ?

A. 4 hours
B. 5 hours
C. 6 hours
D. 7 hours
E. 8 hours

Plugging numbers is less time consuming than the algebraic method for me. Let's start with C since it's the middle value.

$$\frac{1}{6+8} + \frac{2}{12+9} =? \frac{1}{6}$$

$$\frac{2}{28} + \frac{2}{21} =? \frac{1}{6}$$

$$\frac{2}{7}(\frac{1}{4} + \frac{1}{3}) =? \frac{1}{6}$$

$$\frac{2}{7} * \frac{7}{12} =? \frac{1}{6}$$

$$\frac{2}{12} = \frac{1}{6}.$$ So, this is the answer. Ans - C.
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Manager
Joined: 24 Jun 2017
Posts: 122
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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28 Aug 2017, 14:00
1
1/(t+8) + 1/(t+4.5) = 1/t
t(2t + 12.5) = (t +8)(t+4.5)
t^2 = 36
t = 6
Manager
Joined: 17 May 2015
Posts: 249
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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28 Aug 2017, 19:17
1
Hi,

The general framework of the above problem:

Total time taken to complete the work (working together) = $$x$$ hrs

Time taken by A to complete the work (working alone) = $$a$$ hrs more than combined time, i.e. $$(x+a)$$ hrs

Time taken by B to complete the work (working alone) = $$b$$ hrs more than combined time, i.e. $$(x+b)$$ hrs

Then $$x = \sqrt{a \times b}$$

In the above problem $$a = 8$$ hrs, and $$b = 4.5$$ hrs. Hence $$x = \sqrt{a \times b} = \sqrt{8 \times 4.5} = \sqrt{36} = 6$$ hrs.

Thanks.
Intern
Joined: 22 May 2018
Posts: 9
Location: Korea, Republic of
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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30 Sep 2018, 02:28
ganand wrote:
Hi,

The general framework of the above problem:

Total time taken to complete the work (working together) = $$x$$ hrs

Time taken by A to complete the work (working alone) = $$a$$ hrs more than combined time, i.e. $$(x+a)$$ hrs

Time taken by B to complete the work (working alone) = $$b$$ hrs more than combined time, i.e. $$(x+b)$$ hrs

Then $$x = \sqrt{a \times b}$$

In the above problem $$a = 8$$ hrs, and $$b = 4.5$$ hrs. Hence $$x = \sqrt{a \times b} = \sqrt{8 \times 4.5} = \sqrt{36} = 6$$ hrs.

Thanks.

I am not quite sure how you came up with this solution.

Thank you!
Manager
Joined: 17 May 2015
Posts: 249
Re: Two workers A and B are engaged to do a work. A working alone takes 8  [#permalink]

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30 Sep 2018, 03:37
1
blueblac92 wrote:
ganand wrote:
Hi,

The general framework of the above problem:

Total time taken to complete the work (working together) = $$x$$ hrs

Time taken by A to complete the work (working alone) = $$a$$ hrs more than combined time, i.e. $$(x+a)$$ hrs

Time taken by B to complete the work (working alone) = $$b$$ hrs more than combined time, i.e. $$(x+b)$$ hrs

Then $$x = \sqrt{a \times b}$$

In the above problem $$a = 8$$ hrs, and $$b = 4.5$$ hrs. Hence $$x = \sqrt{a \times b} = \sqrt{8 \times 4.5} = \sqrt{36} = 6$$ hrs.

Thanks.

I am not quite sure how you came up with this solution.

Thank you!

Hi @blueblac9,

Ok. I'll derive the relation.

Let A works at the rate of $$p$$ units/hr. Let B works at the rate of $$q$$ units/hr.

Total time taken to complete the work (working together) = $$x$$ hrs.

Time taken by A to complete the work (working alone) = $$a$$ hrs more than combined time, i.e. $$(x+a)$$ hrs.

$$\Rightarrow p(x+a) = (p+q)x \Rightarrow px + pa = px + qx \Rightarrow pa = qx \Rightarrow p = \frac{qx}{a}$$ --- (1)

Time taken by B to complete the work (working alone) = $$b$$ hrs more than combined time, i.e. $$(x+b)$$ hrs.

$$\Rightarrow q(x+b) = (p+q)x \Rightarrow qx + qb = px + qx \Rightarrow qb = px \Rightarrow p = \frac{qb}{x}$$ --- (2)

From (1) and (2), we have following:

$$\frac{qx}{a} = \frac{qb}{x} \Rightarrow x^2 = ab \Rightarrow x = \sqrt{ab}$$.

Hope this helps.

Thanks.
Re: Two workers A and B are engaged to do a work. A working alone takes 8   [#permalink] 30 Sep 2018, 03:37
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