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# If Adam can do a job in 10 days and Ben's speed is twice of Adam's spe

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If Adam can do a job in 10 days and Ben's speed is twice of Adam's spe  [#permalink]

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04 Jul 2018, 21:52
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Difficulty:

45% (medium)

Question Stats:

56% (02:13) correct 44% (01:55) wrong based on 46 sessions

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If Adam can do a job in 10 days and Ben's speed is twice of Adam's speed and both work one after another, what % of the work should be done by Adam to complete the work in exact 7 days?
(A) 67%
(B) 60%
(C) 50%
(D) 40%
(E) 33%

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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
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If Adam can do a job in 10 days and Ben's speed is twice of Adam's spe  [#permalink]

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04 Jul 2018, 22:25
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chetan2u wrote:
If Adam can do a job in 10 days and Ben's speed is twice of Adam's speed and both work one after another, what % of the work should be done by Adam to complete the work in exact 7 days?
(A) 67%
(B) 60%
(C) 50%
(D) 40%
(E) 33%

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Doesn't matter in what sequence or pattern they work, we are interested in the % of work completed in the end of 7 days.
And to finish the work in 7 days, only one solution is possible

The shortest method..

1) weighted average method

So if works can be completed in 5 and 10 days and we are looking for 7 days
5--7---10
So Adam who does in 10 days will do $$\frac{7-5}{10-5}=\frac{2}{5}$$ of the work and 2/5 in % = $$100*\frac{2}{5}=40%$$

2) work/rate problem

Let A works for x days, and his x day work will be $$\frac{x}{10}$$
so B will work for 7-x days, and his 7-x days work $$=\frac{(7-x)}{5}$$

$$\frac{x}{10} + \frac{(7-x)}{5} = 1.........x+14-2x=10......x=4$$
So A works for 4 days in total .....% of work completed in 3 days =$$100*\frac{4}{10}=40%$$
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If Adam can do a job in 10 days and Ben's speed is twice of Adam's spe  [#permalink]

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04 Jul 2018, 22:42
2
chetan2u wrote:
If Adam can do a job in 10 days and Ben's speed is twice of Adam's speed and both work one after another, what % of the work should be done by Adam to complete the work in exact 7 days?
(A) 67%
(B) 60%
(C) 50%
(D) 40%
(E) 33%

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Adam does the work in 10 days and since Ben's speed is twice Adam's speed,
Ben does the same job in 5 days. Let's assume the total work to be 100 units.

Now, the individual rates are Adam - 10 units/day and Ben - 20 units/day.

We need to find how much work does Adam need to do such that the work is completed in
7 days. If $$x$$ is the number of days Adam works, then $$7-x$$ is the number of days Ben works

$$10x + 20(7-x) = 100$$ -> $$10x + 140 - 20x = 100$$ -> $$10x = 40$$ -> $$x = \frac{40}{10} = 4$$

Therefore, Adam does $$40(4*10)$$ units of the total $$100$$ units, which is $$\frac{40}{100} = 40$$% (Option D) of the work.
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Re: If Adam can do a job in 10 days and Ben's speed is twice of Adam's spe  [#permalink]

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04 Jul 2018, 23:06
chetan2u wrote:
If Adam can do a job in 10 days and Ben's speed is twice of Adam's speed and both work one after another, what % of the work should be done by Adam to complete the work in exact 7 days?
(A) 67%
(B) 60%
(C) 50%
(D) 40%
(E) 33%

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Question stem, $$\frac{W_{A}}{W_{Total}}$$*100=?

Given, $$r_{A}$$=$$\frac{1}{10}$$ & $$r_{B}$$=2*$$\frac{1}{10}$$=$$\frac{1}{5}$$
$$t_A+t_B=7$$

We know, work=Rate*Time & Total work done=1

So, $$r_A*t_A+r_B*t_B=1$$
Or, $$\frac{1}{10}*t_A+\frac{1}{5}*(7-t_A)=1$$
Or, $$14-t_A=10$$
Or, $$t_A=4$$

Now, $$\frac{W_{A}}{W_{Total}}$$*100=$$\frac{r_A*t_A}{1}$$*100=$$\frac{1}{10}$$*4*100=40%

Ans. (D)
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Re: If Adam can do a job in 10 days and Ben's speed is twice of Adam's spe  [#permalink]

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05 Jul 2018, 01:58
chetan2u wrote:
If Adam can do a job in 10 days and Ben's speed is twice of Adam's speed and both work one after another, what % of the work should be done by Adam to complete the work in exact 7 days?
(A) 67%
(B) 60%
(C) 50%
(D) 40%
(E) 33%

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Both work one after another??? both work one after another for 1 day each or 2 day each?? no clarity
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Joined: 02 Aug 2009
Posts: 7106
Re: If Adam can do a job in 10 days and Ben's speed is twice of Adam's spe  [#permalink]

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05 Jul 2018, 02:05
praveenkuragodi wrote:
chetan2u wrote:
If Adam can do a job in 10 days and Ben's speed is twice of Adam's speed and both work one after another, what % of the work should be done by Adam to complete the work in exact 7 days?
(A) 67%
(B) 60%
(C) 50%
(D) 40%
(E) 33%

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Both work one after another??? both work one after another for 1 day each or 2 day each?? no clarity

That doesn't matter at all..
Since we have to ensure that the work has to be completed in 7 days and both have different speed, IT is possible only if they do certain% of work ..
The answer will not change if A works for 10 h and then B works for 1hr or there is NO pattern in their working. Finally the work jaa to be completed in exact 7 daya
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

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If Adam can do a job in 10 days and Ben's speed is twice of Adam's spe  [#permalink]

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05 Jul 2018, 02:35
Quote:
both work one after another

A reasonable test-taker might interpret this information as follows:
If Adam works one day, then Ben must work the next day.
Earlier posts suggest that a different interpretation is intended.
I believe that the following clarifies the intent of the problem:

Quote:
Adam can do a job in 10 workdays. Ben's speed is twice Adam's speed. Every workday either Adam or Ben works but not both. If the job in completed in exactly 7 workdays, what percent of the job is produced by Adam?

(A) 67%
(B) 60%
(C) 50%
(D) 40%
(E) 33%

Let the job = 10 units.
Since Adam takes 10 days to complete the 10-unit job, Adam's rate = $$\frac{w}{t} = \frac{10}{10} = 1$$ unit per day.
Since Ben is twice as fast as Adam, Ben's rate = 2 units per day.
For the 10-unit job to be completed in exactly 7 days, only one case is possible:
Adam works at his rate of 1 unit per day for 4 days, producing a total of 4 units.
Ben works at his rate of 2 units per day for 3 days, producing a total of 6 units.
Thus:
$$\frac{Work-by-Adam}{Total-work} = \frac{4}{10} = 40$$%.

An algebraic way to determine the number of days worked by Adam:
Let $$a=$$ Adam's number of days and $$b=$$ Ben's number of days.
Since Adam produces 1 unit per day, Ben produces 2 units per day, and a total of 10 units are produced, we get:
$$a+2b=10$$
Since a total of 7 days are worked, we get:
$$a+b=7$$
$$2a+2b=14$$
Subtracting the red equation from the blue equation, we get:
$$(2a+2b)-(a+2b) = 14-10$$
$$a=4$$
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If Adam can do a job in 10 days and Ben's speed is twice of Adam's spe &nbs [#permalink] 05 Jul 2018, 02:35
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