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Mr. Verma can do a job in 10 days. A helper joins him after 3 days, an

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Mr. Verma can do a job in 10 days. A helper joins him after 3 days, and together they work for 4 days to complete the task. How many days would it take the helper to do the job alone?

a) 3
b) \(5 \frac{5}{7}\)
c) 6
d) 7
e) \(13\frac{1}{3}\)
[Reveal] Spoiler: OA

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Re: How many days would it take the helper to do the job alone? [#permalink]

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Mr Verma can do a job in 10 days. His per day work = 1/10
So in 3 days, he can do alone = 3/10 of the work. Remaining work = 1 - 3/10 = 7/10

Now Mr Verma and helper can do 7/10 of the work in 4 days.
That means together, they can do complete work in = 4*10/7 = 40/7 days
So together, per day work of Mr Verma and helper combined = 7/40

This means work of helper per day = 7/40 - 1/10 (1/10 is per day work of Mr Verma)
= 3/40

This means helper can do complete work in 40/3 days.. Or 13 1/3 days.

Hence E answer

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Re: Mr. Verma can do a job in 10 days. A helper joins him after 3 days, an [#permalink]

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New post 13 Jun 2017, 11:26
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Assume total work to be 40 units, this means work done by Mr Verma per day = 40/10 = 4 units
In 3 days, he will do = 4*3 = 12 units

Remaining work = 40-12 = 28 units
This is done by Mr Verma and helper in 4 days, so their combined per day work = 28/4 = 7 units

But out of these 7 units, Mr Verma does 4 units, so daily work done by helper = 7-4 = 3 units

Since helper does 3 units per day, if he has to alone do this 40 unit work, he will take= 40/3 days or 13 1/3 days

Hence E answer

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Re: Mr. Verma can do a job in 10 days. A helper joins him after 3 days, an [#permalink]

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rosmann wrote:
Mr. Verma can do a job in 10 days. A helper joins him after 3 days, and together they work for 4 days to complete the task. How many days would it take the helper to do the job alone?

a) 3
b) \(5 \frac{5}{7}\)
c) 6
d) 7
e) \(13\frac{1}{3}\)


This can be done with almost no algebra.

So, when a helper joins there are 7 days of work left to be done. If the helper's rate was equal to Mr. Verma's rate, then the job will be done in 3.5 days. Since the actual time (4 days) is longer than that, then the helper's rate must be lower than that of Mr. Verma's. Mr. Verma can do the job in 10 days, so the helper will take longer. Only E fits.

Answer: E.
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Re: Mr. Verma can do a job in 10 days. A helper joins him after 3 days, an [#permalink]

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For me, easiest way to do work/ rate problems is by taking the LCM or multiple of all the given numbers in a question and assume it to be total number of units of work to be done. This eliminates the chances of making error while doing calculation with fractions.

In this question, we are given 3 numbers, 10, 3 and 4. Let's say total units of work to be done is 60 (60 is a multiple of 10, 3 and 4).

Mr. Verma will do 6 units per day (Mr Verma's rate of work) to do the given task in 10 days.
He works for 3 days alone so he'll do 18 units of work. Then helper joins. Work to be done is 60-18 = 42 units.
Both complete the remaining job in 4 days so combined rate is 10.5 units per day.
Work rate of helper will be, 10-5-6 = 4.5 units per day.

At this rate if worker has to do 60 units of work alone, he'll take 60/4.5 = 40/3 days or 13 1/3 days.

Option E is correct.

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Re: Mr. Verma can do a job in 10 days. A helper joins him after 3 days, an [#permalink]

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New post 06 Jul 2017, 17:18
rosmann wrote:
Mr. Verma can do a job in 10 days. A helper joins him after 3 days, and together they work for 4 days to complete the task. How many days would it take the helper to do the job alone?

a) 3
b) \(5 \frac{5}{7}\)
c) 6
d) 7
e) \(13\frac{1}{3}\)


We are given that Mr. Verma can complete a job in 10 days; thus, the rate of Mr. Verma is 1/10.

So, after 3 days, he has completed 3/10 of the job.

When he works with the helper, they complete 7/10 of the job in 4 days, or work at a rate of (7/10)/4 = 7/40.

If we let the rate of the helper = 1/x, then:

1/10 + 1/x = 7/40

Multiplying the entire equation by 40x, we have:

4x + 40 = 7x

40 = 3x

x = 40/3 = 13 ⅓

Answer: E
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New post 06 Jul 2017, 19:51
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Efficiency of Verma = 10%
Work done by Verma in 3 days = 30%

Work done by Verma and helper in 4 days = 70% (remaining work), work done by helper in 4 days = 30% (70%-40%).

Efficiency of worker = 30/4 = 7.5%. So, worker would take 100/7.5 = 13 1/3 days. Ans - E.
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Re: Mr. Verma can do a job in 10 days. A helper joins him after 3 days, an [#permalink]

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New post 12 Aug 2017, 22:12
rosmann wrote:
Mr. Verma can do a job in 10 days. A helper joins him after 3 days, and together they work for 4 days to complete the task. How many days would it take the helper to do the job alone?

a) 3
b) \(5 \frac{5}{7}\)
c) 6
d) 7
e) \(13\frac{1}{3}\)

Mr. Verma can do a job in \(10\) days.

\(1\) day work of Mr. Verma \(= \frac{1}{10}\)

\(3\) days work of Mr. Verma \(= 3(\frac{1}{10}) = \frac{3}{10}\)

Amount of work left \(= 1- \frac{3}{10} = \frac{10-3}{10} = \frac{7}{10}\)

\(\frac{7}{10}\) Work is completed by Mr. Verma and helper together in \(4\) days.

\(1\) day work of Mr. Verma and helper together \(= (\frac{7}{10})(\frac{1}{4}) = \frac{7}{40}\)

Let the \(1\) day work of helper be \(= \frac{1}{x}\)

\(\frac{1}{x} + \frac{1}{10} = \frac{7}{40}\)

\(\frac{1}{x} = \frac{7}{40} - \frac{1}{10} = \frac{7-4}{40} = \frac{3}{40}\)

\(x = \frac{40}{3} = 13\frac{1}{3}\)

Answer (E)...

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Mr. Verma can do a job in 10 days. A helper joins him after 3 days, an [#permalink]

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New post 13 Aug 2017, 00:58
If Mr. Verma can do the work in 10 days,
after three days of working alone, Mr. Verma is joined by the helper and the helper complete the work in 4 days.

The helper does \(\frac{3}{10}\) of the total work in 4 days,
which Mr. Verma would have done in the remaining 3 days had he worked alone on the work.

So, the helper does 3/10 of the work in 4 days and would do \(\frac{\frac{3}{10}}{4}\) or \(\frac{3}{40}\)th of the work in a day.
Hence, it would take the helper \(\frac{1}{\frac{3}{40}}\) or \(13\frac{1}{3}\) to complete the work on his own(Option E)
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Mr. Verma can do a job in 10 days. A helper joins him after 3 days, an   [#permalink] 13 Aug 2017, 00:58
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