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Bunuel
Working together but independently, Scott and Eric can address X envelopes in 18 hours. How long would it take Scott working alone to address X envelopes?

(1) In M minutes, Scott address three times as many envelopes as Eric address in M minutes.

(2) Eric can address X envelopes in 72 hours.


­

Assuming Scott and Eric work at constant rate, during the whole work.

Given \( \frac{1}{s} + \frac{1}{e} = \frac{1}{18}\) ...(i)

(1) In M minutes, Scott address three times as many envelopes as Eric address in M minutes.

\(\frac{3}{e} + \frac{1}{e} = \frac{1}{m} \)

This means Scott's rate is \(3\) times Eric's rate

Thus \( \frac{3}{e} + \frac{1}{e} = \frac{1}{18}\) ... from (i)

SUFF.

(2) Eric can address X envelopes in 72 hours.

\( \frac{1}{s} + \frac{1}{72} = \frac{1}{18}\) ... from (i)

SUFF.

Ans D

7seven7

Hope it helped.
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stne

Assuming Scott and Eric work at constant rate, during the whole work.

Given \( \frac{1}{s} + \frac{1}{e} = \frac{1}{18}\) ...(i)

(1) In M minutes, Scott address three times as many envelopes as Eric address in M minutes.

\(\frac{3}{e} + \frac{1}{e} = \frac{1}{m} \)

This means Scott's rate is \(3\) times Eric's rate

Thus \( \frac{3}{e} + \frac{1}{e} = \frac{1}{18}\) ... from (i)

SUFF.

(2) Eric can address X envelopes in 72 hours.

\( \frac{1}{s} + \frac{1}{72} = \frac{1}{18}\) ... from (i)

SUFF.

Ans D

7seven7

Hope it helped.
Thanks stne, I too would have solved like this but as "constant rate" was not mentioned anywhere, I did not take s=3e due to this sole reason. There is a similar question in DS where answer was Option E as "Constant rate" was not given. And there are many more in time, distance, speed DS questions too where the answer is different due to either constant speed not mentioned or waiting time not known (only speed while driving given).

So, why to consider the rates as constant here (if not mentioned, provided it being a DS question).
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stne

Assuming Scott and Eric work at constant rate, during the whole work.

Given \( \frac{1}{s} + \frac{1}{e} = \frac{1}{18}\) ...(i)

(1) In M minutes, Scott address three times as many envelopes as Eric address in M minutes.

\(\frac{3}{e} + \frac{1}{e} = \frac{1}{m} \)

This means Scott's rate is \(3\) times Eric's rate

Thus \( \frac{3}{e} + \frac{1}{e} = \frac{1}{18}\) ... from (i)

SUFF.

(2) Eric can address X envelopes in 72 hours.

\( \frac{1}{s} + \frac{1}{72} = \frac{1}{18}\) ... from (i)

SUFF.

Ans D

7seven7

Hope it helped.
Thanks stne, I too would have solved like this but as "constant rate" was not mentioned anywhere, I did not take s=3e due to this sole reason. There is a similar question in DS where answer was Option E as "Constant rate" was not given. And there are many more in time, distance, speed DS questions too where the answer is different due to either constant speed not mentioned or waiting time not known (only speed while driving given).

So, why to consider the rates as constant here (if not mentioned, provided it being a DS question).

Can you post an example of a D.S. question, which is of the kind you are talking about. Since this question has not been archived or locked, I am guessing that this is a valid question. Please do share those questions that seem to confuse so that we can discuss them. Thanks.
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