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Working together but independently, Scott and Eric can addre

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Working together but independently, Scott and Eric can addre  [#permalink]

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New post Updated on: 28 Mar 2014, 17:00
7
00:00
A
B
C
D
E

Difficulty:

  45% (medium)

Question Stats:

63% (01:30) correct 37% (01:09) wrong based on 323 sessions

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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.

Originally posted by gkncwb on 28 Mar 2014, 16:52.
Last edited by Bunuel on 28 Mar 2014, 17:00, edited 1 time in total.
RENAMED THE TOPIC.
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Re: Working together but independently, Scott and Eric can addre  [#permalink]

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New post 28 Mar 2014, 17:08
I have solved this way:

E
S
E+S 1.000 18

(1)

E 1.000 M
S 3.000 M
E+S 1.000 18
3.000 / M + 1.000 / M = 1.000 / 18 >>> M = 72 SOLVE

(2)

1.000 / 72 + 1.000 / x = 1.000 / 18 >>>>> x = 24 SOLVE

Answer: D
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Re: Working together but independently, Scott and Eric can addre  [#permalink]

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New post 20 Jan 2016, 10:32
It is clear for me statement 2, but I didn't understand how can I solve the question using statement 1.

Can someone can help me?

Thaaaanks
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Working together but independently, Scott and Eric can addre  [#permalink]

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New post Updated on: 20 Jan 2016, 13:29
pepo wrote:
It is clear for me statement 2, but I didn't understand how can I solve the question using statement 1.

Can someone can help me?

Thaaaanks


Let E and S be the number of envelopes/hour produced by Eric and Scott resp.

Per statement 1, You are given that 3E=S ---> in M minutes, Eric can produce 3 times as many envelopes as produced by Scott. IN 1 hour, they will produce each

--->Eric = EM/60 Scott = 3EM/60 . In 18 hours, they produce,

3EM*18/60+EM*18/60 = X ---> X= 72EM/60

Thus, from above, number of envelopes produced by Eric in 1 hour = EM/60 ---> time for him to produce X (=72EM/60) envelopes =X/(EM/60) = (72EM/60)/(EM/60) = 72 hours.

Hence statement 1 is sufficient.

Hope this helps.

Originally posted by ENGRTOMBA2018 on 20 Jan 2016, 10:56.
Last edited by ENGRTOMBA2018 on 20 Jan 2016, 13:29, edited 1 time in total.
Edited the typos
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Re: Working together but independently, Scott and Eric can addre  [#permalink]

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New post 20 Jan 2016, 13:13
1
Engr2012 wrote:
pepo wrote:
It is clear for me statement 2, but I didn't understand how can I solve the question using statement 1.

Can someone can help me?

Thaaaanks


Let E and S be the number of envelopes/hour produced by Eric and Scott resp.

Per statement 1, You are given that E=3S ---> in M minutes, Eric can produce 3 times as many envelopes as produced by Scott. IN 1 hour, they will produce each

---> Scott = SM/60 and E = 3SM/60. In 18 hours, they produce,

3SM*18/60+SM/60 = X ---> X= 4SM/60

Thus, from above, number of envelopes produced by Eric in 1 hour = 3SM/60 ---> time for him to produce X (=4SM/60) envelopes =X/(3SM/60) = (4SM/60)/(3SM/60) = 4/3 hours.

Hence statement 1 is sufficient.

Hope this helps.


Hi Engr2012,

thanks for your reply!!!

I have just 2 issues:
1) statement says: "In M minutes, Scott address three times as many envelopes as Eric address in M minutes". It should be S=3E, right?
2) I don't understand how you can get X= 4SM/60 from 3SM*18/60+SM/60 = X.
I performed the calculations and I got X=55sm/60.

Best,
Pepo
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Re: Working together but independently, Scott and Eric can addre  [#permalink]

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New post 20 Jan 2016, 13:30
pepo wrote:
Engr2012 wrote:
pepo wrote:
It is clear for me statement 2, but I didn't understand how can I solve the question using statement 1.

Can someone can help me?

Thaaaanks


Let E and S be the number of envelopes/hour produced by Eric and Scott resp.

Per statement 1, You are given that E=3S ---> in M minutes, Eric can produce 3 times as many envelopes as produced by Scott. IN 1 hour, they will produce each

---> Scott = SM/60 and E = 3SM/60. In 18 hours, they produce,

3SM*18/60+SM/60 = X ---> X= 4SM/60

Thus, from above, number of envelopes produced by Eric in 1 hour = 3SM/60 ---> time for him to produce X (=4SM/60) envelopes =X/(3SM/60) = (4SM/60)/(3SM/60) = 4/3 hours.

Hence statement 1 is sufficient.

Hope this helps.


Hi Engr2012,

thanks for your reply!!!

I have just 2 issues:
1) statement says: "In M minutes, Scott address three times as many envelopes as Eric address in M minutes". It should be S=3E, right?
2) I don't understand how you can get X= 4SM/60 from 3SM*18/60+SM/60 = X.
I performed the calculations and I got X=55sm/60.

Best,
Pepo


My bad. Thanks for pointing it out. I have corrected the typos in my solution above although nothing would change as this is a DS question and any unique value will make statement 1 sufficient.
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Re: Working together but independently, Scott and Eric can addre  [#permalink]

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New post 21 Jan 2016, 02:58
2
pepo wrote:
It is clear for me statement 2, but I didn't understand how can I solve the question using statement 1.

Can someone can help me?

Thaaaanks


Hello Pepo, :)
Statement 1 simply presents the rate of doing work for the two people in the question, we have the total time required and using 1 we can calculate ratio of doing work for 2 which is sufficient for making the calculation. Also, M/60 is done to convert the minutes into hour :)

Please consider a kudos if it was helpful
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Working together but independently, Scott and Eric can addre  [#permalink]

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New post 04 Feb 2016, 10:51
Could someone please explain how Statement 1 is sufficient? I went through the responses already here but could not understand as they are rather unclear. Thank you.

Edit 1: I managed to solve it this way, could anyone confirm that this is in fact correct?

S = Rate of Scott
E = Rate of Eric

\(S + E =\frac{x}{18}\) X being the amount of work done, and 18 being the time taken, thus X/18 given the rate for the two together

1. In M minutes, Scott addresses three times as many envelopes as Eric addresses in M minutes.

\(Eric = EM\)
\(Scott = 3EM\)

\(3EM + EM = \frac{x}{18}\)
\(4EM = \frac{x}{18}\)
\(x = 72EM\)

x = 72EM which means it would take Eric 72 hours to complete X amount of work, and since Scott can address three times as many envelopes as Eric, he can complete X amount of work in 24 hours.
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Re: Working together but independently, Scott and Eric can addre  [#permalink]

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New post 08 Feb 2017, 04:21
The formula for the statement in question is (S+E)*18=X. The questions is S*?=X
1.The first statement gives us ratio of Scott's rate of work to Eric's. The ratios is S/E=3r/1r
(3r+1r)*18=X; 72r=X; r=X/72; S=3X/72=X/24. It takes Scott 24 hours to address X envelopes. This statement is sufficient.
2. The second statement gives us Eric's rate: E=X/72; (S+E)*18=72E; 18S=54E; 1S=3E; It takes Eric three times as much time to address X enveloped as it takes Scott. Since Eric can do the work in 72 hours, Scott can do it in 72/3=24 hours. This statement is also sufficient.

Both statements alone are sufficient to answer this question.
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Re: Working together but independently, Scott and Eric can addre  [#permalink]

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New post 12 May 2018, 02:38
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Re: Working together but independently, Scott and Eric can addre &nbs [#permalink] 12 May 2018, 02:38
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