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# Two dressmakers, Sue and Anne, sewed costumes for a local theater prod

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Math Expert
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Two dressmakers, Sue and Anne, sewed costumes for a local theater prod  [#permalink]

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29 Oct 2015, 01:28
13
00:00

Difficulty:

55% (hard)

Question Stats:

61% (01:32) correct 39% (02:08) wrong based on 422 sessions

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Two dressmakers, Sue and Anne, sewed costumes for a local theater production. Sue sewing alone for 10 hours sewed some of the costumes, Anne sewing for 16 hours finished sewing the remaining costumes. How any hours would it have taken Sue alone to sew all of the costumes?

(1) Sue sewed 1 costume every 2 hours.
(2) Sue sewed twice as many costumes working in 10 hours as Anne produced in 16 hours.

Kudos for a correct solution.

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Re: Two dressmakers, Sue and Anne, sewed costumes for a local theater prod  [#permalink]

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29 Oct 2015, 03:04
Is 'C' the correct solution ?
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Re: Two dressmakers, Sue and Anne, sewed costumes for a local theater prod  [#permalink]

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29 Oct 2015, 04:24
Ans is B ,

suppose Anne finished n costumes , then sue would have finished 2n costumes , therefore total costumes become 3n.

Question has asked us that in how much time sue would have finished total costumes.

We know sue takes 10 hrs to finish 2n costumes , therefore by unitary method we can easily find time taken by sue to finish 3n costumes.
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Re: Two dressmakers, Sue and Anne, sewed costumes for a local theater prod  [#permalink]

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29 Oct 2015, 05:09
2
Amount of work done by Sue per hour = s
Amount of work done by Anne per hour = a
Total work done = w
10s + 16a = w

1. Sue sewed costume per hour s=1/2 costume per hour

Not sufficient

2. 10s = 2*(16a)
=> 5s = 16a
Sue will need 5 hours to complete Anne's part of the job .
Sue will complete the entire work in 10+5=15 hours working alone .

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Re: Two dressmakers, Sue and Anne, sewed costumes for a local theater prod  [#permalink]

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30 Oct 2015, 14:09
Hi sachinsnambiar,

In DS questions, you should be able to PROVE whatever solution you think is correct. One of the easiest ways to LOSE points on the GMAT is to not do work (or do very little work 'in your head"). You should elaborate on why you think the answer is C; if you made a logical mistake, then we should be able to fix it.

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Re: Two dressmakers, Sue and Anne, sewed costumes for a local theater prod  [#permalink]

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01 Nov 2015, 05:32
1
Bunuel wrote:
Two dressmakers, Sue and Anne, sewed costumes for a local theater production. Sue sewing alone for 10 hours sewed some of the costumes, Anne sewing for 16 hours finished sewing the remaining costumes. How any hours would it have taken Sue alone to sew all of the costumes?

(1) Sue sewed 1 costume every 2 hours.
(2) Sue sewed twice as many costumes working in 10 hours as Anne produced in 16 hours.

Kudos for a correct solution.

Solution:-

Lets look at the information given in the question:-

Sue sewed some costumes for 10 hrs, lets assume Sew sews x costumes per hour, therefore she sew 10x costumes
Anne sewed teh rest of the costumes in 16 hrs, lets assume Anne sews y costumes per hour, therefore she sews 16y costumes .
Total costumes sown by Sue and Anne = 10x + 16y

The question is asking how much time is would take Sue to sew 10x+16y costumes.

Statement 1 - Sue sews 1 costume for every 2 hr. From here we know that Sue must have sown 5 costumes in 10 hrs. However since we dont know the total number of costumes sown by Anne, and neither do we know the rate at which Anne works we cannot find the total amount of costumes sown by Sue and Anne. Hence we can conclude that we cannot find out how much time Sue will take to complete the entire work, therefore statement 1 is insufficient.

Statement 2 - Sue sewed twice as many costumes in 10 hrs as Anne in 16 hrs. We have assumed earlier Sue Sews 10x costumes in 10 hrs, From Statement 2 we know that Sue sewed twice as many costumes than Anne, therefore Anne must have sewed 5x costumes in 16 hrs, and 16y= 5x.

Replacing 5x into 16y we get that the total work done is equal to 10x + 5x = 15x

Now using unitary method we can calculate that if Sue sewed 10x costumes in 10 hrs she would take 15 hrs to sew 15x costumes. Therefore statement 2 is sufficient to answer the question .

Tip - Understand the question and write down the information given and then devise an approach to answer the question. This allows a systematic way to approach the solution.
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Two dressmakers, Sue and Anne, sewed costumes for a local theater prod  [#permalink]

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01 Aug 2016, 06:42
Bunuel wrote:
Two dressmakers, Sue and Anne, sewed costumes for a local theater production. Sue sewing alone for 10 hours sewed some of the costumes, Anne sewing for 16 hours finished sewing the remaining costumes. How any hours would it have taken Sue alone to sew all of the costumes?

(1) Sue sewed 1 costume every 2 hours.
(2) Sue sewed twice as many costumes working in 10 hours as Anne produced in 16 hours.

Kudos for a correct solution.

(1) Sue sewed 1 costume every 2 hours.
Sue worked ten hours therefore she her rate 10/2 = 5 costumes per hour
Number of costumes produced by Anne is not known, therefore total number of costumes in not known ; therefore time taken by sue to complete the rest of the work cannot be determined
INSIFFICIENT

(2) Sue sewed twice as many costumes working in 10 hours as Anne produced in 16 hours.
From stimulus we know Sue made "SOME" dresses and Anne made the "REAMAINING" dress
This option tell us Sue made 2x dresses and the remaining x dress were made by anne
Total dress made were =("Some + remaining"= all) = 2x+x= 3x
we know sue made 2x dress in 10 hour
therefore sue would made 3x dresses in 10 + 5 = 15 hour
SUFFICIENT

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Re: Two dressmakers, Sue and Anne, sewed costumes for a local theater prod  [#permalink]

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07 Jan 2019, 13:53
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Re: Two dressmakers, Sue and Anne, sewed costumes for a local theater prod   [#permalink] 07 Jan 2019, 13:53
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