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# Alice, Barbara, and Cynia work on identical tasks at differe

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Joined: 21 Mar 2013
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GMAT Date: 03-20-2014

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23 Dec 2013, 08:01
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60% (01:40) correct 40% (01:51) wrong based on 333 sessions

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Alice, Barbara, and Cynia work on identical tasks at different constant rates. Alice, working alone, can complete the task in 21 hours. Is Alice's rate the slowest rate?

(1) Barbara, working alone, can complete the task in 14 hours, and Barbara and Cynia working together can complete the task in approximately 86% of the time taken by Alice and Cynia working together to complete the task.

(2) Barbara and Cynia can complete the task in approximately 71% of the time taken for Alice and Barbara to complete the task.
Magoosh GMAT Instructor
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Re: Alice, Barbara, and Cynia work on identical tasks at differe  [#permalink]

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23 Dec 2013, 09:59
5
idinuv wrote:
Alice, Barbara, and Cynia work on identical tasks at different constant rates. Alice, working alone, can complete the task in 21 hours. Is Alice's rate the slowest rate?

(1) Barbara, working alone, can complete the task in 14 hours, and Barbara and Cynia working together can complete the task in approximately 86% of the time taken by Alice and Cynia working together to complete the task.

(2) Barbara and Cynia can complete the task in approximately 71% of the time taken for Alice and Barbara to complete the task.

Dear idinuv,
I'm happy to help.

http://magoosh.com/gmat/2012/gmat-work-rate-problems/

We know that A takes 21 hours to complete the job alone. From the prompt we know zilch about how she compares to B or C.

Statement #1: Barbara, working alone, can complete the task in 14 hours, and Barbara and Cynia working together can complete the task in approximately 86% of the time taken by Alice and Cynia working together to complete the task.
OK, so first of all, they tell us that B takes only 14 hours, so B definitely works faster than A.

Then we find out that B & C take less time than A & C, which means B & C works faster than A & C. Well, hmm. We already know, from the first part of this statement, that B works faster than A. From casual inspection, it's not clear that this second statement adds more information. We have to investigate it numerically, so we can involve that 86% figure, which is approximately 5/6.

A finishes in 21 hours, so her rate (job/hours) is 1/21. B's rate is 1/14. Let's say C takes c hours, so C's rate is 1/c. For combinations of two people working together, we add rates.
Rate of B & C = 1/14 + 1/c
Rate of A & C = 1/21 + 1/c
B & C take only 5/6 the time of A & C, which means B & C have 6/5 the rate of A & C.

(1/14 + 1/c) = 6/5(1/21 + 1/c)

At this point, we have a single equation we could solve for c. This is GMAT DS. The trap of DS would be to go through all the steps to find the numerical value of c. At this point, we see that we could solve for c. That's enough. If we knew c, we would know who was the fastest and slowest, and we would be able to answer the prompt question. This statement, alone and by itself, is sufficient.

Statement #2: Barbara and Cynia can complete the task in approximately 71% of the time taken for Alice and Barbara to complete the task
Here, we have two unknowns, because we don't know C's rate or time, and we also don't know B's rate or time. Remember, B's rate is something given in statement #1, so we can't use it here. We would be able to set up a single equation, but one equation with two unknowns is not enough to solve, and we would still be left with questions. This statement, alone and by itself, is insufficient.

Does all this make sense?
Mike
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Re: Alice, Barbara, and Cynia work on identical tasks at differe  [#permalink]

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26 Dec 2013, 07:13
mikemcgarry wrote:
idinuv wrote:
Alice, Barbara, and Cynia work on identical tasks at different constant rates. Alice, working alone, can complete the task in 21 hours. Is Alice's rate the slowest rate?

(1) Barbara, working alone, can complete the task in 14 hours, and Barbara and Cynia working together can complete the task in approximately 86% of the time taken by Alice and Cynia working together to complete the task.

(2) Barbara and Cynia can complete the task in approximately 71% of the time taken for Alice and Barbara to complete the task.

Dear idinuv,
I'm happy to help.

http://magoosh.com/gmat/2012/gmat-work-rate-problems/

We know that A takes 21 hours to complete the job alone. From the prompt we know zilch about how she compares to B or C.

Statement #1: Barbara, working alone, can complete the task in 14 hours, and Barbara and Cynia working together can complete the task in approximately 86% of the time taken by Alice and Cynia working together to complete the task.
OK, so first of all, they tell us that B takes only 14 hours, so B definitely works faster than A.

Then we find out that B & C take less time than A & C, which means B & C works faster than A & C. Well, hmm. We already know, from the first part of this statement, that B works faster than A. From casual inspection, it's not clear that this second statement adds more information. We have to investigate it numerically, so we can involve that 86% figure, which is approximately 5/6.

A finishes in 21 hours, so her rate (job/hours) is 1/21. B's rate is 1/14. Let's say C takes c hours, so C's rate is 1/c. For combinations of two people working together, we add rates.
Rate of B & C = 1/14 + 1/c
Rate of A & C = 1/21 + 1/c
B & C take only 5/6 the time of A & C, which means B & C have 6/5 the rate of A & C.

(1/14 + 1/c) = 6/5(1/21 + 1/c)

At this point, we have a single equation we could solve for c. This is GMAT DS. The trap of DS would be to go through all the steps to find the numerical value of c. At this point, we see that we could solve for c. That's enough. If we knew c, we would know who was the fastest and slowest, and we would be able to answer the prompt question. This statement, alone and by itself, is sufficient.

Statement #2: Barbara and Cynia can complete the task in approximately 71% of the time taken for Alice and Barbara to complete the task
Here, we have two unknowns, because we don't know C's rate or time, and we also don't know B's rate or time. Remember, B's rate is something given in statement #1, so we can't use it here. We would be able to set up a single equation, but one equation with two unknowns is not enough to solve, and we would still be left with questions. This statement, alone and by itself, is insufficient.

Does all this make sense?
Mike

Well done Mike
Cheers!
J
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Re: Alice, Barbara, and Cynia work on identical tasks at differe  [#permalink]

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19 Mar 2018, 13:31
1
I may have a quicker way to solve (experts please chime in):

We know A's rate but don't know B's rate and C's rate. Further, we're asked if A is the slowest. To make this determination, we need to find B's rate and C's rate.

(1) B's rate is defined. Then, we're given: (B+C) = (.86)(A+C). We have 3 variables here. However, we know what "B" is, and "C" should be the same value that we are adding to either side. Due to this, we can theoretically get "A".

(2) (B+C) = (.71)(A+B). We don't know what any of the variables are. I suppose we can rationalize that C = (.71)(A). Even so, depending on what "plug" you use for A or C, you can get different answers. D out.

A remains.

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Re: Alice, Barbara, and Cynia work on identical tasks at differe  [#permalink]

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13 Oct 2018, 16:03
idinuv wrote:
Alice, Barbara, and Cynia work on identical tasks at different constant rates. Alice, working alone, can complete the task in 21 hours. Is Alice's rate the slowest rate?

(1) Barbara, working alone, can complete the task in 14 hours, and Barbara and Cynia working together can complete the task in approximately 86% of the time taken by Alice and Cynia working together to complete the task.

(2) Barbara and Cynia can complete the task in approximately 71% of the time taken for Alice and Barbara to complete the task.

Obs.: the term approximately invalidates the question because we know NOTHING about the approximation precision.
(What should be considered "near" 86%, for instance?)
We will consider equality in both cases, with the following (not problematic) aside: the numbers b and c (below) may be non-integers.

Let´s imagine the task is defined by 42 identical units of job (from now on simply "units").

Alice (A) can do 2 units/h (therefore in 21h she will do 2*21 = 42 units, i.e., the task).
Barbara (B) can do (say) b units/h
Cynia (C) can do (say) c units/h

$$2\,\,\mathop < \limits^? \,\,\,\min \left( {b,c} \right)$$

$$\left( 1 \right)\,\,\left\{ \matrix{ b = 3 \hfill \cr \,{{{T_{B \cup C}}} \over {{T_{A \cup C}}}} = {{43} \over {50}}\,\,\,\,\,\mathop \Rightarrow \limits^{W = \,{\rm{work}}\,{\rm{rate}}} \,\,\,\,\,{{3 + c} \over {2 + c}} = {{{W_{B \cup C}}} \over {{W_{A \cup C}}}} = {{50} \over {43}} \hfill \cr} \right.$$

$${{3 + c} \over {2 + c}} = {{50} \over {43}}\,\,\,\, \Rightarrow \,\,\,\,c\,\,{\rm{unique}}\,\,\,\, \Rightarrow \,\,\,\,{\rm{SUFF}}.$$

$$\left( 2 \right)\,\,\,{{{T_{B \cup C}}} \over {{T_{A \cup B}}}} = {{71} \over {100}}\,\,\,\,\,\mathop \Rightarrow \limits^{W = \,{\rm{work}}\,{\rm{rate}}} \,\,\,\,\,{{b + c} \over {2 + b}} = {{100} \over {71}}$$

$$\left\{ \matrix{ \,{\rm{Take}}\,\,\left( {b;c} \right) = \left( {1;{{300} \over {71}} - 1} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr \,{\rm{Take}}\,\,\left( {b;c} \right) = \left( {3;{{500} \over {71}} - 3} \right)\,\,\,\,\, \Rightarrow \,\,\,\,\left\langle {{\rm{YES}}} \right\rangle \hfill \cr} \right.$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
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Re: Alice, Barbara, and Cynia work on identical tasks at differe &nbs [#permalink] 13 Oct 2018, 16:03
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