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

You may find this article on work rate helpful:
https://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.

Answer = (A)

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]
Expert Reply
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 [#permalink]
as per given info-

info-1 : it's clearly says about age of another person. and by give other information it is sufficient.
info-2: Here lack of information to conclude. only total work done percentage is given. not sufficient to answer.

Option A is correct.
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Re: Alice, Barbara, and Cynia work on identical tasks at differe [#permalink]
A takes 21 hrs
B takes x hrs
C takes y hrs
Now,
As per Statement 2:
Time taken by C and B:
(xy)/(x+y)
Time taken by A and B:
21x/(x+y)

Now :
xy/(x+y) = 0.71(21x/(21+x))
x got cancelled
100y(21+x) = 71(21x+21y)
2100y+100xy= 1491x + 1491y
609y+100xy=1491x

As question asks if Alice rate at 21 is the slowest I put y=20
12180 +2000x = 1491 x this makes X negative so X should be less than 20 which means A is not the slowest.

Please suggest where I am wrong.
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Re: Alice, Barbara, and Cynia work on identical tasks at differe [#permalink]
Official explanation:

This is a Yes/No Data Sufficiency question. The question stem provides 21 hours as the time for Alice to complete a task, and asks about the rates of Alice, Barbara, and Cynia. Since the amount of work that comprises the task is not specified, this is a Hidden Plug In. Plug In for the amount of work that comprises the task. Also Plug In for the rates if there is not enough information to solve. The task of a Yes/No Data Sufficiency is to determine whether the information in the statements produces a consistent Yes or No response for any number that satisfies the statement(s). Evaluate the statements one at a time.

Evaluate Statement (1). Statement (1) provides 14 hours as the time for Barbara to complete the task, and says that Barbara and Cynia working together can complete the task in approximately 86% of the time taken by Alice and Cynia. Plug In for the amount of work that comprises the task. Alice completes the task in 21 hours, and Barbara completes the task in 14 hours, so Plug In a number divisible by both 21 and 14. Plug In 42 for the task. Then Alice's rate is 42/21 = 2per hour, and Barbara’s rate is 42/14 = 3 per hour. Statement (1) compares the time that it takes Barbara and Cynia to complete the task to the time that it takes Alice and Cynia to complete the task. Use two Rate Pies to set up an equation. Call Cynia’s rate C and the time for Alice and Cynia to complete the task T. For Barbara and Cynia, the top of the Rate Pie is the total for the task, the time is 0.86T, and the rate is 3 + C, so the equation is 0.86T(3 + C) = 42. For Alice and Cynia, the top of the Rate Pie is the total for the task, the time is T, and the rate is 2 + C, so the equation is T(2 + C) = 42. Because both equations are equal to 42 on the right side, set the two equations equal to each other, yielding 0.86T(3 + C) = T(2 + C). Divide both sides by T to get 0.86(3 + C) = (2 + C). This is one equation with one unknown so it is possible to solve for Cynia’s rate and answer the question of whether Alice’s rate is the slowest. Statement (1) is sufficient. So, write down AD.

Now, evaluate Statement (2). Plug In 42 again for the task. Only Alice’s rate of 2 per hour is known. Statement (2) says that Barbara and Cynia can complete the task in approximately 71% of the time taken by Alice and Barbara. Use two Rate Pies to set up an equation. Call Barbara’s rate B, Cynia’s rate C, and the time for Alice and Barbara to complete the task T. For Barbara and Cynia, the top of the Rate Pie is the total for the task, the time is 0.71T, and the rate is B + C, so the equation is 0.71T(B + C) = 42. For Alice and Barbara, the top of the Rate Pie is the total for the task, the time is T, and the rate is 2 + B, so the equation is T(2 + B) = 42. Because both equations are equal to 42, set the two equations equal to each other, yielding 0.71T(B + C) = T(2 + B). Divide both sides by T to get 0.71(B + C) = (2 + B). This is one equation with two unknowns, so Plug In for the rates to see whether Alice’s rate of 2 has to be the slowest. If B = 1, then 0.71(1 + C) = (2 + 1). Distribute the 0.71 to get 0.71 + 0.71C = 3. Subtract 0.71 from both sides to get 0.71C = 2.29. Finally, divide both sides by 0.71, and C = 2.29/0.71 (approx 3) . In this case, Barbara has the slowest rate, so the answer to the question is “No.” Now, Plug In again, to see whether there is a way to produce a "Yes" answer. If B = 3, then 0.71(3 + C) = (2 + 3). Distribute the 0.71 to get 2.13 + 0.71C = 5. Subtract 2.13 from both sides to get 0.71C = 2.87. Finally, divide both sides by 0.71, and C = 2.87/0.71 (approx 4) . In this case, Alice has the slowest rate, so the answer to the question is “Yes.” When different numbers that satisfy a statement yield different answers to the question, the statement is insufficient. Eliminate choice D.

The correct answer is choice A.
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Re: Alice, Barbara, and Cynia work on identical tasks at differe [#permalink]
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