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

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".
- AD in, BCE out.

(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.

Kudos please if helpful and corrections please if I am incorrect in my thinking.

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.

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.

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.

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.