Data Sufficiency

BTGmoderatorDC 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.

TIME and RATE have a reciprocal relationship.

Statement 1:
The time ratio for Barbara and Alice = (14 hours)/(21 hours).
The rate ratio for Barbara and Alice is equal to the reciprocal of the time ratio:
(B's rate)/(A's rate) = 21/14 = 3/2.
Let B = 3 units per hour and A = 2 units per hour.

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.
The time ratio for B+C and A+C = 86/100 = 43/50.
The rate ratio for B+C and A+C is equal to the reciprocal of the time ratio:
(B+C)/(A+C) = 50/43.
Plugging B=3 and A=2 into the equation above, we get:
(3+C)/(2+C) = 50/43.
Since we can solve for C, we can determine whether Alice has the lowest rate.
SUFFICIENT.

Statement 2:
The time ratio for B+C and B+A= 71/100.
The rate ratio for B+C and B+A is equal to the reciprocal of the time ratio:
(B+C)/(B+A) = 100/71.
Implication:
B+C > B+A
C>A.
No way to determine whether Alice is slower than Barbara.
INSUFFICIENT.

The correct answer is A.

Complete solution for Statement 1:
(3+C)/(2+C) = 50/43
129 + 43C = 100 + 50C
29 = 7C
C = 29/7 = 4 1/7.
Since A=2, B=3 and C = 4 1/7, A's rate is the lowest.

BTGmoderatorDC 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.
Source: Princeton Review

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.