Machines A and B, working together, take t minutes to

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Source: Veritas Prep

Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?

A. 5:8
B. 8:5
C. 25:64
D. 25:39
E. 64:25

The OA is B

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by GMATGuruNY » Sun Jan 27, 2019 4:05 am
BTGmoderatorLU wrote:Source: Veritas Prep

Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?

A. 5:8
B. 8:5
C. 25:64
D. 25:39
E. 64:25
Always keep your eye on the answer choices.

Since A's time alone (t+64) is greater than B's time alone (t+25), the ratio of A's time alone to B's time alone must be GREATER THAN 1.
Eliminate A, C and D.

Since the ratio of A's time to B's time = (t+64)/(t+25), option E implies the following:
(t+64)/(t+25) = 64/25.
The resulting equation is valid only if t=0.
Not possible.
Eliminate E.

The correct answer is B.
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by fskilnik@GMATH » Sun Jan 27, 2019 4:59 am
BTGmoderatorLU wrote:Source: Veritas Prep

Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?

A. 5:8
B. 8:5
C. 25:64
D. 25:39
E. 64:25
$${\rm{times}}\,\,{\rm{A,}}\,B\,\,\,\,\,\, \to \,\,\,\,\,\min $$
$$? = {A \over B}$$

$$\left( * \right)\,\,\,\left\{ \matrix{
A = t + 64 \hfill \cr
B = t + 25 \hfill \cr} \right.$$
$${1 \over A} + {1 \over B} = {1 \over t}\,\,\,\,\left( {{\rm{stem}}} \right)\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,{{1 \cdot B} \over {A \cdot B}} + {{1 \cdot A} \over {B \cdot A}} = {1 \over t}\,\,\,\,\,\mathop \Rightarrow \limits^{\left( * \right)} \,\,\,\,\,{{2t + 89} \over {\left( {t + 64} \right)\left( {t + 25} \right)}} = {1 \over t}$$
$$2{t^2} + 89t = {t^2} + 89t + 25 \cdot 64\,\,\,\,\,\,\mathop \Rightarrow \limits^{t\,\, > \,\,0} \,\,\,\,\,t = 5 \cdot 8 = 40$$

$$?\,\,\,\mathop = \limits^{\left( * \right)} \,\,\,{{40 + 64} \over {40 + 25}}\,\, = \,\,{{104:13} \over {65:13}} = {8 \over 5}$$


We follow the notations and rationale taught in the GMATH method.

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by GMATGuruNY » Sun Jan 27, 2019 7:59 am
BTGmoderatorLU wrote:Source: Veritas Prep

Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?

A. 5:8
B. 8:5
C. 25:64
D. 25:39
E. 64:25
One more approach:

Since A+B together complete the job in t minutes:
Job = (A's output in t minutes) + (B's output in t minutes)

If A works alone for t minutes, A can then complete the rest of the job in 64 minutes.
In this case, the rest of the job = B's output in t minutes.
Thus, A's output in 64 minutes is equal to B's output in t minutes:
A/64 = B/t
A/B = 64/t

If B works alone for t minutes, B can then complete the rest of the job in 25 minutes.
In this case, the rest of the job = A's output in t minutes.
Thus, A's output in t minutes is equal to B's output in 25 minutes:
A/t = B/25
A/B = t/25

The expressions in blue are both equal to A/B and thus must be equal to each other:
t/25 = 64/t
t² = 25*64
t = 5*8
t = 40

Thus:
A's time alone = t+64 = 40+64 = 104
B's time alone = t+25 = 40+25 = 65
(A's time alone) : (B's time alone) = 104:65 = (8*13) : (5*13) = 8:5

The correct answer is B.
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by [email protected] » Sun Jan 27, 2019 12:30 pm
Hi All,

We're told that Machines A and B, working together, take T minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than T to complete the same work. Machine B, working alone, takes 25 minutes more than T to complete the same work. We're asked for the ratio of the time taken by machine A to the time taken by machine B to complete this work. While this question is layered, it's ultimately a "Work Formula" question, so you can use the Work Formula - and a bit of Algebra - to solve it.

Work = (A)(B)/(A+B) where A and B are the individual times needed to complete the task

In this prompt, we have:
A = T + 64
B = T + 25

Substituting in those values, we have:

(T+64)(T+25)/(T+64 + T+25) = T

(T^2 + 89T + 1600)/(2T + 89) = T
T^2 + 89T + 1600 = T(2T + 89)
T^2 + 89T + 1600 = 2T^2 + 89T
1600 = T^2
T = +40 or -40

Since we cannot have a "negative time" for T, we know that T MUST be 40 minutes, meaning that the two individuals times for A and B are:
A = 40 + 64 = 104 minutes
B = 40 + 25 = 65 minutes
A/B = 104/65 = 8/5

Final Answer: B

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by Scott@TargetTestPrep » Wed Jan 30, 2019 6:03 pm
BTGmoderatorLU wrote:Source: Veritas Prep

Machines A and B, working together, take t minutes to complete a particular work. Machine A, working alone, takes 64 minutes more than t to complete the same work. Machine B, working alone, takes 25 minutes more than t to complete the same work. What is the ratio of the time taken by machine A to the time taken by machine B to complete this work?

A. 5:8
B. 8:5
C. 25:64
D. 25:39
E. 64:25

The OA is B
We can let a = the number of minutes A can finish the work by itself and b = the number of minutes B can finish the work by itself. Thus, their rates are 1/a and 1/b, respectively. We can create the equations:

t(1/a) + t(1/b) = 1

(t + 64)(1/a) = 1

and

(t + 25)(1/b) = 1

From the second and third equations, we see that 1/a = 1/(t + 64) and 1/b = 1/(t + 25). Substituting these into the first equation, we have:

t(1/(t + 64)) + t(1/(t + 25)) = 1

Multiplying both sides by (t + 64)(t + 25), we have:

t(t + 25) + t(t + 64) = (t + 64)(t + 25)

t^2 + 25t + t^2 + 64t = t^2 + 25t + 64t + 1600

t^2 = 1600

t = ±40

Since t can't be negative, t = 40. Therefore, a = 40 + 64 = 104 and b = 40 + 25 = 65. So the ratio of the time taken by machine A to the time taken by machine B to complete this work is 104:65 = 8:5.

Answer: B

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