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A and B can complete a task in 20 days. B and C can...

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A and B can complete a task in 20 days. B and C can...

by AAPL » Thu Feb 15, 2018 6:24 am

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A and B can complete a task in 20 days. B and C can complete the same task in 30 days. If A and C can complete the task in 40 days, arrange the efficiencies of A, B, and C in descending order.

$$A.\ A>B>C$$
$$B.\ A>C>B$$
$$C.\ B>A>C$$
$$D.\ B>C>A$$
$$E.\ C>A>B$$

The OA is C.

I don't have clear this PS question. I appreciate if any expert explains it to me. Thank you so much.

I think that I can solve it in the following way,

A and B complete the task in 20 days
B and C complete the task in 30 days
A and C complete the task in 40 day

I can see that B is fast than A and C, and also I can see that C is slower than B and A, hence the descending order for them will be, B > A > C.
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by ErikaPrepScholar » Thu Feb 15, 2018 8:21 am
We can also solve this problem algebraically using the combined rate equation:
$$\frac{1}{A}+\frac{1}{B}=\frac{1}{A\ and\ B}$$
Where A and B are the time it would take each person (or machine, pump, printer, etc. depending on the question) to finish a job alone, and A and B is the time it would take both to finish the same job together.
From our equation, we know:
$$\frac{1}{A\ and\ B}=20$$ $$\frac{1}{\ B\ and\ C}=30$$ $$\frac{1}{\ A\ and\ C}=40$$ This means $$\frac{1}{\ A\ and\ B}<\frac{1}{B\ and\ C}<\frac{1}{A\ and\ C}$$
Using the combined rate equation, we can express each term as the sum of the reciprocals of the time it would take each person to complete the task individually:
$$\frac{1}{\ A}+\frac{1}{B}<\frac{1}{B}+\frac{1}{C}<\frac{1}{A}+\frac{1}{C}$$ or separated out $$\frac{1}{\ A}+\frac{1}{B}<\frac{1}{B}+\frac{1}{C}\ \ and\ \ \frac{1}{B}+\frac{1}{C}<\frac{1}{A}+\frac{1}{C}$$ Then subtracting like terms from both sides $$\frac{1}{\ A}<\frac{1}{C}\ \ and\ \ \frac{1}{B}<\frac{1}{A}$$ We can then cross-multiply without changing the direction of the sign, since we know all of the variables are positive (since a task can't be completed in negative time) to give

C < A and A < B

or combined

B > A > C
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by GMATGuruNY » Thu Feb 15, 2018 10:10 am
AAPL wrote:A and B can complete a task in 20 days. B and C can complete the same task in 30 days. If A and C can complete the task in 40 days, arrange the efficiencies of A, B, and C in descending order.

$$A.\ A>B>C$$
$$B.\ A>C>B$$
$$C.\ B>A>C$$
$$D.\ B>C>A$$
$$E.\ C>A>B$$
Let the task = the LCM of 20, 30, and 40 = 120 units.

Since A+B take 20 days to complete the task, the rate for A+B = 120/20 = 6 units per day.
Since B+C take 30 days to complete the task, the rate for B+C = 120/30 = 4 units per day.
Since A+C take 40 days to complete the task, the rate for A+C = 120/40 = 3 units per day.

Adding together A+B=6, B+C=4 and A+C=3, we get:
(A+B) + (B+C) + (A+C) = 6+4+3
2A + 2B + 2C = 13
A+B+C = 13/2 = 6.5 units per day.

Since A+B+C = 6.5 units per day, and A+B = 6 units per day, C = 0.5 unit per day.
Since A+B+C = 6.5 units per day, and B+C = 4 units per day, A = 2.5 units per day.
Since A+B+C = 6.5 units per day, and A+C = 3 units per day, B = 3.5 units per day.

The results in blue indicate that B > A > C.

The correct answer is C.
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by Scott@TargetTestPrep » Tue Feb 20, 2018 4:19 pm
AAPL wrote:A and B can complete a task in 20 days. B and C can complete the same task in 30 days. If A and C can complete the task in 40 days, arrange the efficiencies of A, B, and C in descending order.

$$A.\ A>B>C$$
$$B.\ A>C>B$$
$$C.\ B>A>C$$
$$D.\ B>C>A$$
$$E.\ C>A>B$$
Since B is in the two least number of days: 20 and 30, it must mean B is the most efficient. Similarly, since C is in the two most number of days: 30 and 40, it must mean C is the least efficient. That leaves A's efficiency between those of B and C. Thus, their efficiencies in descending order, is: B > A > C.

Answer: C

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by deloitte247 » Fri Mar 09, 2018 4:36 am
Let's express the efficiency in the number of hours used in completing the same task.
from the question,
$$A+B=20\ \ \ \ --\left(i\right)$$
$$B+C=30,.....\ where\ B=30-C----\left(ii\right)$$
$$and\ A+C=40\ \ \ \ --\left(iii\right)$$
put (ii) into (i), we have
$$A+30-C=20$$
$$A-C=-10\ .......\left(iv\right)$$
$$solve\ \left(iii\right)\ and\ \left(iv\right)\ simul\tan eously,\ we\ have$$
$$A+C=40$$
$$A-C=-10$$
2A=30 and A=15
Since $$A+C=40\ ,then\ 15+C=40$$
$$C=40-15=25$$
$$and\ B=30-C...\ Then\ B=30-25=5$$
Since it takes B 5 hours, it takes A 15 hours and it takes C 25 hours to do the same task. Therefore, $$B>A>C\ in\ descending\ order\ of\ efficiency.$$
Hence, option C is very correct
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by swerve » Fri Mar 09, 2018 9:43 am
Hi AAPL,

Also, you can try the following,

A + B --> 20
C + B --> 30
So, A >> C (1)

B + C --> 30
A + C --> 40
So, B >> A (2)

(1) & (2): B >> A >> C

Hence, option C.

Regards!
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