Snowplow S, working alone at its own constant rate, will clear

Question

Snowplow S, working alone at its own constant rate, will clear a certain parking lot in 6 hours. Snowplow S and snowplow T, working simultaneously and independently at their respective constant rates, will clear the parking lot in 2.4 hours. How many hours will it take snowplow T, working alone at its own constant rate, to clear the parking lot?

A. 3
B. 3.6
C. 4
D. 4.2
E. 4.4

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Official Answer

C

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Bunuel · Accepted Answer

Assuming the rates of S and T are s and t lots/hours, respectively, we are given:

s = 1/6
s + t = 1/2.4 = 1/(12/5) = 5/12

Subtracting the first equation from the second gives:

t = 5/12 - 1/6 = 3/12 = 1/4

Since time is the reciprocal of the rate, then T, working alone, would need 4 hours to clear the parking lot.

Answer: C.

gmatophobia · Answer

ext{Time taken Together} = \frac{	ext{Time taken by Snowplow S * Time taken by Snowplow T}}{	ext{Time taken by Snowplow S + Time taken by Snowplow T}}

2.4 = \frac{6*x}{6 + x}

2.4*6 + 2.4*x = 6*x

2.4*6 = x (6 - 2.4)

x = \frac{2.4 * 6 }{ 3.6}

x = 4

Option C

Confide · Answer

R*T=W
(1/6)*(12/5) + (1/t)*(12/5) = 1
2+12t = 5
t=4

simple

NEYR0N · Answer

KarishmaB ,  Bunuel  ,

When dealing with time values, can we simply add one decimal place to the times? For example, if we take 60 and 24, then calculate the difference of their reciprocals 1/24 - 1/60 = 1/4 rate. So the time would be 4. While this approach works in this specific case, is it applicable to other similar problems involving minutes/hours ?

KarishmaB · Answer

No, we cannot do that. We must adjust for the movement of decimal.

1/24 - 1/60 = 1/40, not 1/4.
Time would be 40 here.

We certainly should get rid of the decimal while calculating but we should adjust for it. So 2.4 becomes 24/10

So we get

10/24 - 1/6 = 6/24 = 1/4
Now time is 4.

Here is a discussion on work rate:  https://youtu.be/88NFTttkJmA

DanTheGMATMan · Answer

Classic work rate:

https://www.youtube.com/watch?v=juYRz63pHrY

jaegarbong · Answer

I am using Smart Number method:
Let's say there are 24 units of snow
 S -> 6 hours -> 4 units/hr

S + T -> 2.4 hours -> 10 units /hr

Only T -> 6 units/hr -> 24/6 hours =  4 hours

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Snowplow S, working alone at its own constant rate, will clear

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Snowplow S, working alone at its own constant rate, will clear

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