A generator runs at two speeds, high speed and low speed. Running

Question

A generator runs at two speeds, high speed and low speed. Running only at high speed, the generator uses 1 tank of fuel in 5 hours; running only at low speed, it uses 1 tank of fuel in 10 hours. If the generator uses 1 tank of fuel in 8 hours, what percentage of the time is it running at high speed?

A 12.5
B 20
C 25
D 40
E 75

A 12.5
B 20
C 25
D 40
E 75

TestPrepUnlimited · Accepted Answer

We can set the percentage of high speed as x%, then we know 1 - x% of the time the generator is running at low speed.

At high speed the tank is used up at a rate of  \frac{1}{5} 	ext{ tank/h} . At low speed the tank is used up at a rate of  \frac{1}{10} 	ext{ tank/h} . The current "average" rate is at  \frac{1}{8} 	ext{ tank/h}

Match the percentage with their speed and we have  \frac{1}{5}*x\% + \frac{1}{10}*(1 - x\%) = \frac{1}{8} .

\frac{1}{10}*x\% = \frac{1}{8} - \frac{1}{10} = \frac{1}{40}

x\% = \frac{1}{4} = 25\%

Ans: C

chetan2u · Answer

You make wrong equations and you get a wrong OA, as also given here.  
 
5x+10(1-x)=8 will get you x=2/3, that is 66%
OR
Weighted average method being applied wrongly. 
5…..8..10 will get you answer as 40%

This question is similar to one man doing a work in 5 days and other in 10 days. If they work one after another to finish the work in 8 days, how much % of work is done by faster man.  
Of course, we require to find one hour work to move ahead.

Similarly here, one hour consumption of fuel by higher speed generator is 1/5, while that of slower is 1/10.

Let x hours be run by higher speed generator, so 8-x by slower one. 
 When we add x hour work of H and 8-x hour work of slower, we should get ONE complete work.

\frac{x}{5}+\frac{8-x}{10}=1

2x+8-x=10…………x=2 
So 2 out of 8 hrs are run at higher speed. 
%= 100*\frac{2}{8}=25 %

C

SiddharthR · Answer

Hello  chetan2u  could you please tell me why the weighted average method does not work for this problem like you had mentioned in your post above

https://gmatclub.com/static/posted_from.png     Posted from    GMAT ToolKit

Pooki · Answer

i think the OA should be C
Lets say time run at high speed is H, time run at low speed is L
We have H+L=8
Run at high speed , 1 hour cost 1/5 of tank, run at low speed 1 hour cost 1/10 of tank
so H/5+L/10=1
Solve both equation we have H=2, L=6 so should be =2/8=1/4=> 25%
correct me if i am wrong

SiddharthR · Answer

Hello VeritasKarishma

Could you please tell me why weighted average does not work for this problem like chetan2u had explained.

I went thru this post of yours as well

https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2014/1 ... -averages/

but couldn't understand why this problem is any different from this one

https://gmatclub.com/forum/a-convenienc ... l#p2160221

brains · Answer

Taking the consumption as work, let us assume 40L of Tank capacity.
 Now at Low speed, it takes 10 hours. So, per hour consumption is : 40/10 = 4 l/hr
At high speed, it takes 5 hours. So per hour consumption is 40/5 =8 l/hr
Now generator working at both speeds takes 8 hours. So mean consumption is 40/8 =5 l /hr
Now we can use Weighted average:
So we have three rates: High rate: 8 l/hr
Low rate: 4 l/hr
Mean rate: 5l/hr
Since Mean rate is closer to Low speed, it must be the case that generator ran at low speed more than at high speed.

Let x be the time at low speed, so 8-x will be the time at high speed
So, with Wt average formula we have, 
4(x) + 8(8-x) / (8) = 5
x= 6 hr, which is time for low speed
Hence high speed time is 2 hour.
Hence 2/8 = 25% . The answer is C

RajatGMAT777 · Answer

KarishmaB 
I am also confused why can't we directly take the given figures and solve this as a Weighted Average sum, In that case we get fraction as 3/5 which is 40% of the time?
Thanks for your help

Regor60 · Answer

When using weighted averages it's important that the weighting factor makes sense in terms of the underlying problem.

For example, you could have a car going two different speeds, 40 kph and 60 kph over 20 and 40 kilometers.

You might think that the average speed is a weighting of one speed by 20 and the other by 40, over 60 kilometers.

This would be wrong because you're trying to get total distance over total TIME to arrive at average speed.

Multiplying speed, which is (distance/time) by distance is meaningless.

Multiplying by time, however, recovers the distance traveled because the time dimension cancels.

So, in this example, the distance numbers cannot be used directly but must be used to find the time at each speed, (20/40) = 1/2 hour and (40/60)= 2/3 hours, which are the weighting factors to be used.

The same thing is going on in this problem.

This problem can be analogized as Distance= Rate * Time, where distance equals 1 tank of fuel.

You can see then that you need to weight the RATE (1/time) by time.

Posted from my mobile device

KarishmaB · Answer

Here is how you can do it using weighted averages: Consumption in case of high speed = 1/5th tank/hour Consumption in case of low speed = 1/10th tank/hour Average consumption = 1/8th tank/hour What are the weights w1 and w2? We are averaging consumption. This is given by tank/hour. So weights will be the denominator i.e. hours i.e. time. w1/w2 = (C2 - Cavg)/(Aavg - C1) = (1/10 - 1/8)/(1/8 - 1/5) = 1/3 So for 1 hour of high speed, 3 hrs of low speed is maintained. Hence, it is running at high speed for 25% of the time. Answer (C) Check out this post to get more details: https://anaprep.com/arithmetic-weights- ... d-average/ SiddharthR RajatGMAT777

KarishmaB · Answer

Note that chetan2u does not say that "weighted average does not work for a problem like this". He says that weighted average could be applied incorrectly. Understand the quantity that you are averaging. You are averaging fuel consumption (No of tanks used per hour). That is what is different at different speeds and you combine them to get an average consumption. In the question you mentioned: https://gmatclub.com/forum/a-convenienc ... l#p2160221 what are we averaging? No of ounces per bottle. C1 = 12 ounce/bottle, C2 = 24 ounce/bottle, Cavg = 22 ounce/bottle. We are averaging capacity of each bottle. Average Capacity per bottle = Total ounces/Total no of bottle What will be the weights here? No of Bottles

Talha59 · Answer

High speed generator in 1 hour uses 1/5th tank of fuel
Low speed generator in 1 hour uses 1/10th tank of fuel
The combination in 1 hour uses 1/8th tank of fuel

in spider method

1/5 1/10
. \ /
. 1/8
. / \
. 18/80 3/40

Now the ratio for the amount of tank fuel used in 1 hour is 18/80 : 3/40
So the ratio of total tank fuel used in 8 hours will be 80/18 : 40/3 or  1:3 (high speed : low speed) 
(Its just like those work problems where we need to inverse the value to get the total time needed. Suppose A can do a job in 10 hours, B can do it in 5 hours, so together they do it in (1/10 +1/5)^-1 hours time)

so the percentage of high speed one will be 1/4 * 100 = 25%

A generator runs at two speeds, high speed and low speed. Running

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A generator runs at two speeds, high speed and low speed. Running

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