Tom travels d miles by a car, whose consumption C gallons per hour is

Question

Tom travels d miles by a car, whose consumption C gallons per hour is directly proportional to the square of the speed of the car. If he travels 80% of the distance at the speed of 40 miles/hour and the remaining at the speed of 20 miles/hour, he consumes 45 gallons of fuel. How much fuel will he consume if Tom travels the whole distance d miles at the speed of 30 miles hour?

A. 30
B. 37.5
C. 50
D. 60
E. None of these

A. 30
B. 37.5
C. 50
D. 60
E. None of these

MartyMurray · Accepted Answer

Tom travels d miles by a car, whose consumption C gallons per hour is directly proportional to the square of the speed of the car. If he travels 80% of the distance at the speed of 40 miles/hour and the remaining at the speed of 20 miles/hour, he consumes 45 gallons of fuel. How much fuel will he consume if Tom travels the whole distance d miles at the speed of 30 miles hour?

A. 30
B. 37.5
C. 50
D. 60
E. None of these

Let the constant by which the speed of the car is multiplied to determine the fuel consumption per hour be k.

Also, to keep this simple, we can let the distance be 100 miles, which works nicely with 80% of the distance covered at 40 miles per hour and 20% covered at 20 miles per hour.

We get the following.

Time at 40 miles per hour: 80/40 = 2

Fuel consumption at 40 miles per hour: 40^2 × k = 1600k

Total fuel consumption at 40 miles per hour: 2 × 1600k = 3200k

Time at 20 miles per hour: 20/20 = 1

Fuel consumption at 20 miles per hour: 20^2 × k = 400k

Total fuel consumption at 20 miles per hour: 1 × 400k = 400K

Total fuel consumed at 40 miles per hour for 80% of the trip and 20 miles per hour for 20% of trip: 3200k + 400k = 3600k

Time at 30 miles per hour: 100/30 = 3 1/3

Fuel consumption at 30 miles per hour: 30^2 × k = 900k

Total consumed at 30 miles per hour for the entire trip: 3 1/3 × 900 = 3000k

Total fuel consumed at 30 miles per hour for the entire trip: 45 × 3000k/3600k = 45 × 5/6 = 37.5

A. 30
B. 37.5
C. 50
D. 60
E. None of these

Correct answer:  B

Kinshook · Answer

Blackcrow1972

Tom travels d miles by a car, whose consumption C gallons per hour is directly proportional to the square of the speed of the car. If he travels 80% of the distance at the speed of 40 miles/hour and the remaining at the speed of 20 miles/hour, he consumes 45 gallons of fuel. How much fuel will he consume if Tom travels the whole distance d miles at the speed of 30 miles hour?

A. 30
B. 37.5
C. 50
D. 60
E. None of these

How much fuel will he consume if Tom travels the whole distance d miles at the speed of 30 miles hour?

Let C gallons/hour = kv^2 ; where k is a constant and v is speed in miles/hour

Total distance = d miles

40  miles/hour
Distance = .8d
Time = .8d/40 = .02d hours
Fuel consumption rate C = k(40)^2 = 1600k gallons/hour
Fuel consumption = 1600k*.02d = 32kd

20  miles/hour
Distance = .2d
Time = .2d/20 = .01d
Fuel consumption rate C = k(20)^2 = 400k gallons/hour
Fuel consumption = 400k*.01d = 4kd

Total fuel consumption = 32kd + 4kd = 36kd = 45 gallons
kd = 45/36

30  miles/hour
Distance = d
Time = d/30
Fuel consumption rate C = k(30)^2 = 900k gallons/hour
Fuel consumption = 900k*d/30 = 30kd = 30*45/36 = 37.5 gallons

IMO B

Dbrunik · Answer

let me know if this makes sense or if theres a better way.

DanTheGMATMan · Answer

Beautiful problem. Remember the formula for direct proportion and also that fuel consumed= rate of consumption x time:

https://www.youtube.com/watch?v=1tIk5Li7eC0

Elite097 · Answer

MartyMurray  how can k be the same for each of the speeds?

KarishmaB · Answer

\frac{Consumption}{Speed^2} = Constant C = kS^2 (in gallons/hour) We are given data of 45 gallons which is consumption in the entire journey, not gallons consumed per hour. Hence we need to find for how many hrs this consumption was maintained. Comparing Consumption in 80% of the journey with that in 20% of the journey: If distance travelled is 4 times and speed is twice, time taken will be twice too because Time = Distance/Speed C1 = k*40^2 * 2 = 3200k gallons (Total consumption in 80% of the journey) C2 = k*20^2 * 1 = 400k gallons (Total consumption in 20% of the journey) Total Consumption = 3600k gallons = 45 gallons (given) Comparing full journey consumption with that of 20% journey: If distance is 5 times and speed is 3/2, time taken = 10/3 times C3 = k*30^2*\frac{10}{3}*k = 900 * \frac{10}{3} * \frac{45}{3600} = 37.5 gallons Answer (B) Here is a video on Direct Variation used in this question: https://youtu.be/AT86tjxJ-f0

reyrest · Answer

KarishmaB

could you please help me to figure out why the below approach is wrong?

C=ks^2

I find average speed for d: 100/3 ( d/((0.8/40) + ( 0.2/20)) )

45 = k(100/3)^2

k = 45*9/(100)^2

c = k*30^2 ~ 36.45

Dj06 · Answer

I tried solving it by weighted average and it worked!

Average speed during the given trip = 40(4)+20(1)/5 = 36 miles per hour! (ratio is 8:2 = 4:1)

At 36 miles per hour, the consumption is 45, at 30 the consumption = 30*45/36 = 37.5!

I am not sure why it worked perfectly, I really did this so that I could eliminate options and mark a decent guess

Any insights woule be nice  KarishmaB   MartyMurray

MartyMurray · Answer

It just happened to work.

What you did doesn't really make sense.

For one thing, the average speed during the trip was not 36. It was d/((.8d/40 + .2d/20) = 33.33.

Then, we couldn't find the fuel consumption using the average speed even if we had the average speed because the fuel consumption is proportional to the square of the speed. So, different pairs of speeds that have the same average speed can be associated with different average rates of fuel consumption.

So, basically, it just so happened that the the numbers worked out when you sought to solve it that way.

MartyMurray · Answer

It doesn't work because the total fuel consumed at the average speed would not be the same as the total fuel consumed at the two speeds.

Notice that ( 2 × 1600 K) +  400 K ≠  33.33^2 K.

HarshavardhanR · Answer

https://gmatclub.com/forum/download/file.php?id=87111 --- Harsha GMAT-Club-Forum-hob4qx6g.png

Pantheroy · Answer

0.8d --> 40 m/h --> 1600k g/h
0.2d --> 20 m/h --> 400k g/h

given total consumption in gallons(g), so all hours needs to be cancelled out

thus,
0.8d/40 * 1600 k  +0.2d/20 * 400k   = 45  
simplifying 
32dk+4dk=45
36dk=45
dk=5/4

we're asked for distance d, and 30 m/h
similar thing for consumption in gallons

d/30 * 900k =30 K·d = 30 * 5/4 = 37.5

Does this make sense?
 MartyMurray

MartyMurray · Answer

Yes, that makes complete sense.

Regor60 · Answer

The problem states that consumption is proportional to speed squared.

So it is specifically stating that the only variable that pertains to consumption is velocity.

Proportional means that there is some constant that is a factor multiplied by velocity squared that will allow one to make an equality between consumption and velocity squared.

In short, k is a constant because it is not defined to be a variable.

dhairyajain · Answer

Is the Weighted average approach correct here?

The logic I used was since 80% of the journey is at 40 miles/hr and the remaining is at 20 miles/hr

==> the weighted average speed becomes 36 miles/hr

and since the fuel consumption is proportional to speed, 45/36 is the proportionality constant.

and the fuel consumption at 30 miles/hr = proportionality constant * 30 = 37.5

egmat · Answer

Hi dhairyajain,

Good news: your approach isn't a fluke. It's genuinely valid, and the number you called a "proportionality constant" ( 45 / 36  =  1.25 ) is exactly the same k·d that the longer solutions in this thread arrive at. Let me show you  why  it works, because the reason is subtle.

The key move is turning "per hour" into "per mile."

The rate is C = k·v2 gallons  per hour . But fuel over a stretch of road is rate × time:

- Fuel for a segment of length x at speed v = k·v2 × (x/v) =  k·v·x .

Dividing by v (the time factor) cancels one power of v. So while fuel  per hour  ∝ v2, fuel  per mile  ∝ v - just one power. That single fact is what makes your shortcut legal.

Now watch your "weighted average."  Total fuel over the whole trip is

- k·( 40 · 0.8 d +  20 · 0.2 d) = k·d·( 32  +  4 ) = k·d· 36  =  45 .

The  36  you computed is the  distance-weighted average speed  (weights = the  0.8  and  0.2  distance fractions), and total fuel = k·d × (that average). So k·d =  45 / 36  =  1.25 . At  30  mph over the same distance d: fuel = k·d· 30  =  30  ×  1.25  =  37.5 . Perfectly valid.

One caution:  it works here only because fuel per mile is  linear  in speed  and  the total distance d is the same in both scenarios. Contrast reyrest's attempt earlier in the thread: they found an average speed and plugged it into k·v2. That fails, because you can't square an average - (avg of v)2 ≠ avg of v2.

Quick check to feel the "per mile" idea for  1 mile :
- At  40  mph:  1600 k × ( 1 / 40 ) =  40k 
- At  20  mph:  400 k × ( 1 / 20 ) =  20k

Double the speed, double the per-mile fuel - linear in v, exactly as your method assumes.

Answer: B

Tom travels d miles by a car, whose consumption C gallons per hour is

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Tom travels d miles by a car, whose consumption C gallons per hour is

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

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