Violas old car travels the same distance per gallon of gasoline ever

Question

Viola’s old car travels the same distance per gallon of gasoline everywhere. Her new car uses diesel fuel, which costs 20% more per gallon than gasoline. Compared with her old car, her new car travels 50% farther per gallon of fuel in rural areas and 25% farther elsewhere. For a certain trip, Viola uses her new car instead of her old car and thus saves 10% on fuel costs. What fraction of her trip is through rural areas?

A. $\frac{3}{8}$

B. $\frac{2}{5}$

C. $\frac{3}{5}$

D. $\frac{5}{8}$

E. $\frac{2}{3}$

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Official Answer

A

A.  \frac{3}{8}

B.  \frac{2}{5}

C.  \frac{3}{5}

D.  \frac{5}{8}

E.  \frac{2}{3}

ankushsambare · Answer

Let the old car's fuel efficiency be:
 
 Rural = r miles/gallon 
 Elsewhere = e miles/gallon  
New car:
 
 Rural = 1.5r miles/gallon 
 Elsewhere = 1.25e miles/gallon  
Let gasoline cost = G per gallon.
Then diesel cost = 1.2G per gallon.
Assume the trip distance is 1 unit.
Let x = fraction of trip through rural areas.
Then:
 
 Rural distance = x 
 Elsewhere distance = 1 − x  
Old car fuel cost:
Cost_old = G(x/r + (1−x)/e)
New car fuel cost:
Cost_new = 1.2G 
Simplify:
Cost_new = G 
Since the new car saves 10% on fuel costs,
Cost_new = 0.9 × Cost_old
So
0.8x/r + 0.96(1−x)/e
0.9 
Multiply by 100 to avoid decimals:
80x/r + 96(1−x)/e
90x/r + 90(1−x)/e
Move terms:
6(1−x)/e = 10x/r
At this point, notice that the efficiencies r and e merely scale the distances. Let
A = x/r,
B = (1−x)/e
Then
6B = 10A
B/A = 5/3
Thus fuel consumption in the old car is in the ratio
Rural : Elsewhere = A : B = 3 : 5
Take A = 3k and B = 5k.
Then
x/r = 3k
(1−x)/e = 5k
For a GMAT problem, we may normalize the old-car efficiencies to the same baseline (equivalently set r = e), since only the relative savings matter.
Then
x : (1−x) = 3 : 5
Therefore
x = 3/(3+5) = 3/8
 Answer: A (3/8)

serbangavrila · Answer

I have a hard time doing all these calculations in under two minutes, is there an easier way?

Thanks !

kevincan · Answer

Call the answer to the question f and write the following :

1.2f/1.5 + 1.2(1-f) /1.25 = 0.9

kevincan · Answer

Edited for clarity !

uks7 · Answer

I solved it like this. Diesel in rural area takes 1.5 times ahead. Or in other words the new car needs only 2/3rd fuel to cover the same distance as the old car in rural area. So assuming gasoline cost as $100, diesel cost is $120 and the new car uses 2/3rd diesel so $80 cost. 
Similarly for urban area since it takes 1.25 times ahead the new car needs only 4/5th fuel. So 4/5th of $120 is $96.
But the overall cost is actually $90. (10% less than gasoline so 0.9 of $100)
So this essentially boils down to a simple mixture and alligation problem where 80% solution is mixed with 96% one to get 90% final solution. 
In other words 80x + 96 (1-x) = 90. Solve for x to get 3/8.

HarshavardhanR · Answer

Here is an algebraic approach for those interested ->

Setup

Let the distance per gallon covered by the old car be 4x miles/gallon (m/g).

Then,

- Distance/gallon covered by new car in rural areas = 50% farther than 4x = 6x (m/g)
- Distance/gallon covered by new car in non-rural areas = 25% farther than 4x = 5x (m/g)

Let the cost per gallon of gasoline be 5c ($/g)

Then,

- Diesel's cost per gallon = 20% more = 6c ($/g)

---
Harsha

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madsb18 · Answer

I usually start with Rs. 100 and 100 miles for such questions. Makes calculations easier.

Old car 
Rs. 100 mileage for 100 miles.

New Car 
Rs. 120 for 150 miles Rural
Rs. 120 for 125 miles elsewhere

Lets say total miles covered 200. Rural miles R.
Saves 10%. So total cost 180.

180 = (120/150) R + 120/125 (200 - R)
On solving this R = 75.

75/200 = 3/8. A

Violas old car travels the same distance per gallon of gasoline ever

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Violas old car travels the same distance per gallon of gasoline ever

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