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Problem Solving

by RiyaR » Wed Sep 17, 2014 9:26 am
A car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city. If the car traveled 6 fewer miles per gallon in the city than on the highway, how many miles per gallon did the car travel in the city?
(A) 14
(B) 16
(C) 21
(D) 22
(E) 27
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by GMATGuruNY » Wed Sep 17, 2014 9:49 am
RiyaR wrote:A car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city. If the car traveled 6 fewer miles per gallon in the city than on the highway, how many miles per gallon did the car travel in the city?
(A) 14
(B) 16
(C) 21
(D) 22
(E) 27
We can PLUG IN THE ANSWERS, which represent the miles per gallon in the city.
When the correct answer choice is plugged in, the same amount of gas -- in other words, ONE TANKFUL -- will be required to travel 336 miles in the city and 462 miles on the highway.

Since all of the values in the problem are INTEGERS, the correct answer choice must divide evenly into the distance traveled in the city (336).
336 = 2*3*7*8.
Eliminate D (2*11) and E (3*3*3), neither of which divide evenly into 2*3*7*8.

Since on the highway 6 more miles per gallon are traveled, 6 more than the correct answer choice must divide evenly into the distance traveled on the highway (462).
Adding 6 to each of the remaining answer choices, we get:
A: 14+6 = 20 = 2*2*5.
B: 16+6 = 22 = 2*11.
C: 21+6 = 27 = 3*3*3.
Since 462 = 2*3*7*11, only B (2*11) divides evenly into the distance traveled on the highway.

The correct answer is B.

Answer choice B: 16 miles per gallon in the city, 22 miles per gallon on the highway
At a rate of 16 miles per gallon, the amount of gas required to travel 336 miles in the city = 336/16 = 21 gallons.
At a rate of 22 miles per gallon, the amount of gas required to travel 462 miles on the highway = 462/22 = 21 gallons.
Success!
The same amount of gas -- 21 gallons -- is sufficient to travel 336 miles in the city and 462 miles on the highway.
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by GMATGuruNY » Wed Sep 17, 2014 10:08 am
RiyaR wrote:A car traveled 462 miles per tankful of gasoline on the highway and 336 miles per tankful of gasoline in the city. If the car traveled 6 fewer miles per gallon in the city than on the highway, how many miles per gallon did the car travel in the city?
(A) 14
(B) 16
(C) 21
(D) 22
(E) 27
Algebraic solution:

Let x = miles per gallon in the city.
At a rate of x miles per gallon, the amount of gas required to travel 336 miles in the city = 336/x.

Since 6 more miles per gallon are traveled on the highway, miles per gallon on the highway = x+6.
At a rate of x+6 miles per gallon, the amount of gas required to travel 462 miles on the highway = 462/(x+6).

Since the same amount of gas -- one tankful -- is required in each case, the expressions in red are EQUAL:
336/x = 462/(x+6)
336x + 2016 = 462x
2016 = 126x
x = 16.

The correct answer is B.
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by [email protected] » Wed Sep 17, 2014 7:45 pm
Hi RiyaR,

I'm a big fan of Mitch's approach to TEST THE ANSWERS (if you're comfortable with the basic arithmetic, it's a pretty fast way to solve the problem).

The "math" approach can actually be done in a number of ways, depending on how you want to set up the algebra. Here's another way:

Since a "tankful" of gasoline is the same number of gallons of gas whether driving on the highway or driving in the city, we can use a variation of the Distance Formula to create 2 equations:

462 = H(G) where H = miles/gallon on the highway and G = # of gallons

336 = C(G) where C = miles/gallon in the city and G = # of gallons

Right now, we have 3 variables and 2 equations. The last sentence gives us one more equation to work with: "the car traveled 6 fewer miles per gallon in the city than on the highway." This translates into:

C = H - 6

So now we have a "system" of equations (3 variables and 3 unique equations means that we CAN solve for all 3 variables). We're trying to solve for C.....

462 = H(G)
336 = C(G)
C = H - 6

H = C + 6 plug this into the first equation....

462 = (C+6)(G)
462 = CG + 6G

336 = CG plug this into the prior equation....

462 = 336 + 6G
126 = 6G
21 = G

**REMINDER: This is the value of G. We want the value of C.**

Plug G=21 into 336 = CG

336 = C(21)
16 = C

Final Answer: B

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