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Re: A car traveled 462 miles per tankful of gasoline on the hig [#permalink]
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336/G = K ---> eq 1
462/G = K + 6 ---> eq 2

Divide both equations

336/462 = K/K + 6
K = 16
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Re: A car traveled 462 miles per tankful of gasoline on the hig [#permalink]
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We can straight away plug in the options directly.
A) 14=city implies 20=highway but 462 not divisible by 20
B) 16=city implies 22=highway. 462 divisible by 22 & 336 divisible by 16 => Ans
C) 21=city but 336 not divisible by 21
D) 22=city but 336 not divisible by 22
E) 27=city but 336 not divisible by 27
Hence, ans is B.
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Re: A car traveled 462 miles per tankful of gasoline on the hig [#permalink]
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grenico wrote:
Asif123 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



So we're looking for:

\(\frac{miles}{gallon}\)

and we're given:

\(\frac{miles}{tank}\), where the tank consists of a fixed maximum number of gallons.


To find out the number of gallons in the tank, we can think of it as a ratio:

\(\frac{gallons}{tank}\)


Writing this out, we get:

\(\frac{462 miles}{tank}\) on the highway.

\(\frac{336 miles}{tank}\) in the city.

We also know that the car in the city drove 6 miles less than the car did on the highway. Let \(x\) be the number of miles driven on the highway. It follows that \(x-6\) miles we're driven in the city. This can be written as:

\(\frac{(x)miles}{gallons}\) on the highway.

\(\frac{(x-6)miles}{gallons}\) in the city.


To find the gallons per tank we can simply divide miles/tank by miles/gallons to eliminate the miles:

\(\frac{462 miles}{tank} / \frac{(x) miles}{gallons}\)

\(\frac{336 miles}{tank} / \frac{(x-6) miles}{gallons}\)

This simplifies to:

\((\frac{462 miles}{tank}) * (\frac{gallons}{(x) miles})\)

\(\frac{462 miles}{(x) miles}\) \(\frac{gallons}{tank}\)


And:

\(\frac{336 miles}{tank} * \frac{gallons}{(x-6) miles}\)

\(\frac{336 miles}{(x-6) miles}\) \(\frac{gallons}{tank}\)

The miles cancel and we're left with:

\(\frac{462}{x}\) \(\frac{gallons}{tank}\)

\(\frac{336}{(x-6)}\) \(\frac{gallons}{tank}\)

Now we can simply equal them to eachother to solve for \(x\):

\(\frac{462}{x}\) \(=\) \(\frac{336}{(x-6)}\)

\(462x - 2772 = 336x\)

\(-2772 = -126x\)

\(x = 22\)

Since we're looking for the number of miles per gallon incurred in the city, we would plug this into the miles per gallon equation written above for the city:

\(\frac{(x-6)miles}{gallon}\) in the city.

\(\frac{(22-6)miles}{gallon}\) in the city.

\(\frac{16miles}{gallon}\) in the city.

So the answer is B


Why did you equate the two eqns
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Re: A car traveled 462 miles per tankful of gasoline on the hig [#permalink]
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