Test the answers. The equation for this one was taking too long. Backsolving was quick.

(Price per gallon P) * (# of gallons Q) = Total cost

Q = Total cost/price per gallon

1) Find original price from given dollar increase

2) Find original # of gallons by dividing $600 by old price per gallon

3) Find new price
= (1.2)($old), OR
($Old price + added $ increase)

4) Find new # gallons. Divide 600 by new price

Answer C) $10

$10 is both the dollar increase in price and a .20 increase of original price, x

Original price:

$10 = .2x = \(\frac{10}{.2} = $50\)

Original number of gallons:

\(\frac{$600}{$50PerGal} = 12\) gallons

New price: $50 + $10 (or 1.2 * $50) = $60

New number of gallons

\(\frac{$600}{$60per} = 10\) gallons

12 - 10 = 2 fewer gallons now. Too small. That means the dollar price increase is too large. To increase quantity, decrease price.

Try B) $5

$5 = .2x

Original price:

\(\frac{$5}{.2} = $25\)

Original quantity:

\(\frac{$600}{$25per} = 24\) gallons

New price: (1.2)($25) or ($25 + $5) = $30

New quantity:

\(\frac{$600}{$30per} = 20\) gallons

24 - 20 = 4 gallons. Correct

Answer B
_________________

To live is the rarest thing in the world.
Most people just exist.
—Oscar Wilde

Can we solve this using a logical approach??

Apart from algebra you could solve it using percentage.

Initial amount = 100
Now price increased by 20% = 120
Since the price increased , you need to decrease consumption by 20/120 * 100 = 16.67% or 1/6 to keep your expenditure the same.

This 16.67% decrease = 4 units of gasoline
Therefore , 100% = 24 units of gasoline
Initial quality = 24 units ; Initial price = 600/24 = 25$
Final quantity = 20 units ; Final Price = 600/20 = 30$

Increase = 5$