Problem Solving

Car A gets 25% more miles per gallon of gasoline than car b does. However, car A requires special gas that is costs 10% more than the gas for car b uses. If the two cars are driven equal distances, what percent less than the money spent on gasoline for car b is the money spen on gas for car a.

a.22.5 %
b.17.5%
c.15%
d.12%
e.10%






qa is D, I am having some good troubles with tranlsating the equations out of here. How can I get better at this? THis shouldnt be this difficult!

Setup Grid

Assume B does 100 miles/gallon and cost of fuel for B is 100$/gallon

Car || Miles/Gallon || Cost/Gallon || Cost/Mile
B || 100 || 100 || 1
A || 125 || 110 || 110/125


Percentage difference in costs is (B-A)/B * 100

solve and get 12.

Excuse the 'grid' it doesnt really display nice here. But hopefully you get the idea.

Enginpasa1 wrote:Car A gets 25% more miles per gallon of gasoline than car b does. However, car A requires special gas that is costs 10% more than the gas for car b uses. If the two cars are driven equal distances, what percent less than the money spent on gasoline for car b is the money spen on gas for car a.

a.22.5 %
b.17.5%
c.15%
d.12%
e.10%

Percent question with unspecified values - perfect for picking numbers.

On most % questions, we want to pick 100. To keep things as simple as possible, let's let 100 be the unmodified quantities.

So:

b gets 100 mpg
a gets 125 mpg (25% more than b)

1 gallon of b gas costs $100
1 gallon of a gas costs $110 (10% more than b)

Let's pick a number of miles that works easily with both a and b: the LCM of 100 and 125 is 500, so that's a great choice.

Let d = 500

at 100mpg, it takes car b 5 gallons of gas and it takes car a 4 gallons of gas.

b pays $100 per gallon, so b pays $500
a pays $110 per gallon, so a pays $440

The question is what % less does a pay than b, so we want:

(amt b - amt a)/(amt b) = 60/500 = 12/100 = 12%: choose (d).

I saw where I went wrong. I made things way more complicated than they needed to be. With your advice here is a way that I would approach it next time. I do best when I simply logically answer the question. It is much easier, faster and accurate than setting up lengthy complicated compuations. I am just sticking to the golden simple rule taught me by kaplan and manhattan gmat.

heres the new approach:

Pick number:
A=125 mpg B=100 mpg A gas= $11 and B gas = $10
Distance is set to 500 miles

So, gallons of gas needed for A= 500/125= 4
gallons of gas needed B= 500/100=5


Cost of trip: A:4 X $11= 44 and cost of trip B: 5x10=$50

now simply stick into percent change formula



(50-44)/50= .12 qa is d

thank you, engin