A reduction in the price of petrol by 10% enables a motorist

Thanks a lot! How did you quickly arrive at this short method (which I was looking to!) 9/10 decrease will lead to increase of 10/9. This seems to be the key and short cut! Little more explanation or any hint or link within posts would be helpful. Awesome approach!

initially he got \(x\) gallons for $180 so 1 gallon = \(\frac{180}{x}\)-- old price per gallon

after reduction of price
he got 5 gallons more , \(x+5\) gallons for $180 so 1 gallon = \(\frac{180}{x+5}\)----new price per gallon

now old price decreased by 10%, so 90% of old price = new price

\((\frac{90}{100})(\frac{180}{x}) = \frac{180}{x+5}\) --> simplifying \(\frac{162}{x} = \frac{180}{x+5}\)
now solving for x we get x= 45 ,so initially he got 45 gallons for 180 hence price of 1gallon \(\frac{180}{45} = 4\)

Here is a short-cut:

Since price decreased by 10% ( i e 1/10) so for same 180$,
quantity increase will be 11.11% (i e 1/9).
So the increase in quantity will be 1/9 of original which is given to be 5.
So original quantity was 5*9 = 45. Thus original price was 180$/45 = 4 $.

Concept explanation:

When ever there are problems involving change (like increase decrease) the numerator is either added or substracted from the denominator to change base. Ex. SP = 600$. Profit % = 20%. Now, Profit in $ ? Profit % is calculated on CP.
Profit% = (P*100/CP) but wee are given SP. SO we change the base from CP to SP. 20% of CP = 1/5 of CP = 1/6 of SP = 1/6 of 600 = 100 $.
We have added numerator to denominator because new base required is SP, which is larger in case of profit. If it were loss, we would have subtracted.
(P/CP )*100% is on base CP (denominator). which can be base changed to SP. SP = P+CP. thus, Profit % can also be shown like (P/(P+CP))*100%

I will be posting so more examples soon... hope this helps and add value to this wonderfull site!!!

Some kudos please.. I need 50.. am working hard to get them!!!

I have a slightly different approach where we can directly arrive at the original unit price.

Let x be the unit price. We know that we get 5 gallons more after the discount. So the difference is 5.
This difference can be expressed as below

180/.9x minus 180/x = 5 where 180/.9x is the no. of gallons we get after discount and 180/x is the no. of gallons we get before discount

solving for x gives you 180/45 = 4

Hope this approach helps.
thanks,
Mohan

NP=180
0.9P(N+5)=180

0.9P(N+5)=NP
N=45

P=180/45=4

This is more of a word problem, but anyways

One gets (0.9P)(Q+5)=180

Now, once we have this it is better to backsolve. Obviously, D is total out.
I'll start with E since answer choices are not in order and I'm guessing GMAT expects me to work in order

180/4 = 45

Replacing in original equaiton one gets (9/10*4)(50) = 180

It matches so answer is E

Hope it helps
Gimme Kuddos if it does ok?

Cheers!
J

We can let n = the number of gallons normally purchased, and p = the normal price and create the equations:

p x n = 180

n = 180/p

And

0.9p x (n + 5) = 180

0.9p x (180/p + 5) = 180

162 + 4.5p = 180

4.5p = 18

p = 4

Answer: E

Hell Bunuel,

Can you please explain how did we derive to 45?

Due to the 10% price reduction, the amount of petrol one could buy increased by a factor of 10/9. If we assume the original amount of petrol purchased was 'x' gallons, then after the price reduction, the amount becomes 'x * 10/9' gallons.

The difference between these two amounts, which is 'x * 10/9 - x', gives us the additional 5 gallons.

When we solve this, we determine that 'x', the original number of gallons bought, was 45.