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I think this is the way to solve it not really sure

Let X be no of litres of petrol & n be the price per litre

Eqn 1 : x*n = 180

Eqn 2 : (x+5) * (0.9n) = 180

Simplifying eqn 2

0.9xn + 4.5n = 180

Substituting eqn 1 in eqn 2

0.9 ( 180) + 4.5n = 180

n = 18/4.5 = 4

Hence E

PS : Thanxs LM for pointing out my error

Correct and systematic approach. I was just wondering whether it can be solved within a minute or so without making efforts of too much calculations. Any fast method to solve percentages in such kind of questions?

A reduction in the price of petrol by 10% enables a motorist to buy 5 gallons more for $180. Find the original price of petrol?

A. $11 B. $5 C. $45 D. $400 E. $4

Price decreased by 10%, so 9/10 times, which means that original gallons bought increased 10/9 times. Since this increase equals to 5 gallons then 45 gallons were bought originally (45*10/9=50 --> increase 5 gallons). Hence original price was 180/45=$4.

Answer: E.

P.S. One can also use backsolving to get the answer.

A reduction in the price of petrol by 10% enables a motorist to buy 5 gallons more for $180. Find the original price of petrol?

A. $11 B. $5 C. $45 D. $400 E. $4

Price decreased by 10%, so 9/10 times, which means that original gallons bought increased 10/9 times. Since this increase equals to 5 gallons then 45 gallons were bought originally (45*10/9=50 --> increase 5 gallons). Hence original price was 180/45=$4.

Answer: E.

P.S. One can also use backsolving to get the answer.

Hope it's clear.

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!

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!

Consider this: original price of an item is $10, decrease the price by 50%: new price $5 ($10*1/2=$5). Now, for some fixed amount of money, say for $100, you'll be able to buy twice as many items for $5 (100/$5=20) than for $10 (100/$10=10): reduction in price (1/2) reciprocal of increase in quantity (2) (or simply for half of the price you'll buy twice as many in quantity). The same for the original question.

Re: A reduction in the price of petrol by 10% enables a motorist [#permalink]

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04 Sep 2012, 07:16

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LM wrote:

A reduction in the price of petrol by 10% enables a motorist to buy 5 gallons more for $180. Find the original price of petrol?

A. $11 B. $5 C. $45 D. $400 E. $4

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 \(\frac{180}{45} = 4\)
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Re: A reduction in the price of petrol by 10% enables a motorist [#permalink]

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05 Jan 2013, 02:25

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LM wrote:

A reduction in the price of petrol by 10% enables a motorist to buy 5 gallons more for $180. Find the original price of petrol?

A. $11 B. $5 C. $45 D. $400 E. $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!!!
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Re: A reduction in the price of petrol by 10% enables a motorist [#permalink]

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03 Jan 2014, 06:45

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LM wrote:

A reduction in the price of petrol by 10% enables a motorist to buy 5 gallons more for $180. Find the original price of petrol?

A. $11 B. $5 C. $45 D. $400 E. $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

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19 Apr 2015, 04:03

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04 Nov 2016, 00:42

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