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

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

17 Feb 2012, 21:29

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59% (02:21) correct

41% (01:52) wrong

based on 2 sessions

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

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Last edited by LM on 17 Feb 2012, 21:44, edited 1 time in total.

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Re: Percentages-PS [#permalink]

17 Feb 2012, 21:41

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 :-D

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Last edited by boomtangboy on 17 Feb 2012, 21:54, edited 1 time in total.

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Re: Percentages-PS [#permalink]

17 Feb 2012, 21:44

boomtangboy wrote:

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

$ 100 - > 100 Litre
$ 90 -> 105 Litre
$180 - > ?

=> 180 * 105 /90 => 210

180/210 => 6/7 => $ 0.14 / Litre

=> 210 Litres

Please check again, what the question is asking.

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Re: Percentages-PS [#permalink]

17 Feb 2012, 22:00

boomtangboy wrote:

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?

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Re: Percentages-PS [#permalink]

17 Feb 2012, 22:16

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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

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.
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Re: Percentages-PS [#permalink]

17 Feb 2012, 23:12

Bunuel wrote:

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

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!

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Re: Percentages-PS [#permalink]

17 Feb 2012, 23:29

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LM wrote:

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.

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

04 Sep 2012, 08:16

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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]

05 Jan 2013, 03:25

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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!!!

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

10 Jan 2013, 12:54

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

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

27 Jan 2013, 04:55

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

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

P=180/45=4
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Re: A reduction in the price of petrol by 10% enables a motorist [#permalink] 27 Jan 2013, 04:55

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