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After the price of chocolates was discounted by 40%, a

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After the price of chocolates was discounted by 40%, a  [#permalink]

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New post Updated on: 30 Oct 2012, 23:05
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Question Stats:

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After the price of chocolates was discounted by 40%, a shopper can buy 4 more chocolates for $12 than he could buy at the original price. How many chocolates could be bought for $12 at the original price?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 16

Originally posted by actleader on 30 Oct 2012, 22:49.
Last edited by Bunuel on 30 Oct 2012, 23:05, edited 1 time in total.
Edited the tags.
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Re: After the price of chocolates was discounted by 40%, a shopp  [#permalink]

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New post 30 Oct 2012, 23:21
12
3
Price per chocolate = P, No. Of choclotaes = N
PN = 12
0.6P(N+4) = 12
P(N+4) = 20
4P = 8
P=2
N=6

ANSWER IS A

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Re: After the price of chocolates was discounted by 40%, a  [#permalink]

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New post 30 Oct 2012, 23:12
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actleader wrote:
After the price of chocolates was discounted by 40%, a shopper can buy 4 more chocolates for $12 than he could buy at the original price. How many chocolates could be bought for $12 at the original price?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 16


Say for $12 at the original price a shopper can buy \(x\) chocolates, then, the original price is \(\frac{12}{x}\) and discounted price is \(0.6*\frac{12}{x}\).

For $12 at the discounted price a shopper can buy \(\frac{12}{(\frac{0.6*12}{x})}=\frac{x}{0.6}=\frac{10x}{6}\) chocolates.

Given that \(x+4=\frac{10x}{6}\) --> \(x=6\).

Answer: A.

Hope it's clear.
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Re: After the price of chocolates was discounted by 40%, a  [#permalink]

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New post 31 Oct 2012, 12:31
Also we can solve it as following:
[12][/P] - number of bars of chocolate can be bought originally, where P is price of one bar of chocolate
0.6*P - new, discounted price of bar of chocolate and
12/0.6*P - number of bars of chocolate we can buy or [20][/P].

We are at the middle of the task

[12][/6] - [20][/P] = 4 and the price of 1 bar of chocolate = 2

so originally we can buy [12][/2] = 6

the answer is A
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Re: After the price of chocolates was discounted by 40%, a  [#permalink]

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New post 04 Nov 2012, 12:30
(Q0 + 4) * 0.6* P0 = Q0 * P0 ==> Q0 = 6
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Re: After the price of chocolates was discounted by 40%, a  [#permalink]

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New post 03 Jan 2014, 08:49
actleader wrote:
After the price of chocolates was discounted by 40%, a shopper can buy 4 more chocolates for $12 than he could buy at the original price. How many chocolates could be bought for $12 at the original price?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 16


What about this approach

Since 0.6 P means a decrease of 40% then Q has to increase 40%

4 = 40%x then X is 10

Now since originally price was P then P will have to increase 40%
Then we need to decrease Q again 40%
So 0.6 Q = 6

Answer is A

Bunuel, is this approach valid?

Cheers
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Re: After the price of chocolates was discounted by 40%, a  [#permalink]

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New post 03 Jan 2014, 09:00
jlgdr wrote:
actleader wrote:
After the price of chocolates was discounted by 40%, a shopper can buy 4 more chocolates for $12 than he could buy at the original price. How many chocolates could be bought for $12 at the original price?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 16


What about this approach

Since 0.6 P means a decrease of 40% then Q has to increase 40%

4 = 40%x then X is 10

Now since originally price was P then P will have to increase 40%
Then we need to decrease Q again 40%
So 0.6 Q = 6

Answer is A

Bunuel, is this approach valid?

Cheers
J :)


The price decreases 3/5 times, thus the quantity increases by 5/3 times --> x+4=5/3*x --> x=6.
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Re: After the price of chocolates was discounted by 40%, a  [#permalink]

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New post 19 May 2016, 07:03
actleader wrote:
After the price of chocolates was discounted by 40%, a shopper can buy 4 more chocolates for $12 than he could buy at the original price. How many chocolates could be bought for $12 at the original price?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 16


Let the Price of Chocolate before discount is x and after reduction it is .6x so

Total no of chocolates which can be brought before discount is 12/x and after discount 12/.6x

No of Chocolates brought after discount are 4 more than the number of chocolates brought before discount Hebce

12/.6x - 12/x = 4 ===> 20/x-12/x =4 ===> (20-12)/x = 4

x=2 so Total number of chocolates brought before the price reduction is 12/2 = 6
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Re: After the price of chocolates was discounted by 40%, a  [#permalink]

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New post 19 May 2016, 09:27
Let the number of chocolates bought in $12= n
Price for each chocolate= x

nx= 12

After price reduction to .60x one can buy 4 more chocolate= n+4

xn=.6x(n+4)

.4n=2.4
n=6
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Re: After the price of chocolates was discounted by 40%, a  [#permalink]

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New post 19 May 2016, 09:38
1
actleader wrote:
After the price of chocolates was discounted by 40%, a shopper can buy 4 more chocolates for $12 than he could buy at the original price. How many chocolates could be bought for $12 at the original price?

(A) 6
(B) 8
(C) 10
(D) 12
(E) 16



Hi,
another method-

If we read the info and the price involved is 12 in original and final scenario, it tells us that NOW you can buy 4 chocolates for 40% price,
so for 100%, one can buy 4*100/40 = 10...
But earlier he could buy 4 less so answer = 10-4 = 6
A
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Re: After the price of chocolates was discounted by 40%, a  [#permalink]

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New post 18 Oct 2016, 18:11
From the prompt we can derive the following equations:

PQ=12
(P-0.40P)(Q+4) = 12

(0.60P)(Q+4)= 12
0.60PQ+2.4P= 12
7.2 + 2.4P = 12
2.4P = 4.8
P = 2 (original price)

12/2 = 6 (original quantity)
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After the price of chocolates was discounted by 40%, a  [#permalink]

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New post 13 Feb 2017, 12:39
To continue my correction spree:

"an buy 4 more chocolates for $12 than he could"

IMO the "for" induces the test taker in error. It indicates that each chocolate costs 12$.

The correct word is "with" 12 dollars to clearly indicate that 12 refers to the total budget and not to the individual chocolate price.

Am I the only one here? :)
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Re: After the price of chocolates was discounted by 40%, a  [#permalink]

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Re: After the price of chocolates was discounted by 40%, a   [#permalink] 04 Aug 2019, 08:10
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