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

Question

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

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

MacFauz · Accepted Answer

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

Kudos Please... If my post helped.

Bunuel · Answer

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.

actleader · Answer

Also we can solve it as following:
   - 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   .

We are at the middle of the task

-    = 4 and the price of 1 bar of chocolate = 2

so originally we can buy    = 6

the answer is A

Ousmane · Answer

(Q0 + 4) * 0.6* P0 = Q0 * P0 ==> Q0 = 6
Brother Karamazov

jlgdr · Answer

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

Bunuel · Answer

The price decreases 3/5 times, thus the quantity increases by 5/3 times --> x+4=5/3*x --> x=6.

mohitmohit11 · Answer

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

Divyadisha · Answer

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

chetan2u · Answer

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

law258 · Answer

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)

iliavko · Answer

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?

rvgmat12 · Answer

12/0.6x-12/x = 4

8/x = 4

x = 2

No of chocolates for 12 = 12/2 = 6

bumpbot · Answer

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

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