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In a certain candy store, 22% of the customers are caught sa

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In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 11 Oct 2013, 02:09
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In a certain candy store, 22% of the customers are caught sampling the candy and are charged a small fine, but 12% of the customers who sample the candy are not caught. What is the total percent of all customers who sample candy?

A) 22%
B) 23%
C) 24%
D) 25%
E) 34%
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 11 Oct 2013, 02:17
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tk1tez7777 wrote:
In a certain candy store, 22% of the customers are caught sampling the candy and are charged a small fine, but 12% of the customers who sample the candy are not caught. What is the total percent of all customers who sample candy?

A) 22%
B) 23%
C) 24%
D) 25%
E) 34%


Since 12% of the customers who sample the candy are not caught, then 88% of the customers who sample the candy are caught:

{% of customers who sample candy}*0.88 = 0.22;
{% of customers who sample candy} = 0.25.

Answer: D.
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In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post Updated on: 02 Apr 2016, 04:51
What is the best way to calculate this question?
I tried to aproximate and i ended up with a wrong answer.

Originally posted by eladavid on 13 Oct 2013, 03:43.
Last edited by eladavid on 02 Apr 2016, 04:51, edited 1 time in total.
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 13 Oct 2013, 04:07
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 09 Apr 2014, 20:06
I am little bit confused in this question.Would you please elaborate the explanation more clearly.

Lets say we have 100 customer in store. 12% sample the candy but didn't caught.
So, 88% customer caught sampling the candy. I didn't get after this.

Please explain.
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 10 Apr 2014, 00:54
tk1tez7777 wrote:
In a certain candy store, 22% of the customers are caught sampling the candy and are charged a small fine, but 12% of the customers who sample the candy are not caught. What is the total percent of all customers who sample candy?

A) 22%
B) 23%
C) 24%
D) 25%
E) 34%


One can solve by two equations:
x-y=22
y=0.12x

we get x-0.12x=22 => 0.88x=22 => x=25
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 10 Apr 2014, 01:23
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vikrantgulia wrote:
I am little bit confused in this question.Would you please elaborate the explanation more clearly.

Lets say we have 100 customer in store. 12% sample the candy but didn't caught.
So, 88% customer caught sampling the candy. I didn't get after this.

Please explain.


Hi Vikrant,

lets say there are 100 customers in store, out of with x sample the candy
12% of this x was not caught, 88% of this x was caught

this 88% of x = 22% of the total customers (it is given that 22% of the total customers were found sampling)

so x = 22/88 = 0.25

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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 14 Apr 2014, 02:08
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vikrantgulia wrote:
I am little bit confused in this question.Would you please elaborate the explanation more clearly.

Lets say we have 100 customer in store. 12% sample the candy but didn't caught.
So, 88% customer caught sampling the candy. I didn't get after this.

Please explain.



Lets say Total = 100
Sample................................... .............................. Dont Sample
x......................................................................... 100-x

\(\frac{12x}{100} =\)Not Caught, means

\(\frac{88x}{100} =\)Caught Sampling

Given that 22% of the customers are caught sampling the candy, i.e

\(\frac{88x}{100} = \frac{22}{100} * 100\)

x = 25%

Answer = D
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 23 Jun 2015, 10:34
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Let's say the no of customers who sample candy = S

Now { 0.12 S }+ {22} = S

>> S = 25
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 23 Jun 2015, 20:09
Hi All,

The presentation in this question is relatively rare on Test Day - the logic most often appears in Overlapping Sets questions (but the wording is a bit more straight-forward). The idea here is essentially about "groups within groups"...

From the prompt we know that there are 3 different types of people....

Those who do NOT sample candy
Those who SAMPLE AND are CAUGHT
Those who SAMPLE but are NOT CAUGHT

The percents that we're given refer to the latter 2 groups....we can TEST VALUES to clarify the logic:

22% of customers are caught sampling candy - this refers to 22% of ALL CUSTOMERS....

IF Total customers = 100
22%(100) = 22 customers SAMPLE AND are CAUGHT.

Next, we're told that 12% of those WHO SAMPLE are NOT CAUGHT. This is NOT 12% of everyone; it's 12% of those who SAMPLE.

TOTAL WHO SAMPLE = X = (those CAUGHT) + (those NOT CAUGHT)

X = 22 + .12(X)

From here, we can combine like terms and solve for X....

.88X = 22
X = 22/.88 = 1/.04 = 25

Thus, 25 out of the original 100 SAMPLE candy. Since that is what the question is focused on, 25% is the answer.

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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 09 Mar 2016, 20:50
i used the total number of clients = 100
now..22 sampling and fined
suppose X is the number of clients who sample
0.12x are not fined.
it means that 22 = 0.88x
now..
x=22*100/88 = 25

so 25 people are sampling.
since we have 100 customers, 25 of them would represent 25%
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 20 May 2016, 08:14
Tricky one .. good question ... 25% it is :)
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 20 May 2016, 08:19
Let 100 be the total number of persons.

22 sample candy and get caught.

Let total number of persons who sample candy be x

12% (.12x) are not caught while 22 (88%) are caught
.88x= 22
x=25%
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 07 Jul 2016, 19:45
Good question. I made the mistake of sorting information too mechanically by dividing the customers into Sample/Did not sample and Caught/ Not Caught categories. I completely missed the fact that there was no one who would did not sample but got caught.. Once I saw it, I realized that the "22% of the customers are caught sampling the candy and are charged a small fine" is in fact 22% of total customers.

Because "12% of the customers who sample the candy are not caught", 88% of customers who sample are caught. That 88% is 22% of total customers, so 100% of people who sample is 22%*100%/88%= 25% of total customers.
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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New post 09 Aug 2017, 07:33
tk1tez7777 wrote:
In a certain candy store, 22% of the customers are caught sampling the candy and are charged a small fine, but 12% of the customers who sample the candy are not caught. What is the total percent of all customers who sample candy?

A) 22%
B) 23%
C) 24%
D) 25%
E) 34%



22% of the overall customers are caught sampling candy and are fined.
12% of the customers who sample candy are not caught.
So, 88% of the customers who sample candy are caught.

So, total percentage of all customers of sample candy = 22/.88 = 22*100/88 = 25 %

Answer D
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Re: In a certain candy store, 22% of the customers are caught sa  [#permalink]

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