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Bunuel
In a certain math class, one-tenth of the students cheated on a test and were caught but one-fifth of the students who cheated did not get caught. What fraction of the class cheated on the test?

(A) 21/100
(B) 1/10
(C) 1/8
(D) 1/12
(E) 3/25


Are You Up For the Challenge: 700 Level Questions: 700 Level Questions

let total students = 100
who cheated and got caught= 1/10 * 100 = 10
1/5th of students who cheated did not get caught so 4/5 of students cheated and got caught ; i.e 4x/5 =10 ; x= 50/4 students got caught
fraction ; 50/4* 1/100 = 1/8
IMO C
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let the total number students = 40

lets number of students who cheated = x

1/10*40 = cheated+caught= 4 students

4/5(x)= 4

therefore total number of students who cheated= 5

ratio= 5/40= 1/8

option C is the correct option!
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lets say total no of students is 100
total nos of studetns who cheated is x
cheated and caught=1/10=10
cheated and not caught=x/5

therefore, 10+x/5=x, solving for x=12.5=1/8.Ans C
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Bunuel
In a certain math class, one-tenth of the students cheated on a test and were caught but one-fifth of the students who cheated did not get caught. What fraction of the class cheated on the test?

(A) 21/100
(B) 1/10
(C) 1/8
(D) 1/12
(E) 3/25


Are You Up For the Challenge: 700 Level Questions: 700 Level Questions

Let's assume there are 100x students.
Now, 10x cheated and were caught.
Now 10x = \(\frac{4}{5}*cheated\).
so cheated = 12.5x
and the proportion of cheated in class = \(\frac{12.5x}{100x}\) = \(\frac{1}{8}\)
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Let number of students = s
Let number of cheaters = c

Given: No. of cheaters who got caught = \(\frac{s}{10}\)
And, No. of cheaters who did not get caught = \(\frac{c}{5}\)

Therefore,
\(\frac{s}{10} + \frac{c}{5} = c,\)
\(\frac{s}{10} = \frac{4c}{5},\)
\(\frac{c}{s} = \frac{5}{40} = \frac{1}{8} \)

Hope this helps.
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2 x 2 grid method would be best for this problem.
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Bunuel
In a certain math class, one-tenth of the students cheated on a test and were caught but one-fifth of the students who cheated did not get caught. What fraction of the class cheated on the test?

(A) 21/100
(B) 1/10
(C) 1/8
(D) 1/12
(E) 3/25


Are You Up For the Challenge: 700 Level Questions: 700 Level Questions

Always read word problems slowly

Assume 100 students
one tenth cheated and caught 10 no
one fifth of those cheated did not get caught so 4/5 th get caught
so four fifth caught equals to 10 no
so all of those cheated equals to 12.5 no ( fraction is ok as i have assumed 100 )

percentage of cheated = 12.5/100 =1/8
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Since the fractions have 10&5 as denominators let's assume with the LCM 10as the number of total students. Then 1/10=1 is the number who cheated and got caught.
Let's assume C as the number of cheaters. Since 1/5 were not caught 4/5of C were caught and we know from the earlier step that this is 1
So 4/5(C)=1or C=5/4
So 5/4 out of total 10 students, cheated.
5/(4*10)=5/40=1/8
Hence C

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A better trick is to assume the number of people that did not get caught
Assume 1/5 that did not get caught as 2.
People who cheated and got caught=4*2=8
People who cheated and got caught were 1/10 of the students so Total students=10*8=80
Total no. of students who had cheated=10/80=1/8 (D)
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Bunuel
In a certain math class, one-tenth of the students cheated on a test and were caught but one-fifth of the students who cheated did not get caught. What fraction of the class cheated on the test?

(A) 21/100
(B) 1/10
(C) 1/8
(D) 1/12
(E) 3/25


Are You Up For the Challenge: 700 Level Questions: 700 Level Questions

1/5th of cheated did not get caught = 4/5th of cheated got caught.

4/5th of cheated = 1/10th of Total

Cheated = 1/8th of Total

Answer (C)
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Let C = # of students who cheated

Let N = # of students who did not cheat

C + N = 100 students


C is further broken down into: (Cheated + Caught) and (Cheated + NOT Caught)


"1/10 of the students cheated AND were Caught"

thus: (Cheated + Caught) = (1/10) * 100 = 10

"1/5 of the students who cheated did NOT get caught"

C = # of students who cheated ------> thus: (1/5)C = (Cheated + NOT Caught)



adding the 2 groups together to make up all the people who cheated = C

10 + (1/5)C = C

10 = (4/5)C

C = 25/2


What fraction of the total students Cheated?

or

what is: C / T = ?


(25/2) / 100 = 25 / 200 = 1/8


answer C: 1/8
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