In a school, 600 students took a common entrance test. Out of those

Question

In a school, 600 students took a common entrance test. Out of those who took the test, 75% of girls and 60% of boys cleared the test. The total pass percentage of students clearing the test is 70%. How many boys attempted the test?

A)100

B)200

C)250

D)350

E)500

A. 100
B. 200
C. 250
D. 350
E. 500

rajatchopra1994 · Answer

Explanation:
let boys is x
girls be y.

(x+y=600 --- Eqn1 ) * 0.6

0.6x + 0.75y = 420 --- Eqn2

Solving Eqn 1 & 2:
0.15y = 60
y=400
x=200

IMO-B

IanStewart · Answer

This is just a standard weighted average, where one group has a 60% average, one a 75% average, and combined we have a 70% average. Using 'alligation', since 70 is twice as far from 60 as from 75, the ratio of the two groups must be 2 to 1, and since we have more girls (the overall average is closer to the girls' average), then 1/3 of the students are boys, and we have 200 boys.

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Kinshook · Answer

Given: In a school, 600 students took a common entrance test. Out of those who took the test, 75% of girls and 60% of boys cleared the test. The total pass percentage of students clearing the test is 70%.

Asked: How many boys attempted the test?

Let the number of boys who attempted the test be x.
Number of girls who attempted the tests = 600 - x

Total students who cleared the test = 75%(600-x) + 60%x = 70%*600
450 - 75%x + 60%x = 420
30 = 15%x
x = 30/15% = 3000/15 = 200

IMO B

Archit3110 · Answer

given total b+g = 600
and .75*g+.60b = .7*600
.75g+.6b= 420
.25g+.2b=140
plug in values of B from given options only option B ; 200 suffices as we get G = 400
OPTION B

MathRevolution · Answer

600 = G + B

\frac{70}{100}  * 60 =  \frac{75}{100}  * G +  \frac{60}{100}  * B

=> 420 =  \frac{{3G} }{ {4}}  +  \frac{{3B} }{ {5}}

=> 420 * 20 = 15G + 12B

=> 2800 = 5G + 4B

=> 2800 = G + 4G + 4B

=> 2800 = G + 4(B + G)

=> 2800 = G + 4(600)

=> 2800 - 2400 = G

=> G = 400 and B = 600 - G = 600 - 400 = 200

Answer B

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In a school, 600 students took a common entrance test. Out of those

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