in a exam 80% students are passed in English,85% are passed in math ,while 75% are passed in both .if 45 students are failed find the total no of students
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Hy there,
Here's your answer,
let total no. of student be x.
then,student passed in english=80% of x=80/100*x=4x/5.
student passed in mathematics=85%0f x=17x/20
student passed in both subjeect=75%ofx=3x/4
total no.of student passed in either subject =4x/5+17x/20–3x/4=9x/10
total no of student failed in both subject=total no. of student-total no.of student passed in either subject.
this implies, x-9x/10 =45 (since it is given total no. of student failed=45)
solving this we get x=450.
so the total no. Of students should be 450.
Hope it helps.
Thanks.
Mark it brainliest if it helped you.
Here's your answer,
let total no. of student be x.
then,student passed in english=80% of x=80/100*x=4x/5.
student passed in mathematics=85%0f x=17x/20
student passed in both subjeect=75%ofx=3x/4
total no.of student passed in either subject =4x/5+17x/20–3x/4=9x/10
total no of student failed in both subject=total no. of student-total no.of student passed in either subject.
this implies, x-9x/10 =45 (since it is given total no. of student failed=45)
solving this we get x=450.
so the total no. Of students should be 450.
Hope it helps.
Thanks.
Mark it brainliest if it helped you.
Answered by
0
80% pass Eng
85%pass mat
75% failed both
now, failed percentage
100%-80%=20% in Eng
100%-85%=15% in Mat
100%-75%=25% in both
n(eUm)=n(e)+n(m)-n(enm)
=20+15-25
=10%
total no. of failed students= 45
10% of x=45
10/100*x=45
1/10*x=45
x=450
85%pass mat
75% failed both
now, failed percentage
100%-80%=20% in Eng
100%-85%=15% in Mat
100%-75%=25% in both
n(eUm)=n(e)+n(m)-n(enm)
=20+15-25
=10%
total no. of failed students= 45
10% of x=45
10/100*x=45
1/10*x=45
x=450
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