Question

The ratio of boys and girls in a class is 5 : 3. If 16% of boys and 8% of girls failed in an examination, then find the percentage of passed students.

Answer 1

Let number of boys be ‘b’ and number of girls be ‘g’

Ratio of boys and girls is given as 5 : 3

b : g = 5 : 3

⇒ `"b"/"g" = 5/3` ...(A)

Failure in boys = 16% = `16/100 xx "b" = (16"b")/100`

Failure in girls = 8% = `8/100 xx "g" = (8"g")/100`

Pass in boys = 100 – 16% = 84% = `84/100 "b"`  ...(1)

Pass in girls = 100 – 8% = 92% = `92/100 "g"` ...(2)

From A, we have `"b"/"g" = 5/3`, adding 1 on both sides, we get

`"b"/"g" + 1 = 5/3 + 1`

`("b" + "g")/"g" = (5 + 3)/3 = 8/3`

∴ g = `3/8("b" + "g")` ...(3)

Similarly b = `5/8 ("b" + "g")` ...(4)

Total pass = pass in girls + pass in boys

= (1) + (2) 

= `84/100 "b" + 92/100 "g"`

Total pass percentage = `"Total pass"/"Total students" xx 100`

Total pass = boys passed + girls passed

= `((84/100 "b" + 92/100 "g")/("b" + "g")) xx 100`

Substituting (3) and (4) in the above, we get

= `((84/100 "b" + 92/100 "g")/("b" + "g")) xx 100`

= `[(84/100 xx 5/8 ("b" + "g"))/(("b" + "g")) + (92/100 xx 3/8 ("b" + "g"))/(("b" + "g"))] xx 100`

= `84/100 xx 5/8 + 92/100 xx 3/8`

= `[420/800 + 276/800] xx 100`

= `696/800 xx 100`

= 87%