Question

In an examination, the ratio of passes to failures was 4 : 1. Had 30 less appeared and 20 less passed, the ratio of passes to failures would have been 5 : 1 Find the number of students who appeared for the examination.

Suppose x candidates passed and y failed;
therefore, $\frac{x}{4}=\frac{4}{1}\dots \left(I\right)$

In the second case: no. of students appeared = x + y - 30 and no. of those who passed = x - 20:

$\therefore$ No. of failed = $x+y-30-(x-20)=y-10$

Given: $\frac{x-20}{y-10}=\frac{5}{1}\dots \left(II\right)$

Answer 1

let



p = the number of passes

f = the number of failures

p + f = the number of the people that appeared at the examination



the ratio of passes to failures was 4:1



P : f = 4 : 1

p = 4f



if 30 less had appeared and 20 less passed,the ratio of passes to failures would have been 5:1



30 less had appeared: p + f - 30



20 less passed: p - 20



number of failures: p + f - 30 - (p - 20) = p + f - 30 - p + 20 = f - 10



the ratio of passes to failures would have been 5:1



(p - 20) : (f - 10) = 5 : 1



p - 20 = 5(f - 10)



by solving the system of equations



p = 4f

p - 20 = 5(f - 10)



we find



p = 120 passes



f = 30 failures
total number of canditates = 120 +30 = 150