In an examination, the ratio of passes to failures was 4 : 1. Had 30 l

1.	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, x4=41 … (I) In the second case: no. of students appeared = x + y - 30 and no. of those who passed = x - 20: ∴ No. of failed = x+y−30−(x−20)=y−10 Given: x−20y−10=51 … (II)
Answer» 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, $\frac{x}{4} = \frac{4}{1} \dots (I)$ In the second case: no. of students appeared = x + y - 30 and no. of those who passed = x - 20: $∴$ No. of failed = $x + y - 30 - (x - 20) = y - 10$ Given: $\frac{x - 20}{y - 10} = \frac{5}{1} \dots (I I)$

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, x4=41 … (I) In the second case: no. of students appeared = x + y - 30 and no. of those who passed = x - 20: ∴ No. of failed = x+y−30−(x−20)=y−10 Given: x−20y−10=51 … (II)

Answer»

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, $\frac{x}{4} = \frac{4}{1} \dots (I)$

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

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

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

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