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.

260

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112

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2166

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150

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Solution

The correct option is D $150$
$L e t t h e p a s s : f a i l = 4 : 1 I f f a i l s t u d e n t s = x, p a s s s t u d e n t s = 4 x T o t a l s t u d e n t s = 5 x I f 30 l e s s a p p e a r e d a n d 20 l e s s p a s s e d . . S t u d e n t s a p p e a r i n g = 5 x - 30 S t u d e n t s p a s s i n g = 4 x - 20 S t u d e n t s f a i l i n g = 5 x - 30 - (4 x - 20) = x - 10 4 x - 20 : x - 10 :: 5 : 1 4 x - 20 = 5 (x - 10) 4 x - 20 = 5 x - 50 x = 30 S t u d e n t s f a i l i n g = 30 S t u d e n t s p a s s i n g = 4 x = 120 T o t l a s t u d e n t s = 5 x = 150$

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Q. 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.

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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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