In examination, the ratio of passes to failures was $4 : 1$ . Had $30$ less appeared and $20$ less passed the ratio passes to failures would have same. Find the no of total students.

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Solution

Let passed $= 4 x$ Failure $= x$ Total $= 4 x + x = 5 x$

$30$ less appeared $= 5 x - 30$

$20$ less passed $= 4 x - 20$

According to question

$\frac{5 x - 30}{4 x - 20} = \frac{4}{1} 16 x - 80 = 5 x - 30 11 x = - 30 + 80 11 x = 50 x = \frac{50}{11}$
total students= 23

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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.

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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In examination, the ratio of passes to failures was 4:1. Had 30 less appeared and 20 less passed the ratio passes to failures would have same. Find the no of total students.

Let passed =4x Failure=x Total =4x+x=5x 30 less appeared =5x−30 20 less passed =4x−20 According to question 5x−304x−20=4116x−80=5x−3011x=−30+8011x=50x=5011total students= 23

In examination, the ratio of passes to failures was $4 : 1$ . Had $30$ less appeared and $20$ less passed the ratio passes to failures would have same. Find the no of total students.

Let passed $= 4 x$ Failure $= x$ Total $= 4 x + x = 5 x$

$30$ less appeared $= 5 x - 30$

$20$ less passed $= 4 x - 20$

According to question

$\frac{5 x - 30}{4 x - 20} = \frac{4}{1} 16 x - 80 = 5 x - 30 11 x = - 30 + 80 11 x = 50 x = \frac{50}{11}$
total students= 23