Question

on an examination , the ratio of passes to failures was 4:1. had 30 losses appeared and 20 less passed the ratio of passes to failures would have been 5:1 . find the no of students appeared for the examination​

Answer 1

Answer :

›»› The number of students appeared for the examination are 150 students.

Given :

• On an examination, the ratio of passes to failures was 4:1. Had 30 losses appeared and 20 less passed the ratio of passes to failures would have been 5:1.

To Find :

• The number of students appeared for the examination.

Solution :

Let us assume that, the number of passes students is "4x" and the number of failure students is "x" respectively.

The total number of students appeared for the examination,

→ 4x + x

→ 5x

In the second case, number of students appeared for the examination,

→ 5x - 30

And number of passes students appeared for the examination,

→ 4x - 20

Number of failure students :

→ (5x - 30) - (4x - 20)

→ 5x - 30 - 4x + 20

→ (5 - 4)x + (-30 + 20)

→ 1x + (-30 + 20)

→ x + (-30 + 20)

→ x + (-10)

→ x - 10

According to the given condition,

New ratio of the number of students passed to fail students is 5:1.

→ 4x - 20/x - 10 = 5/1

→ 5x - 50 = 4x - 20

→ 5x - 4x = -20 + 50

→ x = -20 + 50

→ x = 30

Therefore,

The total number of students appeared for the examination.

→ 5x

→ 5 * 30

→ 150

Hence the number of students appeared for the examination are 150 students.

Answer 2

Answer

$\:\:$

Given

$\:\:$

• The ratio of passes to failures was 4:1

$\:\:$

• If 30 less appeared and 20 less passed the ratio of passes to failures would have been 5:1

Find

$\:\:$

• No. of students appeared for the examination

$\:\:$

Solution

$\:\:$

• Let "x" candidates have passed

• Let "y" candidates have failed

$\:\:$

$\underline{\bold{\texttt{Original Condition :}}}$

$\:\:$

$\bf \dag \ \ \dfrac { x } { y } = \dfrac { 4 } { 1 }$

$\:\:$

$\sf \longmapsto x = 4y$ ------(1)

$\:\:$

$\underline{\bold{\texttt{New condition :}}}$

$\:\:$

• Total number of students = x + y - 30

• Number of students passed = x - 20

• Number of students failed = x + y - 30 - (x - 20) = y - 10

$\:\:$

$\underline{\bold{\texttt{As per new condition -}}}$

$\:\:$

$\bf \dag \ \ \dfrac { (x-20) } { (y-10) } = \dfrac { 5 } { 1 }$

$\:\:$

$\sf \longmapsto x - 20 = 5y - 50$

$\:\:$

$\sf \longmapsto 5y - x = 30$ ------(2)

$\:\:$

$\underline{\bold{\texttt{Putting the value of "x" from (1) to (2) }}}$

$\:\:$

$\sf \longmapsto 5y - 4y = 30$

$\:\:$

$\purple\longrightarrow$ $\bf y = 30$

$\:\:$

$\rm \dashrightarrow x = 4 \times 30$

$\:\:$

$\purple\longrightarrow$ $\bf x = 120$

$\:\:$

• Hence, number of students appeared for the exam is 120 + 30 = 150