JIPMAT 2024 (QA) - A candidate scoring x \% marks in an examination fails by ' a ' marks while another candidate who scores y \% marks, gets ' b ' marks more than the minimum required pass marks. What is the maximum marks for the examination ? | PYQs + Solutions

JIPMAT 2024

Arithmetic

Percentages

Medium

A candidate scoring $x \%$ marks in an examination fails by ' $a$ ' marks while another candidate who scores $y \%$ marks, gets ' $b$ ' marks more than the minimum required pass marks. What is the maximum marks for the examination ?

\frac{100(a+b)}{(x-y)}

\frac{100(a-b)}{(x+y)}

\frac{100(a+b)}{(y-x)}

\frac{100(a-b)}{(x-y)}

Correct Option: 3

1) Let's say maximum marks is

M

\newline

- Then

x\%

M = \frac{xM}{100}

2) Let pass marks be

P

3) For first candidate:

\newline

- Score =

\frac{xM}{100}

\newline

- Fails by

a

marks

\newline

- Therefore:

\frac{xM}{100} = P - a

4) For second candidate:

\newline

- Score =

\frac{yM}{100}

\newline

- Exceeds by

b

marks

\newline

- Therefore:

\frac{yM}{100} = P + b

5) From equations:

\newline

\frac{xM}{100} = P - a

...(1)

\frac{yM}{100} = P + b

...(2)

6) Subtracting (1) from (2):

\frac{(y-x)M}{100} = a+b

M = \frac{100(a+b)}{y-x}

Therefore, maximum marks =

\frac{100(a+b)}{y-x}