# Difficult Percentages Solved QuestionAptitude Discussion

 Q. A candidate who gets $20\%$ marks fails by 10 marks but another candidate who gets $42\%$ marks gets $12\%$ more than the passing marks. Find the maximum marks.
 ✖ A. 50 ✔ B. 100 ✖ C. 150 ✖ D. 200

Solution:
Option(B) is correct

From the given statement pass percentage is $42\% - 12\% = 30\%$

By hypothesis, $30\%$ of $x – 20\%$ of $x = 10$ (marks)

i.e., $10\%$ of $x = 10$

Therefore,  $x$ = 100 marks.

Edit: Thank you, Suzie, for explaining in detail in the comments.

Edit 2: For yet another alternative solution, check comment by PULOK KHAN.

## (11) Comment(s)

Shiva
()

Total marks = t

Passing marks=p

Score of first candidate be x

Score of second candidate be y

So, x + 10 = p......(1)

Also x=20%of t

t=5x

Now, y - 12%p=p.......(2)

Also y=42%t

y=(42/100)×(5x) as t=5x

y=2.1x

Subsitute in ......(2)

We get, 2.1x-.12p=p

x=(11.2/21)p

Substitute in.....(1)

(11.2/21)p +10= p

p= 2100/98 = 55.8

#essjay

Pvaish
()

Let maximum marks =x

From the question

20%of x + 10=42%of x - 12%of x

=> 20x + 1000= 42x- 30x

=> x= 100

Akshay
()

There is a flaw in the question. It has to be 12 marks more than the passing marks not 12% more than the passing marks. What do u think..??

Xyz
()

If I take 'T' as max marks and 'x' as passing marks ...

The equation that I am getting is -

0.2T + 10 = 0.42T - 0.12( 0.2T +10 )

as it is given in the question that 42% is 12% more than PASSING MARKS and not TOTAL marks...

n here I get stuck

Poonam
()

Let passmark be X

Let maximum mark be Y

.20Y = X-10 ------- (1)

.42Y = X+.12Y

.42Y-.12Y=X

.30Y=X ---------- (2)

Substitute X of (2) equation with (1)

.20Y = .30Y-10

.10Y = 10

Y= 100 (Maximum mark)

.30X-10=

PULOK KHAN
()

Difference between percentage $(42\%-20\%)=22\%$

Total mark difference between these two $(10+12)=22$

so, $22\%=22$

$\Rightarrow 100\%=100$

Kumar
()

It is 12% and not 12 marks!

SundararajC
()

i am getting different answer as follows.

0.2max=passmark-10

0.4max=passmark+0.12passmark

to solve above equations i got answer maximum mark around 58.can u explain what i did wrong?

Priya
()

I believe in your second equation you need to change last term to $0.12 \times \text{ max marks}$ .

Gaur
()

Suzie
()

Let's say total number of marks $=x$

Now, as per the question, candidate who gets 42% marks gets 12% more than the passing marks.

So, $30$ out of $x$ is passing marks, i.e. Passing Percentage

$=42\%-12\%$

$=30\%$

Now, again from the question a candidate who gets 20% marks fails by 10 marks.

Thus,

$\text{20% marks}+10=\text{passing % marks}$

$\Rightarrow \left(\dfrac{20}{100}\times x \right)+10=\left(\dfrac{30}{100}\times x \right)$

$\Rightarrow 10=\left(\dfrac{30-20}{100}\times x \right)$

$\Rightarrow x=100$

Hope this helps.