Question

Ram obtain 40% of the marks in a paper of 200 marks. Shyam is ahead of ram by 25% of Ram's marks, while bhuvan is ahead of shyam by one ninth of his own marks. How many marks does bhuvan got?

Answer 1

Ram secured 40 % of 200 marks
(i.e) 40*200/100 = 80 marksShyam’s Mark is 25 % more than Ram(i.e) 80+(80*25)/100= 100 MarksLet Bhuvan’s Mark be XX=Shyam’s Mark + X/9X-X/9=100
8X/9=100
8X=900
X=900/8 = 112.5 Ans: Bhuvan got 112.5 Marks

Answer 2

Marks obtained by Bhuvan are 112.5

GIVEN

Maximum marks = 200

Marks obtained by Ram = 40% of 200.

Marks of shyaam = 25% more than marks of Ram.

Marks of Bhuvan = marks of shyaam + (1/9)th of marks of Bhuvan.

TO FIND

Marks of Bhuvan.

SOLUTION

We can simply solve the above problem as follows-

It is given

Marks of Ram = 25% of The total marks

Marks of Ram =

$\frac{40}{100} \times 200$

= 80.

Marks of Shyaam

It is given,

Marks of shyaam are 25% more than the marks of Ram.

if we take,

marks of Ram = 100%

Marks of shyaam = 100%+25% = 125%.

Let the marks of shyaam be M

Now if take the quotient of both the marks we get.

$\frac{125}{100} = \frac{M}{80}$

Solving for M,

$M = \frac{125 \times 80}{100}$

M = 100.

Therefore, Marks of shyaam = 100.

Marks Of Bhuvan.

Number of marks obtained by Bhuvan be, B.

It is given,

Marks obtained by Bhuvan = marks of Shyaam + (1/9)th of marks of bhuvan.

Putting the values,

$B = 100 + \frac{B}{9}$

simplifying the above equation we get,

$B - \frac{B}{9} = 100$

This can be rewritten as -

$\frac{9B -B }{9} = 100$

8B = 100× 9

8B = 900

B = 900/8

B = 112.5

Hence, Marks obtained by Bhuvan are 112.5

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