Question

Two students appeared for an examination. One of them got 9 marks more than the other. His marks was also equal to 56% of the sum of their marks. What are their marks?

• A. 39, 30
• B. 42, 33
• C. 43, 34
• D. 45, 54

Answer 1

Answer:

B. 42, 33

marks are A and A+9

take

$A+9=B$ ---------------- (1)

$(A+B)\times \frac{56}{100}=B$

$56A+56B=100B$

$56A=44B$-----------------(2)

put (1) in (2)

$56A=44\times(A+9)$

$56A=44A+(44\times9$)

$56A-44A=44\times9$

$12A=44\times9$

$A=\frac{44\times9}{12}$

$A=33$

$B=33+9=42$