Question

Two students appeared at an examination. One of them secured 9 marks more than the other and his marks was \(56%\) of the sum of their marks. The marks obtained by them are:

• A. 39, 30
• B. 41, 32
• C. 42, 33
• D. 43, 34

Hint:

For “one score is \(k%\) of the sum” problems, set scores as \(x\) and \(x\pm d\). Form the sum \(S\), equate the given score to \(\frac{k{100S\), solve, and finally verify numerically.

Answer 1

The Correct Option is C

Step 1 (Set variables).
Let the lower score be \(x\). Then the higher score is \(x+9\).
Sum \(S = x + (x+9) = 2x+9.\)
Step 2 (Translate the percentage statement).
“The higher score equals \(56%\) of the sum” \(\) \[ x+9 = 0.56\,(2x+9). \] Step 3 (Solve the linear equation).
\(x+9 = 1.12x + 5.04\)
\( 9-5.04 = 1.12x - x\)
\( 3.96 = 0.12x\)
\( x = \dfrac{3.96}{0.12} = 33.\)
Hence the two marks are \(x=33\) and \(x+9=42\). Step 4 (Verify against the condition).
Sum \(= 42+33=75\). \(56%\) of the sum \(= 0.56\times 75=42\) \(\) matches the higher mark. Correct.
Step 5 (Check options quickly).
Only option (c) lists \(42,33\); others fail either the “difference \(=9\)” or the “\(56%\) of sum” test. \[ \boxed{\text{42 and 33 (Option (c)}} \]