Question

In an examination, Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali. After a review, the score of each of them increased by 6. The revised scores of Anjali, Mohan, and Rama were in the ratio 11:10:3. Then Anjali's score exceeded Rama's score by

• A. 24
• B. 26
• C. 32
• D. 35

Answer 1

The Correct Option is C

Let the original scores of Anjali, Mohan, and Rama be denoted by \( A \), \( M \), and \( R \) respectively. According to the problem, Rama's score was one-twelfth of the sum of Mohan's and Anjali's scores. Therefore, we can write the equation:

\[ R = \frac{1}{12}(M + A) \]

After the review, each score increased by 6. Therefore, the revised scores are \( A + 6 \), \( M + 6 \), and \( R + 6 \). Given that these scores are in the ratio 11:10:3, we can express this as:

\[ \frac{A+6}{11} = \frac{M+6}{10} = \frac{R+6}{3} = k \]

From these equations, we can deduce:

\[ A + 6 = 11k, \quad M + 6 = 10k, \quad R + 6 = 3k \]

Subtracting 6 from both sides yields:

\[ A = 11k - 6, \quad M = 10k - 6, \quad R = 3k - 6 \]

Substituting back into the original condition:

\[ 3k - 6 = \frac{1}{12}(10k - 6 + 11k - 6) \]

Simplify the equation:

\[ 3k - 6 = \frac{21k - 12}{12} \]

Cross-multiplying gives:

\[ 36k - 72 = 21k - 12 \]

Solving for \( k \):

\[ 36k - 21k = 72 - 12 \]

\[ 15k = 60 \]

\[ k = 4 \]

Using the value of \( k \), calculate Anjali's revised score:

\[ A + 6 = 11 \times 4 = 44 \]

Thus, \( A = 44 - 6 = 38 \).

Calculate Rama's revised score:

\[ R + 6 = 3 \times 4 = 12 \]

Thus, \( R = 12 - 6 = 6 \).

The difference between Anjali's and Rama's scores is:

\[ A - R = 38 - 6 = 32 \]

Hence, Anjali's score exceeded Rama's score by 32.