Understanding the Exam Score Problem

This question asks us to determine how many individuals out of three (Ram and his two friends) scored more than 70% in an exam, given the ratio of their obtained scores and the relationship between the sum of their scores and the sum of the maximum marks.

Setting up the Problem with Ratios

We are given that the ratio of the obtained marks for the three individuals is 4 ∶ 5 ∶ 6. Let the obtained scores be $S_1$, $S_2$, and $S_3$ for Ram, his first friend, and his second friend, respectively. Based on the ratio, we can represent their scores using a common multiplier, say $x$.

• Ram's score ($S_1$) = $4x$
• Friend 1's score ($S_2$) = $5x$
• Friend 2's score ($S_3$) = $6x$

The sum of their obtained scores is $S_1 + S_2 + S_3 = 4x + 5x + 6x = 15x$.

Relating Total Scores to Maximum Marks

Let's assume the maximum marks for the exam are the same for each of the three individuals. Let this maximum mark be $M$. The total maximum marks for all three individuals would then be $M + M + M = 3M$.

According to the problem, the sum of the obtained scores is 60% of the sum of the maximum marks for the exam for all three of them.

Mathematically, this can be written as:

\( \text{Sum of obtained scores} = 60\% \text{ of } (\text{Sum of maximum marks}) \)

\( 15x = 0.60 \times (3M) \)

\( 15x = 1.8M \)

Now, we can solve for $x$ in terms of $M$:

\( x = \frac{1.8M}{15} = \frac{18M}{150} = \frac{3M}{25} \)

Calculating Individual Scores in terms of Maximum Marks

Now we can substitute the value of $x$ back into the expressions for individual scores:

• Ram's score ($S_1$) = \( 4x = 4 \times \frac{3M}{25} = \frac{12M}{25} \)
• Friend 1's score ($S_2$) = \( 5x = 5 \times \frac{3M}{25} = \frac{15M}{25} \)
• Friend 2's score ($S_3$) = \( 6x = 6 \times \frac{3M}{25} = \frac{18M}{25} \)

Determining the Percentage Scores

To find the percentage score for each individual, we divide their obtained score by the maximum marks ($M$) and multiply by 100%.

• Ram's percentage = \( \left( \frac{S_1}{M} \right) \times 100\% = \left( \frac{12M/25}{M} \right) \times 100\% = \left( \frac{12}{25} \right) \times 100\% \)

\( \left( \frac{12}{25} \right) \times 100\% = 12 \times 4\% = 48\% \)

• Friend 1's percentage = \( \left( \frac{S_2}{M} \right) \times 100\% = \left( \frac{15M/25}{M} \right) \times 100\% = \left( \frac{15}{25} \right) \times 100\% \)

\( \left( \frac{15}{25} \right) \times 100\% = \left( \frac{3}{5} \right) \times 100\% = 0.6 \times 100\% = 60\% \)

• Friend 2's percentage = \( \left( \frac{S_3}{M} \right) \times 100\% = \left( \frac{18M/25}{M} \right) \times 100\% = \left( \frac{18}{25} \right) \times 100\% \)

\( \left( \frac{18}{25} \right) \times 100\% = 18 \times 4\% = 72\% \)

Counting Those Who Scored More Than 70%

Now we compare each individual's percentage score to 70%:

• Ram: 48% (Not > 70%)
• Friend 1: 60% (Not > 70%)
• Friend 2: 72% (Yes, > 70%)

Only one person (Friend 2) scored more than 70% marks in the exam.

Therefore, exactly one of them scored more than 70%.

Revision Table: Key Concepts in Exam Score Problems

Additional Information: Solving Ratio and Percentage Problems

Problems involving both ratios and percentages are common in competitive exams. The key is to use the ratio to represent individual quantities in terms of a common variable ($x$ in this case) and then use the percentage information to form an equation. Once the equation is solved, you can find the specific values or percentages required.

In this problem, the percentage related the sum of obtained scores to the sum of maximum marks. If the percentage had been given for individual scores (e.g., Ram scored 60%), the approach would be slightly different, perhaps directly relating $4x$ to 60% of $M$. Assuming uniform maximum marks for everyone is often a reasonable step in such problems unless stated otherwise.