Solving Mathematical Problems Using Algebraic Equations for Exam Scores

Mathematics is a subject that challenges students to think logically and solve problems in a structured manner. Today, we will walk through a series of problems involving exam scores and algebraic equations. These examples will help you understand how to use algebraic methods to solve real-world problems.

Problem 1: Exam Scores with a 6 Point Difference

Two students appeared in an exam. One of the students scored 6 marks more than the other, and the marks of the stronger student were 52% of the total sum of the marks of both the students. What were the marks scored by the weaker student?

Solution: Let the marks scored by the weaker student be x. Therefore, the marks scored by the stronger student will be x 6. According to the problem, the marks scored by the stronger student, which is 52% of the total marks, can be expressed as:

x 6 0.52(2x 6)

x 6 1.04x 3.12

Rearranging the equation to isolate x:

x - 1.04x 3.12 - 6

-0.04x -2.88

x -2.88 / -0.04 72

Therefore, the marks scored by the weaker student are 72.

Verification:

The weaker student's marks: x 72

The stronger student's marks: 72 6 78

Total marks: 72 78 150

52% of the total marks: 0.52 * 150 78

This confirms that the stronger student's score is indeed 78. Thus, the solution is verified.

The weaker student scored 72.

Problem 2: Solving for Exam Scores Proportional to 57%

Let the marks scored by the first student be X. Then, the marks scored by the second student is X/9. The sum of their marks is X X/9 2X/9. According to the problem, the marks scored by the first student, which is 57% of the total marks, can be expressed as:

X/9 57/100 * 2X/9

1/9 57/50 * X/9

1 - 57 5/54

54 - 2850 5

49 2850

X 2850/490 27.64 or 28

The marks scored by the first student are 28, and the marks scored by the second student are 28/9 37.

Problem 3: Direct 56% Proportion

Let the marks scored by student 1 be x. According to the given condition, student 2 scored 9 marks more, so the marks scored by student 2 will be x 9. According to the condition 2, x 9 is 56% of x.

x 9 56/100 * x

10 900 56x

44x 900

x 900/44 33

Therefore, student 1 scored 33 marks, and student 2 scored 42 marks (33 9).

Problem 4: Solving with a 60% Benchmark

Let the sum of the marks be x. The marks of one student are 60% of x, which is 3/5x. The marks of the second student will be x - 3/5x 2/5x. The difference between the marks of the two students is 10:

3/5x - 2/5x 10

x/5 10

x 50

The marks of the first student are 30 (3/5 * 50), and the marks of the second student are 20 (50 - 30).

Problem 5: Another 60% Benchmark Problem

Let the marks obtained by the one student be X. Therefore, the marks obtained by the other student will be X 10. The sum of their marks is X X 10 2X 10. According to the problem, 60% of the sum of their marks is equal to the marks of the second student:

X 10 0.6(2X 10)

X 10 1.2X 6

Rearranging the equation to isolate X:

X - 1.2X 6 - 10

-0.2X -4

X -4 / -0.2 20

Therefore, the marks obtained by the first student are 20, and the marks obtained by the second student are 30 (20 10).

The sum of their marks is 50.