Question

In a school, the number of students in each class, from Class I to X, in that order, are in an arithmetic progression. The total number of students from Class I to V is twice the total number of students from Class VI to X.
If the total number of students from Class I to IV is 462, how many students are there in Class VI?

• A. 93
• B. . 88
• C. 83
• D. 77
• E. None of the remaining options is correct

Answer 1

The Correct Option is D

Let the number of students in Class I be a, and the common difference of the arithmetic progression be d.

Step 1: Express the total students in terms of a and d. The total number of students from Class I to IV is:

S₄ = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

We are given S₄ = 462.

4a + 6d = 462

Simplify:

2a + 3d = 231 (1)

The total students from Class I to V is:

S₅ = a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d) = 5a + 10d

The total students from Class VI to X is:

S_5-10 = (a + 5d) + (a + 6d) + (a + 7d) + (a + 8d) + (a + 9d) = 5a + 35d

We are given S₅ = 2S_5-10:

5a + 10d = 2(5a + 35d)

Simplify:

5a + 10d = 10a + 70d

5a = −60d = =\(>\) a = −12d (2)

Step 2: Solve the equations. Substitute a = −12d from (2) into (1):

2(−12d) + 3d = 231

−24d + 3d = 231 = =\(>\) −21d = 231 = =\(>\) d = −11

Substitute d = −11 into a = −12d:

a = −12(−11) = 132

Step 3: Find the number of students in Class VI. The number of students in Class VI is:

a + 5d = 132 + 5(−11) = 132 − 55 = 88

Answer: 88