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

Answer: (D)

77

Answer 2

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 aand d. The total number of students from Class I to IV is:

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

We are given S 4 = 462.

4 a + 6 d = 462

Simplify:

2 a + 3 d = 231 (1)

The total students from Class I to V is:

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

The total students from Class VI to X is:

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

We are given S 5 = 2 S 5-10:

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

Simplify:

5 a + 10 d = 10 a + 70 d

5 a = −60 d = =$>$ a = −12 d (2)

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

2(−12 d) + 3 d = 231

−24 d + 3 d = 231 = =$>$ −21 d = 231 = =$>$ d = −11

Substitute d = −11 into a = −12 d :

a = −12(−11) = 132

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

a + 5 d = 132 + 5(−11) = 132 − 55 = 88

Answer: 88