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

To solve this problem, we will break it down using the properties of an arithmetic progression (AP).

Let's denote the number of students in Class I as \(a\) and the common difference of the AP as \(d\). The sequence for the number of students will be:

• Class I: \(a\)
• Class II: \(a + d\)
• Class III: \(a + 2d\)
• Class IV: \(a + 3d\)
• Class V: \(a + 4d\)
• Class VI: \(a + 5d\)
• Class VII: \(a + 6d\)
• Class VIII: \(a + 7d\)
• Class IX: \(a + 8d\)
• Class X: \(a + 9d\)

The total number of students from Class I to IV is given as 462:

\(a + (a + d) + (a + 2d) + (a + 3d) = 462\)

Simplifying this gives:

\(4a + 6d = 462\)

\(2a + 3d = 231 \quad \text{(Equation 1)}\)

The problem states the sum of students from Class I to V is twice the sum from Class VI to X:

\((a + (a + d) + (a + 2d) + (a + 3d) + (a + 4d)) = 2((a + 5d) + (a + 6d) + (a + 7d) + (a + 8d) + (a + 9d))\)

Simplifying, we get:

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

\(5a + 10d = 10a + 70d\)

\(5a + 60d = 0\)

\(a + 12d = 0 \quad \text{(Equation 2)}\)

By solving Equation 1 and Equation 2, we can find values for \(a\) and \(d\).

Substitute \(a = -12d\) from Equation 2 into Equation 1:

\(2(-12d) + 3d = 231\)

\(-24d + 3d = 231\)

\(-21d = 231\)

\(d = -11\)

Using the value of \(d\), find \(a\):

\(a = -12 \times (-11) = 132\)

Now, calculate the number of students in Class VI:

\(a + 5d = 132 + 5 \times (-11) = 132 - 55 = 77\)

Therefore, the number of students in Class VI is 77.

The correct answer is: 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 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 = 77

Answer: 77