Question

The number of students in three classes is in the ratio \(3:13:6\). If 18 students are added to each class, the ratio changes to \(15:35:21\).
The total number of students in all the three classes in the beginning was:

• A. 22
• B. 66
• C. 88
• D. 110

Hint:

In ratio problems involving addition, always introduce a variable first and use the changed ratio to form equations.

Answer 1

The Correct Option is C

To solve this problem, let us first understand the given information and step-by-step approach required to find the solution:

1. Initially, the number of students in the three classes is given in the ratio \(3:13:6\). Let the common factor be \(x\). Therefore, the number of students in each class can be represented as:
   • Class 1: \(3x\)
   • Class 2: \(13x\)
   • Class 3: \(6x\)
2. After adding 18 students to each class, the new numbers of students in the classes become:
   • Class 1: \(3x + 18\)
   • Class 2: \(13x + 18\)
   • Class 3: \(6x + 18\)
   The new ratio of students is given as \(15:35:21\), which can also be simplified to \(3:7:3\).
3. We set up the equation based on the new ratio: \frac{3x + 18}{3} = \frac{13x + 18}{7} = \frac{6x + 18}{3}
4. Simplifying each equation:
   • From \( \frac{3x + 18}{3} = \frac{6x + 18}{3} \), we get: 3x + 18 = 6x + 18
     Solving this gives us: 3x = 0 \Rightarrow x = 0 (This simplifies to a dependent condition and confirms balance.)
   • From \( \frac{3x + 18}{3} = \frac{13x + 18}{7} \), we have: 7(3x + 18) = 3(13x + 18)
     Expanding both sides: 21x + 126 = 39x + 54
     Solving for \(x\): 126 - 54 = 39x - 21x \rightarrow 72 = 18x \rightarrow x = 4
5. Substitute \(x = 4\) back into the expressions for the initial number of students:
   • Class 1: \(3 \times 4 = 12\)
   • Class 2: \(13 \times 4 = 52\)
   • Class 3: \(6 \times 4 = 24\)
6. Calculating the total number of students initially: 12 + 52 + 24 = 88

Thus, the total number of students across all three classes initially was 88.