The students in three classes are in the ratio 2 : 3 : 5. After 20 students are added to each class, the ratio becomes 4 : 5 : 7. What was the total number of students before the increase?

Introduction / Context:
When equal numbers are added to each term of a ratio trio, the overall ratio changes in a predictable way. Setting the original classes as multiples of a common factor allows solving with simple equations using two of the ratio comparisons.

Given Data / Assumptions:

• Original classes = 2k, 3k, 5k
• After adding 20 each: (2k + 20) : (3k + 20) : (5k + 20) = 4 : 5 : 7

Concept / Approach:
Form a proportion from any two adjacent classes and solve for k. Then compute the original total as 2k + 3k + 5k = 10k.

Step-by-Step Solution:
(2k + 20)/(3k + 20) = 4/55(2k + 20) = 4(3k + 20) ⇒ 10k + 100 = 12k + 80 ⇒ 2k = 20 ⇒ k = 10Original total = 10k = 100

Verification / Alternative check:
Check second proportion: (3k + 20)/(5k + 20) with k = 10 ⇒ (50)/(70) = 5/7, confirming the final ratio.

Why Other Options Are Wrong:
10 and 90 are too small; “None of these” is invalid because 100 satisfies both ratio conditions.

Common Pitfalls:
Adding 20 to the ratio numbers directly; ratios are not absolute counts. Always return to the k-based actual numbers.

Final Answer:
100