3.16 Miscellaneous Solved Examples (Combined Concepts)

These three examples deliberately combine two or more ideas from this chapter, exactly the way SSC CGL Tier-II and RRB NTPC CBT-2 questions are typically framed — as a check on whether you can chain concepts rather than apply a single formula in isolation.

Example 1 (Successive change + reverse percentage): After a 20% increase followed by a 25% decrease, the price of an article is ₹960. Find the original price. Net multiplying factor = 1.20 × 0.75 = 0.90 (a 10% net decrease, which also follows from the successive formula: 20 − 25 + (20×−25)/100 = −5 −5 = −10%). Original price = 960 ÷ 0.90 = ₹1,066.67 (approximately ₹1,066.67, or exactly ₹9,600/9).

Example 2 (Ratio + percentage + population-style compounding): The number of men and women in a town is in the ratio 5 : 4. If the number of men increases by 10% and the number of women increases by 20% next year, find the percentage increase in the total population, given the town’s current population is 9,000. Men = 5/9 × 9,000 = 5,000; Women = 4/9 × 9,000 = 4,000. New men = 5,000 × 1.10 = 5,500; new women = 4,000 × 1.20 = 4,800. New total = 5,500 + 4,800 = 10,300. Increase = 10,300 − 9,000 = 1,300. % increase = (1,300/9,000) × 100 = 14.44% (approximately).

Example 3 (Election + percentage error style trap): In an election with two candidates, the winner got 56% of the valid votes and won by 2,400 votes. If 12% of the total votes polled were invalid, find the total number of votes polled (i.e., total voters). Winner − Loser (as % of valid votes) = 56% − 44% = 12% of valid votes = 2,400. Valid votes = 2,400 × (100/12) = 20,000. Since valid votes = 88% of total voters (as 12% were invalid): Total voters = 20,000 ÷ 0.88 = 22,727 (approximately). This example is a common trap because the “12%” appears twice with two completely different meanings — once as the vote-share gap between candidates, and once as the invalid-vote percentage — and confusing the two leads directly to a wrong setup.

Example 4 (Successive change + depreciation + reverse percentage, three ideas in one problem): A machine depreciates every year at the rate of 10% of its value at the beginning of that year. It was purchased for a certain price, and after two years of depreciation, its value is ₹16,200. In the same two-year period, a second machine, purchased at the same original price, appreciated in value by 10% in the first year and then depreciated by 10% in the second year. Find the value of the second machine at the end of two years. First, find the original price using the first machine: Original × (0.9)² = 16,200 → Original × 0.81 = 16,200 → Original = 16,200/0.81 = ₹20,000. Now apply the second machine’s changes to this same original price: a = +10, b = −10 → net change = 10 − 10 + (10×−10)/100 = −1%. Value of second machine = 20,000 × 0.99 = ₹19,800. This question is designed to test whether you can use one part of a problem (reverse percentage on the first machine) purely to unlock a number needed for an unrelated second calculation — a structure increasingly common in SSC CGL Tier-II and RRB NTPC CBT-2 percentage questions.