In 2015 the populations of City X and City Y were equal. From 2010 to 2015, the population of City X increased by 20% and the population of City Y decreased by 10%. If the population of City X was 120,000 in 2010, what was the population of City Y in 2010?

Correct answer: C. 160,000

  • A. 60,000
  • B. 90,000
  • C. 160,000
  • D. 240,000

Explanation

To solve the problem, start with City X. It had a population of 120,000 in 2010, and with a 20% increase, its population in 2015 was 120,000 × 1.20 = 144,000. For City Y, let y represent the population in 2010. A 10% decrease means the 2015 population is y × 0.90. Since both populations are equal in 2015: y × 0.90 = 144,000. Solving this gives y = 144,000 / 0.90 = 160,000. Thus, City Y's population in 2010 was 160,000. The other options do not match this outcome when recalculated with the given percentage changes.

Last updated

About Quantitative Reasoning

Quantitative reasoning applies arithmetic and basic algebra to numerical relationships, including percentages, ratios, averages, fractions, proportions, profit and loss, rates, time, work and measurement. Questions focus on translating words into equations, choosing an efficient method and checking whether the result is numerically reasonable.

Practise Quantitative Reasoning

505 free Quantitative Reasoning MCQs from Analytical and Logical Reasoning, each with the correct answer and an explanation. Unlimited attempts, no account needed.

Exams that ask Analytical and Logical Reasoning questions like this

Analytical and Logical Reasoning is on 24 papers prepared for on TestUstad, and all of them draw the same bank, so this question is worth knowing for every one of them.

Related questions