If 50 one-cent coins were stacked on top of each other in a column, the column would be approximately 3 (7/8) inches tall. At this rate, which of the following is closest to the number of one-cent coins it would take to make an 8-inch-tall column?
Correct answer: B. 100
- A. 75
- B. 100
- C. 200
- D. 390
Explanation
Choice B is correct. A column of 50 stacked one-cent coins is about 3(7/8) inches tall, which is slightly less than 4 inches tall. Therefore a column of stacked one-cent coins that is 4 inches tall would contain slightly more than 50 one-cent coins. It can then be reasoned that because 8 inches is twice 4 inches, a column of stacked one-cent coins that is 8 inches tall would contain slightly more than twice as many coins; that is, slightly more than 100 one-cent coins. An alternate approach is to set up a proportion comparing the column height to the number of one-cent coins, or 3(7/8) inches / 50 coins = 8 inches / x coins, where x is the number of coins in an 8-inch-tall column. Multiplying each side of the proportion by 50x gives 3(7/8) x = 400. Solving for x gives x = 400 x 8 / 31, which is approximately 103. Therefore, of the given choices, 100 is closest to the number of one-cent coins it would take to build an 8-inch-tall column.
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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.
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