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Question

Difficulty: HardDirect Proportions

A specialized construction 33 D printer uses 2.42.4 metric tons of concrete to print 1515 linear feet of a wall. At this constant rate, how many kilograms of concrete are required to print a wall that is 2525 yards long? (11 metric ton = 1,0001,000 kilograms; 11 yard = 33 feet)

  1. A
    4,0004,000
  2. B
    1,2001,200
  3. 12,00012,000Answer
  4. D
    480480

Answer

The correct answer is 12,00012,000.
The correct answer is found by harmonizing the length units, determining the scaled proportion, and harmonizing the final weight units. Converting 2525 yards to 7575 feet reveals that the target length is 55 times the initial length (75÷15=575 \div 15 = 5). Multiplying the initial concrete requirement by this scale factor yields 1212 metric tons (2.4×5=122.4 \times 5 = 12). Finally, converting 1212 metric tons to kilograms yields 12,00012,000 kilograms.

Step-by-Step Solution

1
Convert the target length from yards to feet.
25 yards=25×3=75 feet25 \text{ yards} = 25 \times 3 = 75 \text{ feet}.
The units of length must match the given rate (1515 linear feet) before setting up a proportion.
2
Set up a direct proportion to find the metric tons of concrete needed.
2.415=x75\frac{2.4}{15} = \frac{x}{75}
The amount of concrete used is directly proportional to the length of the wall printed.
3
Solve the proportion for xx.
x=2.4×7515=2.4×5=12 metric tonsx = \frac{2.4 \times 75}{15} = 2.4 \times 5 = 12 \text{ metric tons}.
Isolating xx provides the required amount of concrete in metric tons.
4
Convert metric tons to kilograms.
12 metric tons=12×1,000=12,000 kilograms12 \text{ metric tons} = 12 \times 1,000 = 12,000 \text{ kilograms}.
The question specifically asks for the final answer in kilograms.

Key Concept

Direct Proportions with Unit Conversions
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Topics

Direct Proportions

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