Examkin

Question

Difficulty: MediumRatios and Proportions

A civil engineering project requires a specific concrete mixture consisting of cement, sand, and gravel in a ratio of 2:3:62:3:6 by weight. If a batch of this concrete mixture weighs a total of 6,600 lbs6,600\text{ lbs}, how many more pounds of gravel are used than cement?

  1. A
    1,8001,800
  2. 2,4002,400Answer
  3. C
    3,3003,300
  4. D
    3,6003,600

Answer

The concrete mixture requires 2,4002,400 more pounds of gravel than cement.
The entire mixture consists of 1111 equal parts (22 cement +3+ 3 sand +6+ 6 gravel). Since the total mixture weighs 6,600 lbs6,600\text{ lbs}, one part weighs 600 lbs600\text{ lbs} (6,600÷116,600 \div 11). The difference between the amount of gravel (66 parts) and cement (22 parts) is 44 parts. Multiplying this difference by the weight of one part yields 4×600=2,400 lbs4 \times 600 = 2,400\text{ lbs}.

Step-by-Step Solution

1
Find the total number of parts in the ratio.
2+3+6=11 parts2 + 3 + 6 = 11\text{ parts}.
The total given weight of the concrete mixture must be distributed across all the ratio parts combined.
2
Calculate the weight of a single proportional part.
6,600÷11=600 lbs6,600 \div 11 = 600\text{ lbs}.
Dividing the total mixture weight by the total number of parts determines the exact value of one unit.
3
Determine the difference in parts between the gravel and the cement.
6−2=4 parts6 - 2 = 4\text{ parts}.
The question asks 'how many more pounds' of gravel are used than cement, which corresponds to the difference in their ratio shares.
4
Calculate the final difference in weight.
4×600=2,400 lbs4 \times 600 = 2,400\text{ lbs}.
Multiplying the part difference by the weight per part gives the final amount.

Key Concept

Solving part-to-whole ratio scenarios and calculating the difference between specific components.
Estimated Time:1m 0s
Rate this question

Topics

Ratios and Proportions

More questions from this topic