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Difficulty: HardRatios, Rates, and Proportions

An agricultural irrigation system draws water from a main tank at a constant rate to supply three fields: Field X, Field Y, and Field Z. The ratio of the water volume distributed daily to Field X and Field Y is 3:43:4, and the ratio of the water volume distributed daily to Field Y and Field Z is 3:53:5. If Field Z receives 2,2002,200 gallons more water daily than Field X, what is the total number of gallons of water distributed daily to all three fields combined?

  1. A
    4,0004,000
  2. B
    5,8005,800
  3. 8,2008,200Answer
  4. D
    13,20013,200
  5. E
    11,27511,275

Answer

8,2008,200 gallons
The correct answer is 8,2008,200 gallons. By finding a common multiple for the Field Y term in the two ratios, we align the proportions to get a joint ratio of 9:12:209:12:20 for Field X, Field Y, and Field Z. Let the volumes be 9k9k, 12k12k, and 20k20k. The difference between Field Z and Field X is 20k−9k=11k=2,20020k - 9k = 11k = 2,200 gallons, giving k=200k = 200. The combined volume is 9k+12k+20k=41k=41×200=8,2009k + 12k + 20k = 41k = 41 \times 200 = 8,200 gallons.

Step-by-Step Solution

1
Find a common unit of comparison for the ratio of water distributed to all three fields.
The joint ratio is 9:12:209:12:20.
We are given Field X to Field Y is 3:43:4 and Field Y to Field Z is 3:53:5. To combine these ratios, we find the least common multiple of the parts representing Field Y, which is LCM(4,3)=12\text{LCM}(4, 3) = 12. Multiplying the first ratio by 3 gives 9:129:12, and multiplying the second ratio by 4 gives 12:2012:20.
2
Set up an equation using the given difference between the water volume of Field Z and Field X.
k=200k = 200
Let the daily volumes of Field X, Field Y, and Field Z be represented by 9k9k, 12k12k, and 20k20k respectively. The difference between Field Z and Field X is 20k−9k=11k20k - 9k = 11k. We are given that this difference is 2,2002,200 gallons, so 11k=2,20011k = 2,200, which simplifies to k=200k = 200.
3
Calculate the combined daily volume for all three fields.
8,2008,200 gallons
The total volume distributed to all three fields is 9k+12k+20k=41k9k + 12k + 20k = 41k. Substituting k=200k = 200 gives 41×200=8,20041 \times 200 = 8,200 gallons.

Key Concept

Solving multi-part ratio word problems by establishing a common ratio and setting up algebraic equations.
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Topics

Ratios, Rates, and Proportions

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