AMC 8 Daily Practice Round 8

Complete problem set with solutions and individual problem pages

Problem 30 Hard

In a certain place, there are two ponds, Pond A and Pond B. Due to planned construction work, the water in both ponds must be pumped out, even during the rainy season. Pond A has an area of 2 acres. Using 3 water pumps, it can be emptied in 2 days; if 2 water pumps are used instead, it will take 4 days to empty it. Pond B has an area 3 times that of Pond A. If it is to be emptied in 6 days, how many water pumps will be needed?

  • A.

    1

  • B.

    2

  • C.

    3

  • D.

    5

  • E.

    10

Answer:D

Let the original amount of water in Pond A be v, and let the daily rainfall into Pond A be x.

Then \frac{v + 2x}{3 \times 2} \quad \text{and} \quad \frac{v + 4x}{2 \times 4} both represent the amount of water a single pump can remove from Pond A in one day.

That is: \frac{v + 2x}{6} = \frac{v + 4x}{8} which gives v = 4x, so x = \frac{v}{4}.

Therefore, the amount of water one pump can remove from Pond A in one day is: \frac{v + 2x}{6} = \frac{v + \frac{v}{2}}{6} = \frac{v}{4}.

From this and the above, we know that the rainfall into Pond A in one day is exactly the amount one pump can remove in one day.

Since Pond B’s area is 3 times that of Pond A, removing Pond B’s daily rainfall requires 3 pumps.

The original amount of water in Pond B is 3v.

Since one pump can remove \frac{v}{4} from Pond A’s original water in one day, it would take: \frac{3v}{\frac{v}{4}} = 12 \ \text{days}for one pump to remove Pond B’s original water.

If we want to remove it in 6 days, we need 2 pumps.

Therefore, to remove both the daily rainfall and the original water from Pond B in 6 days, we need: 3 + 2 = 5 pumps.

 

Arithmetic method

(1) For Pond A (including rainfall), using 3 pumps takes 2 days to empty it; to empty it in 1 day requires 6 pumps.

Similarly, using 2 pumps takes 4 days to empty it; to empty it in 1 day requires 8 pumps.

The extra 2 days mean 2 extra days of rainfall, which requires 8 - 6 = 2 extra pumps.

Therefore, the daily rainfall into Pond A requires 1 pump to remove.

If we remove only the original water in Pond A, using 3 - 1 = 2 pumps takes 2 days; using 1 pump takes 4 days.

Thus, each pump removes \frac14 of Pond A’s original water per day.

(2) Since Pond B is 3 times the size of Pond A, removing its daily rainfall requires 3 pumps.

Also, one pump removes \frac14 \div 3 = \frac{1}{12} of Pond B’s original water per day.

To remove it in 6 days, we need: \frac{1/6}{1/12} = 2 pumps.

Therefore, to remove both the rainfall and the original water from Pond B in 6 days requires: 3 + 2 = 5 pumps.

Answer: \text{D}.