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The Eight Word Game

milagros317 said:
I recently solved a math puzzle which made me happy. 😀
Click to expand...
The puzzle and its solution are attached below.

A man named Zeke had a problem. He had a swimming pool that needed to be drained so that it could be renovated but he didn't own a water pump. Fortunately, he had three friends willing to come over to his backyard with their water pumps and drain his pool.

Albert, Bob, and Cassandra, the three friends, all came over and set up their pumps. All three pumps were started at the same moment. All three pumps pumped consistently, at the same rate, until the pool was drained.

When the job was done, each of the three friends made a statement and all three statements were true.

Albert said, "If only my pump ran twice as fast, while your two pumps were unchanged, then we would have had the pool drained 4 hours sooner."

Bob said, "If only my pump ran three times as fast, while your two pumps were unchanged, then we would have had the pool drained 16 hours sooner."

Cassandra said, "If only my pump ran four times as fast, while your two pumps were unchanged, then we would have had the pool drained 25 hours sooner."

How long did it actually take them to drain the pool?
Click to expand...
A solution is below.
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x
x

A solution:

Let x be the answer, the number of hours that it actually took to drain the pool.
Let a be the number of hours it would have taken Albert's pump to drain the pool, working alone.
Let b be the number of hours it would have taken Bob's pump to drain the pool, working alone.
Let c be the number of hours it would have taken Cassandra's pump to drain the pool, working alone.

Then we have:
1/a + 1/b + 1/c = 1/x (equation 1)

From Albert's statement we have:
2/a + 1/b + 1/c = 1/(x-4) (equation 2)

From Bob's statement we have:
1/a + 3/b + 1/c = 1/(x-16) (equation 3)

From Cassandra's statement we have:
1/a + 1/b + 4/c = 1/(x-25) (equation 4)

Equations 1 and 2 give us:
1/a = 1/(x-4) - 1/x (equation 5)

Equations 1 and 3 give us:
2/b = 1/(x-16) - 1/x (equation 6)

Equations 1 and 4 give us:
3/c = 1/(x-25) - 1/x (equation 7)

Now, multiply equation 1 by the constant 6 to obtain:
6/a + 6/b + 6/c = 6/x (equation 8)

Next, use equations 5, 6, and 7 to substitute their right hand sides into equation 8 to obtain:
6/(x-4) - 6/x + 3/(x-16) - 3/x + 2/(x-25) -2/x = 6/x (equation 9)

A little algebra on equation 9 yields:
0 = 17/x - 6/(x-4) - 3/(x-16) - 2/(x-25) (equation 10)

A lot of algebra now needs to be done on equation 10. We need to multiply both sides by the quartic polynomial x(x-4)(x-16)(x-25), then multiply all the addends out, and then collect terms. The result will be this cubic:

0 = 6x^3 - 392x^2 + 6,760x - 27,200 (equation 11)

Dividing by the greatest common factor of the coefficients, 2, we get:

0 = 3x^3 - 196x^2 + 3,380x - 13,600 (equation 12)

Remark: From the terms of the problem, we must have x>25 because it must have been possible for the job to have been finished in 25 fewer hours.

While it is, in general, tedious to solve a cubic equation, we can test for rational roots using the Rational Root Theorem for polynomial equations with integer coefficients. We are interested only in positive solutions (indeed, only solutions greater than 25). Thus, we can use the Rational Root Theorem to say that any positive rational root, R, has R=n/k where n is a positive factor of 13,600 and k=1 or k=3. There are quite a lot of factors of 13,600, namely 1, 2, 4, 5, 8, 10, 16, 17, 20, 25, 32, 34, 40, 50, 68, 80, 85, 100, 136, 160, 170, 200, 272, 340, 400, 425, 544, 680, 800, 850, 1360, 1700, 2720, 3400, 6800, and 13600.

Depending on how good your intuition is about the likely size of the solution, it will take you a few minutes or a few days with a calculator to discover that R = 40/1 = 40 is a rational root.

Having found a rational root, it is easy to factor the right hand side of equation 12:

0 = (x-40)(3x^2 - 76x + 340) (equation 13)

Then, the other two roots of the cubic are:

x = (38±2 ✔106)/3 which can be approximated as x≈19.53 and x≈5.80

With both of those below 25, the only solution is x=40.

This implies that a=360, b=120, and c=72.

Answer: 40 hours.
Click to expand...
 
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