2465.99 – The Rope

A piece of rope is hanging over a pulley. A weight is at one end of the rope and a monkey is hanging at the other end, scratching himself. The monkey and the weight are exactly level with each other. The monkey's age and the age of his mother total seven years. The monkey weighs as many pounds as his mother is years old. Four feet of rope weigh one pound. The monkey's mother is one third again as old as the monkey would be if the monkey's mother were half as old as the monkey will be when the monkey is three times as old as the monkey's mother was when the monkey's mother was three times as old as the monkey was then.

The weight of the weight and the rope together come to twice the difference between the sum of the weight of the weight plus the weight of the monkey and the weight of the weight.

How long is the rope?

Solution

Overall observations:

  1. We can find the length of the rope by finding its weight.

  2. The monkey weighs the same as the weight.

  3. To find the monkey's weight, we need to find the mother's age. So we start with ages.

We have four times: the present "is": M=M = mother's age, mm = monkey's age

"would be": M+aM + a, m+am + a

"will be": M+cM + c, m+cm + c

"was: M+d(d<0)M + d (d < 0), m+dm + d.

So here we go:

(1) The monkey's mother is 1/3 again as old as the monkey would be

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(1) M=43(m+a)→7−m=43(m+a)M = \frac{4}{3}(m + a) \rightarrow 7 - m = \frac{4}{3} (m + a)&nbsp&nbsp(n.b. m+M=7,M=7−mm + M = 7, M = 7 - m)

(2) if the monkey's mother were half as old as the monkey will be

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(2) M+a=12(m+c)→7−m+a=12(m+c)M + a = \frac{1}{2}(m + c) \rightarrow 7 - m + a = \frac{1}{2}(m + c)

(3) when the monkey is 3 times as old as his mother was

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(3) m+c=3(M+d)→m+c=3(7−m+d)m + c = 3(M + d) \rightarrow m + c = 3(7 - m + d)

(4) when the monkey's mother was 3 times as old as the monkey was then

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(4)M+d=3(m+d)→7−m+d=3(m+d)M + d = 3(m + d) \rightarrow 7 - m + d = 3(m + d)

We go to work on these:

(1) 7−m=43(m+a)→21−3m=4m+4a→21−7m=4a→a=21−7m4.7 - m = \frac{4}{3}(m + a) \rightarrow 21 - 3m = 4m + 4a \rightarrow 21 - 7m = 4a \rightarrow a = \frac{21 - 7m}{4}.

(2)&(3) M+a=12(m+c)=12(3(M+d))=32(M+d)=32(3(m+d))=92m+92d.M + a = \frac{1}{2}(m + c) = \frac{1}{2}(3(M + d)) = \frac{3}{2}(M + d) = \frac{3}{2}(3(m + d)) = \frac{9}{2}m + \frac{9}{2}d.

&nbsp(We used (2), (3), and (4) to get this.)

Continuing,

M+a=M+aM + a = M + a

7−m+21−7m4=92m+92d7 - m + \frac{21 - 7m}{4} = \frac{9}{2}m + \frac{9}{2}d.

28−4m+21−7m=18m+18d28 - 4m + 21 - 7m = 18m + 18d

49−11m−18m=18d49 - 11m - 18 m = 18 d

49−29m18=d.\frac{49 - 29m}{18} = d. &nbsp &nbsp Now then,

(4) 7−m+d=3m+3d7 - m + d = 3m + 3d

7−m−3m=2d7 - m - 3m = 2d

7−4m=2d→d=72−2m7 - 4m = 2d \rightarrow d = \frac{7}{2} - 2m

So 49−29m18=72−2m→49−29m=63−36m→7m=14→m=2.\frac{49 - 29m}{18} = \frac{7}{2} - 2m \rightarrow 49 - 29m = 63 - 36m \rightarrow 7m = 14 \rightarrow m = 2.

The monkey is 2 years old. We pause to enjoy this.

OK. Onward. The mother is 5 years old, so the monkey weighs 5 pounds.

weight + rope = 2(weight + monkey - weight) →\rightarrow weight + rope = 2 x monkey = 10.

weight + rope = 2 x monkey →\rightarrow 5 + rope = 10 →\rightarrow rope weighs 5 pounds.

At 4 feet per pound, that's 2020 feet of rope. We pause to enjoy this, too.