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Problem 177 Solution

m_g = math's gathering speed
m_s = math's shaking speed
b_g = boy's gathering speed
b_s = boy's shaking speed

The problems tells us that 3*m_g = b_s. Let's arbitrarily define m_g=1. So b_s=3.

The problem also tells us that "Any one of the men can shake apples off the trees 25% faster than the other two men and boy can pick them up." To put this in the form of an equation:

(1) m_s = 1.25 * (2*m_g + b_g) = 1.25 * (2 + b_g).

The problem also tells us that "The same number of apples are gathered regardless of who is the shaker, as long as everybody is always doing something and no excess apples are left on the ground." Assume that a man is designated the shaker. At some point he should switch from shaking to gathering, lest he shake off more apples than can be gathered, which would violate the condition that no apples are left on the group. Let's call the ratio of time the shaker spends shaking r. We can now set up two more equations:

(2) 3 = r*m_s

(3) 3 = 2 + b_g + (1-r)

Rearranging equation (2): r=3/m_s

Substituting this into equation (3):

3 = 2 + b_g + 1 - 3/m_s

3 = 3 + b_g - 3/m_s

(4) b_g = 3/m_s

Now let's combine equations (1) and (4):

m_s = 1.25 * (2 + 3/m_s). Next multiply both sides by 4*m_s

4m_s² = 5 * (2*m_s + 3).

4m_s² = 10*m_s + 15.

4m_s² - 10*m_s - 15 = 0.

m_s = [ 10 + (10² + 4*4*15)^1/2 ] / 2*4

m_s = ( 10 + 340^1/2 ) / 8

m_s = ( 5 + 85^1/2 ) / 4 =~ 3.5549

So the ratio of the man's shaking speed to the boy's shaking speed is 3.5549 to 3 =~ 1.1850. This is also the ratio by which they should get paid. To verify we can also find the ratio of the man's gatering speed to the boy's gathering speed. To do this plug this into equation (1) to get b_g = (4/5)*(m_s - 2.5 ) =~ 0.8439. So the ratio of gathering speeds is 1 to 0.8439 =~ 1.1850

Let call s_m the man's salary and s_b the boy's salary.

s_b = 0.8439 * s_m.

$500 = 3*s_m + s_b

$500 = 3.8439*3*s_m

s_m = $500/3.8439 = $130.08.

s_b = 0.8439 * $130.08 = $109.77

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