Word problem related to ratios and proportions

Question

This is the problem at hand:

Farmer John has 41 hens. If he had one more solid colored hen, then exactly one-third of his hens would be speckled. From years of experience, Farmer John knows that one-half of the speckled hens will lay speckled eggs and that each hen and a half will lay an egg and a half in a day and a half. After how many full days will Farmer John have 11 dozen speckled eggs to sell? Explain your reasoning

I am able to do this problem when I add one more hen so Farmer John has 42 hens. However, he only has 41 and I am just not sure how to go about figuring out this problem. If he had 42 colored hens then exactly 1/3 of them are speckled. He only has 41 though?

Answer 1

The first part of the question states that if Farmer John had 42 hens, exactly one-third or 14 of those hens would be speckled; since the "missing" solid-coloured hen is not speckled, he has 14 speckled hens. 7 of these speckled hens lay speckled eggs, and John needs 132 (11 dozen) speckled eggs.

Each hen and a half lays an egg and a half in a day and a half. Therefore, among the speckled hens that lay speckled eggs:

• one hen lays one egg in a day and a half
• seven hens lay seven eggs in a day and a half
• seven hens lay 132 eggs in $\frac32\times\frac{132}7=\frac{198}7=28\frac27$ days.

Therefore Farmer John needs 29 whole days – a lunar month – to collect 11 dozen speckled eggs for sale.