Word problem

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Rutgers24

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Jul 20, 2019
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Hi, learning Algebra on my own and am using an old math book and having trouble solving this:

A boy was sent for forty cents worth of eggs. On his way home he broke four and the actual cost was thereby six cents more per dozen than the original purchase price. How many eggs did he buy?

I let "x" be number of eggs and my equation is: (x - 4) (6 + x/12) = 40. But it's not solving. I'm having trouble understanding how to translate "cents per dozen" as an equation i think. Help please!?
 
Hello, and welcome to FMH! :)

If \(x\) is the number of eggs the boy originally bought for 40 cents, then we may write:

[MATH]40=C\frac{x}{12}\implies C=\frac{480}{x}[/MATH]
Here, \(C\) is the cost per dozen. And so, given the information about the reduced number of eggs, we may write:

[MATH]\frac{480}{x}+6=\frac{480}{x-4}[/MATH]
Can you proceed?
 
Can you explain your equation?
How to calculate cents per dozen: first calculate how many dozens we have. E.g. how many dozens in 36? Next, divide total cents by the number of dozens.
 
cHere is another way to think about it.

First name EACH numeric quantity that is unknown.

[MATH]a = \text {number of eggs purchased.}[/MATH]
[MATH]b = \text {number of eggs after breakage.}[/MATH]
[MATH]c = \text {price per egg.}[/MATH]
[MATH]d = \text {price per dozen eggs.}[/MATH]
[MATH]e = \text {effective cost per egg after breakage.}[/MATH]
[MATH]f = \text {effective cost per dozen eggs after breakage.}[/MATH]
Second, with six unknown quantities, we need six independent equations to find the unknowns.

[MATH]b = a - 4.[/MATH]
[MATH]a * c = 40.[/MATH]
[MATH]d = 12c.[/MATH]
[MATH]e = \dfrac{40}{b}.[/MATH]
[MATH]f = 12e[/MATH].

[MATH]f = d + 6.[/MATH]
Now finding each of these equations is relatively straightforward. This approach of identifying each unknown minimizes the abstract thinking required, but it does increase the mechanics involved.

[MATH]a * c = 40 \implies c = \dfrac{40}{a} \text { and } d = 12c \implies d = 12 * \dfrac{40}{a} = \dfrac{480}{a}.[/MATH]
[MATH]e = \dfrac{40}{b} \text { and } f = 12e \implies f = 12 * \dfrac{40}{b} = \dfrac{480}{b}.[/MATH]
[MATH]f = d + 6 \implies \dfrac{480}{b} = \dfrac{480}{a} + 6.[/MATH]
It is now easy to get Mark's equation mechanically by replacing b by a - 4.

[MATH]\dfrac{480}{a - 4} = \dfrac{480}{a} + 6.[/MATH]
I am not nearly so clever as Mark, but I get there step by mechanical step.
 
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