Word problem: profit from crates of eggs

Question

A shopkeeper buys a crate of eggs at \$1.50 per dozen. He buys another crate containing 3 dozens more than the first crate, at $2.00 per dozen.

He sells them all at \$2.50 a dozen and obtained a profit of \$15. How many dozens where there in each of the crates?

Answer 1

Suppose there are $x$ dozen in the first crate. Then they cost $3x/2$. There must be $x+3$ in the second crate and they cost $2x+6$. Total cost $7x/2+6$. He sells $2x+3$ dozen for $5x+15/2$. So his profit is $3x/2+3/2=15$. Hence $3x/2=27/2$.

Hence $x=9$. So there were 9 dozen in the first crate and 12 dozen in the second.

Check: he made 1 profit on each dozen in the first crate and 1/2 on each dozen in the second crate, so a total of $9+12/2=15$.

Answer 2

Let there be $x$ eggs in the first crate he bought for $1.50$ a dozen. Therefore he would have to spend $\frac{1.5x}{12}$ on the first crate. Given he buys the next crate containing $x+36$ eggs at the rate of $2.00 $ per dozen, he would have to spend $\frac{x+36}{6}$ on the second crate giving us his total spending as: $$\frac{1.5x}{12} + \frac{x+36}{6}$$ Now as he sells all the eggs for $2.50$ a dozen, he should be earning $$\frac{2x+36}{12} 2.5$$ As the profit he earned is the difference of these two amounts, $$15=\frac{2x+36}{12}2.5 -( \frac{1.5x}{12}+\frac{x+36}{6})$$ On solving the above equation we get $x$=108. $\therefore$ The first crate contains 9 dozens and the second contains 12 dozens of eggs.

Answer 3

He makes $\$1.00$ profit from eggs in the first box, and $\$0.50$ from eggs in the second.

Total profit is $x+(\frac12x+\frac32)=15$, or $\frac32x=13.50$, so $x=9$.