Word problem: profit from crates of eggs 
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?

 A: 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.
A: 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 $9+12/2=15$.
A: 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 $x=9$.
