sat-dog

Choice C is correct. It’s given that store A sells raspberries for $\$5.50$ per pint and blackberries for $\$3.00$ per pint, and a certain purchase of raspberries and blackberries at store A would cost $\$37.00$. It’s also given that store B sells raspberries for $\$6.50$ per pint and blackberries for $\$8.00$ per pint, and this purchase of raspberries and blackberries at store B would cost $\$66.00$. Let $r$ represent the number of pints of raspberries and $b$ represent the number of pints of blackberries in this purchase. The equation $5.50r+3.00b=37.00$ represents this purchase of raspberries and blackberries from store A and the equation $6.50r+8.00b=66.00$ represents this purchase of raspberries and blackberries from store B. Solving the system of equations by elimination gives the value of $r$ and the value of $b$ that make the system of equations true. Multiplying both sides of the equation for store A by $6.5$ yields $\left(5.50r\right)\left(6.5\right)+\left(3.00b\right)\left(6.5\right)=\left(37.00\right)\left(6.5\right)$, or $35.75r+19.5b=240.5$. Multiplying both sides of the equation for store B by $5.5$ yields $\left(6.50r\right)\left(5.5\right)+\left(8.00b\right)\left(5.5\right)=\left(66.00\right)\left(5.5\right)$, or $35.75r+44b=363$. Subtracting both sides of the equation for store A, $35.75r+19.5b=240.5$, from the corresponding sides of the equation for store B, $35.75r+44b=363$, yields $\left(35.75r-35.75r\right)+\left(44b-19.5b\right)=\left(363-240.5\right)$, or $24.5b=122.5$. Dividing both sides of this equationby $24.5$ yields $b=5$. Thus, $5$ pints of blackberries are in
this purchase.

Choices A and B are incorrect and may result from conceptual or calculation errors. Choice D is incorrect. This is the number of pints of raspberries, not blackberries, in the purchase.