Question

Calvin owns a toy store. He can spend at most $200 on restocking cars and dolls. A doll costs $6.50, and a car costs $8.00. Let x represent the number of cars, and let y represent the number of dolls. Identify an inequality for the number of toys he can buy. Then identify the number of dolls Calvin can buy if he buys 10 cars.

Answer 1

The inequality expression is 6.5y + 8x ≤ 200 and the value is no more than 19

How to determine the inequality expression and the value

From the question, we have the following parameters that can be used in our computation:

Spendings = At most $200

Cost = $6.50 and $8.00

Represent the number of cars as x and the number of dolls as 'y'.

So, we have

6.5y + 8x ≤ 200

If Calvin buys 10 cars, we can substitute x = 10 into the inequality:

6.5y + 8(10) ≤ 200

Solving for y, we get:

6.5y ≤ 120

y ≤ 18.46

Approximate

y < 19

Hence, the inequality expression is 6.5y + 8x ≤ 200 and the value is no more than 19

Answer 2

Given that;

Calvin owns a toy store. He can spend at most $200 on restocking cars and dolls. A doll costs $6.50, and a car costs $8.00.

Let x represent the number of cars, and let y represent the number of dolls.

The maximum amount Calvin can spend is

`8x+6.50y\leq200`

If he buys 10 cars, the maximum number of dolls is

`\begin{gathered} 8x+6.50y\leq200 \\ 8(10)+6.50y\leq200 \\ 6.50y\leq200-80 \\ 6.50y\leq120 \\ y\leq(120)/(6.50) \\ y\leq18.46153 \\ y\leq18\text{ \lparen nearest whole number\rparen} \end{gathered}`

Hence, the answer is

`8x+6.50y\leq200;\text{ no more than 18 dolls}`