Question

The students at a high school collected $3,800 to buy stuffed animals to give to a children’s hospital. The students bought both stuffed bears and stuffed dogs. Each bear cost $7, and each dog cost $8. The students bought a total of 500 stuffed animals and spent all the money. How many stuffed dogs did the students buy? A. 200 B. 205 C. 295 D. 300​

Answer 1

D

Explanation:

Remark

Let the stuffed bear = x

Let the stuffed dog = y

Equations

x + y = 500 (1)

7x + 8y = 3800 (2)

Solution

There are two ways you could do this. You can use substitution or you could use multiplying equation 1 by either 7 or 8 to get rid of the x or y.

Multiply (1) by 7

7x + 7y = 3500 (3) Write equation (2) below (3) and subtract

7x + 8y = 3800 Subtract

-y = - 300 Multiply by - 1

y = 300

Use the result for y to get x using equation 1

x + y = 500

x + 300 = 500 Subtract 300

x = 500 - 300

x = 200

Stuffed bears = 200

Stuffed dogs = 300

Answer 2

D. 300

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Equality Properties

• Multiplication Property of Equality
• Division Property of Equality
• Addition Property of Equality
• Subtraction Property of Equality

Algebra I

• Solving systems of equations using substitution/elimination

Explanation:

Step 1: Define Variables

Stuffed bears = b

Stuffed dogs = d

Step 2: Define Systems

$7b + $8d = $3800

b + d = 500

Step 3: Rewrite Systems

b + d = 500

1. [Equality Property] Multiply everything by -7: -7b - 7d = -3500

Step 4: Redefine Systems

7b + 8d = 3800

-7b - 7d = -3500

Step 5: Solve for d

Elimination

1. Combine 2 equations: d = 300

Step 6: Define Answer

d = 300 means that the students bought 300 stuffed dogs.