# Algebra Word Problems Worksheet

Algebra Word Problems Worksheet
• Page 1
1.
Smith bought 7 oranges for $3 and $x$ apples at$0.60 each. Which equation represents the total cost($c$)?
 a. $c$ = 2 - 0.5$y$ b. $c$ = 0.6 + 3$x$ c. $c$ + 3 = 0.6$x$ d. $c$ = 3 + 0.6$x$

#### Solution:

Cost of 7 oranges = $3 Cost of 1 apple =$0.60

Cost of x apples = x × 0.60 = 0.6x

Total cost = Cost of 7 oranges + Cost of x apples

c = 3 + 0.6x

Therefore, the equation that represents the total cost is c = 3 + 0.6x.

2.
Justin buys a bat and 5 balls for a total cost of $40. Identify the equation representing this situation if $x$, $y$ are the costs of a bat and a ball in dollars respectively.  a. $x$ - 5$y$ = 40 b. 40 + 5$y$ = $x$ c. $x$ + 5$y$ = 40 d. 40 + $x$ = 5$y$ #### Solution: Cost of one bat and 5 balls =$40

Cost of one bat = x

Cost of one ball = y

Cost of 5 balls = 5 × y = 5y

Cost of one bat + Cost of 5 balls = Total cost

x + 5y = 40

Therefore, the equation representing the situation is x + 5y = 40.

3.
Mr. Smith's wife is 42 years old. His son is 15 years old. The ages of Mr. Smith, his wife, and their son add up to 104. Which equation could be used to determine Mr. Smith's age ($n$)?
 a. 42 + 15 + 104 = $n$ b. $n$ + 42 + 15 = 104 c. 42 + 104 = $n$ + 15 d. 42 + 15 + 104 + $n$ = 0

#### Solution:

Age of Mrs. Smith = 42

Age of Smith's son = 15

Sum of ages of Mr. Smith, his wife, and their son = 104

n + 42 + 15 = 104

Therefore, the equation that can be used to determine Mr. Smith's age is n + 42 + 15 = 104.

4.
A company offers jeans for $24 each and the shipping charges for entire order is$13. If John wants to buy $n$ jeans, then which equation will help to determine the cost of his order ($c$)?
 a. $c$ = 24 + 13$n$ b. $c$ = 24$n$ +13 c. $c$ = (24 - 13)$n$ d. $c$ = (24 + 13) $n$

#### Solution:

Cost of jeans = $24 Shipping charges for entire order =$13

Total cost = cost of each jeans × total number of jeans + shipping charges

c = 24 × n + 13

Therefore, the equation that will help to determine the cost of his order is c = 24n + 13.

5.
Julia had $45 with her. She sold $n$ cards at$5 each and earned some money. Write an equation for the money($m$) she had finally.
 a. $m$ = 49 + $n$ b. $m$ = 5$n$ + 45 c. $m$ = 45$n$ + 5 d. $m$ = 5$n$ - 45

#### Solution:

The money she had finally = money earned by selling n cards + money she had initially

The money earned by selling n cards = n × cost of each card = n × 5 = \$5n

So, the equation for the total money she had finally is m = 5n + 45.

6.
Identify an equation for the algebra tiles and solve the equation for the value of $y$.

 a. 8 b. 6 c. 2 d. 4

#### Solution:

4y + 3 + 1 - 3y = 2y - 3 - 1
[Original equation.]

4y -3y + 3 + 1 = 2y - 3 - 1
[Rearrange the terms.]

y + 4 = 2y - 4
[Combine like terms.]

y + 4 - y = 2y - 4 - y
[Subtract y from each side.]

4 = y - 4
[Combine like terms.]

4 + 4 = y - 4 + 4

8 = y
[Simplify.]

So, the value of y is 8.

7.
Identify an equation for the algebra tiles and solve the equation for the value of $y$.

 a. -3 b. 5 c. 4 d. -4

#### Solution:

4y + 3 - 2y = -3
[Original equation.]

4y - 2y + 3 = -3
[Rearrange the terms.]

2y + 3 = -3
[Combine like terms.]

2y + 3 - 3 = -3 - 3
[Subtract 3 from each side.]

2y = -6
[Simplify.]

2y2 = -62
[Divide by 2 on each side.]

y = -3
[Simplify.]

8.
Find the value of $y$ in the equation $y$ + 2 = 2, by using algebra tiles.
 a. 6 b. 1 c. 4

#### Solution:

[Model the equation in the form of algebra tiles.]

[Add two negative tiles on both sides.]

[Remove the zero pairs on both sides.]

The value of y is 0.
[No tiles on the right side represent 0.]

9.
Find the value of $y$ in the equation $y$ - 3 = - 7 using algebra tiles.
 a. -4 b. -5 c. -6 d. -9

#### Solution:

[Model the equation in the form of algebra tiles.]

[Isolate the common constant tiles from both sides.]

[Remove isolated tiles from both sides.]

The value of y = -4.

10.
Find the value of $y$ from the algebra tiles.

 a. 6 b. 8 c. 4 d. 2

#### Solution:

In the model, the left hand side contains one variable y-tile and four -1 tiles, the right hand side has two +1 tiles.

The linear equation for the model is y - 4 = 2
[Four -1 tiles represent -4 and two 1-tiles represent 2.]

y - 4 + 4 = 2 + 4
[Add 4 on each side to isolate the variable y.]

y = 6
[Add 4 and 2 to get 6.]

So, the value of the variable y is 6.