Question

1. The Yeoman family spent a total of $26.75 on lunch. They bought 5 drinks and 3 sandwiches. Each drink costs $2.50 less than a sandwich. Which of the following equations could be used to find the cost of each sandwich?

a. 26.75 = 5(2.50) + 3s

b. 26.75 = 3(2.50) + 5s

c. 26.75 = 5s+ 3(s + 2.50)

d. 26.75 = 3s + 5(s - 2.50)

2. Solve the equation for a rounded to the nearest tenth:
3a - 0.3 ( a + 50) = 75

a. 5.6

b. 22.2

c. 27.3

d. 33.3

Answer 1

1. 26.75 = 3s + 5(s - 2.50)

2. 3a - 0.3(a + 50) = 75
3a - 0.3a - 15 = 75
3a - 0.3a = 75 + 15
2.7a = 90
a = 90/2.7
a = 33.3 <===

Answer 2

1) Option d

2) Option b

Explanation:

1) Given : The Yeoman family spent a total of $26.75 on lunch. They bought 5 drinks and 3 sandwiches. Each drink costs $2.50 less than a sandwich.

To find : Which of the following equations could be used to find the cost of each sandwich?

Solution :

Let d represent the cost of a drink and

s represent the cost of a sandwich.

They bought 5 drinks and 3 sandwiches.

The Yeoman family spent a total of $26.75 on lunch.

The total cost is
`26.75=5d+3s`

Each drink costs $2.50 less than a sandwich

i.e.
`d=s-2.50`

Substitute in the cost equation,

`26.75=5(s-2.50)+3s`

Therefore, Equation could be used to find the cost of each sandwich is
`26.75=3s+5(s-2.50)`

So, Option d is correct.

2) Given : Expression
`3a-0.3(a + 50) = 75`

To find : Solve the equation for a rounded to the nearest tenth ?

Solution :

`3a-0.3(a + 50) = 75`

Apply distributive property,
`a(b+c)=ab+ac`

`3a-0.3a+15= 75`

Take the like terms together,

`3a-0.3a=75-15`

Solve,

`2.7a=60`

`a=(60)/(2.7)`

`a=22.22`

Nearest to tenth, a=22.2

Therefore, Option b is correct.