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

Answer:

1) Option d

2) Option b

Step-by-step 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,

$2.7a=60$

$a=\frac{60}{2.7}$

$a=22.22$

Nearest to tenth, a=22.2

Therefore, Option b is correct.