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Question 1 of 5
Sue went shopping on two occasions. On the first occasion she bought 5 5 apples and 2 2 bananas for $2.80. On the second occasion she paid $5.10 for 3 3 apples and 5 5 bananas. What is the cost of each fruit?
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First, let variables to represent apples and bananas.
a = a = cost of each apple
b = b = cost of each banana
Next, write the simultaneous equations being represented in the problem.
5 5 a a + 2 + 2 b b
= =
2.80 2.80
First occasion
3 3 a a + 5 + 5 b b
= =
5.10 5.10
Second occasion
Multiply equation 1 1 by 3 3
( 5 ( 5 a a + 2 + 2 b b ) ) × 3 × 3
= =
2.80 2.80 × 3 × 3
15 15 a a + 6 + 6 b b
= =
8.40 8.40
Multiply equation 2 2 by 5 5
( 3 ( 3 a a + 5 + 5 b b ) ) × 5 × 5
= =
5.10 5.10 × 5 × 5
15 15 a a + 25 + 25 b b
= =
25.50 25.50
Subtract the transformed equations.
15 15 a a + 6 + 6 b b
= =
8.40 8.40
15 15 a a + 25 + 25 b b
= =
25.50 25.50
- 19 b − 19 b
= =
- 17.10 − 17.10
b b
= =
0.90 0.90
Divide both sides by - 19 − 19
Solve for a a , the cost of each apple.
3 3 a a + 5 + 5 b b
= =
5.10 5.10
3 3 a a + 5 + 5 ( 0.90 ) ( 0.90 )
= =
5.10 5.10
Substitute b = 0.90 b = 0.90
3 a + 4.5 3 a + 4.5 - 4.5 − 4.5
= =
5.10 5.10 - 4.5 − 4.5
Subtract 4.5 4.5 from both sides
3 a 3 a
= =
0.60 0.60
Divide both sides by 3 3
a a
= =
0.20 0.20
Apple = $ 0.20 = $ 0.20 , Banana = $ 0.90 = $ 0.90
Question 2 of 5
A customer bought 4 4 drinks and 3 3 pizzas for $ 40.80 $ 40.80 . Another costumer bought 2 2 drinks and 1 1 pizza for $ 15.00 $ 15.00 . Find the price of the following:
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First, let variables represent drinks and pizzas.
d = d = cost of each drink
p = p = cost of each pizza
Next, write the both of the customer’s purchases as simultaneous equations.
To make the solution easier, convert the dollar value into cents first..
4 4 d d + 3 + 3 p p
= =
4080 4080
Equation 1 1
2 2 d d + 1 + 1 p p
= =
1500 1500
Equation 2 2
Next, multiply the values of equation 2 2 by 2 2 and label the product as equation 3 3 .
2 d + 1 p 2 d + 1 p
= =
1500 1500
Equation 2 2
( 2 d + 1 p ) ( 2 d + 1 p ) × 2 × 2
= =
1500 1500 × 2 × 2
Multiply the values of both sides by 2 2
4 d + 2 p 4 d + 2 p
= =
3000 3000
Equation 3 3
Then, subtract equation 3 3 from equation 1 1 .
4 d - 3 p 4 d − 3 p
= =
4080 4080
- − ( 4 d + 2 p ) ( 4 d + 2 p )
= =
3000 3000
p p
= =
1080 1080
4 d - 4 d 4 d − 4 d cancels out
Now, substitute the value of p p into any of the two equations.
2 d + 1 2 d + 1 p p
= =
1500 1500
Equation 2 2
2 d + 1 2 d + 1 ( 1080 ) ( 1080 )
= =
1500 1500
p = 1080 p = 1080
2 d + 1080 2 d + 1080 - 1080 − 1080
= =
1500 1500 - 1080 − 1080
Subtract 1080 1080 from both sides
2 d 2 d ÷ 2 ÷ 2
= =
420 420 ÷ 2 ÷ 2
Divide both sides by 2 2
d d
= =
210 210
Finally, convert the values back to dollars by dividing each them by 100 100
Dollar value of each drink
d d
= =
210 ÷ 100 210 ÷ 100
= =
$ 2.10 $ 2.10
Dollar value of each pizza
p p
= =
1080 ÷ 100 1080 ÷ 100
= =
$ 10.80 $ 10.80
Price of each drink = $ 2.10 Price of each drink = $ 2.10
Price of each pizza = $ 10.80 Price of each pizza = $ 10.80
Question 3 of 5
2 2 oranges and 5 5 apples cost $ 5.75 $ 5.75 and 4 4 oranges and 6 6 apples cost $ 8.10 $ 8.10 . Find the price of each apple and orange
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First, let variables represent apples and oranges.
x = x = cost of each orange
y = y = cost of each apple
Next, write the simultaneous equations being represented in the problem.
2 2 x x + 5 + 5 y y
= =
575 575
First occasion
4 4 x x + 6 + 6 y y
= =
810 810
Second occasion
Multiply equation 1 1 by 2 2
( 2 ( 2 x x + 5 + 5 y y ) ) × 2 × 2
= =
575 575 × 2 × 2
4 4 x x + 10 + 10 y y
= =
1150 1150
Subtract the second equation from the transformed equation.
4 4 x x + 10 + 10 y y
= =
1150 1150
4 4 x x + 6 + 6 y y
= =
810 810
4 y 4 y
= =
340 340
y y
= =
85 85
Divide both sides by 4 4
Solve for x x , the cost of each orange.
2 2 x x + 5 + 5 y y
= =
575 575
2 2 x x + 5 + 5 ( 85 ) ( 85 )
= =
575 575
Substitute y = 85 y = 85
2 x + 425 2 x + 425 - 425 − 425
= =
575 575 - 425 − 425
Subtract 425 425 from both sides
2 x 2 x
= =
150 150
Divide both sides by 2 2
x x
= =
75 75
Orange = 75 cents = 75 cents , Apple = 85 cents = 85 cents
Question 4 of 5
At a theatre, there are 400 400 patrons. If there were 60 60 more women than men, how many men attended the show?
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First, let variables to represent men and women.
x = x = number of women
y = y = number of men
Next, write the simultaneous equations being represented in the problem.
x x + + y y
= =
400 400
There are a total of 400 400 patrons
x x
= =
y y + 60 + 60
60 60 more women than men
Rewrite equation 2 2 such that x x and y y are on the same side.
x x
= =
y y + 60 + 60
x x - y − y
= =
y y + 60 + 60 - y − y
Subtract y y from both sides
x x - − y y
= =
60 60
x x + 4 + 4 y y
= =
400 400
x x - − y y
= =
60 60
2 x 2 x
= =
460 460
x x
= =
230 230
Divide both sides by 2 2
Solve for y y , the number of men.
x x - − y y
= =
60 60
230 230 - − y y
= =
60 60
Substitute x = 230 x = 230
230 - y 230 − y - 230 − 230
= =
60 60 - 230 − 230
- y − y
= =
- 170 − 170
y y
= =
170 170
There were 170 170 men who attended the show.
Question 5 of 5
Find the value of x x and y y
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First, write the simultaneous equations being represented in the problem.
3 3 x x - − y y
= =
12 12
Length
2 2 x x + 3 + 3 y y
= =
19 19
Width
Multiply equation 1 1 by 3 3
( 3 ( 3 x x - − y ) × 3
=
12 × 3
9 x - 3 y
=
36
Add the second equation to the transformed equation.
2 x + 3 y
=
19
9 x - 3 y
=
36
11 x
=
55
x
=
5
Divide both sides by 11
2 x + 3 y
=
19
2 ( 5 ) + 3 y
=
19
Substitute x = 5
10 + 3 y - 10
=
19 - 10
Subtract 10 from both sides
3 y
=
9
Divide both sides by 3
y
=
3