Question

1. Find the percent of change when the price of a radio changes from $40.00 to $44.00.

Answer 1

Answer:

(1)

option-b

(2)

option-d

(3)

option-a

(4)

option-d

(5)

option-d

Step-by-step explanation:

(1)

previous value =$40

new value =$44

so, change = new value - previous value

change =44-40

change =4

percent of change = ((change)/(previous value))*100

percent of change is

$=\frac{4}{40}\times 100$

percent of change is 10%..................Answer

(2)

previous value =$450

new value =$396

so, change = new value - previous value

change =396-450

change =-54

percent of change = ((change)/(previous value))*100

percent of change is

$=\frac{-54}{450}\times 100$

$=-12$

percent of change is -12%..................Answer

(3)

previous value =20

new value =24

so, change = new value - previous value

change =24-20

change =4

percent of change = ((change)/(previous value))*100

percent of change is

$=\frac{4}{20}\times 100$

$=20$

percent of change is 20%..................Answer

(4)

previous value =150 pounds

new value =138 pounds

so, change = new value - previous value

change =138-150

change =-12

percent of change = ((change)/(previous value))*100

percent of change is

$=\frac{-12}{150}\times 100$

$=-8$

percent of change is -8%..................Answer

(5)

previous value =40

new value =60

so, change = new value - previous value

change =60-40

change =20

percent of change = ((change)/(previous value))*100

percent of change is

$=\frac{20}{40}\times 100$

$=50$

percent of change is 50%..................Answer

Answer 2

Answer:

1. b) 10%

2. d) 12%

3. a) 20%

4. d) 8%

5. d) 50%

Step-by-step explanation:

We will use percentage change formula to solve our given problems.

$\text{Percent change}=\frac{\text{Final value - Initial value}}{\text{initial value}}*100$

1.  The percent of change when the price of a radio changes from $40.00 to $44.00.

$\text{Percent change}=\frac{44-40}{40}*100$

$\text{Percent change}=\frac{4}{40}*100$

$\text{Percent change}=\frac{1}{10}*100$

$\text{Percent change}=1*10=10$

Therefore, the price of radio increased by 10% and option b is the correct choice.

2. The percent of change when the price of an oven changes from $450 to $396.

$\text{Percent change}=\frac{396-450}{450}*100$

$\text{Percent change}=\frac{-54}{450}*100$

$\text{Percent change}=-0.12*100$

$\text{Percent change}=-12$

Therefore, the price of oven decreased by 12% and option d is the correct choice.

3. The top player’s scoring average changes from 20 points per game to 24 points per game.

$\text{Percent change}=\frac{24-20}{20}*100$

$\text{Percent change}=\frac{4}{20}*100$

$\text{Percent change}=4*5$

$\text{Percent change}=20$    

Therefore, the player's scoring average increased by 20% and option a is the correct choice.

4.  At the start of the football season, Jorge weighed 150 pounds. At the end of the season, he weighed 138 pounds.

[\text{Percent change}=\frac{138-150}{150}*100[/tex]

$\text{Percent change}=\frac{-12}{150}*100$

$\text{Percent change}=-0.08*100$

$\text{Percent change}=-8$

Therefore, the Jorge's weight decreased by 8% and option d is the correct choice.

5. Sara raised her raw score on a math test from 40 to 60.

$\text{Percent change}=\frac{60-40}{40}*100$

$\text{Percent change}=\frac{20}{40}*100$

$\text{Percent change}=20*2.5$

$\text{Percent change}=50$

Therefore, Sara's score on math test increased by 50% and option d is the correct choice.