SAT Math - Percent of Change

1

The radius of a given circle is increased by 20%. What is the percent increase of the area of the circle.

20%

40%

44%

100%

144%

Explanation

If we plug-in a radius of 5, then a 20% increase would give us a new radius of 6 (which is 1.2 x 5). The area of the new circle is π(6)2 = 36π, and the area of the original circle was π(5)2 = 25π . The numerical increase (or difference) is 36π - 25π = 11π. Next we have to divide this difference by the original area: 11π/25π = .44, which multiplied by 100 gives us a percent increase of 44%. The percent increase = (the numerical increase between the new and original values)/(original value) x 100. The algebraic solution gives us the same answer. If radius r of a certain circle is increased by 20%, then the new radius would be (1.2)r. The area of the new circle would be 1.44 π r2 and the area of the original circle πr2. The difference between the areas is .44 π r2, which divided by the original area, π r2, would give us a percent increase of .44 x 100 = 44%.

2

The radius of a given circle is increased by 20%. What is the percent increase of the area of the circle.

20%

40%

44%

100%

144%

Explanation

If we plug-in a radius of 5, then a 20% increase would give us a new radius of 6 (which is 1.2 x 5). The area of the new circle is π(6)2 = 36π, and the area of the original circle was π(5)2 = 25π . The numerical increase (or difference) is 36π - 25π = 11π. Next we have to divide this difference by the original area: 11π/25π = .44, which multiplied by 100 gives us a percent increase of 44%. The percent increase = (the numerical increase between the new and original values)/(original value) x 100. The algebraic solution gives us the same answer. If radius r of a certain circle is increased by 20%, then the new radius would be (1.2)r. The area of the new circle would be 1.44 π r2 and the area of the original circle πr2. The difference between the areas is .44 π r2, which divided by the original area, π r2, would give us a percent increase of .44 x 100 = 44%.

3

The cost of a shirt in January is dollars. In February, the cost is decreased by 10%. In March, the cost is decreased by another 10%. By what percentage did the shirt decrease in total between January and March?

We must know the original cost to find the answer

Explanation

The best way to answer this question is to plug in a number for n. Since you are working with percentages, it may be easiest to use 100 for n.

We know that in the month of February, the cost of this shirt was decreased by 10%. Because 10% of 100 is \$10, the new cost of the shirt is \$90.

In March, the cost of the shirt decreased another 10%. 10% of 90 is 9, so the cost of the shirt is now \$81.

To find the total percentage decrease, you must divide 81 by 100 and subtract it from 1.

1 – (81/100) = 1 – 0.81 = 0.19

The total decrease was 19%.

4

The cost of a shirt in January is dollars. In February, the cost is decreased by 10%. In March, the cost is decreased by another 10%. By what percentage did the shirt decrease in total between January and March?

We must know the original cost to find the answer

Explanation

The best way to answer this question is to plug in a number for n. Since you are working with percentages, it may be easiest to use 100 for n.

We know that in the month of February, the cost of this shirt was decreased by 10%. Because 10% of 100 is \$10, the new cost of the shirt is \$90.

In March, the cost of the shirt decreased another 10%. 10% of 90 is 9, so the cost of the shirt is now \$81.

To find the total percentage decrease, you must divide 81 by 100 and subtract it from 1.

1 – (81/100) = 1 – 0.81 = 0.19

The total decrease was 19%.

5

If a rectangle's length decreases by fifteen percent, and its width decreases by twenty percent, then by what percent does the rectangle's area decrease?

32

35

40

36

45

Explanation

Let's call the original length and width of the rectangle $l_{1}$ and $w_{1}$ , respectively.

The initial area, $A_{1}$ , of the rectangle is equal to the product of the length and the width. We can represent this with the following equation:

$A_{1}=l_{1}\cdot w_{1}$

Next, let $l_{2}$ and $w_{2}$ represent the length and width, respectively, after they have been decreased. The final area will be equal to $A_{2}$ , which will be equal to the product of the final length and width.

$A_{2}=l_{2}\cdot w_{2}$

We are asked to find the change in the area, which essentially means we want to compare $A_{1}$ and $A_{2}$ . In order to do this, we will need to find an expression for $A_{2}$ in terms of $l_{1}$ and $w_{1}$ . We can rewrite $l_{2}$ and $w_{2}$ in terms of $l_{1}$ and $w_{1}$ .

First, we are told that the length is decreased by fifteen percent. We can think of the full length as 100% of the length. If we take away fifteen percent, we are left with 100 – 15, or 85% of the length. In other words, the final length is 85% of the original length. We can represent 85% as a decimal by moving the decimal two places to the left.

$l_{2}$ = 85% of $l_{1}$ = $0.85l_{1}$

Similarly, if we decrease the width by 20%, we are only left with 80% of the width.

$w_{2}$ = 80% of $w_{1}$ = $0.80w_{1}$

We can now express the final area in terms of $l_{1}$ and $w_{1}$ by substituting the expressions we just found for the final length and width.

$A_{2}=l_{2}\cdot w_{2}$ = ( $0.85l_{1}$ )( $0.80w_{1}$ ) = $0.68l_{1}w_{1}$

Lastly, let's apply the formula for percent of change, which will equal the change in the area divided by the original area. The change in the area is equal to the final area minus the original area.

percent change = $\frac{(A_{2}-A_{1})}{A_{1}}$ (100%)

= $\frac{(0.68l_{1}w_{1}-l_{1}w_{1})}{l_{1}w_{1}}$ (100%)

= $\frac{-0.32l_{1}w_{1}}{l_{1}w_{1}}$ (100%) = –0.32(100%) = –32%

The negative sign indicates that the rectangle's area decreased. The change in the area was a decrease of 32%.

The answer is 32.

6

If a rectangle's length decreases by fifteen percent, and its width decreases by twenty percent, then by what percent does the rectangle's area decrease?

32

35

40

36

45

Explanation

Let's call the original length and width of the rectangle $l_{1}$ and $w_{1}$ , respectively.

The initial area, $A_{1}$ , of the rectangle is equal to the product of the length and the width. We can represent this with the following equation:

$A_{1}=l_{1}\cdot w_{1}$

Next, let $l_{2}$ and $w_{2}$ represent the length and width, respectively, after they have been decreased. The final area will be equal to $A_{2}$ , which will be equal to the product of the final length and width.

$A_{2}=l_{2}\cdot w_{2}$

We are asked to find the change in the area, which essentially means we want to compare $A_{1}$ and $A_{2}$ . In order to do this, we will need to find an expression for $A_{2}$ in terms of $l_{1}$ and $w_{1}$ . We can rewrite $l_{2}$ and $w_{2}$ in terms of $l_{1}$ and $w_{1}$ .

First, we are told that the length is decreased by fifteen percent. We can think of the full length as 100% of the length. If we take away fifteen percent, we are left with 100 – 15, or 85% of the length. In other words, the final length is 85% of the original length. We can represent 85% as a decimal by moving the decimal two places to the left.

$l_{2}$ = 85% of $l_{1}$ = $0.85l_{1}$

Similarly, if we decrease the width by 20%, we are only left with 80% of the width.

$w_{2}$ = 80% of $w_{1}$ = $0.80w_{1}$

We can now express the final area in terms of $l_{1}$ and $w_{1}$ by substituting the expressions we just found for the final length and width.

$A_{2}=l_{2}\cdot w_{2}$ = ( $0.85l_{1}$ )( $0.80w_{1}$ ) = $0.68l_{1}w_{1}$

Lastly, let's apply the formula for percent of change, which will equal the change in the area divided by the original area. The change in the area is equal to the final area minus the original area.

percent change = $\frac{(A_{2}-A_{1})}{A_{1}}$ (100%)

= $\frac{(0.68l_{1}w_{1}-l_{1}w_{1})}{l_{1}w_{1}}$ (100%)

= $\frac{-0.32l_{1}w_{1}}{l_{1}w_{1}}$ (100%) = –0.32(100%) = –32%

The negative sign indicates that the rectangle's area decreased. The change in the area was a decrease of 32%.

The answer is 32.

7

In his most recent film, it was estimated that Joaquin Phoenix was paid of the $\$23,000,000$ budget. In his next roll, he is expected to make less. How much money should Joaquin expect to make for his next film?

$\$977500$

$\$172500$

$\$2300000$

$\$1250000$

$\$652500$

Explanation

First, we find how much Mr. Phoenix made in his most recent movie

$0.05 * 23000000 = \$1,150,000$

Then, we decrease this by fifteen percent according the to the formula:

$\A = P(1 - r) \A = 1150000(1- 0.15) \A = \$977,500$

8

Cindy is running for student body president and is making circular pins for her campaign. She enlarges her campaign image to fit the entire surface of a circular pin. After the image is enlarged, its new diameter is 75 percent larger than the original. By approximately what percentage has the area of the image increased?

225%

100%

75%

200%

20%

Explanation

Pick any number to be the original diameter. 10 is easy to work with. If the diameter is 10, the radius is 5. The area of the original image is A = πr2, so the original area = 25π. Now we increase the diameter by 75%, so the new diameter is 17.5. The radius is then 8.75. The area of the enlarged image is approximately 77π. To find the percentage by which the area has increased, take the difference in areas divided by the original area. (77π - 25π)/25π = 51π/25π = 51/25 = 2.04 or approximately 200%

9

In his most recent film, it was estimated that Joaquin Phoenix was paid of the $\$23,000,000$ budget. In his next roll, he is expected to make less. How much money should Joaquin expect to make for his next film?

$\$977500$

$\$172500$

$\$2300000$

$\$1250000$

$\$652500$

Explanation

First, we find how much Mr. Phoenix made in his most recent movie

$0.05 * 23000000 = \$1,150,000$

Then, we decrease this by fifteen percent according the to the formula:

$\A = P(1 - r) \A = 1150000(1- 0.15) \A = \$977,500$

10

Cindy is running for student body president and is making circular pins for her campaign. She enlarges her campaign image to fit the entire surface of a circular pin. After the image is enlarged, its new diameter is 75 percent larger than the original. By approximately what percentage has the area of the image increased?

225%

100%

75%

200%

20%

Explanation

Pick any number to be the original diameter. 10 is easy to work with. If the diameter is 10, the radius is 5. The area of the original image is A = πr2, so the original area = 25π. Now we increase the diameter by 75%, so the new diameter is 17.5. The radius is then 8.75. The area of the enlarged image is approximately 77π. To find the percentage by which the area has increased, take the difference in areas divided by the original area. (77π - 25π)/25π = 51π/25π = 51/25 = 2.04 or approximately 200%

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