Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
10. Change Expressed as a Percent
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Exercise 41 Page 150

Practice makes perfect
a It is given that the original side length is 4 meters. Since the area of a square is found by squaring its side length, we can find the area of the original square.
A_(original)= 4^2= 16 If the side length grows by 10 % , the square's dimensions change and the area of the square will increase. We will first calculate the change in the side's length by using the percent proportion.
a/b=p/100
a/4=10/100
a/4=0.1
a=0.4
If the square's side length increases by 0.4centimeters, then the new side length becomes 4+0.4=4.4 meters. Now we can calculate the new area. A_(new)= 4.4^2= 19.36 The area of the square increased. To find the amount of increase, we will subtract the original area from the new area. A_(new) -A_(original) &= 19.36- 16 & = 3.36 We now have enough information to calculate the percent increase in the area.
Percent increase=Amount of increase/Original amount
Percent increase=3.36/16
Percent increase=0.21
Percent increase = 21 %
The area of the square is increased by 21 %.
b As in the previous part, we can calculate the area by simply squaring the side length.
A_(original)=6^2= 36 Using the percent proportion, we calculate what value is 10 % of the square's side length.
a/b=p/100
a/6=10/100
a/6=0.1
a=0.6
The new length of the square's side is 6+0.6=6.6 meters. Now we can calculate the new area. A_(new)=6.6^2= 43.56 The area of the square increased. To find the amount of increase, we will subtract the original area from the new area. A_(new) -A_(original) &= 43.56- 36 & = 7.56 We now have enough information to calculate the percent increase in the area.
Percent increase=Amount of increase/Original amount
Percent increase=7.56/36
Percent increase=0.21
Percent increase = 21 %
The area increased by 21 %.
c Instead of doing the same thing again, let's prove that a 10 % increase in the square's side length corresponds to a 21 % increase in its area no matter the side length. 10 % increase in the square's side length ⇓ ? 21 % increase in its area For this purpose, we will call the square's side length s which means the square has an area of s^2. Let's use the percent proportion to find the change in the square's side length.
a/b=p/100
a/s=10/100
a/s=0.1
a=0.1s
The new side length will grow by 0.1s which we will add to the original side length.
s+0.1s
1s+0.1s
1.1s
An expression for the new square's side length is 1.1s. By squaring this expression we get the new area. A=( 1.1s)^2= 1.21s^2 The new area is 1.21s^2. From this expression, we can tell that the new area will be 21 % larger because 1.21 as a percent is 121 %. 1.21 = 121/100= 121 % This corresponds to an increase of 21 %. Regardless of the length’s numeric value, in this case 8 centimeters, anytime the square’s side length is increased by 10 %, then the area of the square will always increase by 21 %. Let's check our answer by repeating what we did previously in Part A and B.
Original New
Side Length (m) 8 8+8* 10 %= 8.8
Area (m^2) 8^2= 64 (8.8)^2= 77.44
Amount of Increase 77.44- 64= 13.44
Percent Increase 13.44/64=0.21= 21 %