Percents: Problems with Solutions

Solution:
Percent derives from the Latin words "per centum",
which means "on 100", which is why to express a fraction to a percent,
the denominator has to be 100.
If it's possible, we amplify the fraction to have the denominator be 100.
In this case, we amplify the fraction by 20
and we get [tex]\frac{40}{100}[/tex] which means 40%.

Solution:
Percent derives from the Latin words "per centum",
which mean "on 100"
which is why to express a fraction as a percent,
the denominator has to be 100.
If it's possible, we amplify the fraction to have the denominator be 100.
In this case, we amplify the fraction by 25
and we get [tex]\frac{25}{100}[/tex] which means 25%

Solution:
Percent derives from the Latin words "per centum",
which mean "on 100",
which is why to express a fraction as a percent,
the denominator has to be 100.
If it's possible, we amplify the fraction to have the denominator be 100.
In this case, we amplify the fraction by 10
and we get [tex]\frac{70}{100}[/tex] which means 70%.

Solution:
[tex]0.15 = \frac {15}{100}= 15\% [/tex]

Solution:
[tex]0.236 = \frac{236}{1000}=\frac{23.6}{100}=23.6\% [/tex]

Solution:
[tex] 2.7 = \frac{27}{10} = \frac{270}{100}= 270\% [/tex]

Solution:
In this case we cannot amplify the fraction to obtain the denominator 100.
We will transform it into a decimal number and then into a percentage.
[tex]\frac{5}{8} = 0.625 = \frac{625}{1000} = \frac{62.5}{100} = 62.5\% [/tex]

Solution:
In this case we cannot amplify the fraction to obtain the denominator 100.
We will transform it into a decimal number and then into a percent.
[tex]\frac{9}{16} = 0.5625 = \frac{5625}{10000}= \frac{56.25}{100} = 56.25\%[/tex]

Solution:
One half is [tex]\frac{1}{2} = \frac{50}{100} = 50\%[/tex]

Solution:
3 quarters is [tex]\frac{3}{4} = \frac{75}{100} = 75\%[/tex]

Solution:
The circle is divided in 4 equal parts.
The red zone occupies [tex]\frac{3}{4}[/tex] of the circle, which means 75%.

Solution:
The blue zone occupies [tex]\frac{2}{5}[/tex] of the rectangle. [tex]\frac{2}{5} = 40\%[/tex]

Solution:
Mary ate [tex]\frac{2}{10}=\frac{20}{100}=20\%[/tex] of the pizza, and John ate [tex]15\%[/tex].
[tex]20\%\gt 15\%[/tex], so Mary ate more.

Solution:
[tex]40\% = \frac{40}{100}[/tex]
[tex]52\% = \frac {52}{100} [/tex]
[tex]\frac{40}{100} < \frac{52}{100}[/tex]
[tex]40\% < 52\%[/tex]

Solution:
[tex]40\% \text{ of } 200 = \frac{40}{100}\cdot 200 = 40 \cdot 2 = 80[/tex]

Solution:
25% of 240 = [tex]\frac{25}{100} \cdot 240 = \frac{1}{4} \cdot 240 = 60[/tex]

Solution:
20% × a = 8
5 × 20% × a = 5 × 8
100% × a = 40
100% = 1
So a = 40

Solution:
Let "a" be the number we seek. We get
45% × a = 36
2 × 45% × a = 2 × 36 = 72
90% × a = 72
10% × a = 72 : 9 = 8
100% × a = 8 × 10 = 80
a = 80

Solution:
Let "a" be the number we seek.
[tex]\frac{24}{100} \cdot a = 96[/tex]
[tex]a = 96 \div \frac{24}{100}=96 \cdot \frac{100}{24}=\frac{96}{24} \cdot 100 = 4 \cdot {100} = 400[/tex]

Solution:
After the discount the price is [tex]100\%-20\%=80\%[/tex] of the original price [tex]x[/tex]:
[tex]0.8x=220 \Rightarrow x=220\div 0.8=275[/tex]
Careful: adding [tex]20\%[/tex] of [tex]220[/tex] gives [tex]264[/tex], which is wrong — the [tex]20\%[/tex] were taken from the original price, not from [tex]220[/tex].

Solution:
The price decreased by [tex]50-42=8[/tex]. We compare this with the original price [tex]50[/tex]:
[tex]\dfrac{8}{50}=\dfrac{16}{100}=16\%[/tex]

Solution:
There are [tex]40[/tex] boys, and the difference is [tex]20[/tex]. In the first question we compare with the number of boys, in the second — with the number of girls:
[tex]\dfrac{20}{40}=50\%[/tex]
[tex]\dfrac{20}{60}=\dfrac{1}{3}=33\dfrac{1}{3}\%[/tex]

Solution:
The dried apples weigh [tex]100\%-84\%=16\%[/tex] of the fresh ones. If [tex]x[/tex] is the weight of the fresh apples:
[tex]0.16x=32 \Rightarrow x=32\div 0.16=200[/tex]

Solution:
[tex]\dfrac{1}{5}\times 175=35[/tex]
[tex]0.85x-50=35 \Rightarrow 0.85x=85 \Rightarrow x=100[/tex]

Solution:
Anton's amount is [tex]100\%+12\%=112\%[/tex] of Ivan's amount [tex]x[/tex]:
[tex]1.12x=56 \Rightarrow x=56\div 1.12=50[/tex]

Solution:
Let the original price be [tex]100[/tex]. After the reduction it is [tex]80[/tex]; [tex]20\%[/tex] of [tex]80[/tex] is [tex]16[/tex], so the final price is [tex]96[/tex]:
[tex]100\to 80\to 96[/tex]
The price decreased by [tex]4\%[/tex]. The increase is smaller than the reduction because it is taken from a smaller amount.

Solution:
[tex]0.2\times 200=40[/tex]
[tex]200-40=160[/tex], [tex]0.5\times 160=80[/tex]
[tex]\dfrac{40+80}{200}=\dfrac{120}{200}=60\%[/tex]
It is not [tex]20\%+50\%=70\%[/tex], because the [tex]50\%[/tex] are taken from the rest, not from the whole book.

Solution:
The soccer players are [tex]25\%[/tex] of the class, so the difference is [tex]75\%-25\%=50\%[/tex] of the class:
[tex]0.5x=16 \Rightarrow x=32[/tex]

Solution:
If the radius is multiplied by [tex]k[/tex], the area [tex]\pi r^2[/tex] is multiplied by [tex]k^2[/tex]:
[tex]\pi (kr)^2=k^2\times \pi r^2=4\times \pi r^2 \Rightarrow k^2=4 \Rightarrow k=2[/tex]
The radius must double, i.e. increase by [tex]100\%[/tex] (not by [tex]300\%[/tex]).