Percents: Difficult Problems with Solutions

Solution:
If the sides were [tex]a[/tex] and [tex]b[/tex], the new sides are [tex]1.2a[/tex] and [tex]0.9b[/tex]:
[tex]1.2a\times 0.9b=1.08ab[/tex]
The area becomes [tex]108\%[/tex] of the original, i.e. it increases by [tex]8\%[/tex] (not by [tex]20\%-10\%=10\%[/tex]).

Solution:
After each reduction by [tex]p\%[/tex] the price is multiplied by [tex]q=1-\frac{p}{100}[/tex]:
[tex]2000q^2=1280 \Rightarrow q^2=0.64 \Rightarrow q=0.8 \Rightarrow p=20[/tex]
Each reduction was [tex]20\%[/tex]: [tex]2000\to 1600\to 1280[/tex]. ([tex]36\%[/tex] is the total reduction.)

Solution:
Money collected = price [tex]\times[/tex] number of spectators. If the price was [tex]p[/tex] and the number was [tex]n[/tex], the new number [tex]n_1[/tex] satisfies:
[tex]0.8p\times n_1=p\times n \Rightarrow n_1=\dfrac{n}{0.8}=1.25n[/tex]
The number of spectators increased by [tex]25\%[/tex].

Solution:
The dry matter does not change. At first it is [tex]100\%-97\%=3\%[/tex] of [tex]210[/tex]:
[tex]0.03\times 210=6.3[/tex]
Later it is [tex]9\%[/tex] of the new weight [tex]x[/tex]:
[tex]0.09x=6.3 \Rightarrow x=70[/tex]
Surprisingly, the weight dropped to one third!

Solution:
The amount of salt does not change:
[tex]0.05\times 10=0.5[/tex]
After adding [tex]x[/tex] gallons it must be [tex]2\%[/tex] of [tex]10+x[/tex]:
[tex]0.02(10+x)=0.5 \Rightarrow 10+x=25 \Rightarrow x=15[/tex]

Solution:
Let [tex]A[/tex] cost [tex]x[/tex]; then [tex]B[/tex] costs [tex]2x[/tex] and the set costs [tex]3x[/tex]. The new price of the set is
[tex]1.4x+1.25\times 2x=1.4x+2.5x=3.9x[/tex]
[tex]\dfrac{3.9x}{3x}=1.3[/tex]
The set becomes [tex]30\%[/tex] more expensive. (Not the average [tex]32.5\%[/tex], because [tex]B[/tex] is the larger part of the price.)

Solution:
Let there be [tex]x[/tex] problems. After each day the following part remains:
[tex]x\to 0.6x\to 0.4\times 0.6x=0.24x\to 0.7\times 0.24x=0.168x[/tex]
[tex]0.168x=21 \Rightarrow x=125[/tex]
Check: [tex]125\to 75\to 30\to 21[/tex]

Solution:
Let Marina have [tex]x[/tex]; then Lina had [tex]200-x[/tex]. Lina kept [tex]40\%[/tex] and Marina [tex]20\%[/tex] of their money:
[tex]0.4(200-x)+0.2x=50 \Rightarrow 80-0.2x=50 \Rightarrow x=150[/tex]
Check: [tex]0.4\times 50+0.2\times 150=20+30=50[/tex]

Solution:
Let there be [tex]n[/tex] students at first, so [tex]\frac{2n}{3}[/tex] girls. After [tex]4[/tex] girls left:
[tex]\dfrac{2n}{3}-4=0.6(n-4)\ \Big|\times 15[/tex]
[tex]10n-60=9n-36 \Rightarrow n=24[/tex]
There were [tex]16[/tex] girls; now there are [tex]12[/tex] girls out of [tex]20[/tex] students, and [tex]\frac{12}{20}=60\%[/tex].

Solution:
Let the side of the [tex]20[/tex]-gon be [tex]a[/tex]; then the side of the octagon is [tex]1.4a[/tex]. The perimeters are:
[tex]8\times 1.4a=11.2a[/tex], [tex]20a[/tex]
[tex]\dfrac{11.2a}{20a}=0.56=56\%[/tex]
The perimeter of the octagon is [tex]100\%-56\%=44\%[/tex] smaller.

Solution:
The interest for one year is [tex]3.5\%[/tex] of the deposit [tex]x[/tex]:
[tex]0.035x=245 \Rightarrow x=245\div 0.035=7000[/tex]

Solution:
Each year the amount is multiplied by [tex]1.04[/tex]:
[tex]2500\times 1.04=2600[/tex]
[tex]2600\times 1.04=2704[/tex]
(With simple interest it would be only [tex]2700[/tex]: in the second year the interest is also paid on the first year's interest.)

Solution:
The sum of the six numbers is [tex]6\times 32=192[/tex]. If the number is [tex]x[/tex], the other five have sum [tex]192-x[/tex]:
[tex]x=1.4\times \dfrac{192-x}{5} \Rightarrow 5x=268.8-1.4x \Rightarrow 6.4x=268.8 \Rightarrow x=42[/tex]
Check: [tex]\dfrac{192-42}{5}=30[/tex], [tex]1.4\times 30=42[/tex]

Solution:
Increasing by [tex]150\%[/tex] means multiplying by [tex]2.5[/tex]. Let the three equal results be [tex]t[/tex]:
[tex]2.5a=t \Rightarrow a=0.4t[/tex], [tex]0.75b=t \Rightarrow b=\dfrac{4t}{3}[/tex], [tex]c-5=t \Rightarrow c=t+5[/tex]
[tex]0.4t+\dfrac{4t}{3}+t+5=87\ \Big|\times 15[/tex]
[tex]6t+20t+15t=1230 \Rightarrow 41t=1230 \Rightarrow t=30[/tex]
[tex]a=12[/tex], [tex]b=40[/tex], [tex]c=35[/tex]

Solution:
[tex]\dfrac{1.4a}{1.25b}=\dfrac{1.4}{1.25}\times \dfrac{a}{b}=1.12\times \dfrac{a}{b}[/tex]
The fraction increases by [tex]12\%[/tex].

Solution:
The total selling price is [tex]105\%[/tex] of the total purchase price:
[tex]210\div 1.05=200[/tex]
Let the first cow cost [tex]x[/tex] and the second [tex]200-x[/tex]:
[tex]0.9x+1.1(200-x)=210 \Rightarrow 220-0.2x=210 \Rightarrow x=50[/tex]
Check: [tex]0.9\times 50+1.1\times 150=45+165=210[/tex]