Percentage: Level 3
1.
If \(x\) is \(0.5\%\) of \(y\), then \(y\) is what percent of \(x\)?
Solution (E):
\(x = \frac{0.5}{100}y = 0.005y\).
Solving for \(y\): \(y = \frac{x}{0.005} = \frac{1000x}{5} = 200x\).
As a percentage: \(200 \times 100\% = 20,000\%\).
2.
If Mary has half as many cents as Nora has dollars, then Nora has what percent more cents than Mary does? (100 cents = 1 dollar)
Solution (D):
Let Nora have \(D\) dollars. In cents, Nora has \(100D\) cents.
Mary has half as many cents as Nora has dollars, so Mary has \(0.5D\) cents.
We compare Nora (\(100D\)) to Mary (\(0.5D\)).
Difference: \(100D – 0.5D = 99.5D\).
Percent more: \(\frac{99.5D}{0.5D} \times 100\% = 199 \times 100\% = 19,900\%\).
3.
A cockroach population doubles every three days. If there were \(c\) cockroaches on June 1st, what was the percent increase in the population on July 1st? (June has 30 days.)
Solution (C):
Days elapsed: 30.
Doubling periods: \(30/3 = 10\).
Final population = \(c \times 2^{10} = 1024c\).
Increase = \(1024c – c = 1023c\).
Percent increase = \(\frac{1023c}{c} \times 100\% = 102,300\%\).
4.
If 35% of a forest was destroyed and the remainder regenerates at a rate of 10% per year, after how many years will the forest exceed its original size?
Solution (C):
Initial size = 100. Destroyed = 35. Remainder = 65.
We need \(65(1.1)^n > 100\).
\(n=1: 71.5\)
\(n=2: 78.65\)
\(n=3: 86.5\)
\(n=4: 95.1\)
\(n=5: 104.6\).
It takes 5 years.
5.
If the area of a circle increases by 44%, by what percent did the radius increase?
Solution (A):
\(A = \pi r^2\). Let original area = 100. New area = 144.
\( \frac{\text{New Area}}{\text{Old Area}} = \frac{144}{100} = 1.44 \).
Since area is proportional to \(r^2\), the ratio of radii squared is 1.44.
Ratio of radii = \(\sqrt{1.44} = 1.2\).
This represents a 20% increase.
6.
If \(x\) is 20% less than \(y\), and \(y\) is 25% less than \(z\), then \(x\) is what percent less than \(z\)?
Solution (B):
Let \(z = 100\).
\(y\) is 25% less than \(z \implies y = 75\).
\(x\) is 20% less than \(y \implies x = 0.8 \times 75 = 60\).
Comparing \(x\) (60) to \(z\) (100):
\(x\) is \(100 – 60 = 40\) less than \(z\).
This is 40%.
7.
The price of a car depreciates by 10% each year. If the car is worth $10,000 today, what will it be worth in 3 years?
Solution (D):
Value = \( 10000 \times (0.9)^3 \).
\( 0.9 \times 0.9 \times 0.9 = 0.729 \).
\( 10000 \times 0.729 = 7290 \).
8.
Fresh grapes contain 90% water, while dried grapes (raisins) contain 20% water. How many kilograms of raisins can be obtained from 20 kg of fresh grapes?
Solution (C):
The “solid” part (pulp) remains constant.
Fresh: 10% solid. \(0.10 \times 20 = 2\) kg solid.
Raisins: 80% solid (since 20% water).
Let \(W\) be the weight of raisins. \(0.80 W = 2\).
\(W = 2 / 0.8 = 20 / 8 = 2.5\) kg.
9.
If the base of a triangle decreases by 20% and its height increases by 25%, how does the area change?
Solution (A):
Area = \(0.5 \times b \times h\).
New Area = \(0.5 \times (0.8b) \times (1.25h)\).
\(0.8 \times 1.25 = 1\).
The new area is \(1 \times (0.5bh)\), which is the same as the original.
10.
A student multiplies a number by \( \frac{3}{5} \) instead of \( \frac{5}{3} \). What is the percentage error?
Solution (B):
Let the number be 15 (LCM of 3 and 5).
Correct answer: \( 15 \times \frac{5}{3} = 25 \).
Student answer: \( 15 \times \frac{3}{5} = 9 \).
Error = \( 25 – 9 = 16 \).
Percent error = \( \frac{16}{25} \times 100\% = 64\% \).
Score: 0 / 0