Set 4: Ratio Proportion (Challenging)
1.
A bacteria culture doubles every 10 minutes. By what percent does the population increase in 40 minutes?
Solution (C/D):
Doubling periods: \(40/10 = 4\).
Growth factor: \(2^4 = 16\).
Start with 1, end with 16.
Increase = \(16 – 1 = 15\).
Percent increase = \(15 \times 100\% = 1,500\%\).
2.
If the radius of a sphere is increased by 20%, by approximately what percent does its volume increase?
Solution (C):
Volume is proportional to \(r^3\).
New volume factor = \((1.2)^3 = 1.728\).
Percent increase = \(1.728 – 1 = 0.728 = 72.8\% \approx 73\%\).
3.
At the start of the year, the ratio of Juniors to Seniors was 3:4. After 10 Juniors left and 20 Seniors joined, the ratio became 1:2. How many Juniors were there originally?
Solution (B):
Original: \(J = 3x, S = 4x\).
New: \(J’ = 3x – 10, S’ = 4x + 20\).
Ratio: \( \frac{3x – 10}{4x + 20} = \frac{1}{2} \).
\( 6x – 20 = 4x + 20 \implies 2x = 40 \implies x = 20 \).
Original Juniors = \(3(20) = 60\).
4.
If \(x\) is 25% greater than \(y\), and \(y\) is 20% less than \(z\), then \(x\) is equal to:
Solution (E):
\(y = 0.8z\).
\(x = 1.25y\).
Substitute \(y\): \(x = 1.25(0.8z) = (\frac{5}{4})(\frac{4}{5})z = 1z\).
So \(x = z\).
5.
If you mix 10 liters of a 20% acid solution with 40 liters of a 50% acid solution, what is the percent acid in the final mixture?
Solution (D):
Acid in sol 1: \(0.2 \times 10 = 2\) L.
Acid in sol 2: \(0.5 \times 40 = 20\) L.
Total Acid = 22 L. Total Volume = 50 L.
Percent = \(\frac{22}{50} \times 100\% = 44\%\).
6.
A number \(N\) is increased by \(p\%\), then the result is decreased by \(p\%\). The final result is 1% less than \(N\). What is \(p\)?
Solution (A):
Formula for change: \( (1 + \frac{p}{100})(1 – \frac{p}{100}) = 1 – (\frac{p}{100})^2 \).
Given that result is 1% less, so \( 1 – 0.01 = 0.99 \).
\( 1 – (\frac{p}{100})^2 = 0.99 \implies (\frac{p}{100})^2 = 0.01 \).
\( \frac{p}{100} = \sqrt{0.01} = 0.1 \).
\( p = 10 \).
7.
In a school election, Candidate A received 20% more votes than Candidate B. If 4400 votes were cast total (and no other candidates), how many votes did Candidate A receive?
Solution (C):
Let B’s votes = \(x\). A’s votes = \(1.2x\).
\(x + 1.2x = 4400 \implies 2.2x = 4400 \implies x = 2000\).
A’s votes = \(1.2(2000) = 2400\).
8.
A sum of money doubles in 5 years at simple interest. What is the annual interest rate?
Solution (B):
To double, Interest = Principal.
\(P = P \times r \times t\). Here \(t = 5\).
\(1 = r \times 5 \implies r = 1/5 = 0.20 = 20\%\).
9.
A shopkeeper sells an item for $240, making a profit of 20%. What was the cost price?
Solution (C):
\(SP = CP \times 1.2\).
\(240 = 1.2 CP \implies CP = \frac{240}{1.2} = 200\).
10.
If 60% of students passed Math, 70% passed Science, and 40% passed both, what percentage passed **neither** subject?
Solution (E):
\( \text{Total} = M + S – \text{Both} + \text{Neither} \).
\( 100 = 60 + 70 – 40 + \text{Neither} \).
\( 100 = 90 + \text{Neither} \).
\( \text{Neither} = 10\% \).
Score: 0 / 0