Interest Practice Questions
Master GMAT Interest with comprehensive practice questions. Build your quantitative reasoning skills through detailed explanations and strategic practice.
3 questions to try, free — no account neededView Explanation
Correct Answer: E
Step 1: Calculate the interest earned after the first 6 months.
\(2\% \times \$10{,}000 = \$200\)
Balance after 6 months: \(\$10{,}000 + \$200 = \$10{,}200\)
Step 2: Calculate the interest earned after the next 6 months (months 7-12).
\(3\% \times \$10{,}200 = \$306\)
Balance after 12 months: \(\$10{,}200 + \$306 = \$10{,}506\)
Step 3: Calculate the interest earned after the final 6 months (months 13-18).
\(4\% \times \$10{,}506 = \$420.24\)
Balance after 18 months: \(\$10{,}506 + \$420.24 = \$10{,}926.24\)
Step 4: Calculate the total interest paid.
Total interest = Final balance - Initial investment = \(\$10{,}926.24 - \$10{,}000 = \$926.24\)
Alternatively: Total interest = \(\$200 + \$306 + \$420.24 = \$926.24\)
The correct answer is E.
View Explanation
Correct Answer: A
Let's call the constant multiple x.
\( 2000(x)(x)(x) = 250,000 \)
\( 2000(x^3) = 250,000 \)
\( x^3 = \frac{250,000}{2,000} = 125 \)
\( x = 5 \)
Therefore, the population gets five times bigger each hour.
At 3 p.m., there were \( 2000(5)(5) = 50,000 \) bacteria.
The correct answer is A.
View Explanation
Correct Answer: D
First, determine the current population:
4 hours ago: \(1{,}000\)
2 hours ago: \(1{,}000 \times 2 = 2{,}000\)
Now: \(2{,}000 \times 2 = 4{,}000\)
Next, track the population forward from now until it exceeds 250,000:
In 2 hours: \(4{,}000 \times 2 = 8{,}000\)
In 4 hours: \(8{,}000 \times 2 = 16{,}000\)
In 6 hours: \(16{,}000 \times 2 = 32{,}000\)
In 8 hours: \(32{,}000 \times 2 = 64{,}000\)
In 10 hours: \(64{,}000 \times 2 = 128{,}000\)
In 12 hours: \(128{,}000 \times 2 = 256{,}000\)
At 12 hours from now, the population reaches 256,000, which exceeds 250,000.
Alternatively, using the geometric progression formula \(a_n = a_1 \times r^{n-1}\), where \(a_1\) is the first term and \(r\) is the common ratio:
Here, \(a_1 = 4{,}000\) (current population) and \(r = 2\). We need to find the smallest \(n\) such that \(4{,}000 \times 2^{n-1} > 250{,}000\).
Dividing both sides by 4,000: \(2^{n-1} > 62.5\)
Since \(2^5 = 32\) and \(2^6 = 64\), we need \(n - 1 = 6\), so \(n = 7\).
This means 6 doubling cycles (each 2 hours) are needed: \(6 \times 2 = 12\) hours.
The correct answer is D.