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SL 1.4—Financial apps – compound interest, annual depreciation - IB Questionbank | RevisionDojo

Practice SL 1.4—Financial apps – compound interest, annual depreciation with authentic IB Mathematics Analysis and Approaches (AA) exam questions for both SL and HL students. This question bank mirrors Paper 1, 2, 3 structure, covering key topics like functions and equations, calculus, complex numbers, sequences and series, and probability and statistics. Get instant solutions, detailed explanations, and build exam confidence with questions in the style of IB examiners.

Paper

Level

Question 1

SL & HLPaper 2

After $7$ years of investing in an account that pays a fixed annual compound interest rate of $4.5\%$ , the total value of the investment has grown to $\$ 9500$.

Calculate the initial amount invested, correct to the nearest dollar. [3]

Question 2

SLPaper 1

A company records annual revenue as $R_n=R_0(1+g)^n$ with growth rate $g$ (could be negative).

If $g>0$ , find $n$ so that $R_n=2R_0$ .

[2]

If $g<0$ (decline), show that the half-life $H$ (time until $R_H=\tfrac12 R_0$ ) equals $H=\dfrac{\log(\tfrac12)}{\log(1+g)}$ . Explain why $H>0$ .

[3]

Prove the law $\log_a(xy)=\log_a x+\log_a y$ from the exponential identity $a^{u+v}=a^u a^v$ using part 1’s reasoning structure.

[3]

Question 3

SL & HLPaper 2

At the start of July, a colony of bacteria has a population of $1500$ .

After $4$ hours, the population of the bacteria is $2100$ .

The population of this colony can be calculated using the formula: $P = a \cdot b^t$

where $P$ is the population of the bacteria $t$ hours after the start of July.

By finding the value of $a$ and the value of $b$ , calculate the population of the bacteria after $7$ hours from the start of July.

Give your answer correct to the nearest integer. [5]

At the start of July, a colony of bacteria has a population of $1500$ .

After $4$ hours, the population of the bacteria is $2100$ .

The population of this colony can be calculated using the formula: $P = a \cdot b^t$

where $P$ is the population of the bacteria $t$ hours after the start of July.

By finding the value of $a$ and the value of $b$ , calculate the population of the bacteria after $7$ hours from the start of July.

Give your answer correct to the nearest integer.

[5]

Question 4

SLPaper 1

Two savings plans for the same horizon $n$ years with annual rate $r>0$ :

Plan A: invest all $P today.
Plan B: invest $\frac{P}{n}$ at each year-end for $n$ years.

State expressions for the future values $V_A$ and $V_B$ , and simplify the expression for $V_B$ .

[3]

Prove that $V_A>V_B$ for $n\ge2$ .

[3]

Show that the relative shortfall $\dfrac{V_A-V_B}{V_A}$ equals $1-\dfrac{(1+r)^n-1}{nr(1+r)^n}$ . Hence, describe the change in the relative shortfall as $r$ decreases, and as $n$ increases.

[3]

Question 5

SLPaper 1

An account pays a fixed annual interest rate $r>0$ compounded once per year.

A lump sum $P$ is invested today. Write the value after $n$ years.

[2]

A target $k>1$ is set so that the balance must reach $kP$ . Find the least integer $n$ required, using logarithms.

[3]

Two different rates $r_1,r_2>0$ are offered for the same period of $n$ years. Prove using log/exponent laws that $P(1+r_1)^n>P(1+r_2)^n$ iff $r_1>r_2$ .

[3]

Question 6

SLPaper 2

Maria is planning for her future and wants to invest in an account that offers a 4% interest rate per year, compounded annually.

Calculate the amount Maria needs to invest today to reach a future value of $20,000 in 10 years. Give your answer to the nearest dollar.

[4]

After investing, Maria finds another fund that offers a 5% interest rate, compounded annually. Calculate how much less she would need to invest today with this higher interest rate to reach the same goal of $20,000 in 10 years.

[3]

Question 7

SL & HLPaper 2

After $7$ years of investing in an account that pays a fixed annual compound interest rate of $4.5\%$ , the total value of the investment has grown to $9500.

Calculate the initial amount invested, correct to the nearest dollar.

[3]

Question 8

SL & HLPaper 2

Sania invests $6000 at a rate of $3.1\%$ per year compound interest.

Work out the interest earned on the investment at the end of $2$ years.

[3]

Question 9

SL & HLPaper 2

An investor has two options:

Option 1: Invest $P$ at an annual compound interest rate of $r_1\%$ . After 12 years, the investment grows to $2P$ .

Option 2: Invest $P$ at an annual compound interest rate of $r_2\%$ , compounded semi-annually.

Determine the value of $r_2$ , correct to 2 decimal places, such that the total value of the investment under Option 2 after 10 years is equal to the total value of the investment under Option 1 after 10 years.

[6]

Question 10

SL & HLPaper 2

David invests $1200 at a rate of $2.5\%$ per year compound interest.

Calculate the amount David has after $8$ years. Give your answer correct to the nearest dollar.

[2]

Paper

Level

Question 1

SL & HLPaper 2

After $7$ years of investing in an account that pays a fixed annual compound interest rate of $4.5\%$ , the total value of the investment has grown to $\$ 9500$.

Calculate the initial amount invested, correct to the nearest dollar. [3]

Question 2

SLPaper 1

A company records annual revenue as $R_n=R_0(1+g)^n$ with growth rate $g$ (could be negative).

If $g>0$ , find $n$ so that $R_n=2R_0$ .

[2]

If $g<0$ (decline), show that the half-life $H$ (time until $R_H=\tfrac12 R_0$ ) equals $H=\dfrac{\log(\tfrac12)}{\log(1+g)}$ . Explain why $H>0$ .

[3]

Prove the law $\log_a(xy)=\log_a x+\log_a y$ from the exponential identity $a^{u+v}=a^u a^v$ using part 1’s reasoning structure.

[3]

Question 3

SL & HLPaper 2

At the start of July, a colony of bacteria has a population of $1500$ .

After $4$ hours, the population of the bacteria is $2100$ .

The population of this colony can be calculated using the formula: $P = a \cdot b^t$

where $P$ is the population of the bacteria $t$ hours after the start of July.

By finding the value of $a$ and the value of $b$ , calculate the population of the bacteria after $7$ hours from the start of July.

Give your answer correct to the nearest integer. [5]

At the start of July, a colony of bacteria has a population of $1500$ .

After $4$ hours, the population of the bacteria is $2100$ .

The population of this colony can be calculated using the formula: $P = a \cdot b^t$

where $P$ is the population of the bacteria $t$ hours after the start of July.

By finding the value of $a$ and the value of $b$ , calculate the population of the bacteria after $7$ hours from the start of July.

Give your answer correct to the nearest integer.

[5]

Question 4

SLPaper 1

Two savings plans for the same horizon $n$ years with annual rate $r>0$ :

Plan A: invest all $P today.
Plan B: invest $\frac{P}{n}$ at each year-end for $n$ years.

State expressions for the future values $V_A$ and $V_B$ , and simplify the expression for $V_B$ .

[3]

Prove that $V_A>V_B$ for $n\ge2$ .

[3]

Show that the relative shortfall $\dfrac{V_A-V_B}{V_A}$ equals $1-\dfrac{(1+r)^n-1}{nr(1+r)^n}$ . Hence, describe the change in the relative shortfall as $r$ decreases, and as $n$ increases.

[3]

Question 5

SLPaper 1

An account pays a fixed annual interest rate $r>0$ compounded once per year.

A lump sum $P$ is invested today. Write the value after $n$ years.

[2]

A target $k>1$ is set so that the balance must reach $kP$ . Find the least integer $n$ required, using logarithms.

[3]

Two different rates $r_1,r_2>0$ are offered for the same period of $n$ years. Prove using log/exponent laws that $P(1+r_1)^n>P(1+r_2)^n$ iff $r_1>r_2$ .

[3]

Question 6

SLPaper 2

Maria is planning for her future and wants to invest in an account that offers a 4% interest rate per year, compounded annually.

Calculate the amount Maria needs to invest today to reach a future value of $20,000 in 10 years. Give your answer to the nearest dollar.

[4]

[3]

Question 7

SL & HLPaper 2

After $7$ years of investing in an account that pays a fixed annual compound interest rate of $4.5\%$ , the total value of the investment has grown to $9500.

Calculate the initial amount invested, correct to the nearest dollar.

[3]

Question 8

SL & HLPaper 2

Sania invests $6000 at a rate of $3.1\%$ per year compound interest.

Work out the interest earned on the investment at the end of $2$ years.

[3]

Question 9

SL & HLPaper 2

An investor has two options:

Option 1: Invest $P$ at an annual compound interest rate of $r_1\%$ . After 12 years, the investment grows to $2P$ .

Option 2: Invest $P$ at an annual compound interest rate of $r_2\%$ , compounded semi-annually.

[6]

Question 10

SL & HLPaper 2

David invests $1200 at a rate of $2.5\%$ per year compound interest.

Calculate the amount David has after $8$ years. Give your answer correct to the nearest dollar.

[2]

Paper

Level

Question 1

SL & HLPaper 2

After $7$ years of investing in an account that pays a fixed annual compound interest rate of $4.5\%$ , the total value of the investment has grown to $\$ 9500$.

Calculate the initial amount invested, correct to the nearest dollar. [3]

Question 2

SLPaper 1

A company records annual revenue as $R_n=R_0(1+g)^n$ with growth rate $g$ (could be negative).

If $g>0$ , find $n$ so that $R_n=2R_0$ .

[2]

If $g<0$ (decline), show that the half-life $H$ (time until $R_H=\tfrac12 R_0$ ) equals $H=\dfrac{\log(\tfrac12)}{\log(1+g)}$ . Explain why $H>0$ .

[3]

Prove the law $\log_a(xy)=\log_a x+\log_a y$ from the exponential identity $a^{u+v}=a^u a^v$ using part 1’s reasoning structure.

[3]

Question 3

SL & HLPaper 2

At the start of July, a colony of bacteria has a population of $1500$ .

After $4$ hours, the population of the bacteria is $2100$ .

The population of this colony can be calculated using the formula: $P = a \cdot b^t$

where $P$ is the population of the bacteria $t$ hours after the start of July.

By finding the value of $a$ and the value of $b$ , calculate the population of the bacteria after $7$ hours from the start of July.

Give your answer correct to the nearest integer. [5]

At the start of July, a colony of bacteria has a population of $1500$ .

After $4$ hours, the population of the bacteria is $2100$ .

The population of this colony can be calculated using the formula: $P = a \cdot b^t$

where $P$ is the population of the bacteria $t$ hours after the start of July.

By finding the value of $a$ and the value of $b$ , calculate the population of the bacteria after $7$ hours from the start of July.

Give your answer correct to the nearest integer.

[5]

Question 4

SLPaper 1

Two savings plans for the same horizon $n$ years with annual rate $r>0$ :

Plan A: invest all $P today.
Plan B: invest $\frac{P}{n}$ at each year-end for $n$ years.

State expressions for the future values $V_A$ and $V_B$ , and simplify the expression for $V_B$ .

[3]

Prove that $V_A>V_B$ for $n\ge2$ .

[3]

Show that the relative shortfall $\dfrac{V_A-V_B}{V_A}$ equals $1-\dfrac{(1+r)^n-1}{nr(1+r)^n}$ . Hence, describe the change in the relative shortfall as $r$ decreases, and as $n$ increases.

[3]

Question 5

SLPaper 1

An account pays a fixed annual interest rate $r>0$ compounded once per year.

A lump sum $P$ is invested today. Write the value after $n$ years.

[2]

A target $k>1$ is set so that the balance must reach $kP$ . Find the least integer $n$ required, using logarithms.

[3]

Two different rates $r_1,r_2>0$ are offered for the same period of $n$ years. Prove using log/exponent laws that $P(1+r_1)^n>P(1+r_2)^n$ iff $r_1>r_2$ .

[3]

Question 6

SLPaper 2

Maria is planning for her future and wants to invest in an account that offers a 4% interest rate per year, compounded annually.

Calculate the amount Maria needs to invest today to reach a future value of $20,000 in 10 years. Give your answer to the nearest dollar.

[4]

[3]

Question 7

SL & HLPaper 2

After $7$ years of investing in an account that pays a fixed annual compound interest rate of $4.5\%$ , the total value of the investment has grown to $9500.

Calculate the initial amount invested, correct to the nearest dollar.

[3]

Question 8

SL & HLPaper 2

Sania invests $6000 at a rate of $3.1\%$ per year compound interest.

Work out the interest earned on the investment at the end of $2$ years.

[3]

Question 9

SL & HLPaper 2

An investor has two options:

Option 1: Invest $P$ at an annual compound interest rate of $r_1\%$ . After 12 years, the investment grows to $2P$ .

Option 2: Invest $P$ at an annual compound interest rate of $r_2\%$ , compounded semi-annually.

[6]

Question 10

SL & HLPaper 2

David invests $1200 at a rate of $2.5\%$ per year compound interest.

Calculate the amount David has after $8$ years. Give your answer correct to the nearest dollar.

[2]

SL 1.4—Financial apps – compound interest, annual depreciation Questionbank