23 QUESTIONS AND ANSWERS ON LUMP SUM AND COMPOUND INTEREST

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A lump sum is a single payment made or received at a particular time. 

When answering lump sum questions, future value, present value, interest rate, compounding frequency, and number of years are crucial factors to consider. 

A lump sum's present value is its current worth, while its future value is how much it will be worth in the future given a specified interest rate.

The interest rate of a lump sum is the interest earned on a lump sum, expressed as a percentage of the principal.

The number of times that compound interest will be applied to a lump sum in a year is known as the compounding frequency.

The number of years is the amount of time needed for a lump sum to eventually reach a specific target amount.

Major formulas in lump sum

1. To calculate the future value of a lump sum, we used the following formula.

$$FV=PV\left(1+\frac{r}{m}\right)^{n × m}$$

2. To calculate the present value of a lump sum, the formula is as follows:

$$PV=\frac{FV}{\left(1+\frac{r}{m}\right)^{n \times m}}$$

3. The interest rate of a lump sum can be calculated as follow:

$$r=m ×\left(\left(\frac{FV}{PV}\right)^{\frac{1}{nm}}-1\right)$$

4. To calculate the number of years for a lump sum, we use this formula.

$$n=\frac{\log \left(\frac{FV}{PV}\right)}{m × \log\left(1+\frac{r}{m}\right)}$$

We would be applying these formulas as we solved these 22 questions and answers on the lump sum.

These 22 lump sum questions and answers are meant to help students better understand lump sum and compound interest.

Question 1

Scott purchased a used automobile with a loan of €7,200 from his uncle. How much did he paid back after 9 months at an annual interest rate of 6.2%?

Solution:

The total amount he will paid is the future of the lump sum after 9 months. The future value of a lump sum with compound interest can be calculated as follow:

$FV=PV(1+i)^n$

Here PV is 7,200, I is 6.2 expressed in percentage which is 0.062. n is 0.75 because 9 month is three-quarters of one year.

$FV=7200(1+0.062)^{0.75}$

$FV=7200(1.062)^{0.75}$

$FV=7532.27$

Question 2

Assume that N20,000 is invested for ten years at 10% annual interest . Find the future value if interest is compound daily. Please assume a 365-day year.

Solution:

Here, interest compound daily, hence, we used the formula for interest that compounds intra-yearly, which is:

$FV=PV(1+\frac{i}{m})^{nm}$

Here, m is 365, hence,

$FV=20,000\left(1+\frac{0.1}{365}\right)^{365}\times 10$

$FV=20,000(1+0.00027397)^{3650}$

$FV=20,000(1.00027397)^{3650}=N54357.67$

Question 3

James and John, two business partners, have decided to contribute an equal amount to their company. John will contribute N10,000 right away.

James intends to make an equivalent contribution in three years time, when he anticipates earning a large sum of money.

How much will he need to invest in three years time to match John's investment now, assuming an interest rate of 6% compounded semiannually?

Solution:

We need to find the future value of N10,000 in three years time. Semi-annually means that m is 2

$FV=10,000\left(1+\frac{0.06}{2}\right)^{3×2}$

$FV=10,000(1.03)^6$

$FV=10,000(1.1940522)$

$FV=11940.522$

James should therefore contribute N11,940.522 in three years time to equal John's N10,000 investment today

Question 4

With €45,000, I opened a savings account. I withdrew €23,000 from the account after three years.

I added another €11,000 to the account six years after it was first opened.

The interest rate for the first five year is 6.5% compounded annually and the interest rate for the remainder of the period is 4.2%.

Determine the balance in my account at the end of the 10th year?

Solution:

The future value for the first three years is

$FV=45,000(1+0.065)^3$

$FV=45,000(1.065)^3$

$FV=45,000(1.207949625)$

$FV=54357.733$

Now, let's subtract the 23,000 withdraw after the third year 

$54357.733-23,000=31,357.733$

Let's now calculate the future value of the balance's over the next two years when the interest rate is 6.5%.

$FV=31,357.733(1+0.065)^2$

$FV=31,357.733(1.065)^2=35,566.721$

The future value of the last year before N11,000 was deposited.

$FV=35,566.721(1+0.042)^1=37,060.524$

After 11,000 was deposited, the balance is

$37,060.524+11,000=48060.524$

Now for the final 4 years, the future value will be:

$FV=48060.524(1+0.042)^4=56657.757$

Question 5

I want to triple my money in 4 years time . At what annual interest rate, compounded monthly, do I need to invest the money to achieve my target.

Solution:

Here, the future value is 3 since we want to triple the money. Present value is assumed to be 1 and n=4, and m=12.

$3=1(1+\frac{r}{12})^{12×4}$

$3=(1+\frac{r}{12})^{48}$

Now, let's divide the power by $\frac{1}{48}$

$3^\frac{1}{48}=1+\frac{r}{12}$

$1.023152-1=\frac{r}{12}$

$0.023152×12=r$

$r=0.2778$

Expressed in percentage, 

$r=27.78\%$

So, I need to invest at a rate of 27.78% to triple my money in four years.

So, in order to triple my money in four years, I must invest at a rate of 27.78%.

Question 6

If the effective interest rate is 5% annually, what is the present value of N6,000 that will be payable in 10 years?

Solution:

The formula to find the present value of a lump sum is:

$PV=\frac{FV}{(1+i)^n}$

$PV=\frac{6,000}{(1.0.5)^{10}}$

$PV=\frac{6,000}{1.62889}$

$PV=N3,683.49$

Question 7

Find the interest rate at which N8000, if compounded quarterly for 8 years, will grow to €11,672.12?

Solution:

The formula for calculating  the interest rate is:

$r=m ×\left(\left(\frac{FV}{PV}\right)^{\frac{1}{nm}}-1\right)$

$r=4 ×\left(\left(\frac{11,672.12}{8000}\right)^{\frac{1}{8×4}}-1\right)$

$r=4 ×\left(\left(1.459015\right)^{\frac{1}{32}}-1\right)$


$r=4 ×\left(1.011875-1\right)$

$r=4 ×\left(0.011875\right)$

$r=0.0475$

Expressed in percentage,

$r=4.75\%$

Question 8

What interest rate must be used to grow €5,000 into €10,000 over a five-year period given that interest is compounded monthly.

Solution:

$r=12\left(\left(\frac{10,000}{5,000}\right)^{\frac{1}{5×12}}-1\right)$

$r=12\left(\left(2\right)^{\frac{1}{60}}-1\right)$

$r=12\left(\left(1.0116194\right)-1\right)$

$r=12\left(0.0116194\right)$

$r=0.1394$

$r=13.94\%$

Question 9

A man invests N1000 at 10% compounded weekly and plans to hold this investment for five years. How much will he have at the end of the holding period?

Solution:

$FV=N1,000\left(1+\frac{0.1}{52}\right)^{5×52}$

$FV=N1,000(1.001923)^{260}$

$FV=N1000(1.16479)$

$FV=N1647.9$

Question 10

A company needs N10,000 in 5 years to replace a piece of equipment. How much must be invested now at interest rate of 8% compounded daily in order to provide for the replacement.

Solution:

10,000 is the future value, 5 years time, interest rate us 0.08 expressed in percentage, Number of years is 5.

$PV=\frac{10,000}{\left(1+\frac{0.08}{365}\right)^{5\times 365}}$

$PV=\frac{10,000}{\left(1+0.000219178\right)^{1825}}$

$PV=\frac{10,000}{\left(10.000219178\right)^{1825}}$

$PV=\frac{10,000}{1.491759}$

$PV=6703.49567$

Question 11

David intends to make a lump-sum investment of €30,000 at a 12% annual interest rate. How long till the amount doubles, to the tune of €60,000.

Solution:

To calculate the number of years of a lump sum, we use the following formula

$n=\frac{\log\left(\frac{FV}{PV}\right)}{m×\log\left(1+\frac{r}{m}\right)}$

m=1, FV=60,000, PV=30,000, r=0.12

$n=\frac{\log\left(\frac{60,000}{30,000}\right)}{1×\log\left(1+\frac{0.12}{1}\right)}$

$n=\frac{\log 2}{\log 1.12}$

$n=\frac{0.301029996}{0.049218023}$

$n=6.12$

Therefore, it will take approximately 6.12 years to double €30,000.

Question 12

Elizabeth invested N30,000 for a period of six years at a rate of 12%. If it compounds monthly, what will the compounded amount be?

Solution:

$FV=N30,000\left(1+\frac{0.12}{12}\right)^{6\times 12}$

$FV=N30,000(1+0.01)^{72}$

$FV=N30,000(1.01)^{72}$

$FV=N30,000(2.047099)$

$FV=N61,412.97$

Question 13

If the lump sum is compounded quarterly, How long will it take €15,000 to amount to €20,000 if its annual interest is 20%?

Solution:

Recalled that

$n=\frac{\log\left(\frac{FV}{PV}\right)}{m×\log\left(1+\frac{r}{m}\right)}$

$n=\frac{\log\left(\frac{20,000}{15,000}\right)}{4×\log\left(1+\frac{0.2}{4}\right)}$

$n=\frac{\log\left(\frac{4}{3}\right)}{4×\log(1+0.05)}$

$n=\frac{\log 1.333333}{4×\log 1.05}$

$n=\frac{0.124928}{0.0847572}$

$n=1.47$ years.

Question 14

How long will it take for a sum of money with compound interest of 6% per annum to double?

Solution:

PV=1 and FV=2 since we want to double the PV.

$n=\frac{\log\left(\frac{2}{1}\right)}{1×\log\left(1+\frac{0.06}{1}\right)}$

$n=\frac{\log 2}{\log(1+0.06)}$

$n=\frac{\log 2}{\log 1.06}$

$n=\frac{0.301029996}{0.025305865}$

$n=11.9$ years.

Question 15

Joshua made a one-time bank investment of €12,000. The payment doubled to €24,000 after five years. How long, starting from the time it was initially invested, will it take for the lump sum to reach €384,013.75 if interest compounds annually?

Solution:

We first need to know the interest rate for the five year before we can determine the number of years.

Remember that 

$r=m ×\left(\left(\frac{FV}{PV}\right)^{\frac{1}{nm}}-1\right)$

Here, m=1, FV=24,000 and PV=12,000

$r=1 ×\left(\left(\frac{24,000}{12}\right)^{\frac{1}{5×1}}-1\right)$

$r=1 ×\left(\left(\frac{24,000}{12}\right)^{\frac{1}{5}}-1\right)$

$r=1((2)^{\frac{1}{5}}-1)$

$r=1(1.1486984-1)$

$r=0.1487$

Expressed in percentage 

$r=14.87\%$

Having gotten the interest rate, we can calculate the interest rate 

$n=\frac{\log\left(\frac{384,013.75}{12,000}\right)}{1\times\log(1+0.1487)}$

$n=\frac{\log(32.00115)}{\log(1.1487)}$

$n=\frac{1.5051656}{0.06021}$

$n=25$ years.

Therefore, it will take 25 years for €12,000 to amount to €384,013.75 if it invested at 14.87 interest per annum.

Question 16 

You borrowed N9,000 from your brother and agreed to return N10,000 to him in five years time.

Assuming annual compound interest, find the interest rate for this loan.

Solution:

Here, FV=10,000, PV=9,000 and number of years is 5, m is 1 because the compounding frequency is annually

$r=1×\left(\left(\frac{10,000}{9,000}\right)^{1/5}-1\right)$

$r=(1.1111)^{\frac{1}{5}}-1$

$r=1.0213-1$

$r=0.0213$

Expressed in percentage

r=2.13% 

Question 17

You agreed to borrow N140,000 from a bank, and after four years, you will pay it back in one lump payment of N161,000. Find the loan's annual interest rate assuming that interest is compounded every three months?

Solution:

Just to be clear, every three months means the interest is compounded quarterly.

$r=4×\left(\left(\frac{161,000}{140,000}\right)^{\frac{1}{4×4}}-1\right)$

$r=4×\left(\left(1.15\right)^{\frac{1}{16}}-1\right)$

$r=4(1.008773-1)$

$r=4(0.008773)$

$r=0.0351$

Expressed in percentage

r=3.51%.

Therefore, the interest rate accruing on the loan is 3.51%.

Question 18

In ten years, James PLC will have the chance to collect N750,000. What is the present value of the money if the company can earn 15% of it?

Solution:

The present value of the lump sum can be calculated as follows:

$PV=\frac{FV}{\left(1+\frac{r}{m}\right)^{n×m}}$

m=1, FV=750,000, n=10, r=0.15

$PV=\frac{750,000}{\left(1+\frac{0.15}{1}\right)^{10×1}}$

$PV=\frac{750,000}{\left(1+0.15\right)^{10}}$

$PV=\frac{750,000}{4.0455577}$

$PV=185,388.53$

Question 19

N10,000 is continuously compounded at 10%. What will the investment be worth at the end of the fifth years?

Solution:

The future value of a lump sum that compounds continuously can be calculated as follows:

$FV=PV×e^{r×n}$

$FV=10,000×e^{0.1×5}$

$FV=10,000×1.648721$

$FV=16,487.21$

Question 20

Find the present value of N12,000 promised to be paid in 8 years if interest is 15% compounding semi-annually.

Solution:

$PV=\frac{FV}{\left(1+\frac{r}{m}\right)^{n×m}}$

m=2 because interest compounds semi-annually.

$PV=\frac{12,000}{\left(1+\frac{0.15}{2}\right)^{8×2}}$

$PV=\frac{12,000}{\left(1.075\right)^{16}}$

$PV=\frac{12,000}{3.18079}$

$PV=N3,772.65$

Question 21

What interest rate, compounded monthly, will cause N50,000 to quadruple to N200,000 over the course of five years?

Solution:

$r=12\left(\left(\frac{200,000}{50,000}\right)^{\frac{1}{60}}-1\right)$

$r=12\left((4)^{\frac{1}{60}}-1\right)$

$r=12(1.02337-1)$

$r=0.2805$

Expressed in percentage, 

$r=28.05\%$

Question 22

When discounted at 8% annually, what is the present value of N50,000 that would be received in 4 years?

Solution:

$PV=\frac{50,000}{(1+0.08)^4}$

$PV=\frac{50,000}{(1.08)^4}=N36,751.49$

Question 23

Suppose Daniel needs N1,000 one year from now. How much should invest today to achieve that goal if the annual interest rate is 5 percent.

Solution:

$PV=\frac{1,000}{(1+0.05)^1}$

$PV=\frac{1,000}{1.05}=N952.38$

Put your knowledge into test by trying out this question.

A bank pays 6% annual interest rate, compounded monthly. How long will it take to double the money in this account






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