## Simple interest questions Compound interest questions with solutions SSC CGL set 79

10 Simple interest questions Compound interest questions to solve in 15 minutes. Verify correctness from answers. Learn to solve quickly from solutions.

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- Compound interest questions.
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### 10 Simple interest questions Compound interest questions for SSC CGL Set 79 - time to solve 15 minutes

**Problem 1.**

A sum of money placed at compound interest doubles itself in 5 years. It will amount to 16 times of itself at the same rate of interest in,

- 20 years
- 15 years
- 10 years
- 12 years

**Problem 2.**

A man gave 50% of his savings of Rs. 84100 to his wife and divided the remaining sum among his two sons A and B aged 15 and 13 years respectively. He divided it in such a way that each of his sons, when they attain the age of 18 years, would receive the same amount at 5% compound interest per annum. The share of B was,

- Rs. 22000
- Rs. 20000
- Rs. 22050
- Rs. 20050

**Problem 3.**

A sum of Rs. 13360 was borrowed at $8\frac{3}{4}$% per annum compound interest and paid back in two years in two equal installments. What was the amount of each installment?

- Rs. 7500
- Rs. 5769
- Rs. 7569
- Rs. 7009

**Problem 4.**

A person deposited a sum of Rs. 6000 in a bank at 5% simple interest. another person deposited Rs. 5000 at 8% per annum compound interest. After two years, the difference of their interests will be,

- Rs. 230
- Rs. 600
- Rs. 832
- Rs. 232

**Problem 5.**

On a certain sum of money, the difference between the compound interest for a year payable half-yearly, and the simple interest for a year is Rs. 180. If the rate of interest in both cases is 10%, then the sum is,

- Rs. 62000
- Rs. 60000
- Rs. 72000
- Rs. 54000

#### Problem 6.

If the compound interest on a certain sum for two years at 12% per annum is Rs. 2544, the simple interest on it at the same rate after 2 years will be,

- Rs. 2400
- Rs. 2500
- Rs. 2440
- Rs. 2480

**Problem 7.**

On a certain principal the compound interest compounded annually for the second year at 10% per annum is 132. The principal is,

- Rs. 1000
- Rs. 1200
- Rs. 1320
- Rs. 1250

**Problem 8.**

On a certain principal of Rs. P, rate of interest is 2r% per annum. The person will then get after 3 years at compound interest a sum of,

- $\text{Rs. }P\left(1+\displaystyle\frac{r}{100}\right)^3$
- $\text{Rs. }P\left(1+\displaystyle\frac{r}{50}\right)^3$
- $\text{Rs. }\displaystyle\frac{6Pr}{100}$
- $\text{Rs. }3P\left(1+\displaystyle\frac{r}{100}\right)^3$

**Problem 9.**

A sum of money amounts to Rs. 850 in 3 years and to Rs. 925 in 4 years at some rate of simple interest. The sum is,

- Rs. 550
- Rs. 625
- Rs. 600
- Rs. 700

**Problem 10.**

What annual installment will discharge a debt of Rs. 6450 due in 4 years at 5% simple interest?

- Rs. 1950
- Rs. 1500
- Rs. 1935
- Rs. 1800

### Answers to the 10 simple interest questions compound interest questions SSC CGL Set 79

**Problem 1.** **Answer: **Option a: 20 years.

**Problem 2.** **Answer: **Option b: Rs. 20000.

**Problem 3.** **Answer: **Option c: Rs. 7569.

**Problem 4. Answer: **Option d: Rs. 232.

**Problem 5.** **Answer: **Option c: Rs. 72000.

**Problem 6.** **Answer: **Option a: Rs. 2400.

**Problem 7. Answer: **Option b: Rs. 1200.

**Problem 8.** **Answer: **Option b: $\text{Rs. }P\left(1+\displaystyle\frac{r}{50}\right)^3$.

**Problem 9.** **Answer: **Option b: Rs. 625.

**Problem 10.** **Answer: **Option d: Rs. 1800.

### Solutions to the simple interest questions compound interest questions SSC CGL set 79 - time to solve was 15 mins

**Problem 1.**

A sum of money placed at compound interest doubles itself in 5 years. It will amount to 16 times of itself at the same rate of interest in,

- 20 years
- 15 years
- 10 years
- 12 years

**Solution 1: Problem analysis and solution**

The growth of principal $P$ at compound interest rate $r$ over $n$ time-periods is governed by the relation,

$P_n=P(1+r)^n$, where $n$ is usually in years.

By first given statement, principal $P$ becomes at the end of 5 years,

$P_5=P(1+r)^5=2P$

If we invest this principal at the same interest rate for 5 more years at the beginning of 6th year, we will get principal $P$ grown in 10 years as,

$P_{10}=P_5(1+r)^5=2P_5=4P$

In other words, with the given rate of interest by the compound growth,

Every 5 years, the principal at the beginning of investment period will get doubled.

As 16 times is equivalent to 4 times doubling, principal $P$ will grow to 16 times its original amount in, $4\times{5}=20$ years.

This happens following the very basic principles of compound growth.

**Answer:** Option a: 20 years.

**Key concepts used:** ** Compound interest **--

**.**

*Compound growth -- Solving in mind*This is solved in mind as the conceptual steps are simple and calculations easy.

**Problem 2.**

A man gave 50% of his savings of Rs. 84100 to his wife and divided the remaining sum among his two sons A and B aged 15 and 13 years respectively. He divided it in such a way that each of his sons, when they attain the age of 18 years, would receive the same amount at 5% compound interest per annum. The share of B was,

- Rs. 22000
- Rs. 20000
- Rs. 22050
- Rs. 20050

**Solution 2: Problem analysis and conceptual solution**

The first thing we evaluate is the amount that was divided between the sons A and B. This is 50% of Rs. 84100, that is, Rs. 42050.

If these two amounts are $P_A$ and $P_B$, the first would have grown for three years and the second for 5 years to become equal when the two sons attain the age of 18 years, the amount growing at the compound interest rate of 5%.

The compound rate of growth is governed by the standard well-known relation,

$P_n=P(1+r)^n$, where $P$ is the beginning principal, $r$ interest rate per annum, $n$ number of years of growth and $P_n$ original principal $P$ grown in $n$ years.

In our problem then,

$P_{A3}=P_A(1.05)^3=P_{B5}=P_B(1.05)^5$, 5% interest is equivalent to its decimal 0.05 when divided by 100,

Or, $P_A=(1.05)^2P_B$, where $P_A$ and $P_B$ are the original shares of A and B,

Or, $P_A+P_B=(1+1.1025)P_B$, $P_B$ is just added to both sides of the equation,

$=2.1025P_B$,

$=42050$, the total share of two sons

So, $P_B=20000$.

**Answer:** Option b: Rs. 20000.

**Key concepts used:** * Compound interest growth* --

**Event sequencing***.*

**-- Solving in mind**Though evaluating $(1.05)^2$ mentally may not be easy, this problem could be solved mentally because of simplicity of the answer.

**Problem 3.**

A sum of Rs. 13360 was borrowed at $8\frac{3}{4}$% per annum compound interest and paid back in two years in two equal installments. What was the amount of each installment?

- Rs. 7500
- Rs. 5769
- Rs. 7569
- Rs. 7009

**Solution 3: Problem analysis and solution**

Let us look at the problem of two equal installments by breaking up the two years into two parts.

**At the end of 1st year:** total amount $A_1$ is, principal Rs. 13360 at beginning of 1st year plus interest earned over a year.

Assuming that $x$ amount of installment is returned to the lender at the end of first year, amount that remained would be $A_1 -x$. This amount must grow to amount $x$ in the second year if invested and returned as second installment to wipe out the dues fully.

The basic mechanism of two equal installments is thus,

After returning the first installment at the beginning of second year, the remaining amount that would have accumulated through the first year if invested, must grow through the second year to equal the first installment.

Let us calculate the values,

$A_1=13360\left(1+\displaystyle\frac{7}{80}\right)$, the interest $8\frac{3}{4}$% is converted to fraction (and dividing it by 100) as it would be easier to deal with fraction calculations,

$=167\times{87}$, we didn't expect fractional rupees and that was another reason for fraction calculation which is faster and can be carried out mentally.

To fulfill equal installment requirement at the end of second year, amount would be,

$A_2=(A_1-x)\left(1+\displaystyle\frac{7}{80}\right)$

$=(167\times{87}-x)\times{\displaystyle\frac{87}{80}}$

$=x$

Or, $167\times{87}\times{87}=(80+87)x=167x$,

Or, $x=87\times{87}=7569$.

**Answer:** Option c: Rs. 7569.

**Key concepts used:** * Event sequencing in mapping the installments over two years and forming the relations* --

*Compound interest -- Installments -- Percentage as fractions for quick calculation.*We didn't solve the problem wholly in mind and wrote down a part of the products, but ultimately the arithmetic calculation was only evaluating $87\times{87}$ that was easy and quick.

#### Alternate Installment formula based solution

The general equal installment mechanism on borrowings is,

Equal installments are paid at the end of each year ending at the end of borrowing period. It is assumed that each such installment would grow up to the end of borrow period at the same rate of interest at which the original amount was borrowed. The total of all these installments with interest rate applied up to the end of borrow period must equal the borrowed amount that would have grown up to the end of borrow period at the same interest rate. This mechanism ensures that the lender does not lose on interests that would have accrued out of his capital if he had invested it up to the end of borrow period instead of lending it.

For example, borrowed principal $P$, compound interest rate $r$% to be paid back in 3 years will then be governed by the following equality if $x$ be the equal installments paid at the end of 1st year, 2nd year and 3rd year,

$x\left(1+\displaystyle\frac{r}{100}\right)^2 +x\left(1+\displaystyle\frac{r}{100}\right)+x$

$=P\left(1+\displaystyle\frac{r}{100}\right)^3$.

This relation can be extended to borrow cases over any number of years $n$ as well as for borrow under simple interest scheme.

Applying this relation to our problem we have then,

$x\left(1+\displaystyle\frac{7}{80}\right)+x=13360\left(1+\displaystyle\frac{7}{80}\right)^2$,

Or, $\displaystyle\frac{167}{80}x=167\times{\displaystyle\frac{87^2}{80}}$

Or, $x=87^2=7569$.

This is a more formal method that can be applied for borrowing over any number of years.

Choose your method.

**Problem 4.**

A person deposited a sum of Rs. 6000 in a bank at 5% simple interest. another person deposited Rs. 5000 at 8% per annum compound interest. After two years, the difference of their interests will be,

- Rs. 230
- Rs. 600
- Rs. 832
- Rs. 232

**Solution 4: Problem analysis and conceptual solution**

The basic concept of compounding of interest is, over every year the interest earned is on not only the capital at the beginning of investment, but also on the total interest earned till that point of time in addition. This additional component of interest on interest earned is the hallmark of compound interest. In contrast under simple interest scheme, interest earned over every year remains fixed on the capital invested at the beginning of the investment period, and not on the capital at the beginning of that year. Because of this additional interest earned, compound interest schemes deliver more earning for same capital and at same interest rate in comparison to simple interest for the same investment period.

In our problem, the first man earns fixed annual simple interest of, $5\text{% of } 6000=300$, and in two years, Rs. 600.

In contrast the second person would have earned fixed annual interest of, $8\text{% of } 5000=400$, and in two years, Rs. 800 if the interest scheme were simple interest. In that case the difference between the two would have been Rs. 200. But as in the second case, the interest is compounded, there would be an additional interest on interest of Rs. 400 to the tune of, $8\text{% of }400=32$ over the second year of investment.

Total difference between the two would then be Rs. 232.

**Answer:** Option d: Rs. 232.

**Key concepts used:** **Simple interest -- Compound interest -- Difference between simple interest and compound interest -- Solving in mind.**

We have intentionally avoided the use of compound interest formula to speed up solution. In many of the compound interest problems, conceptual approach easily yields quick all in mind solution.

**Problem 5.**

On a certain sum of money, the difference between the compound interest for a year payable half-yearly, and the simple interest for a year is Rs. 180. If the rate of interest in both cases is 10%, then the sum is,

- Rs. 62000
- Rs. 60000
- Rs. 72000
- Rs. 54000

#### Solution 5: Problem analysis, problem modelling and solution

We need to be very clear about what is actually happening here.

If the compound interest were for 1 year payable yearly at the same interest rate of 10% as the simple interest, there would have been no difference in interest earning between the two cases. The difference happens because for the compound interest, the compounding effectively happens at the middle of the year, that is half-yearly. At what rate the interest is paid at the midyear? It would be half the 10% annual rate, that is, 5%.

So the mechanism is,

At midyear, 5% of capital is paid as interest, and at the end of the year again this 5% of capital is paid, but in addition to, 5% of 5% interest paid at midyear.

Knowing this crucial mechansim we have,

$5\text{% of } 5\text{% of } x=180$, where $x$ is the capital

Or, $0.0025x=180$, $0.05\times{0.05}=0.0025$, be careful about the decimals

Or, $25x=1800000$,

Or, $x=72000$.

**Answer:** Option c: Rs. 72000.

**Key concepts used:** * Problem model of half-yearly compounding* --

**Compound interest-- Simple interest -- Difference between simple interest and compound interest -- Solving in mind.**Once we are clear of the problem model, it could be solved in mind easily and quickly.

#### Problem 6.

If the compound interest on a certain sum for two years at 12% per annum is Rs. 2544, the simple interest on it at the same rate after 2 years will be,

- Rs. 2400
- Rs. 2500
- Rs. 2440
- Rs. 2480

**Solution 6: Problem analysis and solution: Difference between simple interest and compound interest**

Knowing how simple interest and compound interest differs, we can straightway break-up the compound interest as simple interest plus compounded interest over two years as,

$24\text{% of }x + 12\text{% of }12\text{% of Capital }x = 2544$, because $24\text{% of }x$ will be the simple interest in two years, where x is the capital, and additional compounded interest will be the interest on interest over the second year, $12\text{% of }12\text{% of } x$

The compounded interest is,

$12\text{% of }12\text{% of } x=0.12\times{0.12}\times{x}=0.0144x$.

Thus compound interest in two years is,

$0.2544x=2544$,

Or, $x=10000$.

And simple interest,

$24\text{% of } x=2400$.

**Answer:** Option a: Rs. 2400.

**Key concepts used:** **Compound interest as simple interest plus interest on interest -- Compound interest -- Simple interest.**

To evaluate the simple interest we needed to evaluate the capital first.

**Problem 7.**

On a certain principal the compound interest compounded annually for the second year at 10% per annum is 132. The principal is,

- Rs. 1000
- Rs. 1200
- Rs. 1320
- Rs. 1250

**Solution 7: Problem analysis and solution**

Here compound interest in second year on principal $x$ is,

$0.1x +0.1\times{0.1}x$, simple interest plus compounded interest on interest earned in first year

$=0.11x$

$=132$, as given.

So the capital is,

$x=\displaystyle\frac{13200}{11}=1200$

**Answer:** Option b: Rs. 1200.

**Key concepts used:** * Compound interest* --

**Interest on interest****-- Solving in mind.**The problem could easily be solved in mind.

**Problem 8.**

On a certain principal of Rs. P, rate of interest is 2r% per annum. The person will then get after 3 years at compound interest a sum of,

- $\text{Rs. }P\left(1+\displaystyle\frac{r}{100}\right)^3$
- $\text{Rs. }P\left(1+\displaystyle\frac{r}{50}\right)^3$
- $\text{Rs. }\displaystyle\frac{6Pr}{100}$
- $\text{Rs. }3P\left(1+\displaystyle\frac{r}{100}\right)^3$

**Solution 8: Problem analysis and solution**

Here we will use the standard formula for growth of principal $C$ at compound interest rate $q$% per annum over $n$ years as,

$C_n=C\left(1+\displaystyle\frac{q}{100}\right)^n$, to convert percentage interest $q$ needed to be divided by 100.

For the given problem then, the principal $P$ would grow in 3 years to be,

$P_3=P\left(1+\displaystyle\frac{2r}{100}\right)^3$

$=P\left(1+\displaystyle\frac{r}{50}\right)^3$.

**Answer:** Option b: $\text{Rs. }P\left(1+\displaystyle\frac{r}{50}\right)^3$.

**Key concepts used:** **Compounded growth of principal**** -- Percentage conversion by division of 100 -- Solving in mind.**

The problem could easily be solved in mind in a few tens of seconds.

**Problem 9.**

A sum of money amounts to Rs. 850 in 3 years and to Rs. 925 in 4 years at some rate of simple interest. The sum is,

- Rs. 550
- Rs. 625
- Rs. 600
- Rs. 700

**Solution 9: Problem analysis and solution**

As simple interest is fixed year to year, for each year the amount invested will grow just by the amount of the fixed simple interest.

In other words, in an investment under simple interest scheme, the difference between the total amounts in two consecutive years will be just the simple interest earned in a year.

On this basis in our problem the simple interest per annum is,

$\text{Rs. } 925-\text{Rs. } 850=\text{Rs. }75$.

Subtracting three times this interest, that is, Rs. 225 from the amount at the end of 3rd year Rs. 850 we get the principal as, Rs. 625.

**Answer:** Option b: Rs. 625.

**Key concepts used:** * Simple interest* --

*--*

**Consecutive year growth in simple interest is the annual simple interest****.**

*Solving in mind***Problem 10.**

What annual installment will discharge a debt of Rs. 6450 due in 4 years at 5% simple interest?

- Rs. 1950
- Rs. 1500
- Rs. 1935
- Rs. 1800

#### Solution 10: Problem analysis and conceptual solution

The basic philosophy working in paying back debt in equal installments is,

the total of each installment with applicable interest up to the end of repayment period must equal total borrowing plus interest over the period. This condition is crucial to ensure that the lender does not lose anything on giving the loan.

The end result should be as if the lender had invested the whole amount at the applicable interest rate for the agreed period of loan.

For example, for a debt of Rs. 6450 at simple interest of 5% to be paid back over 4 years, if $x$ be the equal installment given back at the end of each year, 3 years' interest on 1st installment paid back at the end of 1st year, 2 years' interest on second installment at the end of second year, 1 year's interest on 3rd installment at the end of 3rd year and 0 interest on 4th installment are to be added to $4x$ to equal original debt plus 20% on it as simple interest.

In other words, the equality expression would be,

$4x+\displaystyle\frac{15+10+5}{100}x=6450+\displaystyle\frac{20}{100}\times{6450}$,

Or, $4x+\displaystyle\frac{3}{10}x=6450+1290=7740$,

Or, $x=7740\times{\displaystyle\frac{10}{43}}=1800$.

**Answer:** Option d: Rs. 1800.

**Key concepts used:** * Installment under simple interest* --

**Simple interest -- Borrow period end value.**The total repayment will not be just Rs. 7200, you have to take into account the time of paying each installment and evaluate how much it would become at the end of debt period after applying the interest for the remaining period. Finally,

The total of period end value of each installment should equal the period end value of the amount borrowed at the beginning.

This mechanism holds good both for borrow under simple interest or compound interest.

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