# CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest

**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

Find the C.I. on Rs. 2000 for 3 years at 10% p.a.

- A.
Rs. 562

- B.
Rs. 662

- C.
Rs. 500

- D.
Rs. 516.50

Answer: Option B

**Explanation** :

C.I. = Amount due – Principal

C.I. = P${\left(1+\frac{r}{100}\right)}^{n}-P$

C.I. =$P\left\{{\left(1+\frac{r}{100}\right)}^{n}-1\right\}$

= 2000$\left\{{\left(1+\frac{10}{100}\right)}^{3}-1\right\}=662$

Hence, option (b).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

Find the C.I. on 5000 for 2 years at 4 1/2% p.a. when interest iscompounded annually.

- A.
Rs. 460.125

- B.
Rs. 480

- C.
Rs. 620.14

- D.
Rs. 504.5

Answer: Option A

**Explanation** :

C.I. =5000$\left\{{\left(1+\frac{{\displaystyle \raisebox{1ex}{$9$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{100}\right)}^{2}-1\right\}=230.06$

Hence, option (a).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

Find the C.I. on Rs. 2000 at 12% p.a. for 3 years when the C.I. is reckoned yearly.

- A.
Rs. 881.84

- B.
Rs. 809.86

- C.
Rs. 888.48

- D.
Rs. 848.88

Answer: Option B

**Explanation** :

C.I. = $2000\left[{\left(1+\frac{12}{100}\right)}^{3}-1\right]=202.46$

Hence, option (b).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

What will Rs. 12,500 amount to at C.I. for 3 years at 4% when C.I. is reckoned yearly.

- A.
Rs. 13,600.2

- B.
Rs. 13,801.2

- C.
Rs. 14,200

- D.
Rs. 14,060.80

Answer: Option D

**Explanation** :

$A=P{\left(1+\frac{r}{100}\right)}^{n}=6250{\left(1+\frac{4}{100}\right)}^{3}=14,060.80$

Hence, option (d).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

The compound interest on Rs. 30,000 at 7% per annum is Rs. 4347. The period (in years) is:

- A.
2

- B.
$2\frac{1}{2}$

- C.
3

- D.
4

Answer: Option A

**Explanation** :

Amount = Rs. (30000 + 4347) = Rs. 34347.

Let the time be n years.

Then,

$30000{\left(1+\frac{7}{100}\right)}^{n}=34347\phantom{\rule{0ex}{0ex}}{\left(\frac{107}{100}\right)}^{n}=\frac{34347}{30000}=\frac{11449}{10000}={\left(\frac{107}{100}\right)}^{2}$

∴ n = 2 years.

Hence, option (a).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

At what rate of compound interest per annum will a sum of Rs. 1200 become Rs. 1348.32 in 2 years?

- A.
6%

- B.
6.5%

- C.
7%

- D.
7.5%

Answer: Option A

**Explanation** :

Let the rate be R% p.a.

Then,

$1200{\left(1+\frac{R}{100}\right)}^{2}=1348.32\phantom{\rule{0ex}{0ex}}{\left(1+\frac{R}{100}\right)}^{2}=\frac{134832}{120000}=\frac{11236}{10000}={\left(\frac{106}{100}\right)}^{2}$

$1+\frac{R}{100}=\frac{106}{100}$

∴ R = 6%

Hence, option (a).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

Every year an amount increases by 1/8th of itself, How much will it be after two years if its present value is Rs. 64000?

- A.
Rs. 81000

- B.
Rs. 80000

- C.
Rs. 75000

- D.
None of these

Answer: Option A

**Explanation** :

Here rate =$\frac{1}{8}\times 100\%=12.5\phantom{\rule{0ex}{0ex}}A=64000{\left(1+\frac{12.5}{100}\right)}^{2}=81000$

Hence, option (a).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

A sum of money put out at C.I. amounts to Rs. 57,840 in 2 years and in 3 years to Rs. 61,455. Find the rate of interest.

- A.
$5\frac{3}{4}\%$

- B.
6%

- C.
$5\frac{1}{2}\%$

- D.
$6\frac{1}{4}\%$

Answer: Option D

**Explanation** :

In C.I., amount increases by R% every year.

∴ 61,455 – 57,840 = 3,615

Now, 3,615 is the interest on Rs. 57,840

∴ Rate $=\frac{3,615\times 100}{57,840\times 1}=6\frac{1}{4}\%$

Hence, option (d).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

The least number of complete years in which a sum of money put out at 20% compound interest will be more than doubled is:

- A.
3

- B.
4

- C.
5

- D.
6

Answer: Option B

**Explanation** :

$P{\left(1+\frac{20}{100}\right)}^{n}>2P\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(\frac{6}{5}\right)}^{n}2$

Now, $\left(\frac{6}{5}\times \frac{6}{5}\times \frac{6}{5}\times \frac{6}{5}\right)>2$

So, n = 4 years.

Hence, option (b).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

Find the amount due on Rs. 2000 for 2.5 years at 10% p.a., compounded yearly.

Answer: 2541

**Explanation** :

Amount due = $P{\left(1+\frac{r}{100}\right)}^{n}$

When we have incomplete years, we calculate the amount due for complete years on CI basis and then for the incomplete year we calculate amount due on SI basis.

Amount due = $2000{\left(1+\frac{10}{100}\right)}^{2}\times \left(1+\frac{10\times {\displaystyle \frac{1}{2}}}{100}\right)=2541$

Hence, 2541.

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

What is the compound interest on Rs. 6950 for 3 years if interest is payable half-yearly, at the rate of 6% p.a. for the first two years and at the rate of 9% p.a. for the third year

- A.
Rs. 1590

- B.
Rs. 1502

- C.
Rs. 1482

- D.
Rs. 1615

Answer: Option A

**Explanation** :

C.I = $6950\left\{{\left(1+\frac{3}{100}\right)}^{4}{\left(1+\frac{4.5}{100}\right)}^{2}\right\}=1590$

Hence, option (a).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

Determine the C.I. on Rs. 2400 at 10% p.a. for $1\frac{1}{2}$ years when C.I. is compounded half yearly.

- A.
Rs. 367.20

- B.
Rs. 390

- C.
Rs. 387.10

- D.
Rs. 378.30

Answer: Option D

**Explanation** :

Rate of interest for half year = 10/2 = 5%

Number of compounding periods = 2 × 1.5 = 3.

C.I. = $2400\left\{{\left(1+\frac{5}{100}\right)}^{3}-1\right\}=378.30$

Hence, option (d).

Workspace:

**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

A bank offers 5% compound interest calculated on half-yearly basis. A customer deposits Rs. 1600 each on 1st January and 1st July of a year. At the end of the year, the amount he would have gained by way of interest is:

- A.
Rs. 120

- B.
Rs. 121

- C.
Rs.122

- D.
Rs.123

Answer: Option B

**Explanation** :

Amount =$\left[1600\times {\left(1+\frac{5}{2\times 100}\right)}^{2}+1600\times \left(1+\frac{5}{2\times 100}\right)\right]=Rs.3321$

∴ C.I. = Rs. (3321 - 3200) = Rs. 121

Hence, option (b).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

What is the difference between the compound interests on Rs. 5000 for 1 years at 4% per annum compounded yearly and half-yearly?

- A.
Rs.2.04

- B.
Rs. 3.06

- C.
Rs.4.80

- D.
Rs.8.30

Answer: Option A

**Explanation** :

C.I. when interest compounded yearly =$\left[5000\times \left(1+\frac{4}{100}\right)\times \left(1+\frac{{\displaystyle \frac{1}{2}}\times 4}{100}\right)\right]=\left(5000\times \frac{26}{25}\times \frac{51}{50}\right)=5304$

C.I. when interest is compounded half-yearly =

$\left[5000\times {\left(1+\frac{2}{100}\right)}^{3}\right]=\left(5000\times \frac{51}{50}\times \frac{51}{50}\times \frac{51}{50}\right)=5306.04$

∴ Difference = Rs. (5306.04 - 5304) = Rs. 2.04

Hence, option (a).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

The effective annual rate of interest corresponding to a nominal rate of 6% per annum payable half-yearly is:

- A.
6.06%

- B.
6.07%

- C.
6.08%

- D.
6.09%

Answer: Option D

**Explanation** :

Amount of Rs. 100 for 1 year when compounded half-yearly = Rs.$\left[100\times {\left(1+\frac{3}{100}\right)}^{2}\right]=Rs.106.09$

∴ Effective rate = (106.09 - 100)% = 6.09%

.

Hence, option (d)

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

The effective annual rate of interest corresponding to a nominal rate of 12% per annum payable every 4 months is:

- A.
1.78%

- B.
12.49%

- C.
12%

- D.
13.26%

Answer: Option B

**Explanation** :

Amount of Rs. 100 for 1 year when compounded every 4 months = Rs

$\left[100\times {\left(1+\frac{{\displaystyle \raisebox{1ex}{$12$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}}{100}\right)}^{3}\right]=Rs.112.49$

∴ Effective rate = (112.49 - 100)% = 12.49%

Hence, option (b).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

An amount lent at CI doubles in 2 years. How long will it take for the amount to become 8 times when lent at same rate of interest?

- A.
15 years

- B.
21 years

- C.
6 years

- D.
Can’t be determined

Answer: Option C

**Explanation** :

Amount due $=P{\left(1+\frac{r}{100}\right)}^{t}$

$\Rightarrow 2P=P{\left(1+\frac{r}{100}\right)}^{2}\phantom{\rule{0ex}{0ex}}\Rightarrow 2={\left(1+\frac{r}{100}\right)}^{2}$

Now, the amount has to quadruple at same rate of interest.

$\Rightarrow 8P=P{\left(1+\frac{r}{100}\right)}^{t}\phantom{\rule{0ex}{0ex}}\Rightarrow 8={\left(1+\frac{r}{100}\right)}^{t}\phantom{\rule{0ex}{0ex}}\Rightarrow {2}^{3}={\left(1+\frac{r}{100}\right)}^{t}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left[{\left(1+\frac{r}{100}\right)}^{2}\right]}^{3}={\left(1+\frac{r}{100}\right)}^{t}\phantom{\rule{0ex}{0ex}}\Rightarrow {\left(1+\frac{r}{100}\right)}^{6}={\left(1+\frac{r}{100}\right)}^{t}$

⇒ t = 6 years.

**Alternately,**

For CI, since amount doubles in 2 years, it means the amount will double every 2 years.

For the amount to become 8 (= 2³) times, it has to double 3 times. Hence, time required = 2 + 2 + 2 = 6 years.

Hence, option (c).

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

An amount lent at CI becomes 5 times in 7 years. How long will it take for the amount to become 125 times when lent at same rate of interest?

Answer: 21

**Explanation** :

For CI, since amount becomes 5 times in 7 years, it means the amount will become 5 times every 7 years.

For the amount to become 125 (= 53) times, it has to become 5× 3 times. Hence, time required = 7 + 7 + 7 = 21 years.

Hence, 21.

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**CRE 2 - Compound Interest | Arithmetic - Simple & Compound Interest**

Find the amount due on Rs. 1000 for 2 1/3 years at 30% p.a., compounded yearly.

Answer: 1859

**Explanation** :

Amount due = $P{\left(1+\frac{r}{100}\right)}^{n}$

When we have incomplete years, we calculate the amount due for complete years on CI basis and then for the incomplete year we calculate amount due on SI basis.

Amount due = $1000{\left(1+\frac{30}{100}\right)}^{2}\times \left(1+\frac{30\times {\displaystyle \frac{1}{3}}}{100}\right)=1859$

Hence, 1859.

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