Compound Interest Questions and Answers

Q1 :

Find the amount of Rs. 800 at compound interest in 2122frac{1}{2} years at 5 p.c.

A
  

Rs. 904.5

B
  

Rs. 805.5

C
  

Rs. 104.5

D
  

Rs. 804.5

View Answer
Correct Answer: A

Rs. 904.5

Description:

Amount= Rs. 800×(1+5100)2×(1+5200)text{Amount} = text{ Rs. } 800 times bigg(1 + frac{5}{100} bigg)^2 times bigg(1 + frac{5}{200}bigg)
= Rs. 904.05

Q2 :

At what annual percent rate, the compound interest will Rs. 80,000 amout to Rs. 88,220 in 2 years?

A
  

5%

B
  

8%

C
  

10%

D
  

12%

View Answer
Correct Answer: A

5%

Description:

88200=80000(1+x100)2text{88200} = 80000 Big(1 + frac{x}{100}Big)^2

∴(1+x100)2=441400=(2120)2therefore Big(1 + frac{x}{100} Big)^2 = frac{441}{400} = Big( frac{21}{20}Big)^2

x=5%x = 5%

Q3 :

What principal will amount to Rs. 1352 in 2 years at 4 p.c. compound interest?

A
  

1250

B
  

1300

C
  

1350

D
  

1500

View Answer
Correct Answer: A

1250

Description:

Principal =1352(1+4100)2text{Principal } = frac{1352}{left( 1 + frac{4}{100}right)^2 }

=1352×625676=Rs. 1250= 1352 times frac{625}{676} = text{Rs. } 1250

Q4 :

Find rate percent at which compound interest does a sum of money become four fold in 2 years?

A
  

100%

B
  

10%

C
  

25%

D
  

50%

View Answer
Correct Answer: A

100%

Description:

4x=x(1+r100)24x = xBig(1 + frac {r}{100} Big)^2

∴(1+r100)2=40 or 1+r100=2therefore Big(1 + frac{r}{100} Big)^2 = 40 text{ or } 1 + frac{r}{100} = 2

or r=100%text{or } r =100%

Q5 :

Find the compound interest of Rs. 800 for 3 years at 5 p.c.

A
  

Rs. 126.10

B
  

Rs. 826.10

C
  

Rs. 820.10

D
  

Rs. 850.00

View Answer
Correct Answer: A

Rs. 126.10

Description:

Amount=800×(1+5100)3text{Amount} = 800 times Big(1 + frac{5}{100}Big)^3

=800×92618000=926.10= 800 times frac{9261}{8000} = 926.10

∴ Interest=Rs.926.10−Rs.800=Rs.126.10therefore text{ Interest} = text{Rs.} 926.10 - text{Rs.} 800 = text{Rs.} 126.10

Q6 :

If a sum of Rs. 10,000 is deposited for 1 year at 10% compounded half yearly, then the interest will be:

A
  

Rs. 1,000

B
  

Rs. 1,025

C
  

Rs. 975

D
  

Rs. 925

View Answer
Correct Answer: B

Rs. 1,025

Description:

∵because Compounded half yearly

∴therefore 1 year = 2 half year

and rate 10% annually = 102%frac{10}{2}% half yearly

= 5%

∵ C.I. =P [(1+r100)n−1]=10,000because text{ C.I. } = text{P }Big[big(1+frac{r}{100}big)^n -1 Big] = 10,000

[(1+5100)2−1]=10,000×41400=1,025= Big[big(1+frac{5}{100}big)^2 -1 Big] = 10,000 times frac{41}{400} = 1,025

Q7 :

In three years, a certain sum of money amounts to Rs. 9261 at 5% per annum compound interest. The sum is

A
  

Rs. 5,000

B
  

Rs. 60,000

C
  

Rs. 70,000

D
  

Rs. 8,000

View Answer
Correct Answer: D

Rs. 8,000

Description:

 Required amount =9261×20×20×2021×21×21=8,000text{ Required amount } = frac{9261 times 20 times 20 times 20 }{21 times 21 times 21} = 8,000

Q8 :

A sum of money amounts to Rs. 1000 in 10 years and to Rs. 1250 in 15 years. Find is sum.

A
  

Rs. 750

B
  

Rs. 400

C
  

Rs. 600

D
  

Rs. 500

View Answer
Correct Answer: D

Rs. 500

Description:

P=1(1+tr100)=1000(1+10r100)text{P} = frac{1}{Big(1 + frac{tr}{100}Big)} = frac{1000}{Big(1 + frac{10r}{100}Big)}
=1000010+r...(I)= frac{10000}{10 + r} quad ... (text{I})


Again

P =1250(1+15r100)=1250×2020+3r...(II)text{P } = frac{1250}{Big(1 + frac{15r}{100}Big)} = frac{1250 times 20}{20 + 3r} quad ... (text{II})

By equation I and II

1000010+r=1250×2020+3rfrac{10000}{10 + r} = frac{1250 times 20}{ 20 + 3r}

∴20+3r10+r=1250×2010000therefore quad frac{20 + 3r}{10 + r} = frac{1250 times 20 }{10000}

∴r=10quad therefore quad r = 10

∴therefore quad Now on putting the value of rr in eq. I we get

P =1000010+10=500text{P } = frac{10000}{10 + 10 } = 500

Q9 :

If PP be the principal amount and the rate of interest be rr% per annum and the compound interest is calculated kk times in a year, then what is the amount at the end of nn yr?

A
  

P(1+r100k)n/kPbigg( 1 + dfrac{r}{100k}bigg)^{n/k}

B
  

P(1+kr100)nkPbigg( 1 + dfrac{kr}{100}bigg)^{nk}

C
  

P(1+kr100)n/kPbigg( 1 + dfrac{kr}{100}bigg)^{n/k}

D
  

P(1+kr100k)n/kPbigg( 1 + dfrac{kr}{100k}bigg)^{n/k}

View Answer
Correct Answer: A

P(1+r100k)n/kPbigg( 1 + dfrac{r}{100k}bigg)^{n/k}

Description:

Given, principal amount = PP
Rate of interest, R=rk%R = frac{r}{k} % per annum and
Time, T=nkT = nk

∵A=P(1+R100)T[Rule 1. (i)] ∴A=P(1+r100k)nkbegin{aligned} & because A = P bigg(1 + frac{R}{100} bigg)^T hspace{0.5cm} [text{Rule 1. (i)}] \ & therefore A = P bigg(1 + frac{r}{100k} bigg)^{nk} end{aligned}

Q10 :

A person borrowed ₹ 7500 at 16% compound interest. How much does he have to pay after end of 2 year to close the loan?

A
  

₹ 9900

B
  

₹ 10092

C
  

₹ 11000

D
  

₹ 11052

View Answer
Correct Answer: B

₹ 10092

Description:

Given, PP = ₹ 7500, RR = 16% and nn = 2 yr

∵A=P(1+R100)n[Rule 1. (i)]=7500(1+16100)2=7500(116100)2=7500×2925×2925=12×841=10092begin{aligned} because A &= P bigg ( 1 + frac{R}{100} bigg) ^n hspace{0.5cm} [text{Rule 1. (i)}] \ & = 7500 bigg ( 1 + frac{16}{100} bigg) ^ 2 \ & = 7500 bigg (frac{116}{100} bigg) ^ 2 \ & = 7500 times frac{29}{25} times frac{29}{25} \ & = 12 times 841 = 10092 end{aligned}

Q11 :

At compound interest, if a certain sum of money doubles in nn yr, then the amount will be four fold in

A
  

2n2 yr 2n^2 text{ yr }

B
  

n2 yr n^2 text{ yr }

C
  

2n yr 2n text{ yr }

D
  

4n yr 4n text{ yr }

View Answer
Correct Answer: C

2n yr 2n text{ yr }

Description:

Let the principal be ₹ xx.
∴therefore  Rate = ₹ RR, Amount = ₹ 2x2x
and Time = nn yr

∵A=P(1+R100)n[Rule 1. (i)]∴x(1+R100)n=2x⇒(1+R100)n=2...(i)begin{aligned} because A &= P bigg ( 1 + frac{R}{100} bigg) ^n hspace{0.2cm} [text{Rule 1. (i)}] &therefore x bigg(1 + frac{R}{100} bigg)^n = 2x &Rightarrow bigg( 1 + frac{R}{100} bigg)^n = 2 hspace{0.4cm} text{...(i)} end{aligned}

Let it becomes four fold in NN yr.
Then, x(1+R100)N=4xxbigg(1 + frac{R}{100}bigg)^N = 4x

⇒(1+R100)N=4⇒22=(1+R100)N⇒(1+R100)2n=(1+R100)N[from Eq. (i)]∴N=2n yrbegin{aligned} Rightarrow bigg( 1 + frac{R}{100} bigg)^N = 4 \ Rightarrow 2^2 = bigg( 1 + frac{R}{100} bigg)^N \ Rightarrow bigg( 1 + frac{R}{100} bigg)^{2n} \ = bigg( 1 + frac{R}{100} bigg)^N [text{from Eq. (i)}] \ therefore N = 2n text{ yr} end{aligned}

Q12 :

The simple interest on a fix sum of money for 3 year at the rate of 8% per annual is half the compound interest on rupees 4000 for 2 year at rate of 10% per annual. What is the amount placed on simple interest?

A
  

₹ 1550

B
  

₹ 1650

C
  

₹ 1750

D
  

₹ 2000

View Answer
Correct Answer: C

₹ 1750

Description:

Let the principal amount be ₹ PP.
By given condition, SI = 12frac{1}{2} CI

⇒P××3100=12[400(1+10100)2−4000][Rule 1 (ii)]begin{aligned} &Rightarrow frac{P times times 3}{100} = frac{1}{2} & Bigg[400 bigg( 1 + frac{10}{100} bigg) ^2 - 4000 Bigg] [text{Rule 1 (ii)}] end{aligned}

⇒24P100=12[4000×121100−4000]=12[4840−4000]⇒24P100=420∴P=420×10024=1750begin{aligned} & Rightarrow frac{24P}{100} = frac{1}{2} Bigg[ 4000 times frac{121}{100} - 4000 Bigg] & = frac{1}{2} [4840 - 4000] & Rightarrow frac{24P}{100} = 420 &therefore P = frac{420 times 100}{24} = 1750 end{aligned}

Q13 :

The difference between the simple interest and the compound interest (compounded annually) on ₹ 1250 for 2 yr at 8% per annum will be

A
  

18

B
  

2

C
  

8

D
  

4

View Answer
Correct Answer: C

8

Description:

Given, PP = ₹ 1250, RR = 8% per annum and nn = 2yr

∵Simple interest =P×R×T100∴Simple interest =1250×8×2100begin{aligned} because text{Simple interest } = frac{P times R times T}{100} therefore text{Simple interest } = frac{1250 times 8 times 2}{100} end{aligned}

= ₹ 200

∵because  Compound Interest

=P[(1+R100)n][Rule 1. (ii)]∵CI =1250(1+8100)2−1250 =1250×(108100)2−1250 =1458−1250=208begin{aligned} &= P Bigg[ bigg(1 + frac{R}{100} bigg)^n Bigg] [text{Rule 1. (ii)}] because text{CI } &= 1250 big(1 + frac{8}{100} big)^2 - 1250 ~ &= 1250 times big(frac{108}{100} big)^2 - 1250 ~ &= 1458 - 1250 = 208 end{aligned}

∴therefore Difference in SI and CI = ₹ 208 − ₹ 200
= ₹ 8

Q14 :

A bank offers interest that is compounded on an annual basis. Raju deposited Rs. 100 and received Rs. 121 at the end of second year. Rate of compound interest per annually is

A
  

10%

B
  

12%

C
  

14%

D
  

16%

View Answer
Correct Answer: A

10%

Description:

Given, principal (P) = Rs. 100
Amount (A) received after 2 yr = Rs. 121
Let rate of interest = R% per annum

∵A=P(1+R100)n[Rule 1. (i)]∴121=100(1+R100)2 ⇒121100=(100+R100)2 ⇒(1110)2=(100+R100)2 ⇒1110=100+R100 ⇒100+R=11×10010⇒100+R=110∴R=110−100∴Rate of interest=10%begin{aligned} & because A = P bigg(1 + frac{R}{100} bigg)^n hspace{0.2cm} [text {Rule 1. (i)}] & therefore 121 = 100 bigg(1 + frac{R}{100} bigg)^2 ~ & Rightarrow frac{121}{100} = bigg(frac{100+R}{100} bigg)^2 ~ & Rightarrow bigg(frac{11}{10} bigg)^2 = bigg(frac{100+R}{100} bigg)^2 ~ & Rightarrow frac{11}{10} = frac{100+R}{100} ~ & Rightarrow 100+R = frac {11 times 100}{10} & Rightarrow 100+R = 110 & therefore R = 110-100 & therefore text{Rate of interest} = 10% end{aligned}