# Compound Interest Questions and Answers updated daily – Aptitude

Compound Interest Questions: Solved 161 Compound Interest Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Compound Interest Questions

21. Two equal amounts lend on S.I. and C.I. for two years at same rates. If amount after two years is in the ratio of 520 : 529 and difference between interest earned in two years is Rs. 18, then find the ratio of interests earned from S.I to C.I in 3 years if total amount invested in both scheme is Rs. 1600.

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Correct Ans:400 : 463

Explanation:

Given:

Amounts that lend on S.I. and C.I. after 2 yrs is in the ratio of 520 : 529.

Let the amount received after 2 yrs be 520X and 529X.

Given that difference between interest earned in two years is Rs.18.

529X - 520X = 18

9X = 18

X = 2

Amount received after 2 yrs on SI = 520X = 520(2) = Rs. 1040

Amount received after 2 yrs on CI = 529X = 529(2) = Rs. 1058

Amount invested in both scheme for 3yrs = Rs. 1600

Given that amount invested and rates on each scheme is equal.

Principal invested in each scheme = 1600/2 = Rs. 800

WKT,

Where, SI = 1040 - 800 = Rs. 240

R = (240 x 100)/(800 x 2)

R = 15%

SI after 3 yrs = (800 x 3 x 15)/100 = Rs. 360

CI after 3 yrs =

= 800[(1 + 15/100)

= 800[(23/20)

= 800[(12167 - 8000)/8000]

= Rs. 416.7

Required ratio = 360/416.7 = 400/463.

Amounts that lend on S.I. and C.I. after 2 yrs is in the ratio of 520 : 529.

Let the amount received after 2 yrs be 520X and 529X.

Given that difference between interest earned in two years is Rs.18.

529X - 520X = 18

9X = 18

X = 2

Amount received after 2 yrs on SI = 520X = 520(2) = Rs. 1040

Amount received after 2 yrs on CI = 529X = 529(2) = Rs. 1058

Amount invested in both scheme for 3yrs = Rs. 1600

Given that amount invested and rates on each scheme is equal.

Principal invested in each scheme = 1600/2 = Rs. 800

WKT,

**SI = PNR/100;****R = (SI x 100)/PN**Where, SI = 1040 - 800 = Rs. 240

R = (240 x 100)/(800 x 2)

R = 15%

SI after 3 yrs = (800 x 3 x 15)/100 = Rs. 360

CI after 3 yrs =

**P [(1 + r/100)**^{n}- 1]= 800[(1 + 15/100)

^{3}- 1]= 800[(23/20)

^{3}- 1]= 800[(12167 - 8000)/8000]

= Rs. 416.7

Required ratio = 360/416.7 = 400/463.

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22. Sanjana invested Rs. 15000 in SI at the rate of 2x% per annum for two years and the same amount is invested in CI at the same rate of interest if she received Rs.150 more interest than S.I, then find the rate of interest per annum?

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Correct Ans:10%

Explanation:

CI - SI = 150

P(1 + r/100)

15000[(1 + (2x/100))

15000[((100 + 2x)/100)

15000[(10000 + 400x + 4x

15000[(10000 + 400x + 4x

15000[(400x + 4x

3/2(400x + 4x

(3*4/2)(100x + x

600x + 6x

6x

x

x = 5

The rate of interest per annum = 2x% = (2*5)% = 10%

P(1 + r/100)

^{n}- Pnr/100 = 15015000[(1 + (2x/100))

^{2}- 1] - (15000*2x*2)/100 = 15015000[((100 + 2x)/100)

^{2}- 1] - 600x = 15015000[(10000 + 400x + 4x

^{2})/10000 - 1] - 600x = 15015000[(10000 + 400x + 4x

^{2}- 10000)/10000] - 600x = 15015000[(400x + 4x

^{2})/10000] - 600x = 1503/2(400x + 4x

^{2}) - 600x = 150(3*4/2)(100x + x

^{2}) - 600x = 150600x + 6x

^{2}- 600x = 1506x

^{2}= 150x

^{2}= 25x = 5

The rate of interest per annum = 2x% = (2*5)% = 10%

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23. A person took a loan of Rs. 6000 for 3 years, at 5% per annum compound interest. He repaid Rs. 2100 in each of the first 2 years. The amount he should pay at the end of 3rd to clear all his debts is:

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Correct Ans:Rs. 2425.50

Explanation:

Given, Principal, P = Rs. 6000

No. of years, n = 3

Rate of interest, r = 5% per annum.

Amount repaid at first year = Rs. 2100

Amount repaid at second year = Rs. 2100

For 1st year, Compound Interest,

---> C.I = 6000 * {[1 + (5/100)]

= 6000 * {[1 + (1/20)] - 1}

= 6000 * {[21/20] - 1}

= 6000 * {1/20)}

=

At the end of 1st year,

= 6000 + 300 =

As the person paid Rs. 2100 at first year, the principal for 2nd year = 6300 - 2100 = 4200

For

= 4200 * {1/20)}

= 210

At the end of 2nd year,

= 4200 + 210

=

As the person paid Rs. 2100 at second year, the principal for 3rd year = 4410 - 2100 = 2310

For

= 2310 * {1/20)}

= 115.5

So,

= 2310 + 115.5

=

No. of years, n = 3

Rate of interest, r = 5% per annum.

Amount repaid at first year = Rs. 2100

Amount repaid at second year = Rs. 2100

For 1st year, Compound Interest,

**C.I = P {[1 + (r/100)]**^{n}- 1}---> C.I = 6000 * {[1 + (5/100)]

^{1}- 1}= 6000 * {[1 + (1/20)] - 1}

= 6000 * {[21/20] - 1}

= 6000 * {1/20)}

=

**300**At the end of 1st year,

**Amount payable = Principal + C.I**= 6000 + 300 =

**6300**As the person paid Rs. 2100 at first year, the principal for 2nd year = 6300 - 2100 = 4200

For

**2nd year**, Compound Interest, C.I = 4200 * {[1 + (5/100)]^{1}- 1}= 4200 * {1/20)}

= 210

At the end of 2nd year,

**Amount payable**= Principal (for 2nd year) + C.I= 4200 + 210

=

**4410**As the person paid Rs. 2100 at second year, the principal for 3rd year = 4410 - 2100 = 2310

For

**3rd year**, Compound Interest, C.I = 2310 * {[1 + (5/100)]^{1}- 1}= 2310 * {1/20)}

= 115.5

So,

**The amount that the person should pay at the end of 3rd to clear all his debts**= Principal (for 3rd year) + C.I= 2310 + 115.5

=

**Rs. 2425.5**
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24. Simple interest on a certain sum of money for 5 years at 10 % per annum is five sixth of the compound interest on Rs. 12,000 for 2 years at 5 % per annum. The sum placed on simple interest is?

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Correct Ans:Rs. 2050

Explanation:

Given:

CI: P = Rs. 12,000; N = 2 yrs; R = 5%

WKT,

CI for 1st yr = 12000[1 + (5/100)] - 12000

= 12000[(21/20) - 1]

= 12000[(21- 20)/20]

= Rs. 600

Principal for 2nd yr = 12000 + 600 = Rs. 12, 600

CI for 2nd yr = 12600[1 + (5/100)] - 12600

= 12600[(21/20) - 1]

= 12600[(21- 20)/20]

= Rs. 630

CI for 2 yrs = 600 + 630 = Rs. 1,230

Given: SI is 5/6 of CI,

SI = (5/6)*1230

SI = Rs. 1025

WKT,

1025 = [P x 5 x 10]/100

P = [1025 x 100]/[5 x 10]

P = Rs. 2050.

CI: P = Rs. 12,000; N = 2 yrs; R = 5%

WKT,

**C.I. = P[(1+R/100)**^{N}] – PCI for 1st yr = 12000[1 + (5/100)] - 12000

= 12000[(21/20) - 1]

= 12000[(21- 20)/20]

= Rs. 600

Principal for 2nd yr = 12000 + 600 = Rs. 12, 600

CI for 2nd yr = 12600[1 + (5/100)] - 12600

= 12600[(21/20) - 1]

= 12600[(21- 20)/20]

= Rs. 630

CI for 2 yrs = 600 + 630 = Rs. 1,230

Given: SI is 5/6 of CI,

SI = (5/6)*1230

SI = Rs. 1025

WKT,

**SI = PNR/100**1025 = [P x 5 x 10]/100

P = [1025 x 100]/[5 x 10]

P = Rs. 2050.

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25. Raghu invested Rs. A on SI for 2 years at the rate of 6% p.a. and Rs. 1500 on CI for 2 years at the rate of 20% p.a. If the ratio of SI and CI after 2 years is 6: 11, find the value of A.

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Correct Ans:Rs. 3000

Explanation:

Given:

In SI: P - Rs. A; N - 2 yr; R - 6%

In CI: P - Rs.1500; N - 2 yr; R - 20%

WKT,

CI = 1500{[1 + (20/100)]

= 1500{(6/5)

= 1500[(36 - 25)/25]

= 1500(11/25)

= Rs. 660

If the ratio of SI and CI after 2 years is 6: 11,

(A x 2 x 6)/(660 x 100) = 6/11

A = (6 x 660 x 100)/(2 x 6 x 11)

A = Rs. 3000.

In SI: P - Rs. A; N - 2 yr; R - 6%

In CI: P - Rs.1500; N - 2 yr; R - 20%

WKT,

**CI = P{[1 + (R/100)]**^{N}- 1}CI = 1500{[1 + (20/100)]

^{2}- 1}= 1500{(6/5)

^{2}- 1}= 1500[(36 - 25)/25]

= 1500(11/25)

= Rs. 660

If the ratio of SI and CI after 2 years is 6: 11,

**SI/CI = 6/11****(PNR/100)/P{[1 + (R/100)]**^{N}- 1} = 6/11(A x 2 x 6)/(660 x 100) = 6/11

A = (6 x 660 x 100)/(2 x 6 x 11)

A = Rs. 3000.

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26. Dev invested Rs. X at the rate of 15% p.a. on CI for two years and gets total interest of Rs. 5805. If Dev invested Rs. (X + 14000) for another 2 yrs at an additional rate of 15%, then what will be the CI on that investment?

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Correct Ans:Rs. 22,080

Explanation:

Given:

Dev invested Rs. X ; N = 2 yrs; R = 15% and CI = Rs. 5,805

WKT,

5805 = X[1 + (15/100)]

5805 = X[23/20]

5805 = X[(529 - 400)/400]

X = 5805*(400/129)

X = Rs. 18,000

If Dev invested Rs. (X + 14000); N = 2 yrs; R = 30%(additional rate of 15%)

CI = (18000 + 14000)[1 + (30/100)]

CI = 32000[13/10]

CI = 32000[(169 - 100)/100]

CI = 32000[69/100]

CI = Rs. 22,080.

Dev invested Rs. X ; N = 2 yrs; R = 15% and CI = Rs. 5,805

WKT,

**CI = P[1 + (R/100)]**^{N}- P5805 = X[1 + (15/100)]

^{2}- X5805 = X[23/20]

^{2}- X5805 = X[(529 - 400)/400]

X = 5805*(400/129)

X = Rs. 18,000

If Dev invested Rs. (X + 14000); N = 2 yrs; R = 30%(additional rate of 15%)

CI = (18000 + 14000)[1 + (30/100)]

^{2}- (18000 + 14000)CI = 32000[13/10]

^{2}- 32000CI = 32000[(169 - 100)/100]

CI = 32000[69/100]

CI = Rs. 22,080.

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27. A certain sum is invested for 2 years in scheme A at 20% p.a. compound interest compounded annually. Same sum is also invested for the same period in scheme B at x%p.a. to a simple interest earned from scheme A is twice of that earned from scheme B. What is the value of x?

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Correct Ans:11%

Explanation:

Let sum invested in both schemes = Rs. P

Let interest earned in scheme A = Rs. 2x

In scheme A, time = 2 years and rate = 20% under compound interest.

=> C.I. = P [(1 + r/100)

=> 2x = P [(1 + 20/100)

=> 2x = P [(6/5)

=> 2x = P [(36/35) - 1] = 11P/25

=> x = 11P/50 ---------- (i)

Now, in scheme B, interest earned = Rs. x

Time = 2 years and rate of interest = x% under simple interest

=> S.I. = Pnr/100

=> x = (P*x*2)/100

Using, equation(i) we get:

=> 11P/50 = (P*x)/50

=> x = 11%

Let interest earned in scheme A = Rs. 2x

In scheme A, time = 2 years and rate = 20% under compound interest.

=> C.I. = P [(1 + r/100)

^{n}- 1]=> 2x = P [(1 + 20/100)

^{2}- 1]=> 2x = P [(6/5)

^{2}- 1]=> 2x = P [(36/35) - 1] = 11P/25

=> x = 11P/50 ---------- (i)

Now, in scheme B, interest earned = Rs. x

Time = 2 years and rate of interest = x% under simple interest

=> S.I. = Pnr/100

=> x = (P*x*2)/100

Using, equation(i) we get:

=> 11P/50 = (P*x)/50

=> x = 11%

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28. A person invested sum of the amount at the rate of 15% SI per annum for two years and received total amount of Rs.19500. He invested same sum at the rate x% per annum compounded annually for two years and he received interest Rs. 2100 more as compared to the Simple Interest, then find the value of 'x'

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Correct Ans:20%

Explanation:

Let us take sum be y

WKT,

A = P (1 + (rn)/100)

19500 = y (1 + 30/100)

19500 = y (130/100)

y = 19500 * (10/13) = 15000

SI = A - P = 19500 - 15000

= 4500

Given,

CI = SI + 2100 = 4500 + 2100

= 6600

Total amount = P + I = 15000 + 6600

= 21600

A = P (1 + r/100)

21600 = 15000 (1 + x/100)

(1 + x/100)

1 + x/100 = 6/5

x/100 = 1/5

x = 20%

WKT,

A = P (1 + (rn)/100)

19500 = y (1 + 30/100)

19500 = y (130/100)

y = 19500 * (10/13) = 15000

SI = A - P = 19500 - 15000

= 4500

Given,

CI = SI + 2100 = 4500 + 2100

= 6600

Total amount = P + I = 15000 + 6600

= 21600

A = P (1 + r/100)

^{n}21600 = 15000 (1 + x/100)

^{2}(1 + x/100)

^{2}= (6/5)^{2}1 + x/100 = 6/5

x/100 = 1/5

x = 20%

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29. A sum of cash 4 times itself at compound interest in 20 years. In how many years will it become 16 times?

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Correct Ans:40

Explanation:

By using given condition,

P (1 + R /100)

(1 + R/100)

(1 + R/100)

Let P (1 + R/100)

(1 + R/100)

Using (1)

(1+R/ 100)

n = 40

Thus required time 40 years.

P (1 + R /100)

^{20}= 4 P(1 + R/100)

^{20}= 4(1 + R/100)

^{20}= 2^{2}-------- (1)Let P (1 + R/100)

^{n}= 16 P(1 + R/100)

^{n}= 16 = 2^{4}---------- (2)Using (1)

(1+R/ 100)

^{n}= (1+R /100)^{40}n = 40

Thus required time 40 years.

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30. A tree increases annually by 1â„5 th of its height. If its height today is 50 cm, what will be the height after 3 years?

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Correct Ans:86.4 cm

Explanation:

This problem can be solved by compound interest method.

Where R = (1/5) of its height = (1/5)*100 = 20 %

Time period n = 3 years so,

Height after 3 years = P (1 + (R/100))

= 50 * (1 + (20/100))

= 50 * (1 + (1/5))

= 50 * ((6*6*6) / (5*5*5))

= 86.4 cm.

Where R = (1/5) of its height = (1/5)*100 = 20 %

Time period n = 3 years so,

Height after 3 years = P (1 + (R/100))

^{n}= 50 * (1 + (20/100))

^{3}= 50 * (1 + (1/5))

^{3}= 50 * ((6*6*6) / (5*5*5))

= 86.4 cm.

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31. The difference between simple and compound interest compounded annually on a certain sum of money for 3 years at 4% per annum is Rs. 1. Find the sum.

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Correct Ans:Rs. 205.6

Explanation:

The difference between simple interest and compound interest for 3years given by

1 = x [(4/100)² * ((4/100) + 3)]

1 = x * (1/25)² * (76/25)

1 = 76x/15625

x = 205.59.

Hence the required sum is Rs. 205.6

**C.I - S.I = P[(R/100)² * ((R/100) + 3)]**1 = x [(4/100)² * ((4/100) + 3)]

1 = x * (1/25)² * (76/25)

1 = 76x/15625

x = 205.59.

Hence the required sum is Rs. 205.6

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32. At what percent per annum will Rs. 1000 amounts to Rs. 1728 in 3 years if interest compounded annually?

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Correct Ans:20%

Explanation:

P = 1000, A = 1728, n = 3 years

The amount is given by

(1 + (r/100))

(1 + (r/100))

(1 + (r/100))

Here both cubes cancels each other.

1 + (r/100) = 12/10

r/100 = (12/10) - 1

r/100 = 1/5

Therefore rate of interest r = 20%.

The amount is given by

**A = P (1 + (r/100))**^{n}(1 + (r/100))

^{3}= A/P(1 + (r/100))

^{3}= (1728/1000)(1 + (r/100))

^{3}= (12/10)^{3}Here both cubes cancels each other.

1 + (r/100) = 12/10

r/100 = (12/10) - 1

r/100 = 1/5

Therefore rate of interest r = 20%.

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33. A man borrows Rs. 4000 at 20% compound rate of interest. At the end of each year he pays back Rs. 1500. How much amount should he pay at the end of the third year to clear all his dues ?

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Correct Ans:Rs. 2952

Explanation:

Amount = 4000

20% of 4000 = 800

At the end of 1st year = 4000 + 800 = 4800

Pays back = 1500

Amount = 4800 -1500 = 3300

20% of 3300 = 660

At the end of 2nd year = 3300 + 660 = 3960

Pays back = 1500

Amount = 3960 - 1500 = 2460

20% of 2460 = 492

Amount to be paid at the end of third year = 2460 + 492 = 2952

20% of 4000 = 800

At the end of 1st year = 4000 + 800 = 4800

Pays back = 1500

Amount = 4800 -1500 = 3300

20% of 3300 = 660

At the end of 2nd year = 3300 + 660 = 3960

Pays back = 1500

Amount = 3960 - 1500 = 2460

20% of 2460 = 492

Amount to be paid at the end of third year = 2460 + 492 = 2952

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34. Some amount of money grows upto Rs. 8000 in 3 yr and upto Rs. 10000 in 4 yr on compound interest. What is the sum?

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Correct Ans:Rs. 4096

Explanation:

Here, sum = A

Where, A

Sum = 8000(8000/10000)

= Rs. 4096

_{1}(A_{1}/A_{2})^{n}Where, A

_{1}= Rs. 8000, A_{2}= Rs. 10000, n = 3 yrSum = 8000(8000/10000)

^{3}= Rs. 4096

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35. The compound interest on Rs 7500 in 2 years when the successive rate of interest on successive years is 8% and 10% respectively:

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Correct Ans:Rs 1410

Explanation:

amount at the end of 2nd year

= Rs 7500 (1 + 8/100) (1 + 10/100)

= Rs 7500 Ã— 1.08 Ã— 1.10

= Rs 8910

Thus C.I. for two years = amount â€“ principal

= Rs 8910 â€“ Rs 7500 = Rs 1410

= Rs 7500 (1 + 8/100) (1 + 10/100)

= Rs 7500 Ã— 1.08 Ã— 1.10

= Rs 8910

Thus C.I. for two years = amount â€“ principal

= Rs 8910 â€“ Rs 7500 = Rs 1410

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36. A lent an amount of Rs. 1100 to B. This is to be paid back to A in two instalments. If the rate of interest, which A charges to B, be 20% compounded annually, then what is the value of each instalment ?

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Correct Ans:Rs. 720

Explanation:

Let x = equal instalment at the end of each year

Now 1st year,

P = Rs. 1100

Interest at the end of 1st year = (1100 Ã— 20 Ã— 1)/100 = Rs. 22

Now, at the beginning of 2nd year,

P = Rs. (1100 + 220 â€“ x) = Rs. (1320 â€“ x)

Interest at the end of 2nd year

= ((1320 â€“ x) Ã— 20 Ã— 1)/100 = 264 â€“ x/5

Now amount remaining after 2 years

= (1320 â€“ x) + (264 â€“ x/5) â€“ x = 0

â‡’ 2x + x/5 = 1320 + 264

â‡’ 11x/5 = 1584

â‡’ x = 720

Now 1st year,

P = Rs. 1100

Interest at the end of 1st year = (1100 Ã— 20 Ã— 1)/100 = Rs. 22

Now, at the beginning of 2nd year,

P = Rs. (1100 + 220 â€“ x) = Rs. (1320 â€“ x)

Interest at the end of 2nd year

= ((1320 â€“ x) Ã— 20 Ã— 1)/100 = 264 â€“ x/5

Now amount remaining after 2 years

= (1320 â€“ x) + (264 â€“ x/5) â€“ x = 0

â‡’ 2x + x/5 = 1320 + 264

â‡’ 11x/5 = 1584

â‡’ x = 720

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37. A certain amount of money is lent out at compound interest at the rate of 20% per annum for two years, compounded annually. It would give Rs. 482 more if the amount is compounded half yearly. Find the principle.

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Correct Ans:None of these

Explanation:

we can apply the net % effect formula

x + y + xy/100%

Compounded annually at rate 20% per annum for 2 years, we get

= 20 + 20 + (20 Ã— 20)/100 = 44% ----- for 2 years

Similarly, compounded half yearly at rate 10%, we get

= 10 + 10 + (10 Ã— 10)/100 = 21% --------- for 2 years

And, 21 + 10 + (21 Ã— 10)/100 = 33.1% --------- for 3rd year

And, 33.1 + 10 + (33.1 Ã— 10)/100 = 46.41% ------- for 4th year

Now as per the question,

Difference between compound interest yearly and half yearly = 46.41 â€“ 44 = 2.41%

Given, 2.41% â‰¡ 482

100% â‰¡ x

â‡’ x = (482 Ã— 100)/2.41 = 20,000

x + y + xy/100%

Compounded annually at rate 20% per annum for 2 years, we get

= 20 + 20 + (20 Ã— 20)/100 = 44% ----- for 2 years

Similarly, compounded half yearly at rate 10%, we get

= 10 + 10 + (10 Ã— 10)/100 = 21% --------- for 2 years

And, 21 + 10 + (21 Ã— 10)/100 = 33.1% --------- for 3rd year

And, 33.1 + 10 + (33.1 Ã— 10)/100 = 46.41% ------- for 4th year

Now as per the question,

Difference between compound interest yearly and half yearly = 46.41 â€“ 44 = 2.41%

Given, 2.41% â‰¡ 482

100% â‰¡ x

â‡’ x = (482 Ã— 100)/2.41 = 20,000

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38. Aruna borrows Rs. 6250 from geetha at 10% CI. At the end of every year she pays Rs. 1000 as part repayment. How much does she still over after 3 such installments?

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Correct Ans:5008.75

Explanation:

6250*(10/100) = 625

(6250+625) -1000 = 5875

5875*(10/100) = 587.5

(5875+587.5) -1000 = 5462.5

5462.5*(10/100) = 546.25

(5462.5+546.25)-1000 = 5008.75

(6250+625) -1000 = 5875

5875*(10/100) = 587.5

(5875+587.5) -1000 = 5462.5

5462.5*(10/100) = 546.25

(5462.5+546.25)-1000 = 5008.75

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39. Rs. 6100 was partly invested in Scheme A at 10% pa compound interest (compounded annually) for 2 years and partly in Scheme B at 10% pa simple interest for 4 years. Both the schemes earn equal interests. How much was invested in Scheme A ?

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Correct Ans:Rs. 4000

Explanation:

Let the amount invested in Scheme A is Rs. x.

Then, the amount invested in Scheme B be Rs. (6100 â€“ x)

Now, according to the question,

x(1 + 10/100)

x(121/100 - 1) = (6100-x)*40 / 100

21x/100 = (6100-x)*40 / 100

21x = (6100-x)*40 - 40x

61x = 6100*40

x = 6100*40 / 61 = Rs. 4000

âˆ´ The amount invested in Scheme A is Rs. 4000.

Hence, option C is correct.

Then, the amount invested in Scheme B be Rs. (6100 â€“ x)

Now, according to the question,

x(1 + 10/100)

^{2}- x = (6100-x)*10*4 / 100x(121/100 - 1) = (6100-x)*40 / 100

21x/100 = (6100-x)*40 / 100

21x = (6100-x)*40 - 40x

61x = 6100*40

x = 6100*40 / 61 = Rs. 4000

âˆ´ The amount invested in Scheme A is Rs. 4000.

Hence, option C is correct.

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40. A person takes a loan of Rs. 200 at 5% p.a compound interest. He returns Rs. 100 at the end of one year. How much amount he would require to pay at the end of 2nd year in order to clear his dues.

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Correct Ans:Rs.115.50

Explanation:

Amount to be paid in 1st year = 200 + (200*5*1)/100 = 210

Amount left = 210-100 = 110

Required amount = 110 + (110*5*1)/100= 115.50

Amount left = 210-100 = 110

Required amount = 110 + (110*5*1)/100= 115.50

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