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Class 8 RD Sharma Solutions – Chapter 14 Compound Interest – Exercise 14.3 | Set 2

  • Last Updated : 08 Apr, 2021

Chapter 14 Compound Interest – Exercise 14.3 | Set 1

Question 15. Find the rate percent per annum, if Rs. 2000 amount to Rs. 2315.25 in a year and a half, interest being compounded six monthly.

Solution:

We have,

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Principal = Rs 2000



Amount = Rs 2315.25

Time = 1 ½ years = 3/2 years

Let rate be = R % per annum

By using the formula,

A = P (1 + \frac{R}{100}  )n

Substituting the values, we have

2315.25 = 2000 (1 + \frac{R}{100}  )3/2

(1 + \frac{R}{100}  )3/2 = 2315.25/2000



(1 + \frac{R}{100}  )3/2 = (1.1576)

(1 + \frac{R}{100}  ) = 1.1025

\frac{R}{100}   = 1.1025 – 1

= 0.1025 × 100

= 10.25

Therefore,

Required Rate is 10.25% per annum.

Question 16. Find the rate at which a sum of money will double itself in 3 years, if the interest is compounded annually.

Solution:

We have,

Time = 3 years



Let rate be = R %

Also principal be = P

So, amount becomes = 2P

By using the formula,

A = P (1 + \frac{R}{100}  )n

Substituting the values, we have

2P = P (1 + \frac{R}{100}  )3

(1 + \frac{R}{100}  )3 = 2

(1 + \frac{R}{100}  ) = 21/3

1 + \frac{R}{100}   = 1.2599



\frac{R}{100}   = 1.2599-1

= 0.2599

R = 0.2599 × 100

= 25.99

Therefore,

Required Rate is 25.99% per annum.

Question 17. Find the rate at which a sum of money will become four times the original amount in 2 years, if the interest is compounded half-yearly

Solution:

We have,

Time = 2 years = 2×2 = 4 half years

Let rate = R % per annum = R/2% half years

Let principal be = P

So, Amount becomes = 4P

By using the formula,

A = P (1 + \frac{R}{100}  )n

Substituting the values, we have

4P = P (1 + \frac{R}{100 \times 2}  )4

(1 + \frac{R}{200}  )4 = 4

(1 + \frac{R}{200}  ) = 41/4

1 + \frac{R}{200}   = 1.4142

\frac{R}{200}   = 1.4142-1



= 0.4142

R = 0.4142 × 200

= 82.84%

Therefore,

Required Rate is 82.84% per annum.

Question 18. A certain sum amounts to Rs. 5832 in 2 years at 8% compounded interest. Find the sum.

Solution:

We have,

Amount = Rs 5832

Time = 2 years

Rate = 8%

Let principal be = P

By using the formula,

A = P (1 + \frac{R}{100}  )n

Substituting the values, we have

5832 = P (1 + \frac{8}{100}  )2

5832 = P (1.1664)

P = 5832/1.1664

= 5000

Therefore,

Required sum is Rs 5000.

Question 19. The difference between the compound interest and simple interest on a certain sum for 2 years at 7.5% per annum is Rs. 360. Find the sum.

Solution:

We have,

Time = 2 years

Rate = 7.5 % per annum

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 360

C.I – S.I = Rs 360

By using the formula,

P [(1 + \frac{R}{100}  )n – 1] – (PTR)/100 = 360

Substituting the values, we have



P [(1 + \frac{7.5}{100}  )2 – 1] – (P(2)(7.5))/100 = 360

P[249/1600] – (3P)/20 = 360

249/1600P – 3/20P = 360

(249P-240P)/1600 = 360

9P = 360 × 1600

P = 576000/9

= 64000

Therefore,

The sum is Rs 64000.

Question 20. The difference in simple interest and compound interest on a certain sum of money at 623 % per annum for 3 years in Rs. 46. Determine the sum.

Solution:

We have,

Time = 3 years

Rate = 6\frac{2}{3}   % per annum = 20/3%

Let principal = Rs P

Compound Interest (CI) – Simple Interest (SI) = Rs 46

C.I – S.I = Rs 46

By using the formula,

P [(1 + \frac{R}{100}  )n – 1] – (PTR)/100 = 46

Substituting the values, we have

P [(1 + \frac{20}{3\times100}  )3 – 1] – (P(3)(20/3))/100 = 46

P[(1 + \frac{20}{300}  )3 – 1] – P/5 = 46

P[721/3375] – P/5 = 46

721/3375P – 1/5P = 46

(721P-675P)/3375 = 46

46P = 46 × 3375

46P = 46 × 3375/46

= 3375

Therefore, 

The sum is Rs 3375.

Question 21. Ishita invested a sum of Rs. 12000 at 5% per annum compound interest. She received an amount of Rs. 13230 after n years. Find the value of n.

Solution:



We have,

Principal = Rs 12000

Amount = Rs 13230

Rate = 5% per annum

Let time = T years

By using the formula,

A = P (1 + \frac{R}{100}  )n

Substituting the values, we have

13230 = 12000 (1 + \frac{5}{100}  )T

13230 = 12000 (\frac{105}{100}  )T

(21/20)T = 13230/12000

(21/20)T = 441/400

(21/20)T = (21/20)2

So on comparing both the sides, n = T = 2

Therefore, 

Time required is 2 years.

Question 22. At what rate percent per annum will a sum of Rs. 4000 yield compound interest of Rs. 410 in 2 years?

Solution:

We have,

Principal = Rs 4000

Time = 2 years

CI = Rs 410

Rate be = R% per annum

By using the formula,

CI = P [(1 + \frac{R}{100}  )n – 1]

Substituting the values, we have

410 = 4000 [(1 + \frac{R}{100}  )2 – 1]

410 = 4000 (1 + \frac{R}{100}  )2 – 4000

410 + 4000 = 4000 (1 + \frac{R}{100}  )2

(1 + \frac{R}{100}  )2 = 4410/4000

(1 + \frac{R}{100}  )2 = 441/400



(1 + \frac{R}{100}  )2 = (21/20)2

By canceling the powers on both the sides,

1 + \frac{R}{100}   = 21/20

\frac{R}{100}   = 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Required Rate is 5% per annum.

Question 23. A sum of money deposited at 2% per annum compounded annually becomes Rs. 10404 at the end of 2 years. Find the sum deposited.

Solution:

We have,

Time = 2years

Amount = Rs 10404

Rate be = 2% per annum

Let principal be = Rs P

By using the formula,

A = P [(1 + \frac{R}{100}  )n

Substituting the values, we have

10404 = P [(1 + \frac{2}{100}  )2]

10404 = P [1.0404]

P = 10404/1.0404

= 10000

Therefore, 

Required sum is Rs 10000.

Question 24. In how much time will a sum of Rs. 1600 amount to Rs. 1852.20 at 5% per annum compound interest?

Solution:

We have,

Principal = Rs 1600

Amount = Rs 1852.20

Rate = 5% per annum

Let time = T years

By using the formula,

A = P (1 + \frac{R}{100}  )n

Substituting the values, we have

1852.20 = 1600 (1 + \frac{5}{100}  )T

1852.20 = 1600 (\frac{105}{100}  )T

(21/20)T = 1852.20/1600

(21/20)T = 9261/8000

(21/20)T = (21/20)3

So on comparing both the sides, n = T = 3



Therefore,

Time required is 3 years.

Question 25. At what rate percent will a sum of Rs. 1000 amount to Rs. 1102.50 in 2 years at compound interest?

Solution:

We have,

Principal = Rs 1000

Amount = Rs 1102.50

Rate = R% per annum

Let time = 2 years

By using the formula,

A = P (1 + \frac{R}{100}  )n

Substituting the values, we have

1102.50 = 1000 (1 + \frac{R}{100}  )2

(1 + \frac{R}{100}  )2 = 1102.50/1000

(1 + \frac{R}{100}  )2 = 4410/4000

(1 +\frac{R}{100}  )2 = (21/20)2

1 + \frac{R}{100}   = 21/20

\frac{R}{100}  = 21/20 – 1

= (21-20)/20

= 1/20

R = 100/20

= 5

Therefore,

Required Rate is 5%.

Question 26. The compound interest on Rs. 1800 at 10% per annum for a certain period of time is Rs. 378. Find the time in years.

Solution:

We have,

Principal = Rs 1800

CI = Rs 378

Rate = 10% per annum

Let time = T years

By using the formula,

CI = P [(1 + \frac{R}{100}  )n – 1]

Substituting the values, we have

378 = 1800 [(1 + \frac{10}{100}  )T – 1]

378 = 1800 [(\frac{110}{100}  )T – 1]

378 = 1800 [(\frac{11}{10}  )T – 1800

378 + 1800 = 1800 [(\frac{11}{10}  )T

(11/10)T = 2178/1800

(11/10)T = 726/600

(11/10)T = 121/100

(11/10)T = (11/10)2



So on comparing both the sides, n = T = 2

Therefore,

Time required is 2 years.

Question 27. What sum of money will amount to Rs. 45582.25 at 6 ¾ % per annum in two years, interest being compounded annually

Solution:

We have,

Time = 2years

Amount = Rs 45582.25

Rate be = 6 ¾ % per annum = 27/4%

Let principal be = Rs P

By using the formula,

A = P [(1 + \frac{R}{100}  )n

Substituting the values, we have

45582.25 = P [(1 + 27/4×100)2]

45582.25 = P (1 + \frac{27}{100}  )2

45582.25 = P (\frac{427}{400}  )2

45582.25 = P × 427/400 × 427/400

P = (45582.25 × 400 × 400) / (427×427)

P = 7293160000/182329

= 40000

Therefore,

Required sum is Rs 40000.

Question 28. Sum of money amounts to Rs. 453690 in 2 years at 6.5% per annum compounded annually. Find the sum.

Solution:

We have,

Time = 2years

Amount = Rs 453690

Rate be = 6.5 % per annum

Let principal be = Rs P

By using the formula,

A = P [(1 + \frac{R}{100}  )n

Substituting the values, we have

453690 = P [(1 + \frac{6.5}{100}  )2]

453690 = P (\frac{106.5}{100}  )2

453690 = P × 106.5/100 × 106.5/100

P = (453690 × 100 × 100) / (106.5×106.5)

P = 4536900000/11342.25

= 400000

Therefore,

Required sum is Rs 400000.




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