Compound Interest Class 8 RS Aggarwal Exe-8C Goyal Brothers ICSE Maths Solutions Ch-8. We provide step by step Solutions of council prescribe textbook / publication to develop skill and confidence. Visit official Website **CISCE** for detail information about ICSE Board Class-8 Mathematics.

## Compound Interest Class 8 RS Aggarwal Exe-8C Goyal Brothers ICSE Maths Solutions Ch-8

Board | ICSE |

Publications | Goyal Brothers Prakashan |

Subject | Maths |

Class | 8th |

writer | RS Aggarwal |

Book Name | Foundation |

Ch-8 | Simple Interest and Compound Interest |

Exe-8C | Compound Interest using Formula |

Academic Session | 2024-2025 |

### How to Solve Compound Interest Problems Using Formula

The formula for finding the amount on compound interest is given by: A = P [1 +(R/100) ]^{n}

This is the amount when interest is compounded annually.

Compound interest (CI) = Amount – Original Principal

### How to Solve Growth and Depreciation Problems Easily

The problems on growth and depreciations can be easily solve by using compound interest formula.

Final (Later) Value must be taken as amount While Current (Previous Value) must be taken as principal

If Depreciation then use formula A = P [1 – (R/100) ]^{n } and if Growth then use formula A = P [1 + (R/100) ]^{n}

**Exercise- 8C**

( Compound Interest Class 8 RS Aggarwal Exe-8C Goyal Brothers ICSE Maths Solutions Ch-8 )

**Que-1: Mr Dubey borrows Rs100000 from a bank at 10% per annum Compound interest. What amount will he have to pay to the bank after 3 years ?**

**Sol: **A = P(1 + R/100)^T

A = 100,000(1 + 10/100)³

A = 100,000(11/10)³

A = 133,100

**Que-2: Find the amount and the compound interest on Rs10240 for 3 years at 12*(1/2)% per annum compounded annually.**

**Sol: **Principal (P) = Rs. 10240

Rate (R) = 12*(1/2)% = 25/2% p.a

.Period (n) = 3 years

∴ Amount (A) = P (1+R100)^n

= Rs. 10240 (1+{25/(2×100)})³

= Rs. 10240×(9/8)³

= Rs. 10240×(9/8)×(9/8)×(9/8)

= Rs. 14580

∴ C.I. = A – P

= Rs. 14580 – Rs. 10240

= Rs. 4340

**Que-3: The present population of a town is 176400. If it increases at the rate of 5% per annum, what will be its population after 2 years?**

**Sol: **Initial population of the town = 176400

Annual rate of increase in the population = 5%

Increased Population = (1 + R/100)^{n} x Initial Population

Total population after two years = (1 + 5/100)^{2} x 176400

[(100 + 5)/100] × [(100 + 5)/100] × 176400

= 194481

**Que-4: Amit started a shop by investing Rs50000. He gained 4% during first year, 5% during second year and 10% during 3rd year. What will be his capital amount after these three years?**

**Sol: **Principal = Rs50000

Rate of gain = 4%, 5%, 10%

Time = 3 year

Amount = P[1+(R1/100)][1+(R2/100)][1+(R3/100)]

A = 50000[1+(4/100)][1+(5/100)][1+(10/100)]

A = 50000(104/100)(105/100)(110/100)

A= 52×105×11

A = 60060.

**Que-5: A mango tree of height 125 cm was planted 3 years ago. If it increases at the rate of 20% per annum, what is its present height?**

**Sol: **Mango tree’s original height = 125 cm

Height after 1 year = 125 + 20/100 * 125 = 125 + 25 = 150cm

Height after second year = 150 + 20/100 * 150 = 150 + 30 = 180cm

Height after third year = 180 + 20/100 * 180 = 180 + 36 = 216 cm

**Que-6: Two years ago, the population of a town was 10000. During first year, it increased at the rate of 5% per annum and during second year, it increased at the rate of 6% per annum. What is its present population?**

**Sol: **We are given that two years ago the population of the town was 10000

During first year it increased at the rate of 5% per annum

So, Population after 1st year = 10000 + [(5/100)10000] = 10500

During second year it increased at the rate of 6% per annum

So, Population after 2nd year

= 10500 + 6%(10500)

= 10500 + [(6/100)10500]

= 11130

**Que-7: Mahesh borrowed Rs16000 at 7*(1/2)% per annum simple interest. On the same day, he lent it to Gagan at the same rate but compounded annually. What does he gain at the end of 2 years?**

**Sol: **P = 16000

R = 7*(1/2)% = 15/2%.

T = 2 years

We know that SI = PRT/100

= (16000 * 15 * 2)/2 * 100

= (16000 * 15)/100

= 2400

We know that A = P + I

= 16000 + 2400

= 18400

Given that he lent it to Gagan at the same rate.

Given R = 15/2.

n = 2 years.

P = 16000.

We know that A = P(1 + r/100)^n

= 16000(1 + 15/200)^2

= 16000(215/200)^2

= 16000(43/40)^2

= 16000 * 1849/1600

= 10 * 1849

= 18490

Therefore the gain = 18490 – 18400

= 90

**Que-8: A machine is purchased for Rs625000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years?**

**Sol: **Value of machine (P) = Rs. 625000

Rate of depreciation (R) = 8% p.a.

Period (n) = 2 years

∴ Value after 2 years = P(1−R100)^n

= Rs. 625000(1−(8/100))^2

= Rs. 625000(23/25)^2

= Rs. 625000×(23/25)×(23/25)

= Rs.529000

**Que-9: A car is purchased for Rs348000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years?**

**Sol: **Cost of car = Rs. 348000

Rate of depreciation(R1) = 10% p.a. for first year

∴ (R2) = 20% p.a. for second year

∴ Value after 2 years = P(1−(R1/100))(1−(R2/100))

= Rs. 348000 (1−(10/100))(1−(20/100))

= Rs. 348000×(9/10)×(4/5)

= Rs. 250560

**–: **Compound Interest Class 8 RS Aggarwal Exe-8C Goyal Brothers ICSE Maths Ch-8 Solutions** :–**

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