ML Aggarwal Compound Interest Chapter Test Class 9 ICSE Maths Solutions. We Provide Step by Step Answer of Chapter Test Questions for Compound Interest council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-9.

## ML Aggarwal Compound Interest Chapter Test Class 9 ICSE Maths Solutions

Board | ICSE |

Subject | Maths |

Class | 9th |

Chapter-2 | Compound Interest |

Topics | Solution of Ch-Test Questions |

Edition | 2024-2025 |

**Chapter Test on Compound Interest **

**Question 1. ₹ 10000 was lent for one year at 10% per annum. By how much more will the interest be, if the sum was lent at 10% per annum, interest being compounded half yearly?**

**Answer :**

It is given that

Principal = ₹ 10000

Rate of interest (r) = 10% p.a.

Period = 1 year

We know that

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

Substituting the values

= 10000 (1 + 10/100)^{1}

By further calculation

= 10000 × 11/10

= ₹ 11000

Here

Interest = A – P

Substituting the values

= 11000 – 10000

= ₹ 1000

In case 2,

Rate (r) = 10% p.a. or 5% half-yearly

Period (n) = 1 year or 2 half-years

We know that

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

Substituting the values

= 10000 (1 + 5/100)^{2}

By further calculation

= 10000 × 21/20 × 21/20

= ₹ 11025

Here

Interest = A – P

Substituting the values

= 11025 – 10000

= ₹ 1025

So the difference between the two interests = 1025 – 1000 = ₹ 25

**Question 2. A man invests ₹ 3072 for two years at compound interest. After one year the money amounts to ₹ 3264. Find the rate of interest and the amount due at the end of 2nd year.**

**Answer :**

It is given that

Principal (P) = ₹ 3072

Amount (A) = ₹ 3264

Period (n) = 1 year

We know that

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

Substituting the values

3264/3072 = (1 + r/100)^{1}

By further calculation

1 + r/100 = 17/16

r/100 = 17/16 – 1 = 1/16

By cross multiplication

r = 100 × 1/16 = 25/4 = 6 ¼

Hence, the rate of interest is 6 ¼%.

Here

Amount after 2 years = 3072 (1 + 25/ (4 × 100))^{2}

By further calculation

= 3072 (1 + 1/16)^{2}

So we get

= 3072 × 17/16 × 17/16

= ₹ 3468

Hence, the amount due at the end of 2 years is ₹ 3468.

**Question 3. What sum will amount to ₹ 28090 in two years at 6% per annum compound interest? Also find the compound interest.**

**Answer :**

It is given that

Amount (A) = ₹ 28090

Rate (r) = 6% p.a.

Period (n) = 2 years

We know that

P = A ÷ (1 + r/100)^{n}

Substituting the values

= 28090 ÷ (1 + 6/100)^{2}

By further calculation

= 28090 ÷ (53/50)^{2}

So we get

= 28090 × 50/53 × 50/53

= ₹ 25000

Here

Amount of CI = A – P

Substituting the values

= 28090 – 25000

= ₹ 3090

**Question 4. Two equal sums were lent at 5% and 6% per annum compound interest for 2 years. If the difference in the compound interest was ₹ 422, find:**

**(i) the equal sums**

**(ii) compound interest for each sum.**

**Answer :**

Consider ₹ 100 as each equal sum

Case I –

Rate (r) = 5%

Period (n) = 2 years

We know that

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

Substituting the values

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

It can be written as

= 100 × 21/20 × 21/20

= ₹ 441/4

Here

CI = A – P

Substituting the values

= 441/4 – 100

= ₹ 41/4

Case II –

Rate of interest (R) = 6^{n}

Period (n) = 2 years

We know that

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

Substituting the values

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

It can be written as

= 100 × 53/50 × 53/50

= ₹ 2809/25

Here

CI = A – P

Substituting the values

= 2809/25 – 100

= ₹ 309/25

So the difference between the two interests = 309/25 – 41/4

Taking LCM

= (1236 – 1025)/ 100

= ₹ 211/100

If the difference is ₹ 211/100, then equal sum = ₹ 100

If the difference is ₹ 422, then equal sum = (100 × 422 × 100)/ 211 = ₹ 20000

Here, Amount in first case = 20000 (1 + 5/100)^{2}

So we get

= 20000 × (21/20)^{2}

It can be written as

= 20000 × 21/20 × 21/20

So we get

= 44100/2

= ₹ 22050

CI = 22050 – 20000 = ₹ 2050

Amount in second case = 20000 (1 + 6/100)^{2}

It can be written as

= 20000 × 53/50 × 53/50

= ₹ 22472

CI = 22472 – 20000 = ₹ 2472

**Question 5. The compound interest on a sum of money for 2 years is ₹ 1331.20 and the simple interest on the same sum for the same period at the same rate is ₹ 1280. Find the sum and the rate of interest per annum.**

**Answer :**

It is given that

CI for 2 years = ₹ 1331.20

SI for 2 years = ₹ 1280

So the difference = 1331.20 – 1280 = ₹ 51.20

Here ₹ 51.20 is the simple interest on 1280/2 = ₹ 640 for one year

We know that

Rate = (SI × 100)/ (P × t)

Substituting the values

= (51.20 × 100)/ (640 × 1)

Multiplying and dividing by 100

= (5120 × 100)/ (100 × 640)

= 8% p.a.

So the SI for two years at the rate of 8% pa

Sum = (SI × 100)/ (r × t)

Substituting the values

= (1280 × 100)/ (8 × 2)

= ₹ 8000

**Question 6. On what sum will the difference between the simple and compound interest for 3 years if the rate of interest is 10% p.a. is ₹ 232.50?**

**Answer :**

Consider sum (P) = ₹ 100

Rate (r) = 10% p.a.

Period (n) = 3 years

We know that

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

Substituting the values

= 100 (1 + 10/100)^{3}

By further calculation

= 100 × 11/10 × 11/10 × 11/10

= ₹ 133.10

Here, CI = A – P

Substituting the values

= 133.10 – 100

= ₹ 33.10

So the simple interest = PRT/100

Substituting the values

= (100 × 10 × 3)/ 100

= ₹ 30

Difference = 33.10 – 30 = ₹ 3.10

Here if the difference is ₹ 3.10 then sum = ₹ 100

If the difference is ₹ 232.50 then sum = (100 × 232.50)/ 3.10

Multiplying and dividing by 100

= (100 × 23250)/ 310

= ₹ 7500

**Question 7. The simple interest on a certain sum for 3 years is ₹ 1080 and the compound interest on the same sum at the same rate for 2 years is ₹ 741.60. Find:**

**(i) the rate of interest**

**(ii) the principal.**

**Answer :**

It is given that

SI for 3 years = ₹ 1080

SI for 2 years = (1080 × 2)/ 3 = ₹ 720

CI for 2 years = ₹ 741.60

So the difference = 741.60 – 720 = ₹ 21.60

Here ₹ 21.60 is the SI on 720/2 = ₹ 360 for one year

##### (i) We know that

Rate = (SI × 100)/ (P × t)

Substituting the values

= (21.60 × 100)/ (360 × 1)

Multiply and divide by 100

= (2160 × 100)/ (100 × 360 × 1)

= 6%

##### (ii) ₹ 1080 is SI for 3 years at the rate of 6% p.a.

So the principal = (SI × 100)/ (r × t)

Substituting the values

= (1080 × 100)/ (6 × 3)

= ₹ 6000

**Question 8. In what time will ₹ 2400 amount to ₹ 2646 at 10% p.a. compounded semi-annually?**

**Answer :**

It is given that

Amount (A) = ₹ 2646

Principal (P) = ₹ 2400

Rate (r) = 10% p.a. or 5% semi-annually

Consider Period = n half-years

We know that

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

Substituting the values

2646/2400 = (1 + 5/100)^{n}

By further calculation

(21/20)^{n} = 441/400 = (21/20)^{2}

n = 2

Therefore, the time period is 2 half years or 1 year.

**Question 9. Sudarshan invested ₹ 60000 in a finance company and received ₹ 79860 after 1 ½ years. Find the rate of interest per annum compounded half-yearly.**

**Answer :**

It is given that

Principal (P) = ₹ 60000

Amount (A) = ₹ 79860

Period (n) = 1 ½ years = 3 half-years

We know that

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

Substituting the values

79860/60000 = (1 + r/100)^{3}

By further calculation

(1 + r/100)^{3} = 1331/1000 = (11/10)^{3}

We get

1 + r/100 = 11/10

r/100 = 11/10 – 1 = 1/10

By cross multiplication

r = 1/10 × 100 = 10% half-yearly

r = 10 × 2 = 20% p.a.

Therefore, the rate of interest per annum compounded half-yearly is 20%.

**Question 10. The population of a city is 320000. If the annual birth rate is 9.2% and the annual death rate is 1.7%, calculate the population of the town after 3 years.**

**Answer :**

It is given that

Birth rate = 9.2%

Death rate = 1.7%

So the net growth rate = 9.2 – 1.7 = 7.5%

Present population (P) = 320000

Period (n) = 3 years

We know that

Population after 3 years (A) = P (1 + r/100)^{n}

Substituting the values

= 320000 (1 + 7.5/100)^{3}

By further calculation

= 320000 (1 + 3/40)^{3}

= 320000 × (43/40)^{3}

So we get

= 320000 × 43/40 × 43/40 × 43/40

= 397535

**Question 11. The cost of a car, purchased 2 years ago, depreciates at the rate of 20% every year. If the present value of the car is ₹ 315600 find:**

**(i) its purchase price**

**(ii) its value after 3 years**

**Answer:**

It is given that

Present value of car = ₹ 315600

Rate of depreciation (r) = 20%

(i) We know that

Purchase price = A ÷ (1 – r/100)^{n}

Substituting the values

= 315600 ÷ (1 – 20/100)^{2}

By further calculation

= 315600 × 5/4 × 5/4

= ₹ 493125

(ii) We know that

Value after 3 years = 315600 × (1 – 20/100)^{3}

By further calculation

= 315600 × 4/5 × 4/5 × 4/5

= ₹ 161587.20

**Question 12. Amar Singh started a business with an initial investment of ₹ 400000. In the first year he incurred a loss of 4%. However, during the second year, he earned a profit of 5% which in the third year rose to 10%. Calculate his net profit for the entire period of 3 years.**

**Answer :**

It is given that

Investment (P) = ₹ 400000

Loss in the first year = 4%

Profit in the second year = 5%

Profit in the third year = 10%

We know that

Total amount after 3 years = P (1 + r/100)^{n}

Substituting the values

= 400000 (1 – 4/100) (1 + 5/100) (1 + 10/100)

By further calculation

= 400000 × 24/25 × 21/20 × 11/10

= ₹ 443520

So the net profit after 3 years = 443520 – 400000 = ₹ 43520

— : End of ML Aggarwal Compound Interest Chapter Test Class 9 ICSE Maths Solutions :–

Return to :- ** ML Aggarawal Maths Solutions for ICSE Class-9**

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