Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Foundation Maths Solutions. In this article you will learn how to solve Multiple Choice Questions on Compound Interest very easily**.** Visit official Website **CISCE** for detail information about ICSE Board Class-9 Mathematics.

## Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions

Board | ICSE |

Publications | Goyal brothers Prakshan |

Subject | Maths |

Class | 9th |

Chapter-2 | Compound Interest |

Writer | RS Aggrawal |

Book Name | Foundation |

Topics | Solution of MCQs |

Academic Session | 2024-2025 |

### How to Solve MCQs on Compound Interest Easily

To solve mcqs on Compound Interest use formula or without formula and calculate unknown value Also focus on growth and deprecitaion formula in some cases. At last check the option available which is most suitable after calculation.

**Multiple Choice Questions **

Compound Interest Class 9 RS Aggarwal Goyal Brothers ICSE Foundation Maths Solutions.

**Page- 41,42**

**Que-1: The compound interest on Rs 3750 for 2 years at 8% p.a., compounded annually is :**

**(a) Rs604 (b) Rs614 (c) Rs624 (d) Rs642**

**Solution- **(c) Rs624

**Reason: **A = P * (1+R/100)^n

A = 3750 * (1+8/100)^2

CI = A- Principal = 624Rs.

**Que-2: A man invests Rs46875 at 4% p.a. compound interest for 3 years. The interest for the first year will be :**

**(a) Rs1785 (b) Rs1587 (c) Rs1875 (d) Rs1758**

**Solution- **(c) Rs1875

**Reason: **It is given that

Principal= 46875

Rate of interest = 4% p.a.

(i) Interest for the first year= Prt/100

Substituting the values

=(46875×4×1)/100

= 1875.

**Que-3: A man deposits Rs10000 in a cooperative bank for 3 years at 9% p.a. If interest is compounded annually, then the amount he will get from the bank after 3 years is :**

**(a) Rs12950.29 (b) Rs12905.29 (c) Rs12059.29 (d) Rs12095.29**

**Solution- **(a) Rs12950.29

**Reason: **Principal amount (P) is Rs.10,000

rate (r) is 9% or 0.09

years (t) is 3

A = 10000(1 + 0.09/1)^(1*3)

= 10000(1 + 0.09)^3

= 10000(1.09)^3

≈ 10000(1.295029)

≈ Rs.12,950.29.

**Que-4: Rs16000 is deposited in a bank for three years. The rate of compound interest for first and second year are 8% and 12% respectively. At the end of third year the amount becomes Rs21384. The rate of interest for third year will be :**

**(a) 7% (b) 10% (c) 11% (d) 12%**

**Solution- **(b) 10%

**Reason: **Initial deposit (P) = Rs16,000

Interest rate for the first year = 8%

Interest rate for the second year = 12%

Final amount after three years = Rs. 21,384

A1 = P(1+r1/100)

where r1 is the interest rate for the first year.

𝐴1 = 16000(1+8/100)

𝐴1 = 16000(1+0.08)

𝐴1 = 16000×1.08

A1 = 17280

A2 = A1(1+r2/100)

where r2 is the interest rate for the second year.

𝐴2 = 17280(1+12/100)

𝐴2 = 17280(1+0.12)

𝐴2 = 17280×1.12

𝐴2 = 19353.60

A3 = A2(1+r3/100)

We know that A3 = 21384.

So, 21384 = 19353.60(1+𝑟3/100)

To find 1+(𝑟3/100):

1+(𝑟3/100) = 21384/19353.60

1+(𝑟3/100) = 1.105

Solving for r3:

𝑟3/100 = 1.105 − 1

𝑟3/100 = 0.105

𝑟3 = 0.105 × 100

𝑟3 = 10%.

**Que-5: A man borrows Rs5000 at 12% compound interest p.a., interest payable every six months. He pays back Rs1800 at the end of every six months. The third payment he has to make at the end of 18 months in order to clear the entire loan will be :**

**(a) Rs2024.60 (b) Rs2204.60 (c) Rs2240.60 (d) Rs2402.60**

**Solution- **(a) Rs2024.60

**Reason: **For first year,

P = 5000

R =12%

T = 1/2 year

I = PTR/100

I = 5000×12×1/100×1/2

A = P + I

A = 5000+ 300

A = 5300

Man repaid = 5300 – 1800

= 3500

For next six month,

P = 3500

R = 12 %

T = 1/2

I = PTR/100

= 3500×12×1/100×1/2

= 210

Now we know that

A = P + I

= 3500 + 210

= 3710

Man repaid = 3710- 1800

= 1910

For last six months ,

P = 1910

R = 12

T = 1/2

I = PTR/100

= 114.6

A = P + I

= 1910+ 114.6

= 2024.6

**Que-6: The** **compound interest for the second year on Rs8000 invested for 3 years at 10% p.a. is :**

**(a) Rs780 (b) Rs880 (c) Rs980 (d) Rs1080**

**Solution- **(b) Rs880

**Reason:**For the first year

P = Rs8,000

N = 1year

R = 10 %

We have S.I. = PNR/100 = (8,000×1×10)/100 = Rs800

And Amount at the end of first year P+S.I. = Rs8,000 + Rs800 = Rs8,800

Now, for the second year

P = Rs8,800

N = 1year

R = 10 %

We have S.I. = PNR/100 = (8,800×1×10)/100 = Rs880

Thus, Compound interest for the second year = Rs880.

**Que-7: A person took a loan of Rs6000 from a bank and agreed to pay back the amount along with interest in 2 years. If the rate of compound interest for the first year is 10% and second year is 12%, the amount he had to pay after 2 years will be :**

**(a) Rs 7329 (b) Rs 7932 (c) Rs 7292 (d) Rs 7392**

**Solution- **(d) Rs7392

**Reason: **Principal amount (P) = Rs. 6000

Interest rate for the first year (r1) = 10%

𝐴1 = 6000(1+10/100)

𝐴1 = 6000(1+0.10)

𝐴1 = 6000×1.10 = 6600

Amount at the end of the first year (A1) = Rs. 6600

Interest rate for the second year (r2) = 12%

𝐴2 = 6600(1+12/100)

𝐴2 = 6600(1+0.12)

𝐴2 = 6600×1.12

𝐴2 = 7392.

**Que-8: Nikita invests Rs 6000 for 2 years at a certain rate of interest compounded annually. At the end of the first year, it amounts to Rs6720. The rate of interest p.a. is :**

**(a) 8% (b) 10% (c) 12% (d) 14%**

**Solution- **(c) 12%

**Reason: **P = Rs. 6000, Amount at the end of the first year = Rs. 6720

S.I. for first year = Rs. (6720 – 6000) = Rs. 720

Let r% be the rate of interest p.a.

720 = (6000×𝑟×1)/100

r = 720/60

r = 12%.

**Que-9: The compound interest on Rs8640 for 3 years at 8% p.a. is :**

**(a) Rs2345 (b) Rs3245 (c) Rs3425 (d) Rs3452**

**Solution- **(a) Rs2345

**Reason: **

**Que-10: If the interest is compounded half-yearly, then, C.I. when the principal is Rs7400, the rate of interest is 5% p.a. and the duration is one year, is :**

(a) Rs373.63 (b) Rs374.63 (c) Rs373.36 (d) Rs373

**Solution- **(b) Rs374.63

**Reason:**It is given that

Principal (P) = 7400

Rate of interest (r) = 5%

Period (n) = 1 year

We know that

A = P(1+r/(2×100))^2×n

Substituting the values

= 7400(1+5/200)^2

By further calculation

= 7400 × 205/200 × 205/200

= 7774.63

C.I. = A – P

C.I. = 7774.63 – 7400

C.I. = 374.63.

**Que-11: The simple interest on a sum of money for 2 years at 4% per annum is Rs340. The compound interest on this sum for one year payable half-yearly at the same rate is :**

**(a) Rs170.70 (b) Rs107.70 (c) Rs171.70 (d) Rs270.70**

**Solution- **(c) Rs171.70

**Reason: **Given : I = Rs340, T = 2 Years and R = 4%

∴P = (I×100)/(R×T)

= (Rs340×100)/(4×2) = Rs.4250

C.I. = P(1+r/(2×100))^n×2 − P

= Rs.4250(1+4/(2×100))^1×2 − Rs.4250

= Rs.4421.70 − Rs.4250

= Rs.171.70

**Que-12: The compound interest on a certain sum of money at 5% p.a. for two years is Rs246. The simple interest on the same sum for three years at 6% p.a. will be :**

**(a) Rs432 (b) Rs430.50 (c) Rs432.75 (d) Rs431.75**

**Solution- **(a) Rs432

**Reason: **CI = Rs246, R = 5%, T = 2 years

CI = A – P

246 = P[1+5/100]^2 – P

246 = P|(21/20)^2 – 1|

246 = P[41/400]

P = (246×400)/41

= Rs2400

Now, P = Rs2400, R = 6%, T = 3 years

SI = (2400×6×3)/100

= Rs432.

**Que-13: Ramesh wants to get Rs6050 from a bank after 2 years. If the bank gives 10% p.a. compound interest, then the amount of money he has to keep now in the bank is :**

**(a) Rs5500 (b) Rs5000 (c) Rs5600 (d) Rs5800**

**Solution- **(b) Rs5000

**Reason: **Amount = Rs. 6,050

Time = 2 years

Rate of interest = 10% p.a.

Let assume that Principal = x

So,

[(100 + Rate of interest)/100]Time = (Amount/Principal)

⇒ [(100 + 10)/100]^{2} = (6050/ x)

⇒ (110/100) × (110/100) = (6050/ x)

⇒ (121/100) = (6050/x)

By cross multiplying

⇒ x = 6050 × (100/121)

⇒ x = 5,000

**Que-14: The difference between the compound interest and simple interest on a certain sum deposited for 2 years at 5% p.a. is Rs12. The sum will be :**

**(a) Rs4500 (b) Rs4600 (c) Rs4800 (d) Rs5000**

**Solution- **(c) Rs4800

**Reason:**N = 2years

R = 5 %

We have S.I. = (PNR)/100 = (P×2×5)/100 = 10P/100 = 0.1P

And on interest being compounded for 2 years and R = 5 %,

Amount = P(1+R/100)^N

= P(1+5/100)²

= P×(1.05)²

= 1.1025P

So, C.I. = A−P = 1.1025P − P = 0.1025P

Given, C.I. − S.I = Rs12

= 0.1025P−0.1P = Rs12

= 0.0025P = Rs12

P = Rs4,800.

**Que-15: At what rate of compound interest p.a. will Rs20000 amount to Rs26620 in 3 years ?**

**(a) 4% (b) 6% (c) 8% (d) 10%**

**Solution- **(d) 10%

**Reason: **Amount = Principal (1 + r/100__)^n__

= 26620 = 20000 (1+r/100)³

( 1+ r%)³ = 26620/20000

= ( 1 +r%)³ = 1331 / 1000

= ( 1+ r %) = ³√ 1331 / 1000

= ( 1+ r%) = 11/10

= ( 1 + r/100 ) = 11/10

(100+r)/100 = 11 /10

= (100 + r) = (__11__×100)/10

= 100 + r = 110

= rate = 110 – 100

= 10%

**Que-16: In what time will Rs5000 amount to Rs5832 at 8% rate of compound interest p.a. ?**

**(a) 2 years (b) 4 years (c) 6 years (d) 8 years**

**Solution- **(a) 2 years

**Reason: **Principal = Rs. 5,000

Amount = Rs. 5,832

Rate = 8% p.a.

Amount = Principal × (1 + 𝑅/100)^n

⇒ 5832 = 5000 × (1 + 8/100)^n

⇒ 5832/5000 = (108/100)^n

⇒ (729/625) = (27/25)^n

⇒ (27/25)² = (27/25)^n

⇒ Time = 2 years

**Que-17: A machine depreciates at the rate of 10% of its value at the beginning of the year. If the present value of the machine is Rs8000, its value after three years will be :**

**(a) Rs5382 (b) Rs5832 (c) Rs5238 (d) Rs5638**

**Solution- **(b) Rs5832

**Reason: **machine value = ₹8000

machine depreciation every year 10%

first year -> 10% of 8000 = 800

value is = 8000 – 800 = ₹7200

second year again 10% decrease

value is = 10% of 7200 = 720

so 7200 – 720 = 6480

final third year also 10% decrease

value is = 10% of 6480 = 648

final value is = 6480 – 648 = ₹5832

**Que-18: The present population of a town is 200000. The population will increase by 10% in the first year and 15% in the second year. The population of the town after two years will be :**

**(a) Rs253000 (b) Rs235000 (c) Rs203500 (d) Rs352000**

**Solution- **(a) Rs253000

**Reason: **Population after n years = Present population x (1+𝑟/100)^𝑛

Present population = 2,00,000

After first year, population = = 2,00,000 x (1+10/100)^1

= 2,00,000 × 11/10

= 2,20,000

Population after two years = 2,20,000 ×(1+15/100)^1

= 253000

Thus, the population after two years is 2,53,000.

**Que-19: A machine depreciates at the rate of 12% of its value at the beginning of a year. The machine was purchased for Rs10000 and is sold for Rs7744. The number of years, that the machine was used is :**

**(a) 2 (b) 4 (c) 6 (d) 8**

**Solution- **(a) 2

**Reason: **V = 7744

P = 10000

r = 12%

n = ?

V = P(1-r/100)^n

7744 = 10000(1-12/100)^n

7744/10000 = (100-12/100)^n

7744/10000 = (88/100)^n

7744/10000 = 0.88^n

0.7744 = 0.88^n

0.2552^2 = 0.127^n {approx}

n = 2 years.

**Que-20: The value of a machine depreciates every year at a constant rate. If the values of the machine in 2006 and 2008 are Rs 25000 and 19360 respectively, then the annual rate of depreciation is :**

**(a) 8% (b) 10% (c) 12% (d) 14%**

**Solution- **(c) 12%

**Reason: **P = 25000 in 2006

V = 19360 in 2008

n = 2008 – 2006 = 2 years

V = P(1+r/100)^n

19360 = 25000(1+r/100)^2

19360/25000 = (1+r/100)^2

0.7744 = (1+r/100)^2

√0.7744 = 1+r/100

0.88 = 1+r/100

1 – 0.88 = r/100

0.12 = r/100

r = 1.2 x 100

r = 12%.

– : End of Compound Interest Class 9 RS Aggarwal MCQs Goyal Brothers ICSE Maths Solutions : —

Return to :- **RS Aggarwal Solutions for ICSE Class-9 Mathematics (Goyal Brother Prakashan)**

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