Compound Interest Class 9 RS Aggarwal Exe-2A Goyal Brothers ICSE Foundation Mathmatics Solutions. In this article you will learn how to solve compound interest problems easily without formula . Visit official Website **CISCE** for detail information about ICSE Board Class-9 Mathematics.

## Compound Interest Class 9 RS Aggarwal Exe-2A 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 Exe-2A |

Academic Session | 2024-2025 |

### How to Solve Compound Interest Problems without Formula

- Solving compound interest problems without formula is as same as simple interest but becomes borius if T (time) is in large numeric value. follow simle step given
- Calculate S I of first year only as simple interest calculates
- add this interest to Principal to get either amount of first year or Principal of second year
- Again calculate S I using 1st year amount as 2nd year pricipal
- Again add this second year interst to second year principal to get either amont of second year or principal for third year
- proceed the same step unless number of year complete to get final amount
- At last to calculate compund Interest subtract original principal from final amount

**Exercise- 2A**

Compound Interest Class 9 RS Aggarwal Exe-2A Goyal Brothers ICSE Foundation Mathmatics Solutions

**Page- 29,30**

**Que-1: Calculate the amount and the compound interest on Rs25000 for 2 years at 8% per annum, compound annually.**

**Solution- **The principal ( P ) is Rs. 25,000**
**The rate (R ) is 8 %

The time ( n ) is 2 years

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

A = 25000{1+8/100}²

A = 25000(27/25)²

A = 25000 x 27/25 x 27/25

A = 40 x 27 x 27

**A = 29160 Ans.**

C.I. = A – P

C.I. = 29160 – 25000

**C.I. = 4160 Ans.**

**Que-2: Rohit borrows Rs62500 from Arun for 2 years at 10% per annum, simple interest. He immediately lends out this sum to Kunal at 10% per annum for the same period, compounded annually. Calculate Rohit’s profit in the transaction at the end of two years.**

**Solution- **Rohit borrowed =62,500, T = 2yrs, R = 10%

S.D = (pxrxt)/100 = (62,500×2×10)/100

⇒ 12,500

Rohit lent = 62,500, T = 2yrs, R = 10%. (Compounded)

Amount = 62,500(1+(10/100))^2 [∵A = P(1+1/100)^n]

= 62,500(1.1)^2

= 75625

C.D = 75625−62500 = 13125

Profit = 13125−12500 **= 625 Rs. Ans.**

**Que-3: A man invests Rs10000 for 3 years at a certain rate of interest, compounded annually. At the end of one year, it amounts to Rs11200. Calculate : (i) the rate of interest per annum; (ii) the interest accrued in the second year: (iii) the amount at the end of third year.**

**Solution-**P = 10000, n = 3 years, A = 11200

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

(i) After 1 year, the amount is Rs 11,200. Using the compound interest formula:

11200 = 10000(1+(𝑟/100))

Solving for r:

11200/10000 = 1+(𝑟/100)

1.12 = 1+(r/100)

1.12−1 = r/100

0.12 = r/100

r = 0.12×100

r=12% Ans.

(ii) Amount after 2 years (n = 2):

𝐴2 = 𝑃(1+(𝑟/100))²

P = 10000 and r = 12:

𝐴2 = 10000(1+(12/100))²

𝐴2 = 10000(1.12)²

𝐴2 = 10000×1.2544

𝐴2 = 12544

The interest accrued in the second year is the amount after 2 years minus the amount after 1 year:

Interest accrued in the second year = A2−A1

= 12544−11200

**= 1344 Ans.**

(iii) Amount after 3 years (n = 3):

𝐴3 = 𝑃(1+(𝑟/100))³

Using P = 10000 and r = 12:

𝐴3 = 10000(1+(12/100))³

𝐴3 = 10000(1.12)³

𝐴3 = 10000×1.404928

**𝐴3 = 14049.28 Ans.**

**Que-4: Sudhakar borrows Rs22500 at 10% per annum, compounded annually. If he repays Rs11250 at the end of first year and Rs12550 at the end of second year, find the amount of loan outstanding against him at the end of the third year.**

**Solution- **P = Rs22500

r = 10%

n = 3 years

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

Amount at the end of the first year

= 𝑃(1+(𝑅/100))¹ = 22500(1+(10/100))¹ = 22500 𝑥 11/10 = ₹24750

Amount paid at the end of the first year ₹11250(given)

Balance after the first year = ₹24750 – ₹11250 = ₹13500

Now, this balance becomes principal for next year’s calculation.

Amount at the end of 2nd year = ₹13500(1+(10/100))¹ = ₹13500 𝑥 11/10 = ₹14850

Amount paid at the end of the second year = ₹12550(given)

Balance after the second year = ₹14850 – ₹12550 = ₹2300

Now, this balance becomes principal for next year’s calculation.

Amount at the end of the third year

= 2300(1+(10/100))¹ = 2300 𝑥 11/10

= ₹2530 Ans.

**Que-5: A man borrows Rs15000 at 12% per annum, compounded annually. If he repays Rs4400 at the end of each year, find the amount outstanding against him at the beginning of the third year.**

**Solution-**Principal = 15000

Rate = 12%

Repays amount = 4400

Calculate the amount due at the end of the first year before repayment:

𝐴1 = 𝑃(1+(𝑟/100)

𝐴1 = 15000(1+12/100)

𝐴1 = 15000×1.12

A1 = 16800

Subtract the repayment made at the end of the first year:

Outstanding after repayment = 16800−4400

Outstanding = 12400

Calculate the amount due at the end of the second year before repayment:

𝐴2 = 12400(1+12/100)

𝐴2 = 12400×1.12

A2 = 13888

Subtract the repayment made at the end of the second year:

Outstanding after repayment = 13888−4400

Outstanding = 9488

So, the amount outstanding against him at the beginning of the third year is 9488 Ans.

**Que-6: Mr. Ravi borrows Rs16000 for 2 years. The rate of interest for two successive years are 10% and 12% respectively. If he repays Rs5600 at the end of first year, find the amount outstanding at the end of second year.**

**Solution- **P = Rs.16,000, T = 2yrs, R = 10% and 12% p.a

For 1st yr:

16000 X 1 X(10 ÷ 100) = 1600

Amount = 16000 + 1600 = 17,600

He repays 5600 at the end of 1st yr

Therefore, 17,600 – 5600 = 12000

New Principal = 12000

For 2nd yr,

Interest = 12000 X 1 X (12 ÷ 100) = 1440

Amount = 12000 + 1440 = 13440

Therefore, the amount outstanding for the 2nd yr is = Rs. 13,440

**Que-7: Calculate the amount of Rs30000 at the end of 2 year 4 months, compounded annually at 10% per annum.**

**Solution- **Interest for the 1st year = Rs ( 30000 x 10 /100 x1 ) = 3000

Amount at the end of 1st year = 30000 + 3000 = 33000 Rs

Principal for the 2nd year = 33000 Rs

Interest for the second year =( 33000 x 10/100 x 1) = 3300

Amount at the end of 2nd year = 33000 + 3300 = 36300

Principal for the 3rd year = 36300

Interest for the 4 months = 36300 x 4/12 x 10 x 1/100 = 1210

Hence, the amount at the end of the 4 months = 36300 + 1210 **= 37510 Ans.**

**Que-8: ****Calculate the amount of Rs31250 at the end of 2*(1/2) years, compounded annually at 8% per annum.**

**Solution- **Principal (P) = Rs 31,250

Annual Interest Rate (r) = 8%

Number of years (n) = 2*(1/2) = 2.5

A2 = P(1+(r/100))²

𝐴2 = 31250(1+(8/100))²

A2 = 31250(1.08)²

A2 = 31250×1.1664

A2 = 36450

For the half-year period, we will use the effective interest rate for half a year, which is:

Half-year interest rate = 8/2 = 4%

So, the amount after the additional half-year is:

𝐴2.5 = 𝐴2(1+4/100)

A2.5 = 36450(1.04)

A2.5 = 36450×1.04

A2.5 = 37908 Ans.

**Que-9: Calculate the amount and the compound interest on Rs15000 for 2 years compounded annually, the rates of interest for the successive years being 8% and 9% per annum respectively.**

**Solution- **Principal (P) = Rs 15,000

Rate of interest for the first year (R1) = 8%

Time (T) = 1 year

𝐴1 = 𝑃(1+(𝑅1/100))

𝐴1 = 15000(1+(8/100))

A1 = 15000(1+0.08)

A1 = 15000×1.08

A1 = 16200

New Principal = Amount at the end of the first year = Rs 16,200

Rate of interest for the second year (R2) = 9%

Time (T) = 1 year

A2 = A1(1+(R2/100))

𝐴2 = 16200(1+(9/100))

A2 = 16200(1+0.09)

A2 =16200×1.09

A2 = 17658

**Thus, the amount at the end of 2 years is Rs 17,658 Ans.**

Compound Interest (CI) = Amount – Principal

Principal (P) = Rs 15,000

Amount (A) = Rs 17,658

𝐶𝐼 =𝐴−𝑃

= 17658−15000 **= 2658 Ans.**

**Que-10: ****Calculate the amount and the compound interest on Rs25000 for 3 years compounded annually, the rates of interest for the successive years being 8%, 9% and 10% per annum respectively.**

**Solution- **Principal (P) = Rs 25,000

Rate of interest for the first year (R1) = 8%

Time (T) = 1 year

𝐴1 = 𝑃(1+(𝑅1/100))

𝐴1 = 25000(1+(8/100))

A1 = 25000(1+0.08)

A1 = 25000×1.08

A1 = 27000

New Principal = Amount at the end of the first year = Rs 27,000

Rate of interest for the second year (R2) = 9%

Time (T) = 1 year

𝐴2 = 𝐴1(1+(𝑅2/100))

𝐴2 = 27000(1+(9/100))

A2 = 27000(1+0.09)

A2 = 27000×1.09

A2 = 29430

New Principal = Amount at the end of the second year = Rs 29,430

Rate of interest for the third year (R3) = 10%

Time (T) = 1 year

A3 = A2(1+(R3/100))

𝐴3 = 29430(1+(10/100))

A3 = 29430(1+0.10)

A3 = 29430×1.10

A3 = 32373**
Thus, the amount at the end of 3 years is Rs 32,373 Ans.
**Compound Interest (CI) = Amount – Principal

Principal (P) = Rs 25,000

Amount (A) = Rs 32,373

𝐶𝐼 = 𝐴−𝑃

= 32373−25000

**= 7373 Ans.**

**Que-11: Peter invested Rs240000 for 2 years at 10% per annum compounded annually. If 20% of the accrued interest at the end of each year is deducted as income tax, find the amount he received at the end of 2 years.**

**Solution- **Amount invested = Principal =𝑅𝑠.2,40,000

Time = 2𝑦𝑟𝑠

Rate =10% 𝑝.𝑎

Interest for 1st year

= 𝑆.𝐼 = (𝑃𝑅T)/100

𝑆.𝐼 = (2,40,000×10×1)/100

𝑆.𝐼 = 𝑅𝑠.24,000

Amount at the end of 1st year = 𝑃+𝑆.𝐼 − (20/100) 𝑜𝑓 𝑆.𝐼.

= 2,40,000 + 24000 − (20/100) × 24000

= 2,40,000 + (80/100) × 24000

= 2,40,000+19200

= 𝑅𝑠.2,59,200

Principal for 2nd year =𝑅𝑠.2,59,200

Interest for 2nd year = (2,59,200×10×1)/100

= 𝑅𝑠.25,920

Amount at the end of 2nd year = 𝑃+𝑆.𝐼−20% 𝑜𝑓 𝑆.𝐼.

= 2,59,200 + 2,5920 − (20/100) 𝑜𝑓 𝑆.𝐼

= 2,59,200 + (80/100) × 25920

= 2,59,200 + 20736

= 𝑅𝑠.2,79,936 Ans.

**Que-12: Find the amount and the compound interest on Rs10000 for 1 year at 12% per annum, compounded half-yearly.**

**Solution- **P = 10000

n = 1 year

r = 12%

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

A = 10000(1+(12/100))²

A = 10000(1+0.6)²

A = 10000(1.06)²

A = 10000 x 1.1236

**A = 11236 Ans.**

C.I. = A – P

C.I. = 11236 – 10000

**C.I. = 1236 Ans.**

**Que-13: ****Find the amount and the compound interest on Rs64000 for**

1*(1/2) year at 15% per annum, compounded half-yearly.

1*(1/2) year at 15% per annum, compounded half-yearly.

**Solution- **Principal (p) = 64000

Time (t) = 1*(1/2) = 3/2

Rate (r) = 15%

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

A = 64000(1+(15/2×100))³

A = 64000 (43/40)³

A = 64000 x (79507/64000)

**A = 79507 Ans.**

C.I. = A – P

C.I. = 79507 – 64000

**C.I. = 15507 Ans.**

**Que-14: The simple interest on the sum of money for 2 years at 10% p.a. is Rs1700. Find: (i) the sum of money; (ii) the compound interest on this sum for 1 year, payable half-yearly at the same rate.**

**Solution- **Here, S.I. = 1700, P = 2 years, R = 10% p.a.

Using simple interest formula, we have

(i) S.I. = (PxNxR)/100

P = (S.I. x 100)/(N x R)

= (1700 x 100)/(2 x 10)

= 8500

Therefore, sum (P) **= Rs. 8500 Ans.**

(ii) A = P(1+ r/2×100)2n

= 8500(1 + 10/200)2

= 8500(21/20)2

= Rs. 9371.25

Compound Interest (C.I.) = A – P = 9371.25 – 8500 **= Rs. 871.25 Ans.**

–: End of Compound Interest Class 9 RS Aggarwal Exe-2A Goyal Brothers ICSE Foundation Mathmatics : —

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

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