# ML Aggarwal Solutions for Chapter 2 Compound Interest Class 9 Maths ICSE

**Exercise 2.1**

**1. Find the amount and the compound interest on ₹ 8000 at 5% per annum for 2 years.**

**Solution**

It is given that

Principal = ₹ 8000

Rate of interest = 5% p.a.

We know that

Interest for the first year = Prt/100

Substituting the values

= (8000×5×1)/100

= ₹ 400

So the amount for the first year or principal for the second year = 8000 + 400 = ₹ 8400

Here,

Interest for the second year = (8400×5 ×1)/100

= ₹ 420

We know that,

Amount after the second year = 8400 + 420 = ₹ 8820

Total compound interest = 400 + 420 = ₹ 820

**2. A man invests ₹ 46875 at 4% per annum compound interest for 3 years. Calculate:**

**(i) the amount standing to his credit at the end of the second year.**

**(ii) the interest for the third year.**

**(iii) the interest for the first year.**

**Solution**

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

So the amount after first year or principal for the second year = 46875 + 1875 = ₹ 48750

Here,

Interest for the second year = (48750×4×1)/100

= ₹ 1950

**(ii) **We know that

Amount at the end of second year = 48750 + 1950

= ₹ 50700

**(iii) **Interest for the third year = (50700×4×1)/100 = ₹ 2028

**3. Calculate the compound interest for the second year on ₹ 8000 for three years at 10% p.a.**

**Also find the sum due at the end of third year.**

**Solution**

It is given that

Principal = ₹ 8000

Rate of interest = 10% p.a.

We know that

Interest for the first year = Prt/100

Substituting the values

= (8000×10×1)/100

= ₹ 800

So the amount after the first year or principal for the second year = 8000 + 800 = ₹ 8800

**(i)** Interest for the second year = (8800×10×1)/100

= ₹ 880

So the amount after second year or principal for the third year = 8800 + 880 = ₹ 9680

Interest for the third year = (9680×10×1)/100

= ₹ 968

**(ii)** Amount due at the end of the third year = 9680 + 968

= ₹ 10648

**4. Ramesh invests ₹ 12800 for three years at the rate of 10% per annum compound interest.**

**Find:**

**(i) the sum due to Ramesh at the end of the first year.**

**(ii) the interest he earns for the second year.**

**(iii) the total amount due to him at the end of three years.**

**Solution**

It is given that

Principal = ₹ 12800

Rate of interest = 10% p.a.

**(i)** We know that

Interest for the first year = (12800×10×1)/100

= ₹ 1280

So the sum due at the end of first year = 12800 + 1280

= ₹ 14080

**(ii) **Principal for second year = ₹ 14080

So the interest for the second year = (14080×10×1)/100

= ₹ 1408

**(iii)** We know that

Sum due at the end of second year = 14080 + 1408

= ₹ 15488

Here

Principal for third year = ₹ 15488

Interest for the third year = (15488×10×1)/100

= ₹ 1548.80

So the total amount due to him at the end of third year = 15488 + 1548.80

= ₹ 17036.80

**5. The simple interest on a sum of money for 2 years at 12% per annum is ₹ 1380. Find:**

**(i) the sum of money.**

**(ii) the compound interest on this sum for one year payable half-yearly at the same rate.**

**Solution**

It is given that

Simple Interest (SI) = ₹ 1380

Rate of interest (R) = 12% p.a.

Period (T) = 2 years

**(i)** We know that

Sum (P) = (SI×100)/(R×T)

Substituting the values

= (1380×100)/ (12×2)

= ₹ 5750

**(ii)** Here

Principal (P) = ₹ 5750

Rate of interest (R) = 12% p.a. or 6% half-yearly

Period (n) = 1 year – 2 half years

So we get

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

Substituting the values

= 5750 (1 + 6/100)^{2}

By further calculation

= 5750 × (53/50)^{2}

So we get

= 5750 × 53/50 × 53/50

= ₹ 6460.70

Here,

Compound Interest = A – P

Substituting the values

= 6460.70 – 5750

= ₹ 710.70

**6. A person invests ₹ 10000 for two years at a certain rate of interest, compounded annually. At the end of one year this sum amounts to ₹ 11200. Calculate:**

**(i) the rate of interest per annum.**

**(ii) the amount at the end of second year.**

**Solution**

It is given that

Principal (P) = ₹ 10,000

Period (T) = 1 year

Sum amount (A) = ₹ 11,200

Rate of interest = ?

(i) We know that

Interest (I) = 11200 – 10000 = ₹ 1200

So the rate of interest

R = (I×100)/(P×T)

Substituting the values

R = (1200×100)/(10000×1)

So we get

R = 12% p.a.

Therefore, the rate of interest per annum is 12% p.a.

**(ii)** We know that

Period (T) = 2 years

Rate of interest (R) = 12% p.a.

Here

A = P (1 + R/100)^{t}

Substituting the values

A = 10000 (1 + 12/100)^{2}

By further calculation

A = 10000 (28/25)^{2}

We can write it as

A = 10000 × 28/25 × 28/25

So we get

A = 16 × 28 × 28

⇒ A = ₹ 12544

Therefore, the amount at the end of second year is ₹ 12544.

**7. Mr. Lalit invested ₹ 75000 at a certain rate of interest, compounded annually for two years. At the end of first year it amounts to ₹ 5325. Calculate**

**(i) the rate of interest.**

**(ii) the amount at the end of second year, to the nearest rupee.**

**Solution**

It is given that

Investment of Mr. Lalit = ₹ 5000

Period (n) = 2 years

**(i)** We know that

Amount after one year = ₹ 5325

So the interest for the first year = A – P

Substituting the values

= 5325 – 5000

= ₹ 325

Here

Rate = (SI × 100)/(P×T)

Substituting the values

= (325 × 100)/(5000×1)

So we get

= 13/2

= 6.5 % p.a.

**(ii)** We know that

Interest for the second year = (5325×13×1)/(100×2)

By further calculation

= (213×13)/(4×2)

So we get

= 2769/8

= ₹ 346.12

So the amount after second year = 5325 + 346.12

We get

= ₹ 5671.12

= ₹ 5671 (to the nearest rupee)

**8. A man invests ₹ 5000 for three years at a certain rate of interest, compounded annually. At the end of one year it amounts to ₹ 5600. Calculate:**

**(i) the rate of interest per annum**

**(ii) the interest accrued in the second year.**

**(iii) the amount at the end of the third year.**

**Solution**

It is given that

Principal = ₹ 5000

Consider r% p.a. as the rate of interest

**(i)** We know that

At the end of one year

Interest = Prt/100

Substituting the values

= (5000 × r × 1)/100

= 50r

Here,

Amount = 5000 + 50r

We can write it as

5000 + 50r = 5600

By further calculation

50r = 5600 – 5000 = 600

So we get

r = 600/50 = 12

Hence, the rate of interest is 12% p.a.

**(ii)** We know that

Interest for the second year = (5600×12×1)/100

= ₹ 672

So the amount at the end of second year = 5600 + 672

= ₹ 6272

**(iii)** We know that

Interest for the third year = (6272×12×1)/100

= ₹ 752.64

So the amount after third year = 6272 + 752.64

= ₹ 7024.64

**9. Find the amount and the compound interest on ₹ 2000 at 10% p.a. for 2 years, compounded annually.**

**Solution**

It is given that

Principal (P) = ₹ 2000

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

Period (n) = 2 ½ years

We know that

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

Substituting the values

= 2000 (1 + 10/100)^{2} ×{1 + 10/(2×100)}

By further calculation

= 2000 × 11/10 × 11/10 × 21/20

So we get

= ₹ 2541

Here,

Interest = A – P

Substituting the values

= 2541 – 2000

= ₹ 541

**10. Find the amount and the compound interest on ₹ 50000 for 1 ½ years at 8% per annum, the interest being compounded semi-annually.**

**Solution**

It is given that

Principal (P) = ₹ 50000

Rate of interest (r) = 8% p.a. = 4% semi-annually

Period (n) = 1 ½ years = 3 semi-annually

We know that

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

Substituting the values

= 50000 (1 + 4/100)^{3}

By further calculation

= 50000 (26/25)^{3}

= 50000 × 26/25 × 26/25 × 26/25

= ₹ 56243.20

Here,

Compound Interest = A – P

Substituting the values

= 56243.20 – 50000

= ₹ 6243.20

**11. Calculate the amount and the compound interest on ₹ 5000 in 2 years when the rate of interest for successive years is 6% and 8%, respectively.**

**Solution**

It is given that

Principal = ₹ 5000

Period = 2 years

Rate of interest for the first year = 6%

Rate of interest for the second year = 8%

We know that

Amount for two years = P (1 + r/100)^{n}

Substituting the values

= 5000 (1 + 6/100) (1 + 8/100)

By further calculation

= 5000 × 53/50 × 27/25

= ₹ 5724

Here

Interest = A – P

Substituting the values

= 5724 – 5000

= ₹ 724

**12. Calculate the amount and the compound interest on ₹ 17000 in 3 years when the rate of interest for successive years is 10%, 10% and 14%, respectively.**

**Solution**

It is given that

Principal = ₹ 17000

Period = 3 years

Rate of interest for 3 successive years = 10%, 10% and 14%

We know that

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

Substituting the values

= 17000 (1 + 10/100) (1 + 10/100) (1 + 14/100)

By further calculation

= 17000 × 11/10 × 11/10 × 57/50

= ₹ 23449.80

Here

Amount of compound interest = A – P

Substituting the values

= 23449.80 – 17000

= ₹ 6449.80

**13. A sum of ₹ 9600 is invested for 3 years at 10% per annum at compound interest.**

**(i) What is the sum due at the end of the first year?**

**(ii) What is the sum due at the end of the second year?**

**(iii) Find the compound interest earned in 2 years.**

**(iv) Find the difference between the answers in (ii) and (i) and find the interest on this sum for one year.**

**(v) Hence, write down the compound interest for the third year.**

**Solution**

It is given that

Principal = ₹ 9600

Rate of interest = 10% p.a.

Period = 3 years

We know that

Interest for the first year = Prt/100

Substituting the values

= (9600×10×1)/100

= ₹ 960

**(i) **Amount after one year = 9600 – 960 = ₹ 10560

So the principal for the second year = ₹ 10560

Here the interest for the second year = (10560×10×1)/100

= ₹ 1056

**(ii)** Amount after two years = 10560 + 1056 = ₹ 11616

**(iii)** Compound interest earned in 2 years = 960 + 10560 = ₹ 2016

**(iv)** Difference between the answers in (ii) and (i) = 11616 – 10560 = ₹ 1056

We know that

Interest on ₹ 1056 for 1 year at the rate of 10% p.a. = (1056×10×1)/100

= ₹ 105.60

**(v)** Here,

Principal for the third year = ₹ 11616

So the interest for the third year = (11616×10×1)/100

= ₹ 1161.60

**14. The simple interest on a certain sum of money for 2 years at 10% p.a. is ₹ 1600. Find the amount due and the compound interest on this sum of money at the same rate after 3 years, interest being reckoned annually.**

**Solution**

It is given that

Period = 2 years

Rate = 10% p.a.

We know that

Sum = (SI×100)/(r×n)

Substituting the values

= (1600×100)/(10×2)

= ₹ 8000

Here

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

Substituting the values

= 8000 (1 + 10/100)^{3}

By further calculation

= 8000 × 11/10 × 11/10 × 11/10

= ₹ 10648

So the compound interest = A – P

Substituting the values

= 10648 – 8000

= ₹ 2648

**15. Vikram borrowed ₹ 20000 from a bank at 10% per annum simple interest. He lent it to his friend Venkat at the same rate but compounded annually. Find his gain after 2 ½ years.**

**Solution**

**First case-**

Principal = ₹ 20000

Rate = 10% p.a.

Period = 2 ½ = 5/2 years

We know that

Simple interest = Prt/100

Substituting the values

= (20000×10×5)/(100×2)

= ₹ 5000

**Second case-**

Principal = ₹ 20000

Rate = 10% p.a.

Period = 2 ½ years at compound interest

We know that

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

Substituting the values

= 20000 (1 + 10/100)^{2} (1 + 10/(2×100))^{2}

By further calculation

= 20000 × 11/10 × 11/10 × 21/20

= ₹ 25410

Here

Compound Interest = A – P

Substituting the values

= 25410 – 20000

= ₹ 5410

So his gain after 2 years = CI – SI

We get

= 5410 – 5000

= ₹ 410

**16. A man borrows ₹ 6000 at 5% compound interest. If he repays ₹ 1200 at the end of each year, find the amount outstanding at the beginning of the third year.**

**Solution**

It is given that

Principal = ₹ 6000

Rate of interest = 5% p.a.

We know that

Interest for the first year = Prt/100

Substituting the values

= (6000×5×1)/100

= ₹ 300

So the amount after one year = 6000 + 300 = ₹ 6300

Principal for the second year = ₹ 6300

Amount paid = ₹ 1200

So the balance = 6300 – 1200 = ₹ 5100

Here

Interest for the second year = (5100×5×1)/100 = ₹ 255

Amount for the second year = 5100 + 255 = ₹ 5355

Amount paid = ₹ 1200

So the balance = 5355 – 1200 = ₹ 4155

**17. Mr. Dubey borrows ₹ 100000 from State Bank of India at 11% per annum compound interest. He repays ₹ 41000 at the end of first year and ₹ 47700 at the end of second year. Find the amount outstanding at the beginning of the third year.**

**Solution**

It is given that

Borrowed money (P) = ₹ 100000

Rate = 11% p.a.

Time = 1 year

We know that

Amount after first year = Prt/100

Substituting the values

= (100000×11×1)/100

By further calculation

= 100000 + 11000

= ₹ 111000

Amount paid at the end of first year = ₹ 41000

So the principal for second year = 111000 – 41000

= ₹ 70000

We know that

Amount after second year = P + (70000×11)/100

By further calculation

= 70000 + 700

= ₹ 77700

So the amount paid at the end of second year = ₹ 47700

Here the amount outstanding at the beginning year = 77700 – 47700

= ₹ 30000

**18. Jaya borrowed ₹ 50000 for 2 years. The rates of interest for two successive years are 12% and 15% respectively. She repays ₹ 33000 at the end of first year. Find the amount she must pay at the end of second year to clear her debt.**

**Solution**

It is given that

Amount borrowed by Jaya = ₹ 50000

Period (n) = 2 years

Rate of interest for two successive years are 12% and 15% respectively

We know that

Interest for the first year = Prt/100

Substituting the values

= (50000×12×1)/100

= ₹ 6000

So the amount after first year = 50000 + 6000 = ₹ 56000

Amount repaid = ₹ 33000

Here

Balance amount for the second year = 56000 – 33000 = ₹ 23000

Rate = 15%

So the interest for the second year = (230000×15×1)/100

= ₹ 3450

Amount paid after second year = 23000 + 3450 = ₹ 26450

**Exercise 2.2**

**1. Find the amount and the compound interest on ₹ 5000 for 2 years at 6% per annum, interest payable yearly.**

**Solution**

It is given that

Principal (P) = ₹ 5000

Rate of interest (r) = 6% p.a.

Period (n) = 2 years

We know that

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

Substituting the values

= 5000 (1 + 6/100)^{2}

By further calculation

= 5000 × 53/50 × 53/50

= ₹ 5618

Here

CI = A – P

Substituting the values

= 5618 – 5000

= ₹ 618

**2. Find the amount and the compound interest on ₹ 8000 for 4 years at 10% per annum interest reckoned yearly.**

**Solution**

It is given that

Principal (P) = ₹ 8000

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

Period (n) = 4 years

We know that

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

Substituting the values

= 8000 (1 + 10/100)^{4}

By further calculation

= 8000 × 11/10 × 11/10 × 11/10 × 11/10

= ₹ 11712.80

Here

CI = A – P

Substituting the values

= 11712.80 – 8000

= ₹ 3712.80

**3. If the interest is compounded half yearly, calculate the amount when the principal is ₹ 7400, the rate of interest is 5% and the duration is one year.**

**Solution**

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

**4. Find the amount and the compound interest on ₹ 5000 at 10% p.a. for 1 ½ years, compound interest reckoned semi-annually.**

**Solution**

It is given that

Principal (P) = ₹ 5000

Rate of interest = 10% p.a. or 5% half-yearly

Period (n) = 1 ½ years or 3 half-years

We know that

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

Substituting the values

= 5000 (1 + 5/100)^{3}

By further calculation

= 5000 × 21/20 × 21/20 × 21/20

= ₹ 5788.12

Here

CI = A – P

Substituting the values

= 5788.12 – 5000

= ₹ 788.12

**5. Find the amount and the compound interest on ₹ 100000 compounded quarterly for 9 months at the rate of 4% p.a.**

**Solution**

It is given that

Principal (P) = ₹ 100000

Rate of interest = 4% p.a. or 1% quarterly

Period (n) = 9 months or 3 quarters

We know that

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

Substituting the values

= 100000 (1 + 1/100)^{3}

By further calculation

= 100000 × 101/100 × 101/100 × 101/100

= ₹ 103030.10

Here

CI = A – P

Substituting the values

= 103030.10 – 100000

= ₹ 3030.10

**6. Find the difference between CI and SI on sum of ₹ 4800 for 2 years at 5% per annum payable yearly.**

**Solution**

It is given that

Principal (P) = ₹ 4800

Rate of interest (r) = 5% p.a.

Period (n) = 2 years

We know that

SI = Prt/100

Substituting the values

= (4800×5×2)/100

= ₹ 480

If compounded yearly

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

Substituting the values

= 4800 (1 + 5/100)^{2}

By further calculation

= 4800 × 21/20 × 21/20

= ₹ 5292

Here

CI = A – P

Substituting the values

= 5292 – 4800

= ₹ 492

So the difference between CI and SI = 492 – 480 = ₹ 12

**7. Find the difference between the simple interest and compound interest on ₹ 2500 for** **2 years at 4% per annum, compound interest being reckoned semi-annually.**

**Solution**

It is given that

Principal (P) = ₹ 2500

Rate of interest (r) = 4% p.a. or 2% half-yearly

Period (n) = 2 years or 4 half-years

We know that

SI = Prt/100

Substituting the values

= (2500×4×2)/100

= ₹ 200

If compounded semi-annually

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

Substituting the values

= 2500 (1 + 2/100)^{4}

By further calculation

= 2500 × 51/50 × 51/50 × 51/50 × 51/50

= ₹ 2706.08

We know that

CI = A – P

Substituting the values

= 2706.08 – 2500

= ₹ 206.08

So the difference between CI and SI = 206.08 – 200 = ₹ 6.08

**8. Find the amount and the compound interest on ₹ 2000 in 2 years if the rate is 4% for the first year and 3% for the second year.**

**Solution**

It is given that

Principal (P) = ₹ 2000

Rate of interest = 4% on the first year and 3% for the second year

Period (n) = 2 years

We know that

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

Substituting the values

= 2000 (1 + 4/100) (1 + 3/100)

By further calculation

= 2000 × 26/25 × 103/100

= ₹ 2142.40

Here

CI = A – P

Substituting the values

= 2142.40 – 2000

= ₹ 142.40

**9. Find the compound interest on ₹ 3125 for 3 years if the rates of interest for the first, second and third year are respectively 4%, 5% and 6% per annum.**

**Solution**

It is given that

Principal (P) = ₹ 3125

Rate of interest for continuous = 4%, 5% and 6%

Period (n) = 3 years

We know that

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

Substituting the values

= 3125 (1 + 4/100) (1 + 5/100) (1 + 6/100)

By further calculation

= 3125 × 26/25 × 21/50 × 53/50

= ₹ 3617.25

Here

CI = A – P

Substituting the values

= 3617.25 – 3125

= ₹ 492.25

**10. What sum of money will amount to ₹ 9261 in 3 years at 5% per annum compound interest?**

**Solution**

It is given that

Amount (A) = ₹ 9261

Rate of interest (r) = 5% per annum

Period (n) = 3 years

We know that

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

Substituting the values

9261 = P (1 + 5/100)^{3}

By further calculation

9261 = P (21/20)^{3}

So we get

P = (9261×20×20×20)/(21×21×21)

P = ₹ 8000

Therefore, the sum of money is ₹ 8000.

**11. What sum invested at 4% per annum compounded semi-annually amounts to ₹ 7803 at the end of one year?**

**Solution**

It is given that

Amount (A) = ₹ 7803

Rate of interest (r) = 4% p.a. or 2% semi-annually

Period (n) = 1 year or 2 half years

We know that

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

Substituting the values

= 7803 + (1 + 2/100)^{2}

By further calculation

= 7803 + (51/20)^{2}

= 7803 × 50/51 × 50/51

= ₹ 7500

Hence, the principal is ₹ 7500.

**12. What sum invested for 1 ½ years compounded half yearly at the rate of 4% p.a. will amount to ₹132651?**

**Solution**

It is given that

Amount (A) = ₹ 132651

Rate of interest (r) = 4% p.a. or 2% half yearly

Period (n) = 1 ½ years or 3 half years

We know that

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

It can be written as

P = A ÷ (1 + r/100)^{n}

Substituting the values

= 132651 ÷ (1 + 2/100)^{3}

By further calculation

= 132651 ÷ (51/50)^{3}

So we get

= 132651 × (50/51)^{3}

= 132651 × 50/51 × 50/51 × 50/51

= ₹ 125000

Hence, the principal amount is ₹ 125000.

**13. On what sum will the compound interest for 2 years at 4% per annum be ₹ 5712?**

**Solution**

It is given that

CI = ₹ 5712

Rate of interest (r) = 4% p.a.

Period (n) = 2 years

We know that

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

It can be written as

CI = A – P = P (1 + r/100)^{n} – P

= P [(1 + r/100)^{n} – 1]

Substituting the values

5712 = P [(1 + 4/100)^{2} – 1]

= P [(26/25)^{2} – 1]

By further calculation

= P [676/625 – 1]

Taking LCM

= P [(676 – 625)/625]

= P × 51/625

Here

P = 5712 × 625/51

= 112 × 625

= ₹ 70000

Hence, the principal amount is ₹ 70000.

**14. A man invests ₹ 1200 for two years at compound interest. After one year the money amounts to ₹ 1275. Find the interest for the second year correct to the nearest rupee.**

**Solution**

It is given that

Principal = ₹ 1200

After one year, the amount = ₹ 1275

So the interest for one year = 1275 – 1200 = ₹ 75

We know that

Rate of interest = (SI×100)/(P×t)

Substituting the values

= (75×100)/(1200×1)

By further calculation

= 75/12

= 25/4

= 6 ¼ % p.a.

Here

Interest for the second year on ₹ 1275 at the rate of 25/4% = Prt/100

Substituting the values

= (1275×25×1)/(100×4)

By further calculation

= 1275/16

= ₹ 79.70

= ₹ 80

**15. At what rate percent per annum compound interest will ₹ 2304 amount to ₹ 2500 in 2 years?**

**Solution**

It is given that

Amount = ₹ 2500

Principal = ₹ 2304

Period (n) = 2 years

Consider r% p.a. as the rate of interest

We know that

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

It can be written as

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

Substituting the values

(1 + r/100)^{2} = 2500/2304

By further calculation

(1 + r/100)^{2} = 625/576 = (25/24)^{2}

So we get

1 + r/100 = 25/24

⇒ r/100 = 25/24 – 1

Taking LCM

r = 100/24 = 25/6 = 4 1/6

Hence, the rate of interest is 4 1/6% p.a.

**16. A sum compounded annually becomes 25/16 time of itself in two years. Determine the rate of interest per annum.**

**Solution**

Consider sum (P) = x

Amount (A) = 25/16x

Period (n) = 2 years

We know that

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

Substituting the values

25x/16x = (1 + r/100)^{2}

By further calculation

(1 + r/100)^{2} = (5/4)^{2}

So we get

1 + r/100 = 5/4

⇒ r/100 = 5/4 – 1/1 = 1/4

By cross multiplication

r = 100 × ¼ = 25

Hence, the rate of interest is 25% p.a.

**17. At what rate percent will ₹ 2000 amount to ₹ 2315.25 in 3 years at compound interest?**

**Solution**

It is given that

Principal (P) = ₹ 2000

Amount (A) = ₹ 2315.25

Period (n) = 3 years

Consider r% p.a. as the rate of interest

We know that

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

Substituting the values

2315.25/2000 = (1 + r/100)^{3}

By further calculation

(1 + r/100)^{3} = 231525/(100×2000) = 9261/8000 = (21/20)^{3}

So we get

1 + r/100 = 21/20

It can be written as

r/100 = 21/20 – 1 = 1/20

⇒ r = 100/20 = 5

Hence, the rate of interest is 5% p.a.

**18. If ₹ 40000 amounts to ₹ 48620.25 in 2 years, compound interest payable half-yearly, find the rate of interest per annum.**

**Solution**

It is given that

Principal (P) = ₹ 40000

Amount (A) = ₹ 48620.25

Period (n) = 2 years = 4 half years

Consider rate of interest = r% p.a. = r/2 % half yearly

We know that

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

Substituting the values

48620.25/40000 = (1 + r/200)^{4}

By further calculation

(1 + r/200)^{4} = 4862025/(100×40000) = 194481/160000

So we get

(1 + r/200)^{4} = (21/20)^{4}

It can be written as

1 + r/200 = 21/20

⇒ r/200 = 21/20 – 1 = 1/20

By cross multiplication

r = 200 × 1/20 = 10

Hence the rate of interest per annum is 10%.

**19. Determine the rate of interest for a sum that becomes 216/125 times of itself in 1 ½ years, compounded semi-annually.**

**Solution**

Consider principal (P) = x

Amount (A) = 216/125 x

Period (n) = 1 ½ years = 3 half years

Take rate percent per year = 2r% and r% half yearly

We know that

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

Substituting the values

216x/125x = (1 + r/100)^{3}

By further calculation

(1 + r/100)^{3} = 216/125 = (6/5)^{3}

So we get

1 + r/100 = 6/5

⇒ r/100 = 6/5 – 1 = 1/5

By cross multiplication

r = 100 × 1/5 = 20%

So the rate percent per year = 2×20 = 40%

**20. At what rate percent p.a. compound interest would ₹ 80000 amounts to ₹ 88200 in two years, interest being compounded yearly. Also find the amount after 3 years at the above rate of compound interest.**

**Solution**

It is given that

Principal (P) = ₹ 80000

Amount (A) = ₹ 88200

Period (n) = 2 years

Consider r% per annum as the rate of interest percent

We know that

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

Substituting the values

88200/80000 = (1 + r/100)^{2}

By further calculation

(1 + r/100)^{2} = 441/400 = (21/20)^{2}

So we get

1 + r/100 = 21/20

⇒ r/100 = 21/20 – 1 = 1/20

By cross multiplication

r = 1/20 × 100 = 5

Hence, the rate of interest is 5% per annum.

**21. A certain sum amounts to ₹ 5292 in 2 years and to ₹ 5556.60 in 3 years at compound interest. Find the rate and the sum.**

**Solution**

It is given that

Amount after 2 years = ₹ 5292

Amount after 3 years = ₹ 5556.60

So the difference = 5556.60 – 5292 = ₹ 264.60

Here ₹ 264.60 is the interest on ₹ 5292 for one year

We know that

Rate % = (SI×100)/(P×t)

Substituting the values

= (264.60×100)/(5292×1)

Multiply and divide by 100

= (26460×100)/(100×5292)

= 5%

Here

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

Substituting the values

5292 = P (1 + 5/100)^{2}

By further calculation

P = 5292 ÷ (1 + 5/100)^{2}

So we get

P = 5292 ÷ (21/20)^{2}

P = 5292 × 21/20 × 21/20

P = ₹ 4800

Hence, the rate is 5% and the sum is ₹ 4800.

**22. A certain sum amounts to ₹ 798.60 after 3 years and ₹ 878.46 after 4 years. Find the interest rate and the sum.**

**Solution**

It is given that

Amount after 3 years = ₹ 798.60

Amount after 4 years = ₹ 878.46

So the difference = 878.46 – 798.60 = ₹ 79.86

Here ₹ 79.86 is the interest on ₹ 798.60 for 1 year.

We know that

Rate = (SI×100)/(P×t)

Substituting the values

= (79.86×100)/(798.60×1)

Multiply and divide by 100

= (7986×100×100)/(79860×100×1)

= 10%

Here

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

It can be written as

P = A ÷ (1 + r/100)^{n}

Substituting the values

P = 798.60 ÷ (1 + 10/100)^{3}

By further calculation

P = 79860/100 × 10/11 × 10/11 × 10/11

P = ₹ 600

**23. In what time will ₹ 15625 amount to ₹ 17576 at 4% per annum compound interest?**

**Solution**

It is given that

Amount (A) = ₹ 17576

Principal (P) = ₹ 15625

Rate = 4% p.a.

Consider n years as the period

We know that

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

Substituting the values

17576/15625 = (1 + 4/100)^{n}

By further calculation

(26/25)^{3} = (26/25)^{n}

So we get

n = 3

**24. (i) In what time will ₹ 1500 yield ₹ 496.50 as compound interest at 10% per annum compounded annually?**

**(ii) Find the time (in years) in which ₹ 12500 will produce ₹ 3246.40 as compound interest at 8% per annum, interest compounded annually.**

**Solution**

**(i)** It is given that

Principal (P) = ₹ 1500

CI = ₹ 496.50

So the amount (A) = P + SI

Substituting the values

= 1500 + 496.50

= ₹ 1996.50

Rate (r) = 10% p.a.

We know that

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

It can be written as

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

Substituting the values

1996.50/1500 = (1 + 10/100)^{n}

By further calculation

199650/(1500 × 100) = (11/10)^{n}

So we get

1331/1000 = (11/10)^{n}

⇒ (11/10)^{3} = (11/10)^{n}

Here Time n = 3 years

**(ii)** It is given that

Principal (P) = ₹ 12500

CI = ₹ 3246.40

So the amount (A) = P + CI

Substituting the values

= 12500 + 3246.40

= ₹ 15746.40

Rate (r) = 8% p.a.

We know that

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

It can be written as

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

Substituting the values

15746.40/12500 = (1 + 8/100)^{n}

Multiply and divide by 100

1574640/(12500×100) = (27/25)^{n}

By further calculation

78732/(12500×5) = (27/25)^{n}

⇒ 19683/(3125×5) = (27/25)^{n}

So we get

19683/15625 = (27/25)^{n}

⇒ (27/25)^{3} = (27/25)^{n}

Here, Period = 3 years

**25. ₹ 16000 invested at 10% p.a., compounded semi-annually, amounts to ₹ 18522, find the time period of investment.**

**Solution**

It is given that

Principal (P) = ₹ 16000

Amount (A) = ₹ 18522

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

Consider period = n half years

We know that

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

Substituting the values

18522/16000 = (1 + 5/100)^{n}

By further calculation

9261/8000 = (21/20)^{n}

So we get

(21/20)^{3} = (21/20)^{n}

n = 3 half years

Here

Time = 3/2 = 1 ½ years

**26. What sum will amount to ₹ 2782.50 in 2 years at compound interest, if the rates are 5% and 6% for the successive years?**

**Solution**

It is given that

Amount (A) = ₹ 2782.50

Rate of interest for two successive years = 5% and 6%

We know that

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

Substituting the values

2782.50 = P (1 + 5/100) (1 + 6/100)

By further calculation

2782.50 = P × 21/20 × 53/50

So we get

P = 2782.50 × 20/21 × 50/53

Multiply and divide by 100

P = 278250/100 × 20/21 × 50/53

P = ₹ 2500

Hence, the principal is ₹ 2500.

**27. A sum of money is invested at compound interest payable annually. The interest in two successive years is ₹ 225 and ₹ 240. Find:**

**(i) the rate of interest**

**(ii) the original sum**

**(iii) the interest earned in the third year.**

**Solution**

It is given that

Interest for the first year = ₹ 225

Interest for the second year = ₹ 240

So the difference = 240 – 225 = ₹ 15

Here ₹ 15 is the interest on ₹ 225 for 1 year

**(i)** Rate = (SI × 100)/ (P × t)

Substituting the values

= (15×100)/(225×1)

So we get

= 20/3

= 6 2/3% p.a.

**(ii)** We know that

Sum = (SI×100)/(R×t)

Substituting the values

= (225×100)/(20/3 × 1)

It can be written as

= (225×100×3)/(20×1)

So we get

= 225 × 15

= ₹ 3375

**(iii)** Here,

Amount after second year = 225 + 240 + 3375 = ₹ 3840

So the interest for the third year = Prt/100

Substituting the values

= (3840×20×1)/(100×3)

= ₹ 256

**28. On what sum of money will the difference between the compound interest and simple interest for 2 years be equal to ₹ 25 if the rate of interest charged for both is 5% p.a.?**

**Solution**

It is given that

Sum (P) = ₹ 100

Rate (R) = 5% p.a.

Period (n) = 2 years

We know that

SI = PRT/100

Substituting the values

= (100×5×2)/100

= ₹ 10

So the amount when interest is compounded annually = P (1 + R/100)^{n}

Substituting the values

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

By further calculation

= 100 × (21/20)^{2}

= 100 × 21/20 × 21/20

So we get

= ₹ 441/4

Here,

CI = A – P

Substituting the values

= 441/4 – 100

= ₹ 41/4

So the difference between CI and SI = 41/4 – 10 = ₹ ¼

If the difference is ₹ ¼ then sum = ₹ 100

If the difference is ₹ 25 then sum = (100×4)/1 × 25 = ₹ 10000

**29. The difference between the compound interest for a year payable half-yearly and the simple interest on a certain sum of money lent out at 10% for a year is ₹ 15. Find the sum of money lent out.**

**Solution**

It is given that

Sum = ₹ 100

Rate = 10% p.a. or 5% half yearly

Period = 1 years or 2 half years

We know that

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

Substituting the values

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

By further calculation

= 100 × 21/20 × 21/20

= ₹ 441/4

Here

CI = A – P

Substituting the values

= 441/4 – 100

= ₹ 41/4

SI = PRT/100

Substituting the values

= (100×10×1)/100

= ₹ 10

So the difference between CI and SI = 41/4 – 10 = ₹ ¼

Here if the difference is ₹ ¼ then sum = ₹ 100

If the difference is ₹ 15 then sum = (100×4×15)/1 = ₹ 6000

**30. The amount at compound interest which is calculated yearly on a certain sum of money is ₹ 1250 in one year and ₹ 1375 after two years. Calculate the rate of interest.**

**Solution**

It is given that

Amount after one year = ₹ 1250

Amount after two years = ₹ 1375

Here the difference = 1375 – 1250 = ₹ 125

So ₹ 125 is the interest on ₹ 1250 for 1 year

We know that

Rate of interest = (SI×100)/(P×t)

Substituting the values

= (125×100)/(1250×1)

= 10%

**31. The simple interest on a certain sum for 3 years is ₹ 225 and the compound interest on the same sum at the same rate for 2 years is ₹ 153. Find the rate of interest and the principal.**

**Solution**

It is given that

SI for 3 years = ₹ 225

SI for 2 years = (225×2)/3 = ₹ 150

CI for 2 years = ₹ 153

So the difference = 153 – 150 = ₹ 3

Here ₹ 3 is interest on one year i.e. ₹ 75 for one year

We know that

Rate = (SI×100)/(P×t)

Substituting the values

= (3×100)/(75×1)

= 4%

SI for 3 years = ₹ 225

Rate = 4% p.a.

So principal = (SI×100)/(R×t)

Substituting the values

= (225×100)/(4×3)

= ₹ 1875

**32. Find the difference between compound interest on ₹ 8000 for 1 ½ years at 10% p.a. when compounded annually and semi-annually.**

**Solution**

It is given that

Principal (P) = ₹ 8000

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

Period = 1 ½ years or 3 half years

Case 1 – When compounded annually

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

Substituting the values

= 8000 (1 + 10/100) (1 + 5/100)

By further calculation

= 8000 × 11/10 × 21/20

= ₹ 9240

We know that

CI = A – P

Substituting the values

= 9240 – 8000

= ₹ 1240

Case 2 – When compounded half-yearly

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

Substituting the values

= 8000 (1 + 5/100)^{3}

By further calculation

= 8000 × 21/20 × 21/20 × 21/20

= ₹ 9261

We know that

CI = A – P

Substituting the values

= 9261 – 8000

= ₹ 1261

Here the difference between two CI = 1261 – 1240 = ₹ 21

**33. A sum of money is lent out at compound interest for two years at 20% p.a., CI being reckoned yearly. If the same sum of money is lent out at compound interest at same rate percent per annum, CI being reckoned half-yearly, it would have fetched ₹ 482 more by way of interest. Calculate the sum of money lent out.**

**Solution**

It is given hat

Sum = ₹ 100

Rate = 20% p.a. or 10% half-yearly

Period = 2 years or 4 half-years

Case 1 – When the interest is reckoned yearly

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

Substituting the values

= 100 (1 + 20/100)^{2}

By further calculation

= 100 × 6/5 × 6/5

= ₹ 144

We know that

CI = A – P

Substituting the values

= 144 – 100

= ₹ 44

Case 2 – When the interest is reckoned half-yearly

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

Substituting the values

= 100 (1 + 10/100)^{4}

By further calculation

= 100 × 11/10 × 11/10 × 11/10 × 11/10

= ₹ 146.41

We know that

CI = A – P

Substituting the values

= 146.41 – 100

= ₹ 46.41

So the difference between two CI = 46.41 – 44 = ₹ 2.41

If the difference is ₹ 2.41 then sum = ₹ 100

If the difference is ₹ 482 then sum = (100×482)/2.41

Multiplying and dividing by 100

= (100×482×100)/241

= ₹ 20000

**34. A sum of money amounts to ₹ 13230 in one year and to ₹ 13891.50 in 1 ½ years at compound interest, compounded semi-annually. Find the sum and the rate of interest per annum.**

**Solution**

It is given that

Amount after one year = ₹ 13230

Amount after 1 ½ years = ₹ 13891.50

So the difference = 13891.50 – 13230 = ₹ 661.50

Here ₹ 661.50 is the interest on ₹ 13230 for ½ years

We know that

Rate = (661.50×100×2)/(13230×1)

Multiplying and dividing by 100

= (66150×100×2)/(13230×1×100)

= 10% p.a.

Here

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

Substituting the values

13891.50 = P (1 + 5/100)^{3}

By further calculation

13891.50 = P × 21/20 × 21/20 × 21/20

So we get

P = 13891.50 × 20/21 × 20/21 × 20/21

P = ₹ 12000

**Exercise 2.3**

**1. The present population of a town is 200000. Its population increases by 10% in the first year and 15% in the second year. Find the population of the town at the end of two years.**

**Solution**

We know that

Population after 2 years = Present population × (1 + r/100)^{n}

Here the present population = 200000

Population after first year = 200000 × (1 + 10/100)^{1}

By further calculation

= 200000 × 11/10

= 220000

Population after two years = 220000 × (1 + 15/100)^{1}

By further calculation

= 220000 × 23/20

= 253000

**2. The present population of a town is 15625. If the population increases at the rate of 4% every year, what will be the increase in the population in next 3 years?**

**Solution**

It is given that

Present population (P) = 15625

Rate of increase (r) = 4% p.a.

Period (n) = 3 years

We know that

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

Substituting the values

= 15625 (1 + 4/100)^{3}

By further calculation

= 15625 × 26/25 × 26/25 × 26/25

= 17576

So the increase = 17576 – 15625 = 1951

**3. The population of a city increase each year by 4% of what it had been at the beginning of each year. If its present population is 6760000, find:**

**(i) its population 2 years hence**

**(ii) its population 2 years ago.**

**Solution**

It is given that

Present population = 6760000

Increase percent = 4% p.a.

**(i)** We know that

Population 2 years hence = P (1 + r/100)^{2}

Substituting the values

= 6760000 (1 + 4/100)^{2}

By further calculation

= 6760000 × 26/25 × 26/25

= 7311616

**(ii)** We know that A = 6760000

Population 2 years ago P = A + (1 + r/100)^{2}

Substituting the values

= 6760000 + (1 + 4/100)^{2}

By further calculation

= 6760000 + (26/25)^{2}

= 6760000 × 25/26 × 25/26

= 6250000

**4. The cost of a refrigerator is ₹ 9000. Its value depreciates at the rate of 5% ever year. Find the total depreciation in its value at the end of 2 years.**

**Solution**

It is given that

Present value (P) = ₹ 9000

Rate of depreciation (r) = 5% p.a.

Period (n) = 2 years

We know that

Value after 2 years = P (1 – r/100)^{n}

Substituting the values

= 9000 (1 – 5/100)^{2}

By further calculation

= 9000 × 19/20 × 19/20

= ₹ 8122.50

So the total depreciation = 9000 – 8122.50 = ₹ 877.50

**5. Dinesh purchased a scooter for ₹ 24000. The value of the scooter is depreciating at the rate of 5% per annum. Calculate its value after 3 years.**

**Solution**

It is given that

Present value of scooter (P) = ₹ 24000

Rate of depreciation (r) = 5%

Period (n) = 3 years

We know that

Value after 3 years = P (1 – r/100)^{n}

Substituting the values

= 24000 (1 – 5/100)^{3}

By further calculation

= 24000 × 19/20 × 19/20 × 19/20

= ₹ 20577

**6. A farmer increases his output of wheat in his farm every year by 8%. This year he produced 2187 quintals of wheat. What was the yearly produce of wheat two years ago?**

**Solution**

It is given that

Present production of wheat = 2187 quintals

Increase in production = 8% p.a.

We know that

Production of wheat 2 years ago = A ÷ (1 + r/100)^{n}

Substituting the values

= 2187 ÷ (1 + 8/100)^{2}

By further calculation

= 2187 ÷ (27/25)^{2}

So we get

= 2187 × 25/27 × 25/27

= 1875 quintals

**7. The value of a property decreases every year at the rate of 5%. If its present value is ₹ 411540, what was its value three years ago?**

**Solution**

It is given that

Present value of property = ₹ 411540

Rate of decrease = 5% p.a.

We know that

Value of property 3 years ago = A ÷ (1 – r/100)^{n}

Substituting the values

= 411540 ÷ (1 – 5/100)^{3}

By further calculation

= 411540 ÷ (19/20)^{3}

So we get

= 411540 × 20/19 × 20/19 × 20/19

= ₹ 480000

**8. Ahmed purchased an old scooter for ₹ 16000. If the cost of the scooter after 2 years depreciates to ₹ 14440, find the rate of depreciation.**

**Solution**

It is given that

Present value = ₹ 16000

Value after 2 years = ₹ 14440

Consider r% p.a. as the rate of depreciation

We know that

A/P = (1 – r/100)^{n}

Substituting the values

14440/16000 = (1 – r/100)^{2}

By further calculation

361/400 = (1 – r/100)^{2}

⇒ (19/20)^{2} = (1 – r/100)^{2}

We can write it as

1 – r/100 = 19/20

So we get

r/100 = 1 – 19/20 = 1/20

By cross multiplication

r = 1/20 × 100 = 5%

Hence, the rate of depreciation is 5%.

**9. A factory increased its production of cars from 80000 in the year 2011-2012 to 92610 in 2014-15. Find the annual rate of growth of production of cars.**

**Solution**

It is given that

Production of cars in 2011-2012 = 80000

Production of cars in 2014-2015 = 92610

Period (n) = 3 years

Consider r% as the rate of increase

We know that

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

Substituting the values

92610/80000 = (1 + r/100)^{3}

By further calculation

(21/20)^{3} = (1 + r/100)^{3}

We can write it as

1 + r/100 = 21/20

⇒ r/100 = 21/20 – 1 = 1/20

By cross multiplication

r = 1/20 × 100 = 5

Hence, the annual rate of growth of production of cars is 5% p.a.

**10. The value of a machine worth ₹ 500000 is depreciating at the rate of 10% every year. In how many years will its value be reduced to ₹ 364500?**

**Solution**

It is given that

Present value = ₹ 500000

Reduced value = ₹ 364500

Rate of depreciation = 10% p.a.

Consider n years as the period

We know that

A/P = (1 – r/100)^{n}

Substituting the values

364500/500000 = (1 – 10/100)^{n}

By further calculation

(9/10)^{n} = 729/1000 = (9/10)^{3}

So we get

n = 3

Therefore, the period in which its value be reduced to ₹ 364500 is 3 years.

**11. Afzal purchased an old motorbike for ₹ 16000. If the value of the motorbike after 2 years is ₹ 14440, find the rate of depreciation.**

**Solution**

It is given that

CP of an old motorbike = ₹ 16000

Price after 2 years = ₹ 14440

Consider r% as the rate of depreciation

We know that

A/P = (1 – r/100)^{n}

Substituting the values

14440/16000 = (1 – r/100)^{2}

By further calculation

361/400 = (1 – r/100)^{2}

⇒ (19/20)^{2} = (1 – r/100)^{2}

So we get,

19/20 = 1 – r/100

⇒ r/100 = 1 – 19/20 = (20 – 19)/20 = 1/20

By cross multiplication

r = 100/20 = 5

Hence, the rate of depreciation is 5%.

**12. Mahindra set up a factory by investing ₹ 2500000. During the first two years, his profits were 5% and 10% respectively. If each year the profit was on previous year’s capital, calculate his total profit.**

**Solution**

It is given that

Investment = ₹ 2500000

Rates of profit during first two years = 5% and 10%

We know that

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

Substituting the values

= 2500000 (1 + 5/100) (1 + 10/100)

By further calculation

= 2500000 × 21/20 × 11/10

= ₹ 2887500

So the net profit = A – P

Substituting the values

= 2887500 – 2500000

= ₹ 387500

**13. The value of a property is increasing at the rate of 25% every year. By what percent will the value of the property increase after 3 years?**

**Solution**

It is given that

Original price of the property (P) = ₹ 100

Rate of increase (r) = 25% p.a.

Period (n) = 3 years

We know that

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

Substituting the values

= 100 (1 + 25/100)^{3}

By further calculation

= 100 × 5/4 × 5/4 × 5/4

= ₹ 3125/16

Here

Increased value = 3125/16 – 100

Taking LCM

= (3125 – 1600)/ 16

= 1525/16

So the percent increase after 3 years = 1525/16 = 95 5/16%

**14. Mr. Durani bought a plot of land for ₹ 180000 and a car for ₹ 320000 at the same time. The value of the plot of land grows uniformly at the rate of 30% p.a.., while the value of the car depreciates by 20% in the first year and by 15% p.a. thereafter. If he sells the plot of land as well as the car after 3 years, what will be his profit or loss?**

**Solution**

It is given that

Price of plot of land = ₹ 180000

Growth rate = 30% p.a.

Period (n) = 3 years

We know that

Amount after 3 years = P (1 + R/100)^{n}

Substituting the values

= 180000 (1 + 30/100)^{3}

By further calculation

= 180000 × (13/10)^{3}

It can be written as

= 180000 × 13/10 × 13/10 × 13/10

= ₹ 395460

Here

Price of car = ₹ 320000

Rate of depreciation = 20% for the first year and 15% for next period

Period (n) = 3 years

We know that

Amount after 3 years = A (1 – R_{1}/100)^{n} × (1 – R_{2}/100)^{2}

Substituting the values

= 320000 (1 – 20/100) (1 – 15/100)^{2}

By further calculation

= 320000 × 4/5 × (17/20)^{2}

So we get

= 320000 × 4/5 × 17/20 × 17/20

= ₹ 184960

Here

Total cost of plot and car = 180000 + 320000 = ₹ 500000

Total sale price of plot and car = 395460 + 184960 = ₹ 580420

We know that

Profit = S.P. – C.P.

Substituting the values

= 580420 – 500000

= ₹ 80420

**Chapter Test**

**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?**

**Solution**

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

**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.**

**Solution**

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.

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

**Solution**

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

**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.**

**Solution**

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

**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.**

**Solution**

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

**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?**

**Solution**

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

**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.**

**Solution**

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

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

**Solution**

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.

**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.**

**Solution**

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%.

**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.**

**Solution**

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

**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**

**Solution**

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

**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.**

**Solution**

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