Given
Principal (P) = Rs 40000
Amount (A) = Rs 46305
Period (n) =1 \frac{1}{2} \text { years } = 3/2 years
So half yearly, 2n = 2 × (3/2) = 3 years.
Let the rate of interest be r% p.a.
WKT
A / P = (1 + r / 100)n
46305 / 40000 = (1 + r / 100)3
(1 + r / 100)3 = 46305 / 40000
On further calculation, we get,
(1 + r / 100)3 = 9261 / 8000
(1 + r / 100)3 = (21 / 20)3
We get,
(1 + r / 100) = (21 / 20)
r / 100 = (21 / 20) – 1
r / 100 = 1 / 20
r = 100 / 20
We get,
r = 5
Therefore, the rate of interest = 5% for half a year.
So, 2 × 5 = 10% per annum
ClearTax offers taxation & financial solutions to individuals, businesses, organizations & chartered accountants in India. ClearTax serves 2.5+ Million happy customers, 20000+ CAs & tax experts & 10000+ businesses across India.
Efiling Income Tax Returns(ITR) is made easy with ClearTax platform. Just upload your form 16, claim your deductions and get your acknowledgment number online. You can efile income tax return on your income from salary, house property, capital gains, business & profession and income from other sources. Further you can also file TDS returns, generate Form-16, use our Tax Calculator software, claim HRA, check refund status and generate rent receipts for Income Tax Filing.
CAs, experts and businesses can get GST ready with ClearTax GST software & certification course. Our GST Software helps CAs, tax experts & business to manage returns & invoices in an easy manner. Our Goods & Services Tax course includes tutorial videos, guides and expert assistance to help you in mastering Goods and Services Tax. ClearTax can also help you in getting your business registered for Goods & Services Tax Law.
Save taxes with ClearTax by investing in tax saving mutual funds (ELSS) online. Our experts suggest the best funds and you can get high returns by investing directly or through SIP. Download ClearTax App to file returns from your mobile phone.
ML Aggarwal Solutions for Class 9 Maths Chapter 2 Compound Interest has problems solved by a set of expert faculty at BYJU’S. In a subject like Mathematics, students should understand the concepts first and start answering the questions wisely. Practising the problems on a daily basis is the main key towards success. Here, the students can refer to ML Aggarwal Solutions for Class 9 Maths Chapter 2 Compound Interest PDF, from the links provided below.
Chapter 2 consists of problems in finding the rate of interest, principal amount and the time period, as per the latest syllabus of the ICSE board. The ML Aggarwal Solutions are available in PDF format so that the students can use them while solving the exercise wise problems from the textbook. It also helps students in improving the time management skills which are important from the exam point of view.
Access ML Aggarwal Solutions for Class 9 Maths Chapter 2: Compound Interest
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
So we get
= ₹ 420
We know that
Amount after the second year = 8400 + 420
= ₹ 8820
Total compound interest = 8820 + 8000
= ₹ 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
So we get
= ₹ 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 – R1/100)n × (1 – R2/100)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) = 6n
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