Q1: The population of a place increased to 54,000 in 2003 at a rate of 5% per annum.
(i) Find the population in 2001.
(ii) What would be its population is 2005?
Sol: Population in 2003 is P = 54000
(i) Let the population in 2001 (i.e. 2 years ago) = P
Since rate of increment in population = 5% p.a.
∴ Present populationThus, the population in 2001 was about 48980.
(ii) Initial population (in 2003), i.e. P = 54000
Rate of increment in population = 5% p.a.
Time = 2 years ⇒n = 2= 59535
Thus, the population in 2005 = 59,535.
Q2: In a laboratory, the count of bacteria in a certain experiment was increasing at the rate of 2.5% per hour. Find the bacteria at the end of 2 hours if the count was initially 5,06,000.
Sol: Initial count of bacteria (P) = 5,06,000
= 531616.25 or 531616 (approx.)
Increasing rate (R) = 2.5% per hour
Time (T) = 2 hour. ⇒ n = 2
Thus, the count of bacteria after 2 hours will be 531616 (approx.).
Q3: A scooter was bought at ₹ 42,000. Its value depreciated at the rate of 8% per annum. Find its value after one year.
Sol: Initial cost (value) of the scooter (P) = Rs 42000
Depreciation rate = 8% p.a.
Time = 1 year ⇒ n = 1= ₹ 420 × 92 = ₹ 38,640
Thus, the value of the scooter after 1 year will be ₹ 38,640.
Q1: Machinery worth Rs 10,500 depreciated by 5%. Find its value after one year.
Ans: Here, P = Rs 10,500, R = –5% p.a., T = 1 year, n = 1
∵ [∵ Depreciation is there, ∴ r = –5%]
∴
= Rs10500 * 19/20 = Rs 525 * 19 = Rs 9975
Thus, machinery value after 1 year = Rs 9975
Q2: Find the population of a city after 2 years, which is at present 12 lakh, if the rate of increase is 4%
Ans: Present population, P = 12 lakh
Rate of increase, R = 4% p.a.
Time, T = 2 years
∴ Population after 2 years =
Thus, the population of the town will be 12,97,920 after 2 years.
Note: For depreciation, we use the formula as
Try These
Q1: Find CI on a sum of Rs 8000 for 2 years at 5% per annum compounded annually. Solution: We have P = Rs 8000, R = 5% p.a., T = 2 years
Sol:
∵
∴
= Rs (20 * 21 * 21) = Rs 8820
Now, compound interest = A – P
= Rs 8820 – Rs 8000
= Rs 820
Remember
- The time period after which the interest is added each time to form a new principal is called the conversion period.
- If the interest is compounded half yearly, the time period becomes twice and rate becomes half of the annual rate.
- If the interest compounded quarterly, the time period becomes 4 times and rate become one-fourth of the annual rate.
- If the conversion period is not specified, then it is taken as one year and the interest is compounded annually.
Q2: Find the time period and rate for each.
Sol:
1. We have an interest rate 8% per annum for year.
∴ Time period half years
Rate (R) = 1/2 (8%) = 4% per half year.
2. We have interest rate 4% per annum for 2 years.
∴ Time period (n) = 2(2) = 4 half years
Rate (R) = 1/2 (4%)= 2% per half year.
Q3: Find the amount to be paid
Sol:
1. We have: P = Rs 2400, R = 5% p.a., T = 2 years
∵ Interest is compounded annually i.e. n = 2
∴
2. Here, interest compounded quarterly.
∴ R = 8% p.a. = 8/4% i.e. 2% per quarter
T = 1 year = 4 * 1, i.e. 4 quarters or n = 4
Now
Q4: Calculate the amount and compound interest on
(a) Rs 10,800 for 3 years at per annum compounded annually.
(b) Rs 18,000 for years at 10% per annum compounded annually.
(c) Rs 62,500 for years at 8% per annum compounded half yearly.
(d) Rs 8,000 for 1 year at 9% per annum compounded half yearly.
(You could use the year by year calculation using SI formula to verify.)
(e) Rs 10,000 for 1 year at 8% per annum compounded half yearly.
Sol: (a ) Here P = Rs 10800, T = 3 years, R =
We have:
[∵ Interest compounded annually,∴ n = 3]
∴ Amount = Rs 15377.34Rs 15377.34
Now, compound interest = Rs 15377.34 – Rs 10800
= Rs 4577.34
(b) Here P = Rs 18000, T = years, R = 10% p.a.
∵ Interest is compounded annually,
= Rs 22869
∴ Amount = Rs 22869
CI = Rs 22869 – Rs 18000 = Rs 4869
(c) Here P = Rs 62500, T = r = 8% p.a.
Compounding half yearly,
R = 8% p.a. = 4% per half year
T = year → n = 3 half years.
∴ Amount
= Rs 4 * 26 * 26 * 26 = Rs 70304
Amount = RS 70304
CI = Rs 70304 – Rs 62500 = Rs 7804
(d) Here P = Rs 8000, T = 1 year, R = 9% p.a.
Interest is compounded half yearly,
∴ T = 1 year = 2 half years
R = 9% p.a = 9/2% half yearly
∴ Amount =
CI = Rs 8736.20 – Rs 8000
= Rs 736.20
(e) Here P = Rs 10000, T = 1 year
R = 8% p.a. compounded half yearly.
∴ R = 8% p.a. = 4% per half yearly
T = 1 year → n = 2 * 1 = 2
Now, amount
= Rs 16 * 26 * 26 = Rs 10816
CI = Rs 10816 – Rs 10000 = Rs 816
Q5: Kamala borrowed RS 26,400 from a Bank to buy a scooter at a rate of 15% p.a. compounded yearly. What amount will she pay at the end of 2 years and 4 months to clear the loan?
(Hint: Find A for 2 years with interest is compounded yearly and then find SI an the 2nd year for 4/12 years.)
Sol:
Note: Here, we shall calculate the amount for 2 years using the CI formula. Then this amount will become the principal for next 4 months, i.e. 4/12 years.
Here, P = Rs 26400, T = 2 years, R = 15% p.a
Again, P = 34914, T = 4 months = 4/12 years, R = 15% p.a.
∴ Using we have
= Rs 1745.70
Amount = S.I. + P
= Rs (1745.70 + 34914) = Rs 36659.70
Thus, the required amount to be paid to the bank after 2 years 4 month = Rs 36659.70
Q6: Fabina borrows Rs 12,500 at 12% per annum for 3 years at simple interest and Radha borrows the same amount for the same time period at 10% per annum, compounded annually. Who pays more interest and by how much?
Sol:
For Fabina
P = Rs 12500
T = 3 years
R = 12% p.a.
= Rs 125 X 3 X 12
For Radha
P = Rs 12500
T = 3 years
R = 10% p.a. (Compounded annually)
= Rs 16637.5
∴ CI = 16637.5 – RS 12500 = 4137.50
Difference = Rs 4500 – Rs 4137.50 = Rs 362.50
Thus, Fabina pays Rs 362.50 more.
Q7: I borrowed Rs 12,000 from Jamshed at 6% per annum simple interest for 2 years. Had I borrowed this sum at 6% per annum compound interest, what extra amount would I have to pay?
Sol:
For SI
Principal = Rs 12000
Time = 2 years
Rate = 6% p.a.
For CI
Principal = Rs 12000
Time = 2 years
Rate = 6% p.a.
Thus, excess amount = Rs 1483.20 – Rs 1440
= Rs 43.20.
Q8: Vasudevan invested Rs 60,000 at an interest rate of 12% per annum compounded half yearly. What amount would he get
(i) after 6 months? (ii) after 1 year?
Sol:
(i) Amount after 6 months
P = Rs 60000
T = 1/2 year , n = 1 [∵ Interest is compounded half yearly.]
R = 12% p.a. = 6% per half year
(ii) Amount after 1 year
P = Rs 60000
T = 1 year; n = 2
R = 12% p.a. = 6% per half year
Thus, amount after 6 months = Rs 63600
and amount after 1 year = Rs 67416
Q9: Arif took a loan of Rs 80,000 from a bank. If the rate of interest is 10% per annum,find the difference in amounts he would be paying after years if the interest is (i) Compounded annually. (ii) Compounded half yearly.
Sol:
(i) Compounded annually
P = Rs 80000
R = 10% p.a.
Amount for 1st year.
= Rs 440 *10 = Rs 4400
∴ Amount = Rs 88000 + Rs 44000
= Rs 92400
(ii) Compounded half yearly
P = Rs 80000
R = 10% p.a. = 5% per half year
= Rs 10 * 21 * 21 * 21
= Rs 92610
Thus, the difference between the two amounts = Rs 92610 – Rs 92400
= Rs 210
Q10: Maria invested Rs 8,000 in a business. She would be paid interest at 5% per annum compounded annually. Find:
(i) The amount credited against her name at the end of the second year.
(ii) The interest for the 3rd year.
Sol: Principal = Rs 8000
Rate = 5% p.a. compounded annually
(i) Time = 2 years ⇒ n = 2
∴ Amount credited against her name at the end of two years = Rs 8820
(ii) Time = 3 years ⇒ n = 3
∴
= Rs (21 * 21 * 21) = Rs 9261
∵ Interest paid during 3rd year
= [Amount at the end of 3rd year] – [Amount at the end of 2nd year]
= Rs 9261 – Rs 8820 = Rs 441
Q11: Find the amount and the compound interest on Rs 10,000 for years at 10% per annum, compounded half yearly. Would this interest be more than the interest he would get if it was compounded annually?
Sol:
Principal = Rs 10,000
Time = %
Rate = 10% p.a.
Case I. Interest on compounded half yearly
We have r = 10 p.a. = 5% per half yearly
∴
∴ Amount = Rs 11576.25
Now CI = Amount – Principal
= Rs 11576.25 – Rs 10,000 = Rs 1576.25
Case II. Interest on compounded annually
We have R = 10% p.a.
Amount for 1 year =
∴ Interest for 1st year = Rs 11000 – Rs 10,000 = Rs 1000
Interest for next 1/2 year on Rs 11000
= Rs 55 * 10 = Rs 550
∴ Total interest = Rs 1000 + Rs 550
= Rs 1550
Since Rs 1576.25 > Rs 1550
∴ Interest would be more in case of it is compounded half yearly.
Q12: Find the amount which Ram will get on Rs 4096, if he gave it for 18 months at per annum, interest being compounded half yearly.
Sol:
We have P = Rs 4096
T = 18 months
∵ Interest is compounded half yearly.
T = 18 months ⇒ n = 18/6 = 3 six months
Now,
Thus, the required amount = Rs 4913
1. What are the key concepts covered in Exercise 7.3 of Comparing Quantities in Old NCERT? |
2. How can I solve problems related to percentage increase and decrease in Exercise 7.3? |
3. Are there any specific formulas to remember for solving questions in Comparing Quantities? |
4. How can I practice effectively for the Comparing Quantities chapter? |
5. What are the common mistakes to avoid while solving Exercise 7.3 questions? |
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