Compound Interest
Class 8th Mathematics RS Aggarwal Solution
- Find the amount and the compound interest on Rs. 2500 for 2 years at 10% per…
- Find the amount and the compound interest on Rs. 15625 for 3 years at 12% per…
- Find the difference between the simple interest and the compound interest on…
- Ratna obtained a loan of Rs. 25000 from the Syndicate Bank to renovate her…
- Harpreet borrowed Rs. 20000 from her friend at 12% per annum simple interest.…
- Manoj deposited a sum of Rs. 64000 in a post office for 3 years, compounded…
- Divakaran deposited a sum of Rs. 6250 in the Allahabad Bank for 1 year,…
- Michael borrowed Rs.16000 from a finance company at 10% per annum, compounded…
- Rs. 6000 for 2 years at 9% per annum compounded annually.
- Rs. 10000 for 2 years at 11% per annum compounded annually.
- Rs. 31250 for 3 years at 8% per annum compounded annually.
- Rs. 10240 for 3 years at 12 1/2 % per annum compounded annually.
- Rs. 62500 for 2 years 6 months at 12% per annum compounded annually.…
- Rs. 9000 for 2 years 4 months at 10% per annum compounded annually.…
- Find the amount of Rs. 8000 for 2 years compounded annually and the rates being…
- Anand obtained a loan of Rs. 125000 from the Allahabad Bank for buying…
- Three years ago, Beeru purchased a buffalo from Surjeet for Rs. 11000. What…
- Shubhalaxmi took a loan of Rs. 18000 from Surya Finance to purchase a TV set.…
- Neha borrowed Rs. 24000 from the State Bank of India to buy a scooter. If the…
- Abhay borrowed Rs. 16000 at 7 1/2 % per annum simple interest. On the same…
- The simple interest on a sum of money for 2 years at 8% per annum is Rs. 2400.…
- The difference between the compound interest and the simple interest on a…
- The difference between the compound interest and the simple interest on a…
- A sum of money amounts to Rs. 10240 in 2 years at 6 2/3 % per annum,…
- What sum of money will amount to Rs. 21296 in 3 years at 10% per annum,…
- At what rate per cent per annum will Rs. 4000 amount to Rs. 4410 in 2 years…
- At what rate per cent per annum will Z 640 amount to Rs. 774.40 in 2 years…
- In how many years will Rs. 1800 amount to Rs. 2178 at 10% per annum when…
- In how many years will Rs. 6250 amount to Rs. 7290 at 8% per annum, compounded…
- The population of a town is 125000. It is increasing at the rate of 2% per…
- Three years ago, the population of a town was 50000. If the annual increase…
- The population of a city was 120000 in the year 2013. During next year it…
- The count of bacteria in a certain experiment was increasing at the rate of 2%…
- The bacteria in a culture grows by 10% in the first hour, decreases by 10% in…
- A machine is purchased for Rs. 625000. Its value depreciates at the rate of 8%…
- A scooter is bought at Rs. 56000. Its value depreciates at the rate of 10% per…
- A car is purchased for Rs. 348000. Its value depreciates at 10% per annum…
- The value of a machine depreciates at the rate of 10% per annum. It was…
- Find the amount and the compound interest on Rs. 8000 for 1 year at 10% per…
- Find the amount and the compound interest on Rs. 31250 for 1 1/2 % years at 8%…
- Find the amount and the compound interest on Rs. 12800 for 1 year at 7 1/2 %…
- Find the amount and the compound interest on Rs. 160000 for 2 years at 10% per…
- Swati borrowed Rs. 40960 from a bank to buy a piece of land. If the bank…
- Mohd. Aslam purchased a house from Avas Vikas Parishad on credit. If the cost…
- Sheela deposited Rs. 20000 in a bank, where the interest is credited…
- Neeraj lent Rs. 65536 for 2 years at 12 1/2 % per annum, compounded annually.…
- Sudershan deposited Rs. 32000 in a bank, where the interest is credited…
- Arun took a loan of Rs. 390625 from Kuber Finance. If the company charges…
- Find the amount and the compound interest on Rs. 3000 for 2 years at 10% per annum.…
- Find the amount of Rs. 10000 after 2 years compounded annually; the rate of interest…
- Find the amount and the compound interest on Rs. 6000 for 1 year at 10% per annum…
- A sum amounts to Rs. 23762 in 2 years at 9% per annum, compounded annually. Find the…
- A scooter is bought for Rs. 32000. Its value depreciates at 10% per annum. What will be…
- The compound interest on Rs. 5000 at 10% per annum for 2 years isA. Rs. 550 B. Rs. 1050…
- The annual rate of growth in population of a town is 5%. If its present population is…
- At what rate per cent per annum will 5000 amount to Rs. 5832 in 2 years, compounded…
- If the simple interest on a sum of money at 10% per annum for 3 years is Rs. 1500, then…
- If the compound interest on a certain sum for 2 years at 10% per annum is Rs. 1050,…
- Fill in the blanks: (i) a = p (1 + l l /100)^n (ii) (Amount) - (Principal) = . (iii)…
- The compound interest on Z 5000 at 8% per annum for 2 years, compounded…
- The compound interest on Z 10000 at 10% per annum for 3 years, compounded…
- The compound interest on Z 10000 at 12% per annum for 1 1/2 years, compounded…
- The compound interest on Rs. 4000 at 10% per annum for 2 years 3 months,…
- A sum of Rs. 25000 was given as loan on compound interest for 3 years…
- The compound interest on Rs. 6250 at 8% per annum for 1 year, compounded half…
- The compound interest on Rs. 40000 at 6% per annum for 6 months, compounded…
- The present population of a town is 24000. If it increases at the rate of 5%…
- The value of a machine depreciates at the rate of 10% per annum. It was…
- The value of a machine depreciates at the rate of 20% per annum. It was…
- The annual rate of growth in population of a town is 10%. If its present…
- If the simple interest on a sum of money at 5% per annum for 3 years is Rs.…
- If the compound interest on a sum for 2 years at 12 1/2 % per annum is Rs.…
- The sum that amounts to Rs.4913 in 3 years at (25/4)% per annum compounded…
- At what rate per cent per annum will a sum of Rs. 7500 amount to Rs. 8427 in 2…
Exercise 11a
Question 1.Find the amount and the compound interest on Rs. 2500 for 2 years at 10% per annum, compounded annually.
Answer:
Present value = Rs.2500
Interest rate = 10% per annum
Time = 2 years
Amount (A) = P (1 + R/100)n
[Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 2500 (1 + 10/100)2
⇒ A = 2500 (11/10)2
⇒ A = 2500 × 121/100
⇒ A = 25 × 121
⇒ A = 3025
∴ Amount = Rs.3025
∴ Compound interest = Rs.(3025 – 2500)
= Rs.525
Question 2.
Find the amount and the compound interest on Rs. 15625 for 3 years at 12% per annum, compounded annually.
Answer:
Present value = Rs.15625
Interest rate = 12% per annum
Time = 3 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 15625 (1 + 12/100)3
⇒ A = 15625 (112/100)3
⇒ A = 15625 (28/25)3
⇒ A = 15625 × 21952/15625
⇒ A = 21952
∴ Amount = Rs.21952
∴ Compound interest = Rs.(21952 – 15625)
= Rs.6327
Question 3.
Find the difference between the simple interest and the compound interest on Rs. 5000 for 2 years at 9% per annum.
Answer:
Present value = Rs.5000
Interest rate = 9% per annum
Time = 2 years
Simple interest (SI) = PRT/100 [where, P = Present value
R = Interest rate, T = Time]
∴ SI = (5000 × 9 × 2)/100
⇒ SI = 50 × 9 × 2
⇒ SI = 900
Now,
Compound interest (CI),
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 5000 (1 + 9/100)2
⇒ A = 5000 (109/100)2
⇒ A = 5000 (1.09)2
⇒ A = 5000 × 1.1881
⇒ A = 5940.5
∴ Amount = Rs.5940.5
∴ Compound interest = Rs.(5940.5 – 5000)
= Rs.940.5
Now,
Difference between the simple interest and the compound interest = (CI – SI)
= (940.5 – 900)
= 40.5
Question 4.
Ratna obtained a loan of Rs. 25000 from the Syndicate Bank to renovate her house. If the rate of interest is 8% per annum, what amount will she have to pay to the bank after 2 years to discharge her debt?
Answer:
Present value = Rs.25000
Interest rate = 8% per annum
Time = 2 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 25000 (1 + 8/100)2
⇒ A = 25000 (108/100)2
⇒ A = 25000 (1.08)2
⇒ A = 25000 × 1.1664
⇒ A = 29160
∴ Amount = Rs.29160
Question 5.
Harpreet borrowed Rs. 20000 from her friend at 12% per annum simple interest. She lent it to Alam at the same rate but compounded annually. Find her gain after 2 years.
Answer:
Present value = Rs.20000
Interest rate = 12% per annum
Time = 2 years
Simple interest (SI) = PRT/100 [where, P = Present value
R = Interest rate, T = Time]
∴ SI = (20000 × 12 × 2)/100
⇒ SI = 200 × 12 × 2
⇒ SI = 4800
Now,
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 20000 (1 + 12/100)2
⇒ A = 20000 (112/100)2
⇒ A = 20000 (1.12)2
⇒ A = 20000 × 1.2544
⇒ A = 25088
∴ Amount = Rs.25088
∴ Compound interest = Rs.(25088 – 20000)
= Rs.5088
Now,
(CI – SI) = 5088 -4800
= Rs.288
∴ The amount of money Harpreet will gain after two years = Rs.288
Question 6.
Manoj deposited a sum of Rs. 64000 in a post office for 3 years, compounded annually at 7
% per annum. What amount will he get on maturity?
Answer:
Present value = Rs.64000
Interest rate = (15/2) % per annum
Time = 3 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 64000 [1 + (15/2 × 1/100)]3
⇒ A = 64000 [1 + 3/40]3
⇒ A = 64000 (43/40)3
⇒ A = 1 × 43 × 43 × 43
⇒ A = 79507
∴ Manoj will get an amount of Rs.79507 after 3 years.
Question 7.
Divakaran deposited a sum of Rs. 6250 in the Allahabad Bank for 1 year, compounded half-yearly at 8% per annum. Find the compound interest he gets.
Answer:
Present value = Rs.6250
Interest rate = 8 % per annum
Time = 1 years
∵ Interest is compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 6250 [1 + (8/2)/100]2
⇒ A = 6250 [1 + 4/100]2
⇒ A = 6250 [26/25]2
⇒ A = 6250 × 26/25 × 26/25
⇒ A = 10 × 26 × 26
⇒ A = 6760
∴ Amount = Rs.6760
∴ Compound interest = Rs.(6760 – 6250)
= Rs.510
∴ Divakaran gets a CI of Rs.510.
Question 8.
Michael borrowed Rs.16000 from a finance company at 10% per annum, compounded half-yearly. What amount of money will discharge his debt after 1years?
Answer:
Present value = Rs.16000
Interest rate = 10% per annum
Time = (3/2) years
∵ Interest is compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 16000 [1 + (10/2)/100]3
⇒ A = 16000 [1 + 5/100]3
⇒ A = 16000 [1 + 1/20]3
⇒ A = 16000 [21/20]3
⇒ A = 16000 × 21/20 × 21/20 × 21/20
⇒ A = 2 × 21 × 21 × 21
⇒ A = 18522
∴ Amount = Rs.18522
Exercise 11b
Question 1.Rs. 6000 for 2 years at 9% per annum compounded annually.
Answer:
Present value = Rs.6000
Interest rate = 9% per annum
Time = 2 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 6000 (1 + 9/100)2
⇒ A = 6000 (109/100)2
⇒ A = 6000 × 109/100 × 109/100
⇒ A = 6 × 109 × 109/10
⇒ A = 7128.6
∴ Amount = Rs.7128.6
∴ Compound interest = Rs.(7128.6 – 6000)
= Rs.1128.6
Question 2.
Rs. 10000 for 2 years at 11% per annum compounded annually.
Answer:
Present value = Rs.10000
Interest rate = 11% per annum
Time = 2 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 10000 (1 + 11/100)2
⇒ A = 10000 (111/100)2
⇒ A = 10000 × 111/100 × 111/100
⇒ A = 1 × 111 × 111
⇒ A = 12321
∴ Amount = Rs.12321
∴ Compound interest = Rs.(12321 – 10000)
= Rs.2321
Question 3.
Rs. 31250 for 3 years at 8% per annum compounded annually.
Answer:
Present value = Rs.31250
Interest rate = 8% per annum
Time = 3 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 31250 (1 + 8/100)3
⇒ A = 31250 (1 + 2/25)3
⇒ A = 31250 (27/25)3
⇒ A = 31250 × 27/25 × 27/25 × 27/25
⇒ A = 31250 × 19683/15625
⇒ A = 2 × 19683
⇒ A = 39366
∴ Amount = Rs.39366
∴ Compound interest = Rs.(39366 – 31250)
= Rs.8116
Question 4.
Rs. 10240 for 3 years at 12% per annum compounded annually.
Answer:
Present value = Rs.10240
Interest rate = (25/2) % per annum
Time = 3 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 10240 [1 + (25/2)/100]3
⇒ A = 10240 [1 + 1/8]3
⇒ A = 10240 [9/8]3
⇒ A = 10240 × 9/8 × 9/8 × 9/8
⇒ A = 10240 × 729/512
⇒ A = 20 × 729
⇒ A = 14580
∴ Amount = Rs.14580
∴ Compound interest = Rs.(14580 – 10240)
= Rs.4340
Question 5.
Rs. 62500 for 2 years 6 months at 12% per annum compounded annually.
Answer:
Present value = Rs.62500
Interest rate = 12 % per annum
Time = 2 years 6 month = (2 + 1/2) years = (5/2) years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 62500 (1 + 12/100)2 × [1 + (1/2 × 12)/100]
⇒ A = 62500 (1 + 3/25)2 × [1 + 6/100]
⇒ A = 62500 (28/25)2 × [106/100]
⇒ A = 62500 × 28/25 × 28/25 × 106/100
⇒ A = 625 × 784/625 × 106
⇒ A = 1 × 784 × 106
⇒ A = 83104
∴ Amount = Rs.83104
∴ Compound interest = Rs.(83104 – 62500)
= Rs.20604
Question 6.
Rs. 9000 for 2 years 4 months at 10% per annum compounded annually.
Answer:
Present value = Rs.9000
Interest rate = 10 % per annum
Time = 2 years 4 month = (2 + 1/3) years = (7/2) years
Amount (A) = P (1 + R/100)n × [1 + (1/3 × R)/100]
[Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 9000 (1 + 10/100)2 × [1 + (1/3 × 10)/100]
⇒ A = 9000 (1 + 1/10)2 × [1 + 1/30]
⇒ A = 9000 (11/10)2 × [31/30]
⇒ A = 9000 × 121/100 × 31/30
⇒ A = 9 × 121 × 31/3
⇒ A = 3 × 121 × 31
⇒ A = 11253
∴ Amount = Rs.11253
∴ Compound interest = Rs.(11253 – 9000)
= Rs.2253
Question 7.
Find the amount of Rs. 8000 for 2 years compounded annually and the rates being 9% per annum during the first year and 10% per annum during the second year.
Answer:
Present value = Rs.8000
Interest rate for 1st year, p = 9 % per annum
Interest rate for 2nd year, q = 10 % per annum
Time = 2 years
Amount (A) = P × (1 + p/100) × (1 + q/100)
A = 8000 × (1 + 9/100) × (1 + 10/100)
= 8000 × (109/100) × (1 + 1/10)
= 8000 × 109/100 × 11/10
= 8 × 109 × 11
= 9592
∴ Amount = Rs.9592
Question 8.
Anand obtained a loan of Rs. 125000 from the Allahabad Bank for buying computers. The bank charges compound interest at 8% per annum, compounded annually. What amount pwill he have to pay after 3 years to clear the debt?
Answer:
Present value = Rs.125000
Interest rate = 8% per annum
Time = 3 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 125000 (1 + 8/100)3
⇒ A = 125000 (108/100)3
⇒ A = 125000 × 108/100 × 108/100 × 108/100
⇒ A = 125000 × 1259712/1000000
⇒ A = 125 × 1259712/1000
⇒ A = 1259712/8
⇒ A = 157464
∴ Amount = Rs.157464
∴ Anand has to pay Rs.157464 after 3 years to clear the debt.
Question 9.
Three years ago, Beeru purchased a buffalo from Surjeet for Rs. 11000. What payment will discharge his debt now, the rate of interest being 10% per annum, compounded annually?
Answer:
Present value = Rs.11000
Interest rate = 10% per annum
Time = 3 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 11000 (1 + 10/100)3
⇒ A = 11000 (1 + 1/10)3
⇒ A = 11000 (11/10)3
⇒ A = 11000 × 11/10 × 11/10 × 11/10
⇒ A = 11000 × 1331/1000
⇒ A = 11 × 1331
⇒ A = 14641
∴ Amount = Rs.14641
∴ Beeru has to pay Rs.14641 to clear the debt.
Question 10.
Shubhalaxmi took a loan of Rs. 18000 from Surya Finance to purchase a TV set. If the company charges compound interest at 12% per annum during the first year and 12
% per annum during the second year, how much will she have to pay after 2 years?
Answer:
Present value = Rs.18000
Interest rate for 1st year, p = 12 % per annum
Interest rate for 2nd year, q = (25/2) % per annum
Time = 2 years
Amount (A) = P × (1 + p/100) × (1 + q/100)
A = 18000 × (1 + 12/100) × [1 + (25/2)/100]
= 18000 × (112/100) × [1 + 25/200]
= 18000 × (112/100) × [1 + 1/8]
= 18000 × 112/100 × 9/8
= 180 × 112 × 9/8
= 180 × 14 × 9
= 22680
∴ Amount = Rs.22680
∴ Shubhlaxmi has to pay Rs.157464 after 2 years.
Question 11.
Neha borrowed Rs. 24000 from the State Bank of India to buy a scooter. If the rate of interest be 10% per annum compounded annually, what payment will she have to make after 2 years 3 months?
Answer:
Present value = Rs.24000
Interest rate = 10 % per annum
Time = 2 years 3 month = (2 + 1/4) years = 2
years.
Amount (A) = P (1 + R/100)n × [1 + (1/4 × R)/100]
[Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 24000 (1 + 10/100)2 × [1 + (1/4 × 10)/100]
⇒ A = 24000 (1 + 1/10)2 × [1 + 1/40]
⇒ A = 24000 (11/10)2 × [41/40]
⇒ A = 24000 × 121/100 × 41/40
⇒ A = 24 × 121 × 41/4
⇒ A = 6 × 121 × 41
⇒ A = 29766
∴ Amount = Rs.29766
∴ Neha should pay Rs. 29766 to the bank after 2 years 3 months.
Question 12.
Abhay borrowed Rs. 16000 at 7
% per annum simple interest. On the same day, he lent it to Gurmeet at the same rate but compounded annually. What does he gain at the end of 2 years?
Answer:
Present value = Rs.16000
Interest rate = 7
% = (15/2) % per annum
Time = 2 years
Simple interest (SI) = PRT/100 [where, P = Present value
R = Interest rate, T = Time]
∴ SI = (16000 × (15/2) × 2)/100
⇒ SI = 160 × 15
⇒ SI = 2400
Now,
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 16000 [1 + (15/2)/100]2
⇒ A = 16000 [1 + 3/40]2
⇒ A = 16000 [43/40]2
⇒ A = 16000 × 1849/1600
⇒ A = 10 × 1849
⇒ A = 18490
∴ Amount = Rs.18490
∴ Compound interest = Rs.(18490 – 16000)
= Rs.2490
Now,
(CI – SI) = 2490 -2400
= Rs.90
∴ Abhay gains Rs.90 at the end of 2 years.
Question 13.
The simple interest on a sum of money for 2 years at 8% per annum is Rs. 2400. What will be the compound interest on that sum at the same rate and for the same period?
Answer:
Simple interest = Rs.2400
Interest rate = 8% per annum
Time = 2 years
Simple interest (SI) = PRT/100 [where, P = Present value
R = Interest rate
∴ 2400 = (P × 8 × 2)/100 T = Time]
⇒ 2400 = P × 16/100
⇒ 2400 = P × 4/25
⇒ P = 2400 × 25/4
⇒ P = 600 × 25
⇒ P = 15000
∴ Sum = Rs.15000
Now,
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 15000 [1 + 8/100]2
⇒ A = 15000 [1 + 2/25]2
⇒ A = 15000 [27/25]2
⇒ A = 15000 × 27/25 × 27/25
⇒ A = 600 × 27 × 27/25
⇒ A = 24 × 27 × 27
⇒ A = 17496
∴ Amount = Rs.17496
∴ Compound interest = Rs.(17496 – 15000)
= Rs.2496
Question 14.
The difference between the compound interest and the simple interest on a certain sum for 2 years at 6% per annum is Rs. 90.
Find the sum.
Answer:
Let sum = P
Interest rate = 6% per annum
Time = 2 years
Simple interest (SI) = PRT/100 [Where, P = Present value
R = Annual interest rate
T = Time in years]
∴ SI = (P × 6 × 2)/100
⇒ SI = 3P/25 __________ (i)
Compound interest (CI) = P (1 + R/100)n - P [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ CI = P (1 + 6/100)2 - P
⇒ CI = P (1 + 3/50)2 - P
⇒ CI = P (53/50)2 - P
⇒ CI = (2809P/2500)- P
⇒ CI = (2809P – 2500P)/2500
⇒ CI = 309P/2500 ______ (ii)
Now,
CI – SI = (309P/2500) – (3P/25)
⇒ 90 = (309P/2500) – (3P/25) [Given, CI – SI = 90]
⇒ 90 = (309P – 300P)/2500
⇒ 90 = 9P/2500
⇒ P = 90 × 2500/9
⇒ P = 10 × 2500
⇒ P = 25000
∴ Sum = Rs.25000
Question 15.
The difference between the compound interest and the simple interest on a certain sum for 3 years at 10% per annum is Rs. 93. Find the sum.
Answer:
Let sum = P
Interest rate = 10% per annum
Time = 3 years
Simple interest (SI) = PRT/100 [Where, P = Present value
R = Annual interest rate
T = Time in years]
∴ SI = (P × 10 × 3)/100
⇒ SI = 3P/10 ______ (i)
Compound interest (CI) = P (1 + R/100)n - P [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ CI = P (1 + 10/100)3 - P
⇒ CI = P (1 + 1/10)3 - P
⇒ CI = P (11/10)3 - P
⇒ CI = (1331P/1000)- P
⇒ CI = (1331P – 1000P)/1000
⇒ CI = 331P/1000 ______ (ii)
Now,
CI – SI = (331P/1000) – (3P/10)
⇒ 93 = (331P/1000) – (3P/10) [Given, CI – SI = 93]
⇒ 93 = (331P – 300P)/1000
⇒ 93 = 31P/1000
⇒ P = 93 × 1000/31
⇒ P = 3 × 1000
⇒ P = 3000
∴ Sum = Rs.3000
Question 16.
A sum of money amounts to Rs. 10240 in 2 years at 6
% per annum, compounded annually. Find the sum.
Answer:
Let sum = P
Interest rate = 6
% = (20/3) % per annum
Time = 2 years
Now,
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = P[1 + (20/3)/100]2
⇒ 10240 = P [1 + 1/15]2
⇒ 10240 = P [16/15]2
⇒ 10240 = P × 256/225
⇒ P = 10240 × 225/256
⇒ P = 40 × 225
⇒ P = 9000
∴ Sum = Rs.9000
Question 17.
What sum of money will amount to Rs. 21296 in 3 years at 10% per annum, compounded annually?
Answer:
Let sum = P
Interest rate = 10% per annum
Time = 3 years
Now,
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = P (1 + 10/100)3
⇒ 21296 = P (1 + 1/10)3
⇒ 21296 = P (11/10)3
⇒ 21296 = P × 1331/1000
⇒ P = 21296 × 1000/1331
⇒ P = 16 × 1000
⇒ P = 16000
∴ Sum = Rs.16000
Question 18.
At what rate per cent per annum will Rs. 4000 amount to Rs. 4410 in 2 years when compounded annually?
Answer:
Let rate = R % per annum
P = Rs.4000
A = Rs.4410
Time = 2 years
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ A = P (1 + R/100)2
⇒ 4410 = 4000 (1 + R/100)2
⇒ (1 + R/100)2 = 4410/4000
⇒ (1 + R/100)2 = 441/400
⇒ (1 + R/100) = √(441/400)
⇒ R/100 = (21/20) - 1
⇒ R/100 = (21 – 20)/20
⇒ R/100 = 1/20
⇒ R = 100/20
⇒ R = 5
∴ Rate = 5% per annuam.
Question 19.
At what rate per cent per annum will Z 640 amount to Rs. 774.40 in 2 years when compounded annually?
Answer:
Let rate = R % per annum
P = Rs.640
A = Rs.774.40
Time = 2 years
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ A = P (1 + R/100)2
⇒ 774.40 = 640 (1 + R/100)2
⇒ (1 + R/100)2 = 774.40/640
⇒ (1 + R/100)2 = 1.21
⇒ (1 + R/100) = √(1.21)
⇒ R/100 = (1.1) - 1
⇒ R/100 = 0.1
⇒ R = 0.1 × 100
⇒ R = 10
∴ Rate = 10% per annuam.
Question 20.
In how many years will Rs. 1800 amount to Rs. 2178 at 10% per annum when compounded annually?
Answer:
Let time = n years
P = Rs.1800
A = Rs.2178
R = 10% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ A = P (1 + R/100)n
⇒ 2178 = 1800 (1 + 10/100)n
⇒ (1 + 1/10)n = 2178/1800
⇒ (11/10)n = 121/100
⇒ (11/10)n = (11/10)2
⇒ n = 2
∴ Time = 2 years.
Question 21.
In how many years will Rs. 6250 amount to Rs. 7290 at 8% per annum, compounded annually?
Answer:
Let time = n years
P = Rs.6250
A = Rs.7290
R = 8% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ A = P (1 + R/100)n
⇒ 7290 = 6250 (1 + 8/100)n
⇒ (1 + 2/25)n = 7290/6250
⇒ (27/25)n = 729/625
⇒ (27/25)n = (27/25)2
⇒ n = 2
∴ Time = 2 years.
Question 22.
The population of a town is 125000. It is increasing at the rate of 2% per annum. What will be its population after 3 years?
Answer:
Population of a town, P = 125000
Time, n = 3 years
Increasing rate, R = 2% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ Population = P (1 + R/100)n
= 125000 (1 + 2/100)3
= 125000 (1 + 1/50)3
= 125000 (51/50)3
= 125000 × 51/50 × 51/50 × 51/50
= 1 × 51 × 51 × 51
= 132651
∴ Population of a town after 3 years is 132651.
Question 23.
Three years ago, the population of a town was 50000. If the annual increase during three successive years be at the rate of 5%, 4% and 3% respectively, what is its present population?
Answer:
Population of a town, P = 50000
Interest rate for 1st year, p = 5%
Interest rate for 2nd year, q = 4%
Interest rate for 3rd year, r = 3%
Time, n = 3 years
Now,
Present population = P × (1 + p/100) × (1 + q/100) × (1 + r/100)
= 50000 × (1 + 5/100) × (1 + 4/100) × (1 + 3/100)
= 50000 × (1 + 1/20) × (1 + 1/25) × (1 + 3/100)
= 50000 × 21/20 × 26/25 × 103/100
= 50 × 21/2 × 26/25 × 103
= 1 × 21 × 26 × 103
= 56238
∴ Present population of a town is 56238.
Question 24.
The population of a city was 120000 in the year 2013. During next year it increased by 6% but due to an epidemic it decreased by 5% in the following year. What was its population in the year 2015?
Answer:
Population of a city in 2013, P = 120000
Time, n = 3 years
Increasing rate, R = 6% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴Population of the city in the year 2014,
∴ Population = P (1 + R/100)n
= 120000 (1 + 6/100)1
= 120000 (1 + 3/50)
= 120000 (53/50)
= 120000 × 53/50
= 2400 × 53
= 127200
∴ Population of a city in 2014 is 127200.
Now,
Decreasing rate = 8%
∴Population of the city in the year 2015,
∴ Population = P (1 - R/100)n
= 127200 (1 - 5/100)1
= 127200 (1 - 1/20)
= 127200 (19/20)
= 127200 × 19/20
= 6360 × 19
= 120840
∴ Population of a city in 2015 is 120840.
Question 25.
The count of bacteria in a certain experiment was increasing at the rate of 2% per hour. Find the bacteria at the end of 2 hours if the count was initially 500000.
Answer:
Count of bacteria, P = 500000
Time, n = 2 hours
Increasing rate, R = 2% per hour
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time]
∴ Count of bacteria = P (1 + R/100)n
= 500000 (1 + 2/100)2
= 500000 (102/100)2
= 500000 × 102/100 × 102/100
= 50 × 102 × 102
= 520200
∴ Count of bacteria at the end of 2 hours is 520200.
Question 26.
The bacteria in a culture grows by 10% in the first hour, decreases by 10% in the second hour and again increases by 10% in the third hour. Find the bacteria at the end of 3 hours if the count was initially 20000.
Answer:
Initial count of bacteria, P = 20000
Time, n = 3 hours
Increasing rate, R = 10% per hour
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time]
∴Count of bacteria at the end of 1st hour,
∴ Count of bacteria = P (1 + R/100)n
= 20000 (1 + 10/100)1
= 20000 (1 + 1/10)
= 20000 (11/10)
= 20000 × 11/10
= 2000 × 11
= 22000
∴ Count of bacteria at the end of 1st hour is 22000.
Now,
Decreasing rate = 10%
∴Count of bacteria at the end of 2nd hour,
∴ Count of bacteria = P (1 + R/100)n
= 22000 (1 - 10/100)1
= 22000 (1 - 1/10)
= 22000 × 9/10
= 2200 × 9
= 19800
∴ Count of bacteria at the end of 2nd hours is 19800.
Now,
Increasing rate = 10%
∴Count of bacteria at the end of 3rd hour,
∴ Count of bacteria = P (1 + R/100)n
= 19800 (1 + 10/100)1
= 19800 (1 + 1/10)
= 19800 (11/10)
= 19800 × 11/10
= 1980 × 11
= 21780
∴ Count of bacteria at the end of 3rd hours is 21780.
Question 27.
A machine is purchased for Rs. 625000. Its value depreciates at the rate of 8% per annum. What will be its value after 2 years?
Answer:
Present value of machine, P = Rs.625000
Time, n = 2 years
Rate of depreciates, R = 8% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ Value = P (1 - R/100)n [∵ Rate decreases]
= 625000 (1 - 8/100)2
= 625000 (1 - 2/25)2
= 625000 (23/25)2
= 625000 × 729/625
= 1000 × 529
= 529000
∴ Value of machine after 2 years will be Rs.529000.
Question 28.
A scooter is bought at Rs. 56000. Its value depreciates at the rate of 10% per annum. What will be its value after 3 years?
Answer:
Present value of scooter, P = Rs.56000
Time, n = 3 years
Rate of depreciates, R = 10% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ Value = P (1 - R/100)n [∵ Rate decreases]
= 56000 (1 - 10/100)3
= 56000 (1 - 1/10)3
= 56000 (9/10)3
= 56000 × 729/1000
= 56 × 729
= 40824
∴ Value of scooter after 3 years will be Rs.40824.
Question 29.
A car is purchased for Rs. 348000. Its value depreciates at 10% per annum during the first year and at 20% per annum during the second year. What will be its value after 2 years?
Answer:
Present value of car, P = Rs.348000
Rate of depreciates for 1st year, p = 10%
Rate of depreciates for 2nd year, q = 20%
Time, n = 2 years
Now,
Value = P × (1 - p/100) × (1 - q/100)
= 348000 × (1 - 10/100) × (1 - 20/100)
= 348000 × (1 - 1/10) × (1 - 1/5)
= 348000 × 9/10 × 4/5
= 34800 × 9 × 4/5
= 6960 × 9 × 4
= 25056
∴ Value of the car after 2 years is Rs.25056.
Question 30.
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago. If its present value is Rs. 291600, for how much was it purchased?
Answer:
Let the 3 years ago machine value = P
Rate of depreciates, R = 10%
Time, n = 3 years
Now,
Value = P (1 + R/100)n [Where,
P = Present value
R = Annual interest rate
n = Time in years]
∴ Value = P (1 - 10/100)n [∵ Rate decreases]
⇒ 291600 = P (1 - 1/10)3
⇒ 291600 = P (9/10)3
⇒ 291600 = P × 729/1000
⇒ P = 291600 × 1000/729
⇒ P = 400 × 1000
⇒ P = 400000
∴ Initial value of machine is Rs.400000.
Exercise 11c
Question 1.Find the amount and the compound interest on Rs. 8000 for 1 year at 10% per annum, compounded half-yearly.
Answer:
Present value, P = Rs.8000
Interest rate, R = 10% per annum
Time, n = 1 years
∵ Compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 8000 [1 + (10/2)/100]2
⇒ A = 8000 [1 + 5/100]2
⇒ A = 8000 [1 + 1/20]2
⇒ A = 8000 [21/20]2
⇒ A = 8000 × 441/400
⇒ A = 20 × 441
⇒ A = 8820
∴ Amount = Rs.8820
∴ Compound interest = Rs.(8820 – 8000) [∵CI = A – P]
= Rs.820
Question 2.
Find the amount and the compound interest on Rs. 31250 for 1
% years at 8% per annum, compounded half-yearly.
Answer:
Present value, P = Rs.31250
Interest rate, R = 8% per annum
Time, n = (3/2) years
∵ Compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 31250 [1 + (8/2)/100]3 [2n = 2 × 3/2]
⇒ A = 31250 [1 + 4/100]3
⇒ A = 31250 [1 + 1/25]3
⇒ A = 31250 [26/25]3
⇒ A = 31250 × 17576/15625
⇒ A = 2 × 17576
⇒ A = 35152
∴ Amount = Rs.35152
∴ Compound interest = Rs.(35152 – 31250) [∵CI = A – P]
= Rs.3902
Question 3.
Find the amount and the compound interest on Rs. 12800 for 1 year at 7% per annum, compounded half-yearly.
Answer:
Present value, P = Rs.12800
Interest rate, R = (15/2)% per annum
Time, n = 1 years
∵ Compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 12800 [1 + (15/4)/100]2
⇒ A = 12800 [1 + 3/80]2
⇒ A = 12800 [83/80]2
⇒ A = 12800 × 6889/6400
⇒ A = 128 × 6889/64
⇒ A = 2 × 6889
⇒ A = 13778
∴ Amount = Rs.13778
∴ Compound interest = Rs.(13778 – 12800) [∵CI = A – P]
= Rs.978
Question 4.
Find the amount and the compound interest on Rs. 160000 for 2 years at 10% per annum, compounded half-yearly.
Answer:
Present value, P = Rs.160000
Interest rate, R = 10% per annum
Time, n = 2 years
∵ Compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 160000 [1 + (10/2)/100]4
⇒ A = 160000 [1 + 5/100]4
⇒ A = 160000 [1 + 1/20]4
⇒ A = 160000 [21/20]4
⇒ A = 160000 × 21/20 × 21/20 × 21/20 × 21/20
⇒ A = 160000 × 194481/160000
⇒ A = 1 × 194481
⇒ A = 194481
∴ Amount = Rs.8820
∴ Compound interest = Rs.(194481 – 160000) [∵CI = A – P]
= Rs.34481
Question 5.
Swati borrowed Rs. 40960 from a bank to buy a piece of land. If the bank charges 12% per annum, compounded half-yearly, what amount will she have to pay after 1years? Also, find the interest paid by her.
Answer:
Present value, P = Rs.40960
Interest rate, R = (25/2)% per annum
Time, n = 3/2 years
∵ Compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 40960 [1 + (25/4)/100]3 [R = 25/2 and n = 3/2 years]
⇒ A = 40960 [1 + 1/16]3
⇒ A = 40960 [17/16]3
⇒ A = 40960 × 4913/4096
⇒ A = 10 × 4913
⇒ A = 49130
∴ Amount = Rs.49130
∴ Compound interest = Rs.(49130 – 40960) [∵CI = A – P]
= Rs.8170
Question 6.
Mohd. Aslam purchased a house from Avas Vikas Parishad on credit. If the cost of the house is Rs. 125000 and the Parishad charges interest at 12% per annum compounded half-yearly, find the interest paid by Aslam after a year and a half.
Answer:
Initial value, P = Rs.125000
Interest rate, R = 12% per annum
Time, n = (1 + 1/2) years = 3/2 years
∵ Compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 125000 [1 + (12/2)/100]3 [n = 3/2 years]
⇒ A = 125000 [1 + 6/100]3
⇒ A = 125000 [1 + 3/50]3
⇒ A = 125000 [53/50]3
⇒ A = 125000 × 53/50 × 53/50 × 53/50
⇒ A = 125000 × 148877/125000
⇒ A = 1 × 148877
⇒ A = 148877
∴ Amount = Rs.148877
∴ Compound interest = Rs.(148877 – 125000) [∵CI = A – P]
= Rs.23877
∴ Rs.23877 interest paid by Aslam after (3/2) years.
Question 7.
Sheela deposited Rs. 20000 in a bank, where the interest is credited half-yearly. If the rate of interest paid by the bank is 6% per annum, what amount will she get after 1 year?
Answer:
Present value, P = Rs.20000
Interest rate, R = 6% per annum
Time, n = 1 years
∵ Compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 20000 [1 + (6/2)/100]2
⇒ A = 20000 [1 + 3/100]2
⇒ A = 20000 [103/100]2
⇒ A = 20000 × 103/100 × 103/100
⇒ A = 2 × 103 × 103
⇒ A = 21218
∴ Amount = Rs.21218
∴ Shella gets Rs.21218 after 1 year.
Question 8.
Neeraj lent Rs. 65536 for 2 years at 12% per annum, compounded annually. How much 2 more could he earn if the interest were compounded half-yearly?
Answer:
Initial value, P = Rs.65536
Interest rate, R = (25/2)% per annum
Time, n = 2 years
∵ Compounded annually.
∴ Amount (A) = P [1 + R/100]n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 65536 [1 +(25/2) /100]2
⇒ A = 65536 [1 + 1/8]2
⇒ A = 65536 [9/8]2
⇒ A = 65536 × 9/8 × 9/8
⇒ A = 65536 × 81/64
⇒ A = 1024 × 81
⇒ A = 82944
∴ Amount = Rs.82944
∴ Compound interest = Rs.(82944 – 65536) [∵CI = A – P]
= Rs.17408
Now,
∵ Compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 65536 [1 + (25/4)/100]4 [R = (25/2)% and n = 2 years]
⇒ A = 65536 [1 + 1/16]4
⇒ A = 65536 [17/16]4
⇒ A = 65536 × 17/16 × 17/16 × 17/16 × 17/16
⇒ A = 65536 × 83521/65536
⇒ A = 1 × 83521
⇒ A = 83521
∴ Amount = Rs.83521
∴ Compound interest = Rs.(83521 – 65536) [∵CI = A – P]
= Rs.17985
Now,
Difference between interests compound half-yearly and yearly,
= Rs.(17985 – 17408)
= Rs.577
Question 9.
Sudershan deposited Rs. 32000 in a bank, where the interest is credited quarterly. If the rate of interest be 5% per annum, what amount will he receive after 6 months?
Answer:
Present value, P = Rs.32000
Interest rate, R = 5% per annum
Time, n = 6 months = (1/2) years
∵ Compounded quarterly,
∴ Amount (A) = P [1 + (R/4)/100]4n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 362000 [1 + (5/4)/100]2 [4n = 4 × 1/2]
⇒ A = 32000 [1 + 1/80]2
⇒ A = 32000 [81/80]2
⇒ A = 32000 × 81/80 × 81/80
⇒ A = 400 × 81 × 81/80
⇒ A = 5 × 81 × 81
⇒ A = 32805
∴ Amount = Rs.32805
∴ Sudershan will receive amount of Rs.32805 after 6 months.
Question 10.
Arun took a loan of Rs. 390625 from Kuber Finance. If the company charges interest at 16% per annum, compounded quarterly, what amount will discharge his debt after one year?
Answer:
Present value, P = Rs.390625
Interest rate, R = 16% per annum
Time, n = 1 year
∵ Compounded quarterly,
∴ Amount (A) = P [1 + (R/4)/100]4n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 390625 [1 + (16/4)/100]4
⇒ A = 390625 [1 + 4/100]4
⇒ A = 390625 [1 + 1/25]4
⇒ A = 390625 [26/25]4
⇒ A = 390625 × 26/25 × 26/25 × 26/25 × 26/25
⇒ A = 390625 × 456976/390625
⇒ A = 1 × 456976
⇒ A = 456976
∴ Amount = Rs.456976
∴ Arun has to pay Rs.45976 after 1 year.
Cce Test Paper-11
Question 1.Find the amount and the compound interest on Rs. 3000 for 2 years at 10% per annum.
Answer:
Present value = Rs.3000
Interest rate = 10% per annum
Time = 2 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 3000 (1 + 10/100)2
⇒ A = 3000 (1 + 1/10)2
⇒ A = 3000 (11/10)2
⇒ A = 3000 × 11/10 × 11/10
⇒ A = 3000 × 121/100
⇒ A = 30 × 121
⇒ A = 3630
∴ Amount = Rs.3630
∴ Compound interest = Rs.(3630 – 3000)
= Rs.630
Question 2.
Find the amount of Rs. 10000 after 2 years compounded annually; the rate of interest being 10% per anum during the first year and 12% per annum during the second year. Also, find the compound interest.
Answer:
Present value = Rs.10000
Interest rate for 1st year, p = 10 % per annum
Interest rate for 2nd year, q = 12 % per annum
Time = 2 years
Amount (A) = P × (1 + p/100) × (1 + q/100)
A = 10000 × (1 + 10/100) × (1 + 12/100)
= 10000 × (1 + 1/10) × (112/100)
= 10000 × 11/10 × 112/100
= 10 × 11 × 112
= 12320
∴ Amount = Rs.12320
∴ Compound interest = Rs.(12320 – 10000)
= Rs.2320
Question 3.
Find the amount and the compound interest on Rs. 6000 for 1 year at 10% per annum compounded half-yearly.
Answer:
Present value = Rs.6000
Interest rate = 10 % per annum
Time = 1 years
∵ Interest is compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 6000 [1 + (10/2)/100]2
⇒ A = 6000 [1 + 5/100]2
⇒ A = 6000 [1 + 1/20]2
⇒ A = 6000 [21/20]2
⇒ A = 6000 × 21/20 × 21/20
⇒ A = 300 × 21 × 21/20
⇒ A = 15 × 21 × 21
⇒ A = 6615
∴ Amount = Rs.6615
∴ Compound interest = Rs.(6615 – 6000)
= Rs.615
Question 4.
A sum amounts to Rs. 23762 in 2 years at 9% per annum, compounded annually. Find the sum.
Answer:
Let sum = P
Amount (A) = Rs.23762
Interest rate = 9 % per annum
Time = 2 years
Now,
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = P(1 + 9/100)22
⇒ 23762 = P (109/100)2
⇒ 23762 = P × 11881/10000
⇒ P = 23762 × 10000/11881
⇒ P = 2 × 10000
⇒ P = 20000
∴ Sum = Rs.20000
Question 5.
A scooter is bought for Rs. 32000. Its value depreciates at 10% per annum. What will be its value after 2 years?
Answer:
Present value, P = Rs.32000
Time, n = 2 years
Rate of depreciates, R = 10% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ Value = P (1 - R/100)n [∵ Rate decreases]
= 32000 (1 - 10/100)2
= 32000 (1 - 1/10)2
= 32000 (9/10)2
= 32000 × 9/10 × 9/10
= 320 × 9 × 9
= 25920
∴ Value of scooter will be Rs.25920 after 2 years.
Question 6.
The compound interest on Rs. 5000 at 10% per annum for 2 years is
A. Rs. 550
B. Rs. 1050
C. Rs. 950
D. Rs. 825
Answer:
Present value = Rs.5000
Interest rate = 10% per annum
Time = 2 years
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 5000 (1 + 10/100)2
⇒ A = 5000 (1 + 1/10)2
⇒ A = 5000 (11/10)2
⇒ A = 5000 × 11/10 × 11/10
⇒ A = 5000 × 121/100
⇒ A = 50 × 121
⇒ A = 6050
∴ Amount = Rs.6050
∴ Compound interest = Rs.(6050 – 5000)
= Rs.1050
Question 7.
The annual rate of growth in population of a town is 5%. If its present population is 4000, what will be its population after 2 years?
A. Rs. 4441
B. Rs. 4400
C. Rs. 4410
D. Rs. 4800
Answer:
Population of a town, P = 4000
Time, n = 2 years
Increasing rate, R = 5% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ Population = P (1 + R/100)n
= 4000 (1 + 5/100)2
= 4000 (1 + 1/20)2
= 4000 (21/20)2
= 4000 × 21/20 × 21/20
= 200 × 21 × 21/20
= 10 × 21 × 21
= 4410
∴ Population of a town after 2 years is 4410.
Question 8.
At what rate per cent per annum will 5000 amount to Rs. 5832 in 2 years, compounded annually?
A. 11%
B. 10%
C. 9%
D. 8%
Answer:
Present value, P = Rs.5000
Amount, A = Rs.5832
Time, n = 2 years
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ Amount (A) = P (1 + R/100)n
⇒ 5832 = 5000 (1 + R/100)2
⇒ (1 + R/100)2 = 5832/5000
⇒ (1 + R/100)2 = 2916/2500
⇒ (1 + R/100)2 = (54/50)2
⇒ 1 + R/100 = 54/50
⇒ R/100 = (54/50) - 1
⇒ R/100 = (54 – 50)/50
⇒ R/100 = 4/50
⇒ R = 400/50
⇒ R = 8
∴ Rate = 8 %.
Question 9.
If the simple interest on a sum of money at 10% per annum for 3 years is Rs. 1500, then the compound interest on the same sum at the same rate for the same period is
A. Rs. 1655
B. Rs. 1155
C. Rs. 1555
D. Rs. 1855
Answer:
Simple interest = Rs.1500
Interest rate = 10% per annum
Time = 3 years
Simple interest (SI) = PRT/100 [where, P = Present value
R = Interest rate
∴ 1500 = (P × 10 × 3)/100 T = Time]
⇒ 1500 = P × 30/100
⇒ 1500 = P × 3/10
⇒ P = 1500 × 10/3
⇒ P = 500 × 10
⇒ P = 5000
∴ Sum = Rs.5000
Now,
Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 5000 [1 + 10/100]3
⇒ A = 5000 [1 + 1/10]3
⇒ A = 5000 [11/10]3
⇒ A = 5000 × 11/10 × 11/10 × 11/10
⇒ A = 5000 × 1331/1000
⇒ A = 5 × 1331
⇒ A = 6655
∴ Amount = Rs.6655
∴ Compound interest = Rs.(6655 – 5000)
= Rs.1655
Question 10.
If the compound interest on a certain sum for 2 years at 10% per annum is Rs. 1050, the sum is
A. Rs. 3000
B. Rs. 4000
C. Rs. 5000
D. Rs. 6000
Answer:
Compound interest, CI = Rs.1050
Interest rate, R = 10% per annum
Time = 2 years
∴ CI = P (1 + R/100)n – P
⇒ 1050 = P (1 + 10/100)2 – P
⇒ 1050 = P (1 + 1/10)2 – P
⇒ 1050 = P (11/10)2 – P
⇒ 1050 = 121P/100 – P
⇒ 1050 = (121P – 100P)/100
⇒ 1050 = 21P/100
⇒ P = 1050 × 100/21
⇒ P = 50 × 100
⇒ P = 5000
∴ Sum = Rs.5000
Question 11.
Fill in the blanks:
(i) 
(ii) (Amount) - (Principal) = ……….
(iii) If the value of a machine is Rs. P and it depreciates at R% per annum, then its value after 2 years is
(iv) If the population P of a town increases at R% per annum, then its population after 5 years is ………..
Answer:
(i) R
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
(ii) Compound interest
Amount (A) = P (1 + R/100)n
And Compound interest = P (1 + R/100)n - P
(iii) Rs. 
∵ Rate decreases.
∴ Value = P (1 - R/100)n
Present value of machine = Rs.P
Interest rate = R% per annum
Time, n = 2
∴ Value = Rs.
(iv) 
Present population of a town = P
Increases rate = R% per annum
Time, n = 5 years
∴ Population = P (1 + R/100)n [Where,
P = Present value
R = Annual interest rate
n = Time in years]
∴ Population after 5 years =
Exercise 11d
Question 1.The compound interest on Z 5000 at 8% per annum for 2 years, compounded annually, is
A. Rs. 800
B. Rs. 825
C. Rs. 832
D. Rs.850
Answer:
Present value, P = Rs.5000
Interest rate, R = 8% per annum
Time, n = 2 years
∴ Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 5000 (1 + 8/100)2
⇒ A = 5000 (1 + 2/25)2
⇒ A = 5000 (27/25)2
⇒ A = 5000 × 27/25 × 27/25
⇒ A = 200 × 27 × 27/25
⇒ A = 8 × 27 × 27
⇒ A = 5832
∴ Amount = Rs.5832
∴ Compound interest = Rs.(5832 – 5000) [∵CI = A – P]
= Rs.832
Question 2.
The compound interest on Z 10000 at 10% per annum for 3 years, compounded annually, is
A. Rs. 1331
B. Rs.3310
C. Rs.3130
D. Rs. 13310
Answer:
Present value, P = Rs.10000
Interest rate, R = 8% per annum
Time, n = 3 years
∴ Amount (A) = P (1 + R/100)n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 10000 (1 + 10/100)3
⇒ A = 10000 (1 + 1/10)3
⇒ A = 10000 (11/10)3
⇒ A = 10000 × 11/10 × 11/10 × 11/10
⇒ A = 10 × 11 × 11 × 11
⇒ A = 13310
∴ Amount = Rs.13310
∴ Compound interest = Rs.(13310 – 10000) [∵CI = A – P]
= Rs.3310
Question 3.
The compound interest on Z 10000 at 12% per annum for 1years, compounded annually, is
A. Rs. 1872
B. Rs. 1720
C. Rs. 1910.16
D. Rs. 1782
Answer:
Present value, P = Rs.10000
Interest rate, R = 12% per annum
Time, n = 1
years
∴ Amount (A) = P (1 + R/100)n × [1 + (R/2)/100] [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 10000 (1 + 12/100)1 × [1 + (12/2)/100]
⇒ A = 10000 (1 + 12/100) × [1 + 6/100]
⇒ A = 10000 (112/100) × [106/100]
⇒ A = 10000 × 112/100 × 106/100
⇒ A = 1 × 112 × 106
⇒ A = 11872
∴ Amount = Rs.11872
∴ Compound interest = Rs.(11872 – 10000) [∵CI = A – P]
= Rs.1872
Question 4.
The compound interest on Rs. 4000 at 10% per annum for 2 years 3 months, compounded annually, is
A. Rs. 916
B. Rs. 900
C. Rs. 961
D. Rs. 896
Answer:
Present value, P = Rs.4000
Interest rate, R = 10% per annum
Time, n =2 years 3 months = (2 + 1/4) years
∴ Amount (A) = P (1 + R/100)n × [1 + (R/4)/100] [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 4000 (1 + 10/100)2 × [1 + (10/4)/100]
⇒ A = 4000 (1 + 1/10)2 × [1 + 1/40]
⇒ A = 4000 (11/10)2 × [41/40]
⇒ A = 4000 × 121/100 × 41/40
⇒ A = 40 × 121 × 41/40
⇒ A = 121 × 41
⇒ A = 4961
∴ Amount = Rs.4961
∴ Compound interest = Rs.(4961 – 4000) [∵CI = A – P]
= Rs.961
Question 5.
A sum of Rs. 25000 was given as loan on compound interest for 3 years compounded annually at 5% per annum during the first year, 6% per annum during the second year and 8% per annum during the third year. The compound interest is
A. Rs. 5035
B. Rs. 5051
C. Rs. 5072
D. Rs. 5150
Answer:
Sum, P = 25000
Interest rate for 1st year, p = 5%
Interest rate for 2nd year, q = 6%
Interest rate for 3rd year, r = 8%
Time, n = 3 years
Now,
Amount (A) = P × (1 + p/100) × (1 + q/100) × (1 + r/100)
= 25000 × (1 + 5/100) × (1 + 6/100) × (1 + 8/100)
= 25000 × (1 + 1/20) × (1 + 3/50) × (1 + 2/25)
= 25000 × 21/20 × 53/50 × 27/25
= 250 × 21/2 × 53/5 × 27/25
= 10 × 21/2 × 53/5 × 27
= 1 × 21 × 53 × 27
= 30051
∴ Compound interest = Rs.(30051 – 25000) [∵CI = A – P]
= Rs.50051
Question 6.
The compound interest on Rs. 6250 at 8% per annum for 1 year, compounded half yearly, is
A. Rs. 500
B. Rs. 510
C. Rs. 550
D. Rs. 512.50
Answer:
Initial value, P = Rs.6250
Interest rate, R = 8% per annum
Time, n = 1 years
∵ Compounded half-yearly.
∴ Amount (A) = P [1 + (R/2)/100]2n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 6250 [1 +(8/2) /100]2
⇒ A = 6250 [1 + 4/100]2
⇒ A = 6250 [1 + 1/25]2
⇒ A = 6250 [26/25]2
⇒ A = 6250 × 26/25 × 26/25
⇒ A = 6250 × 26/25 × 26/25
⇒ A = 250 × 26 × 26/25
⇒ A = 10 × 26 × 26
⇒ A = 6760
∴ Amount = Rs.6760
∴ Compound interest = Rs.(6760 – 6250) [∵CI = A – P]
= Rs.510
Question 7.
The compound interest on Rs. 40000 at 6% per annum for 6 months, compounded quarterly, is
A. Rs. 1209
B. Rs. 1902
C. Rs. 1200
D. Rs. 1306
Answer:
Present value, P = Rs.40000
Interest rate, R = 6% per annum
Time, n =6 months = 1/2 years
∵ Compounded quarterly.
∴ Amount (A) = P [1 + (R/4)/100]4n [Where, P = Present value
R = Annual interest rate
n = Time in years]
∴ A = 40000 [1 +(6/4) /100]2 [4n = 4 × 1/2]
⇒ A = 40000 [1 + 3/200]2
⇒ A = 40000 [1 + 3/200]2
⇒ A = 40000 [203/200]2
⇒ A = 40000 × 203/200 × 203/200
⇒ A = 40000 × 203/200 × 203/200
⇒ A = 200 × 203 × 203/200
⇒ A = 1 × 203 × 203
⇒ A = 41209
∴ Amount = Rs.41209
∴ Compound interest = Rs.(41209 – 40000) [∵CI = A – P]
= Rs.1209
Question 8.
The present population of a town is 24000. If it increases at the rate of 5% per annum, what will be its population after 2 years?
A. 26400
B. 26460
C. 24460
D. 26640
Answer:
Population of a town, P = 24000
Time, n = 2 years
Increasing rate, R = 5% per annum
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ Population = P (1 + R/100)n
= 24000 (1 + 5/100)2
= 24000 (1 + 1/20)2
= 24000 (21/20)2
= 24000 × 21/20 × 21/20
= 240 × 21/2 × 21/2
= 60 × 21 × 21
= 26460
∴ Population of a town is 26460 after 2 years.
Question 9.
The value of a machine depreciates at the rate of 10% per annum. It was purchased 3 years ago for Rs. 60000. What is the present value of the machine?
A. Rs. 53640
B. Rs. 51680
C. Rs. 43740
D. Rs. 43470
Answer:
Value of a machine 3 years ago, P = Rs.60000
Time, n = 3 years
Rate of depreciates, R = 10% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ Value = P (1 - R/100)n [∵ Rate decreases]
= 60000 (1 - 10/100)3
= 60000 (1 - 1/10)3
= 60000 (9/10)3
= 60000 × 9/10 × 9/10 × 9/10
= 60 × 9 × 9 × 9
= 43740
∴ Present value of the machine is Rs.43740.
Question 10.
The value of a machine depreciates at the rate of 20% per annum. It was purchased 2 years ago. If its present value is Rs. 40000 for how much was it purchased?
A. Rs. 56000
B. Rs. 62500
C. Rs. 65200
D. Rs. 56500
Answer:
Let value of a machine 2 years ago, = P
Present value of machine = Rs.40000
Time, n = 2 years
Rate of depreciates, R = 20% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ Value = P (1 - R/100)n [∵ Rate decreases]
⇒ 40000 = P (1 - 20/100)2
⇒ 40000 = P (1 - 1/5)2
⇒ 40000 = P (4/5)2
⇒ 40000 = P × 16/25
⇒ P = 40000 × 25/16
⇒ P = 2500 × 25
⇒ P = 62500
∴ Value of a machine 2 years ago is Rs.62500.
Question 11.
The annual rate of growth in population of a town is 10%. If its present population is 33275, what was it 3 years ago?
A. Rs. 25000
B. Rs. 27500
C. Rs. 30000
D. Rs. 26000
Answer:
Let 3 years ago population = P
Present population = 33275
Time, n = 3 years
Increases rate, R = 10% per annum
Now,
Amount (A) = P (1 + R/100)n [Where, A = Amount with compound interest
P = Present value
R = Annual interest rate
n = Time in years]
∴ Population = P (1 + R/100)n
⇒ 33275 = P (1 + 10/100)3
⇒ 33275 = P (1 + 1/10)3
⇒ 33275 = P (11/10)3
⇒ 33275 = P × 1331/1000
⇒ P = 33275 × 1000/1331
⇒ P = 25 × 1000
⇒ P = 25000
∴ 3 years ago population is 25000.
Question 12.
If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200 then the compound interest on the same sum for the same period at the same rate will be
A. Rs. 1225
B. Rs. 1236
C. Rs. 1248
D. Rs. 1261
Answer:
Interest rate, R = 5% per annum
Time = 3 years
Simple interest = Rs.1200
Simple interest = PRT/100
⇒ 1200 = (P × 5 × 3)/100
⇒ 1200 = P × 15/100
⇒ P = 1200 × 100/15
⇒ P = 8000
Now,
Amount (A) = P (1 + R/100)n
= 8000 (1 + 5/100)3
= 8000 (1 + 1/20)3
= 8000 (21/20)3
= 8000 × 9261/8000
= 9261
∴ Amount = 9261
∴ Compound interest = Rs.(9261 – 8000) [∵CI = A – P]
= Rs.1261
Question 13.
If the compound interest on a sum for 2 years at 12% per annum is Rs. 510, the simple interest on the same sum at the same rate for the same period of time is
A. Rs. 400
B. Rs. 450
C. Rs. 460
D. Rs. 480
Answer:
Compound interest, CI = Rs.510
Interest rate, R = 12% = 25/2% per annum
Time = 2 years
CI = P (1 + R/100)n – P
⇒ 510 = P (1 + (25/2)/100)2 – P
⇒ 510 = P (1 + 1/8)2 – P
⇒ 510 = P (9/8)2 – P
⇒ 510 = 81P/64 – P
⇒ 510 = (81P – 64P)/64
⇒ 510 = 17P/64
⇒ P = 510 × 64/17
⇒ P = 30 × 64
⇒ P = 1920
Now,
SI = PRT/100
= (1920 × 25/2 × 2)/100
= (1920 × 25)/100
= 480
∴ Simple interest = Rs.480
Question 14.
The sum that amounts to Rs.4913 in 3 years at (25/4)% per annum compounded annually, is
A. Rs. 3096
B. Rs. 4076
C. Rs. 4085
D. Rs. 4096
Answer:
Amount, A = Rs.4913
Interest rate, R = (25/4)% per annum
Time = 3 years
Amount (A) = P (1 + R/100)n
⇒ 4913 = P (1 + (25/4)/100)3
⇒ 4913 = P (1 + 1/16)3
⇒ 4913 = P (17/16)3
⇒ 4913 = P × 4913/4096
⇒ P = 4913 × 4096/4913
⇒ P = 4096
∴ Sum = Rs.4096
Question 15.
At what rate per cent per annum will a sum of Rs. 7500 amount to Rs. 8427 in 2 years, compounded annually?
A. 4%
B. 5%
C. 6%
D. 8%
Answer:
Present value, P = Rs.7500
Amount, A = Rs.8427
Time, n = 2 years
Now,
Amount (A) = P (1 + R/100)n
⇒ 8427 = 7500 (1 + R/100)2
⇒ (1 + R/100)2 = 8427/7500
⇒ (1 + R/100)2 = (53/50)2
⇒ (1 + R/100) = (53/50)
⇒ R/100 = 53/50 – 1
⇒ R/100 = (53 – 50)/50
⇒ R = 3/50 × 100
⇒ R = 6
∴ Rate = 6%