Let Us Work Out 6.1
1. I have ₹5000 in my hand. I deposited that money in a bank at the rate of 8.5% compound interest per annum for two years. Let us write by calculating how much money I shall get at the end of 2 years.
Solution :
Principal (P) = ₹5000
Rate (R) = 8.5%
Time (n) = 2 years
Amount = P
(
1 +
R
100
)
n
= 5000
(
1 +
8.5
100
)
²
= 5000
(
108.5
100
)
²
= 5000 × 1.177225
= ₹5886.13
∴ Required amount = ₹5886.13
2. Let us calculate the amount on ₹5000 at the rate of 8% compound interest per annum for 3 years.
Solution :
Principal (P) = ₹5000
Rate (R) = 8%
Time (n) = 3 years
Amount = P
(
1 +
R
100
)
n
= 5000
(
1 +
8
100
)
³
= 5000
(
108
100
)
³
= 5000 × 1.259712
= ₹6298.56
∴ Required amount = ₹6298.56
3. Goutam babu borrowed ₹2000 at the rate of 6% compound interest per annum for 2 years. Let us write by calculating how much compound interest at the end of 2 years he will pay.
Solution :
Principal (P) = ₹2000
Rate (R) = 6%
Time (n) = 2 years
Amount = 2000
(
1 +
6
100
)
²
= 2000
(
106
100
)
²
= 2000 × 1.1236
= ₹2247.20
Compound Interest = Amount − Principal
= 2247.20 − 2000
= ₹247.20
∴ Required compound interest = ₹247.20
4. Let us write by calculating compound interest on ₹30000 at the rate of 9% compound interest per annum for 3 years.
Solution :
Principal (P) = ₹30000
Rate (R) = 9%
Time (n) = 3 years
Amount = 30000
(
1 +
9
100
)
³
= 30000
(
109
100
)
³
= 30000 × 1.295029
= ₹38850.87
Compound Interest = 38850.87 − 30000
= ₹8850.87
∴ Required compound interest = ₹8850.87
5. Let us write by calculating the amount on ₹80000 for 2½ years at the rate of 5% compound interest per annum.
Solution :
Principal (P) = ₹80000
Rate (R) = 5%
Time = 2½ years
Amount after 2 years
= 80000
(
1 +
5
100
)
²
= 80000 × 1.1025
= ₹88200
Interest for next 6 months :
=
88200 × 5 × 1
100 × 2
= ₹2205
Total Amount = 88200 + 2205
= ₹90405
∴ Required amount = ₹90405
6. Chandadevi borrowed some money for 2 years at the rate of 8% compound interest per annum. If the compound interest is ₹2496, then how much money she had borrowed.
Solution :
Rate (R) = 8%
Time (n) = 2 years
Compound Interest = ₹2496
Amount = P
(
1 +
8
100
)
²
= P × 1.1664
Compound Interest = 1.1664P − P
2496 = 0.1664P
P =
2496
0.1664
= ₹15000
∴ Required principal = ₹15000
7. Let us write by calculating the principal whose compound interest becomes ₹2648 after getting 10% compound interest per annum for 3 years.
Solution :
Rate (R) = 10%
Time (n) = 3 years
Compound Interest = ₹2648
Amount = P
(
1 +
10
100
)
³
= P × 1.331
2648 = 1.331P − P
2648 = 0.331P
P =
2648
0.331
= ₹8000
∴ Required principal = ₹8000
8. Rahaman chacha deposited some money in a cooperative bank at the rate of 9% compound interest per annum and he received amount ₹29702.50 after 2 years. Let us calculate how much money Rahaman chacha had deposited in cooperative bank.
Solution :
Amount (A) = ₹29702.50
Rate (R) = 9%
Time (n) = 2 years
A = P
(
1 +
9
100
)
²
29702.50 = P × 1.1881
P =
29702.50
1.1881
= ₹25000
∴ Required deposited money = ₹25000
9. Let us write by calculating what sum of money at the rate of 8% compound interest per annum for 3 years will amount to ₹31492.80.
Solution :
Amount (A) = ₹31492.80
Rate (R) = 8%
Time (n) = 3 years
A = P
(
1 +
8
100
)
³
31492.80 = P × 1.259712
P =
31492.80
1.259712
= ₹25000
∴ Required principal = ₹25000
10. Let us calculate the difference between compound interest and simple interest on ₹12000 for 2 years at 7.5% per annum.
Solution :
Principal (P) = ₹12000
Rate (R) = 7.5%
Time = 2 years
Simple Interest =
12000 × 7.5 × 2
100
= ₹1800
Amount at compound interest
= 12000
(
1 +
7.5
100
)
²
= 12000 × 1.155625
= ₹13867.50
Compound Interest = 13867.50 − 12000
= ₹1867.50
Difference = 1867.50 − 1800
= ₹67.50
∴ Required difference = ₹67.50
11. Let us write by calculating the difference between compound interest and simple interest on ₹10000 for 3 years at 5% per annum.
Solution :
Principal (P) = ₹10000
Rate (R) = 5%
Time = 3 years
Simple Interest =
10000 × 5 × 3
100
= ₹1500
Amount at compound interest
= 10000
(
1 +
5
100
)
³
= 10000 × 1.157625
= ₹11576.25
Compound Interest = 11576.25 − 10000
= ₹1576.25
Difference = 1576.25 − 1500
= ₹76.25
∴ Required difference = ₹76.25
12. Let us write by calculating the sum of money, if the difference between compound interest and simple interest for 2 years at 9% per annum is ₹129.60.
Solution :
Difference between C.I. and S.I. for 2 years
=
P
(
R
100
)
²
129.60 =
P
(
9
100
)
²
129.60 =
P ×
81
10000
P =
129.60 × 10000
81
= ₹16000
∴ Required sum = ₹16000
13. Let us write by calculating the sum of money if the difference between compound interest and simple interest for 3 years becomes ₹930 at the rate of 10% per annum.
Solution :
Difference between C.I. and S.I. for 3 years
=
P
[
(
1 +
10
100
)
³
− 1 −
10 × 3
100
]
930 = P(1.331 − 1 − 0.3)
930 = P × 0.031
P =
930
0.031
= ₹30000
∴ Required principal = ₹30000
14. If the rates of compound interest for the first and the second year are 7% and 8% respectively, let us write by calculating compound interest on ₹6000 for 2 years.
Solution :
Principal (P) = ₹6000
Amount after first year
= 6000 ×
(
1 +
7
100
)
= 6000 × 1.07
= ₹6420
Amount after second year
= 6420 ×
(
1 +
8
100
)
= 6420 × 1.08
= ₹6933.60
Compound Interest = 6933.60 − 6000
= ₹933.60
∴ Required compound interest = ₹933.60
15. If the rates of compound interest for the first and second year are 5% and 6% respectively, let us calculate the compound interest on ₹5000 for 2 years.
Solution :
Principal (P) = ₹5000
Amount after first year
= 5000 × 1.05
= ₹5250
Amount after second year
= 5250 × 1.06
= ₹5565
Compound Interest = 5565 − 5000
= ₹565
∴ Required compound interest = ₹565
16. If simple interest of a certain sum of money for 1 year is ₹50 and compound interest for 2 years is ₹102, let us write by calculating the sum of money and the rate of interest.
Solution :
Simple interest for 1 year = ₹50
PR
100
= 50
PR = 5000
Compound interest for 2 years
= 2 × 50 +
P
(
R
100
)
²
102 = 100 +
P
(
R
100
)
²
2 =
P
(
R
100
)
²
2 =
5000R
10000
R = 4%
P =
5000
4
= ₹1250
∴ Required principal = ₹1250
∴ Required rate = 4%
17. If simple interest and compound interest of a certain sum of money for two years are ₹8400 and ₹8652 respectively, then let us write by calculating the sum of money and the rate of interest.
Solution :
Difference = 8652 − 8400
= ₹252
252 =
P
(
R
100
)
²
Simple Interest for 2 years = ₹8400
PR
2
100
= 8400
PR = 420000
252 =
420000R
10000 × 100
252 × 1000000 = 420000R
R = 6%
P =
420000
6
= ₹70000
∴ Required principal = ₹70000
∴ Required rate = 6%
18. Let us calculate compound interest on ₹6000 for 1 year at the rate of 8% compound interest per annum compounded at the interval of 6 months.
Solution :
Principal (P) = ₹6000
Rate for 6 months = 4%
Number of intervals = 2
Amount = 6000
(
1 +
4
100
)
²
= 6000 × 1.0816
= ₹6489.60
Compound Interest = 6489.60 − 6000
= ₹489.60
∴ Required compound interest = ₹489.60
19. Let us write by calculating compound interest on ₹6250 for 9 months at the rate of 10% compound interest per annum compounded at the interval of 3 months.
Solution :
Principal (P) = ₹6250
Rate for 3 months = 2.5%
Number of intervals = 3
Amount = 6250
(
1 +
2.5
100
)
³
= 6250 × 1.076890625
= ₹6730.57
Compound Interest = 6730.57 − 6250
= ₹480.57
∴ Required compound interest = ₹480.57
20. Let us write by calculating at what rate of interest per annum will ₹6000 amount to ₹6984 in 2 years.
Solution :
Principal (P) = ₹6000
Amount (A) = ₹6984
Time = 2 years
6984 = 6000
(
1 +
R
100
)
²
1.164 =
(
1 +
R
100
)
²
√1.164 = 1 +
R
100
1.08 = 1 +
R
100
R = 8%
∴ Required rate = 8%
21. Let us calculate in how many years will ₹40000 amount to ₹46656 at the rate of 8% compound interest per annum.
Solution :
Principal (P) = ₹40000
Amount (A) = ₹46656
Rate (R) = 8%
A = P
(
1 +
R
100
)
ⁿ
46656 = 40000
(
1 +
8
100
)
ⁿ
46656 = 40000 ×
(
108
100
)
ⁿ
46656 = 40000 ×
(
27
25
)
ⁿ
46656
40000
=
(
27
25
)
ⁿ
1.1664 = (1.08)ⁿ
(1.08)² = 1.1664
n = 2
∴ Required time = 2 years
22. Let us write by calculating at what rate of compound interest per annum, the amount on ₹10000 for 2 years is ₹12100.
Solution :
Principal (P) = ₹10000
Amount (A) = ₹12100
Time (n) = 2 years
12100 = 10000
(
1 +
R
100
)
²
1.21 =
(
1 +
R
100
)
²
√1.21 =
1 +
R
100
1.1 =
1 +
R
100
R
100
= 0.1
R = 10%
∴ Required rate = 10% per annum
23. Let us calculate in how many years will ₹50000 amount to ₹60500 at the rate of 10% compound interest per annum.
Solution :
Principal (P) = ₹50000
Amount (A) = ₹60500
Rate (R) = 10%
60500 = 50000
(
1 +
10
100
)
ⁿ
60500 = 50000 × (1.1)ⁿ
60500
50000
= (1.1)ⁿ
1.21 = (1.1)ⁿ
(1.1)² = 1.21
n = 2
∴ Required time = 2 years
24. Let us write by calculating in how many years will ₹300000 amount to ₹399300 at the rate of 10% compound interest per annum.
Solution :
Principal (P) = ₹300000
Amount (A) = ₹399300
Rate (R) = 10%
399300 = 300000 × (1.1)ⁿ
399300
300000
= (1.1)ⁿ
1.331 = (1.1)ⁿ
(1.1)³ = 1.331
n = 3
∴ Required time = 3 years
25. Let us calculate the compound interest and amount on ₹1600 for 1½ years at the rate of 10% compound interest per annum compounded at the interval of 6 months.
Solution :
Principal (P) = ₹1600
Rate of interest per annum = 10%
Interest for 6 months = 5%
Time = 1½ years = 3 half-years
Amount = 1600
(
1 +
5
100
)
³
= 1600 × (1.05)³
= 1600 × 1.157625
= ₹1852.20
Compound Interest = 1852.20 − 1600
= ₹252.20
∴ Required amount = ₹1852.20
∴ Required compound interest = ₹252.20
Let Us Work Out 6.2
1. At present the population of village of Pahalanpur is 10000; if population is being increased at the rate of 3% every year, let us write by calculating its population after 2 years.
Solution :
Present population of the village = 10000
Rate of increase per year = 3%
Time = 2 years
We know,
A = P
\(
1 +
R
100
\)
ⁿ
= 10000
\(
1 +
3
100
\)
²
= 10000
\(
1 +
0.03
\)
²
= 10000 × (1.03)²
= 10000 × 1.0609
= 10609
∴ Population after 2 years = 10609
2. Rate of increase in population of a state is 2% in a year. The present population is 80000000; let us calculate the population of the state after 3 years.
Solution :
Present population = 80000000
Rate of increase per year = 2%
Time = 3 years
Using compound growth formula,
A = P
\(
1 +
R
100
\)
ⁿ
= 80000000
\(
1 +
2
100
\)
³
= 80000000 × (1.02)³
= 80000000 × 1.061208
= 84896640
∴ Population after 3 years = 84896640
3. The price of a machine in a leather factory depreciates at the rate of 10% every year. If the present price of the machine be 100000, let us calculate what will be the price of that machine after 3 years.
Solution :
Present price of the machine = ₹100000
Rate of depreciation = 10%
Time = 3 years
Since the price decreases every year,
A = P
\(
1 -
R
100
\)
ⁿ
= 100000
\(
1 -
10
100
\)
³
= 100000 × (0.9)³
= 100000 × 0.729
= 72900
∴ Price of the machine after 3 years = ₹72900
4. As a result of Sarva Shiksha Abhiyan, the students leaving the school before completion, are readmitted, so the students in a year is increased by 5% in comparision to its previous year. If the number of such readmitted students in a district be 3528 in the present year. Let us write by calculating, the number of students readmitted 2 years before in this manner.
Solution :
Present number of readmitted students = 3528
Rate of increase = 5%
Time = 2 years
We know,
A = P
\(
1 +
R
100
\)
ⁿ
3528 = P(1.05)²
3528 = P × 1.1025
P =
3528
1.1025
= 3200
∴ Number of students 2 years before = 3200
5. Through the publicity of road-safety programme, the street accidents in Purulia district are decreased by 10% in comparison to its previous year. If the number of street accidents in this year be 8748, let us write by calulating, the number of street accidents 3 years before in the district.
Solution :
Present number of accidents = 8748
Rate of decrease = 10%
Time = 3 years
A = P(1 - R/100)ⁿ
8748 = P(0.9)³
8748 = P × 0.729
P =
8748
0.729
= 12000
∴ Number of accidents 3 years before = 12000
6. A cooperative society of fisherman implemented such an improved plan for the production of fishes that the production in a year will be increased 10% in comparison to its previous year. In the present year if the cooperative society can produce 406 quintals of fishes, let us write by calculating, what will be the production of fishes after 3 years.
Solution :
Present production = 406 quintals
Rate of increase = 10%
Time = 3 years
A = P(1 + R/100)ⁿ
= 406(1.1)³
= 406 × 1.331
= 540.386
≈ 540.39 quintals
∴ Production after 3 years = 540.39 quintals
7. The height of tree increases at the rate of 20% every year. If the present height of tree is 28.8 metre, let us calculate the height of tree 2 years before.
Solution :
Present height of the tree = 28.8 metre
Rate of increase = 20%
Time = 2 years
A = P(1 + R/100)ⁿ
28.8 = P(1.2)²
28.8 = P × 1.44
P =
28.8
1.44
= 20
∴ Height of the tree 2 years before = 20 metre
8. Three years before from today a family had planned to reduce the experditure of electric bill by 5% in comparison to its previous year. 3 years ago, that family had to spend 4000 in a year for electric bill. Let us write by calculating how much amount will the family have to spend to pay the electric bill in the present year.
Solution :
Electric bill 3 years ago = ₹4000
Rate of decrease = 5%
Time = 3 years
A = P(1 - R/100)ⁿ
= 4000(0.95)³
= 4000 × 0.857375
= 3429.50
∴ Present electric bill = ₹3429.50
9. The weight of Savan babu is 80kg. In order to reduces his weight, he started regular morning walk. He decided to reduce his weight every year by 10%. Let us write by calculating, his weight after 3 years.
Solution :
Present weight = 80 kg
Rate of decrease = 10%
Time = 3 years
A = P(1 - R/100)ⁿ
= 80(0.9)³
= 80 × 0.729
= 58.32
∴ Weight after 3 years = 58.32 kg
10. At present the sum of the number of students in all M.S.K in a district is 3993. If the of students increased in a year was 10% of its previous year, let us calculate the sum of the number of students 3 years before in all the M.S.K in the district.
Solution :
Present number of students = 3993
Rate of increase per year = 10%
Time = 3 years
We know,
A = P
\(
1 +
R
100
\)
ⁿ
3993 = P(1.1)³
3993 = P × 1.331
P =
3993
1.331
= 3000
∴ Number of students 3 years before = 3000
11. As the farmers are becoming more alert of the harmful effects of using the chemical fertiliserns and insecticides in agricultural lands, the number of farmers using fertilisers and insectisides in the Rasulpur village decreases by 20% in a year in comparison to its previous year. Three years ago, the number of such farmers was 3000, let us calculate the number of such farmers in that village now.
Solution :
Initial number of farmers = 3000
Rate of decrease per year = 20%
Time = 3 years
We know,
A = P
\(
1 -
R
100
\)
ⁿ
= 3000(1 - 0.20)³
= 3000(0.8)³
= 3000 × 0.512
= 1536
∴ Present number of farmers = 1536
12. The price of a machine of a factory is 180000. The price of that machine decreases by 10% in each year. Let us calculate its price after 3 years.
Solution :
Present price of the machine = ₹180000
Rate of depreciation = 10%
Time = 3 years
A = P(1 - R/100)ⁿ
= 180000(0.9)³
= 180000 × 0.729
= 131220
∴ Price after 3 years = ₹131220
13. For the families having no electricity in their house, a Panchayat samity of Bakultala village accepted a plan to offer electric connections. 1200 families in this village have no electric connection in their house. In comparison to its previous year, it is possible to arrange electricity every year for 75% of the families having no electricity, let us write by calculating, the number of families without electricity after 2 years.
Solution :
Present number of families without electricity = 1200
Rate of decrease per year = 75%
Time = 2 years
A = P(1 - R/100)ⁿ
= 1200(1 - 0.75)²
= 1200(0.25)²
= 1200 × 0.0625
= 75
∴ Number of families without electricity after 2 years = 75
14. As a result of continuous publicity about harmful reactions in the use of cold drinks, the number of users of cold drinks has decreased by 25% every year in comparison to its privious year. 3 ago the before number of users of cold drink in a town was 80000. Let us write by calculating, the number of users of cold drink in the present year.
Solution :
Number of users 3 years ago = 80000
Rate of decrease per year = 25%
Time = 3 years
A = P(1 - R/100)ⁿ
= 80000(0.75)³
= 80000 × 0.421875
= 33750
∴ Present number of users = 33750
15. As a result of the non smoking campaign, the number of smoker has decreased by 67% every year in comparison to its previous year. If the number of smokers at present in a city is 33750, let us write by calculating, the number of smokers in that city 3 years before.
Solution :
Present number of smokers = 33750
Rate of decrease per year = 67%
Time = 3 years
A = P(1 - R/100)ⁿ
33750 = P(0.33)³
33750 = P × 0.035937
P =
33750
0.035937
≈ 939443
∴ Number of smokers 3 years before ≈ 939443
16. Very short answer type questions (V.S.A.)
(A) M.C.Q. :
(i) In case of compound interest, the rate of compound interest per annum is
Ans : (c) both equal or unequal
(ii) In case of compound interest
Ans : (b) Principal changes in each year
(iii) At present the population of a village is p and if increase rate of population per year be 2r%, the population after n years will be
Ans : (b) p
\(
1 +
r
50
\)
ⁿ
(iv) Present price of a machine is 2p and if price of the machine decreases by 2r% in each year, the price of machine after 2n years will be
Ans : (b) ₹2p
\(
1 -
r
50
\)
ⁿ
(v) A person deposited ₹100 in a bank and got the amount ₹121 for two years, the rate of compound interest per annum is
A = P
\(
1 +
R
100
\)
ⁿ
121 = 100
\(
1 +
R
100
\)
²
1.21 =
\(
1 +
R
100
\)
²
1.1 =
1 +
R
100
R
100
= 0.1
R = 10%
Ans : (a) 10%
(B) True or False :
(i) The compound interest will be always less than simple interest for some money at fixed rate of interest for fixed time.
Ans : False
(ii) In case of compound interest, interest is to be added to principal at the fixed time interval i.e. the amount of principal increases continuously.
Ans : True
(C) Fill in the blanks :
(i) The compound interest and simple interest for one year at the fixed rate of interest on fixed sum of money are equal.
(ii) If some things are increased by fixed rate with respect to time, that is uniform growth of compound increase.
(iii) If some things are decreased by fixed rate with respect to time this is uniform rate of depreciation.
17. Short answer (S.A.)
(i) Let us write the rate of compound interest per annum, so that the amount on ₹400 for 2 years becomes ₹441.
Solution :
P = 400
A = 441
n = 2
441 = 400
\(
1 +
R
100
\)
²
441
400
=
\(
1 +
R
100
\)
²
1.1025 =
\(
1 +
R
100
\)
²
1.05 =
1 +
R
100
R
100
= 0.05
R = 5%
∴ Required rate = 5% per annum
(ii) If a sum of money doubles itself at the fixed rate of compound interest per annum in n years, let us write in how many years will it become four times.
Solution :
If the money doubles in n years,
2P = P
\(
1 +
R
100
\)
ⁿ
2 =
\(
1 +
R
100
\)
ⁿ
Again,
4P = P
\(
1 +
R
100
\)
ᵗ
4 =
\(
1 +
R
100
\)
ᵗ
4 = 2²
=
\[
\(
1 +
R
100
\)
ⁿ
\]²
=
\(
1 +
R
100
\)
²ⁿ
t = 2n
∴ Required time = 2n years
(iii) Let us calculate the principal that at the rate of 5% compound interest per annum becomes ₹615 after two years.
Solution :
A = 615
R = 5%
n = 2
615 = P(1.05)²
615 = P × 1.1025
P =
615
1.1025
= 557.82
∴ Required principal = ₹557.82
(iv) The price of a machine depreciates at the rate of r% per annum. If the price of the machine after n years be v, let us find the price of the machine that was n years before.
Solution :
A = P
\(
1 -
R
100
\)
ⁿ
v = P
\(
1 -
r
100
\)
ⁿ
P =
v
\(
1 -
r
100
\)
ⁿ
∴ Price of the machine n years before =
v
\(
1 -
r
100
\)
ⁿ
(v) If the rate of increase in population is r% per year, the population after n years is p; let us find the population that was n years before.
Solution :
A = P
\(
1 +
R
100
\)
ⁿ
p = P
\(
1 +
r
100
\)
ⁿ
P =
p
\(
1 +
r
100
\)
ⁿ
∴ Population n years before =
p
\(
1 +
r
100
\)
ⁿ
