West Bengal Board Class 10 Math Book Solution In English Chapter 6 Compound Interest And Uniform Rate Of Increase Or Decrease Exercise 6.2
Question 1. At present the population of the village of Pahalanpur is 10000; if the population is being increased at the rate of 3% every year, let us write by calculating its population after 2 years.
Solution: Present population = 10000
After one year, the population will be
= 10000 + 3/100 x 10000
=10000 + 300
= 10300.
After 2nd year, population will be = 10300 + 3 / 100 X 10300 + 309
= 10609
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Question 2. The rate of increase in the 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
After one-year population will be
= 8000000 + 2/100 × 80000000
= 80000000 + 1600000 ‘
= 81600000
After 2nd year the population will be
= 83232000+ 83232000 x 2/100
= 81600000+1632000
= 83232000
After 3rd year the population will be
= 83232000+83232000 x 2/100
= 83232000+ 1664640 84896640.
West Bengal Board Class 10 Math Book Solution In English
Question 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 Rs. 100000, let us calculate what will be the price of that machine after 3 years.
Solution: Present price of the machine = Rs. 100000
After 1 year the price will be = Rs. (100000 – 10/100 × 100000 ) = Rs. 90,000
After 2nd year the price will be = Rs. 90000- 10/100 x 90000) = Rs. 81000
∴ After 3rd year the price will be
Rs.(81000-10/100x 81000)
= Rs. (81000 – 8100) Rs. 72900.
Question 4. As a result of Sarba Siksha Abhiyan, the students leaving the school before completion, the students are readmitted, so the number of students in a year is increased by 5% in comparison to the previous year. If the number of such readmit- ted 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: Let at present no. of students = 3528.
Let two years before, no. of students was x & rate = 5%.
∴According to the problem,
X (1+5/100) = 3528
Or X x 105/100×105/100 = 3528
∴ X = 3528 x 100 x 100/105×105
= 3200
∴Two years before no. of students was 3200.
WBBSE Solutions Guide Class 10
Question 5. Through the publicity of the road-safety program, street accidents in the Purulia district decreased by 10% in comparison to the previous year. If the number of street accidents this year is 8748, let us write by calculating the number of street accidents 3 years before in the district.
Solution: No. of street accidents decreases by 10% every year in comparison to the previous year.
No. of street accidents this year is 8748.
Let no. of street accidents 3 years before was x.
According to the problem,
x(1-10/100)³= 8748
Or, X x(9/10)³ = 8748
or, X x 9×9×9/10x10x10 = 8748
∴x = 8748x10x10x10 / 9x9x9
= 12000
∴No. of street accidents 3 years before was 12000
Question 6. A cooperative society of fishermen implemented such an improved plan for the production of fishes that the production in a year will be increased by 10% in comparison to the previous year. In the present year if the cooperative society can produce 406 quintals of fish, let us write by calculating what will be the production of fishes after 3 years.
Solution: Production in a year increases by 10% in comparison to the previous year. In the present year, the production of fish is 400 quintals.
∴ Production of fish after 3 years will be
=400 (1+10/100)³ quintals
= 400 X 11/100×11/100×11/100
=532.4 quintals
WBBSE Solutions Guide Class 10
Question 7. The height of a tree increases at a rate of 20% every year. If the present height of the tree is 28.8 metres, let us calculate the height of the tree 2 years before.
Solution: Let 2 years before the height of the tree was x m.
Now the height of the tree = X (1+20/100)² m = 28.8 m.
X X(6/5)² = 28.8
X x 36/25 = 28.8
X = 28.8×25 / 36 = 20 m.
∴ 2 years before the height of the tree was = 20 m.
Question 8. Three years before today a family had planned to reduce the expenditure of electric bills by 5% in comparison to the previous year. 3 years before, that family had to spend Rs. 4000 a year on electric bills. Let us write by calculating how much amount the family will have to spend to pay the electric bill in the present year.
Solution: The expenditure on the electric bill was reduced by 5% in comparison to the previous year.
3 years before the expenditure for the electric bill was Rs. 4000.
.. The present electric expenditure will be
=Rs. 4000(1-5/100)³
=Rs. 4000 x 95/100 x 95/100 x 95/100
= Rs. 3429.50. Ans.
WBBSE Solutions Guide Class 10
Question 9. Weight of Savan babu is 80 kg. In order to reduce his weight, he started regular morning walks. He decided to reduce his weight every year by 10%. Let us write by calculating his weight after 3 years.
Solution: Present mass of Savan babu is 80 kg.
He decided to reduce his mass every year by 10%.
∴After 3 years his mass will be
= 80 x(1-10/100)³
= 80 x (9/10)³kg
=80 x 9/10 x 9/10 x 9/10 kg
= 58.32 kg.
Question 10. At present, the sum of the number of students in all M.S.K in a district is 399. If the number 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: If the number of students increased in a year by 10% in comparison to the previous year.
The present no. of students = 3993 and let the no. of students 3 years was x.
∴ X x (1+10/100)³ = 3993
Or, X x 11/10 x 11/10 x 11/10 = 3993
∴ X = 3993 x 10 x10x10 / 11x11x11
x = 3 x 1000
= 3000. Ans.
Maths WBBSE Class 10 Solutions
Question 11. As the farmers are becoming more alert to the harmful effects of applying only chemical fertilizers and insecticides in agricultural lands, the number of farmers using fertilizers and insecticides in the village of Rasulpur decreases by 20% in a year in comparison to its previous year. Three years before, the number of such farmers was 3000; let us calculate the number of such farmers in that village now.
Solution: No. of farmers using chemical fertilizer decreases by 20% in the current year in comparison to the previous year.
3 years before no. of such farmers was 3000.
.. No. of farmers will be
=3000 x (1-20/100)³
=3000 x 4/5 x 4/5 x 4/5
= 1536.
Question 12. The price of a machine in the factory is Rs. 18000. The price of that machine decreases by 10% in each year. Let us calculate its price after 3 years.
Solution: The value of a machine in a factory is Rs. 180000.
The value of the machine depreciates at 10% every year.
∴ The value of the machine after 3 years will be
= Rs. 180000 x (1-10/100)³
= Rs. 18000 x (9/10)³
= Rs. 18000 x 729 / 1000
= Rs. 131220.
Maths WBBSE Class 10 Solutions
Question 13. For the families having no electricity in their houses, a Panchayat samiti of village Bakultala accepted a plan to offer electricity connections. 1200 families in. this village have no electric connection in their houses. In comparison to the 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: No. of family = 1200
In comparison to the previous year, it is possible to arrange electricity every year for 75% of the family having no electricity.
.. No. of families without electricity after 2 years
= 1200 x (1-75/100)²
=1200 x (1-3/4)²
= 1200 x 1/7 x 1/4
=75.
Question 14. As a result of continuous publicity on harmful reactions to the use of cold drinks filled, bottles, the number of users of cold drinks is decreased by 25% every year in comparison to the previous year. 3 years before the number of users of cold drinks in a town was 80000. Let us write by calculating the number of users of cold drinks in the present year.
Solution: No. of users of cold drinks decreases by 25% every year in comparison to the previous year.
3 years before the number of users of cold drinks in a town was 80000.
∴No. of users of cold drinks in the present year
= 80000 x (1-25/100)³
= 80000 x (3/4)³
=80000 x 27/64
= 33750.
Maths WBBSE Class 10 Solutions
Question 15. As a result of publicity on smoking, the number of smokers is decreased by 6 1/4 % every year in comparison to the 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: No. of smokers decreases by 6 1/4 % (25/4 %) every year in comparison to the previous year.
Let the no. of smokers 3 years before be x.
According to the problem,
X x (1- 25/4 /100)³
=33750
Or, X x (1-25/100)³
=33750
Or, X x (15/16)³
=33750
Or, X x 15 x 15 x 115 / 16 x 16 x 16
=33750
∴ X =33750 x 16 x 16 x 16 / 15 x 15 x 15
= 40960
∴ 3 years before no.of smokers was 40960
Maths WBBSE Class 10 Solutions Chapter 6 Compound Interest And Uniform Rate Of Increase Or Decrease Exercise 6.2 Multiple Choice Questions
Question 1 In the case of compound interest, the rate of compound interest per annum is–
1. Equal
2. Unequal
3. Both equal or unequal
4. None of these
Answer. 1. Equal
Question 2. In the case of compound interest
1. The principal remains unchanged each year
2. Principal changes in each year
3. The principal may be equal or unequal in each year
Answer. 2. Principal changes in each year
Question 3. At present the population of a village is p and if the rate of increase of population per year is 2r%, the population will be after n years.
1. P (1+r/100) n
2. p (1+r/50) n
3. p (1+r/50)2n
4. P (1-r/50) n
Answer: 2. p (1+r/50) n
Question 4. The present price of a machine is Rs. 2p and if the price of the machine decreases by 2r% each year, the price of the machine will be
1. Rs.p (1-r/50)n
2. Rs.2p (1-r/50)n
3. Rs.p (1-r/100)2n
4. Rs.2p (1-r/100)2n
Answer: Rs.2p (1-r/100)2n
Question 5. A person deposited Rs. 100 in a bank and got the amount Rs. 121 for two years. The rate of compound interest is
1. 10%
2. 20%
3. 5%
4. 10 ½%
Answer. 1. 10%
WBBSE Solutions Guide Class 10 Chapter 6 Compound Interest And Uniform Rate Of Increase Or Decrease Exercise 6.2 True Or False
Question 1. The compound interest will be always less than simple interest for some money at a fixed rate of interest for a fixed time.
Answer: False
Question 2. In the case of compound interest, interest is to be added to the principal at a fixed time interval, i.e., the number of principals increases continuously.
Answer: True
Chapter 6 Compound Interest And Uniform Rate Of Increase Or Decrease Exercise 6.2 Fill In The Blanks
1. The compound interest and simple Interest for one year at the fixed rate of interest on a fixed sum of money are Equal.
2. If some things are increased by a fixed rate with respect to time, that is the same rate.
3. If some things are decreased by a fixed rate with respect to time, this is a uniform rate of Decrease
WBBSE Solutions Guide Class 10 Chapter 6 Compound Interest And Uniform Rate Of Increase Or Decrease Exercise 6.2 Short Answer
Question 1. Let us write the rate of compound interest per annum so that the amount of Rs. 400 for 2 years becomes Rs. 441.
Question 2. If a sum of money doubles itself at compound interest in n years, let us write in how many years It will become four times.
Question 3. Let us calculate the principle that at the rate of 5% compound interest per annum becomes Rs. 615 after two years.
Question 4. The price of a machine depreciates at the rate of r% per annum, let us find the price of the machine that was n years before.
Solution: The value of the machine n years before was Rs.= V/ (1-r/100) n
= V x (1-r/100)-n
Question 5: 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: n years before, no. of population was = P / (1-r/100) n = P x (1+r/100)-n
