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West Bengal Board Class 10 Math Book Solution In English

Chapter 7 Theorems Related To Angles In A Circle Exercise 7.2

 

Question 1. I drew a circle and I drew two angles in the segment in that circle which (a) are formed with the same arc; (b) are not formed with the same arc. 

Question 2. I drew angles at the centre and the angle in the segment which (a) are formed with same arc; (b) are not formed with the same arc.

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(a) In fig (1), the angle at the centre O and the angle in the segment are formed with the same arc. 

(b) In fig (2) angle at the centre & angle in the segment are not formed with the same arc.

West Bengal Board Class 10 Math Book Solution In English

Question 3. Let us look at the picture and let us give an answer (O is the centre of the circle)

 

 

Solution: 

(i) ∠AOB, is the angle at the centre of the arc APB.

(ii) ∠APB in the segment is formed with arc ÁQB

(iii) In fig (iii) ∠ADB is cyclically formed with arc AQB

(iv) In fig (iv) ∠ACB is cyclically formed with arc APB

(v) In fig (v) ∠ADB is not an angle in the segment formed with the arc AQB.

 

 

In fig. (i) two angles∠ACB &∠ADB are in the same arc.

In fig. (ii)∠ACB & ∠ADB are two angles that lie in the segment ADCB.

West Bengal Board Class 10 Math Book Solution In English

Application 1. I drew a circle with centre P and by drawing an angle∠APB at the centre and an angle ∠AQB at the point on the circle by the arc ACB, we shall prove that, ∠APB2∠AQB

Solution: Let on the arc ACB of the circle with centre P, ∠APB is the angle at the centre & ∠AQB is the angle on the circumference.

To prove,∠APB = 2 ∠AQB.

Join Q, P and produce to R.

Proof: In ΔAPQ, AP = PQ (Radii of the same circle)

∴ ∠PAQ = ∠AQP

Again, QP is produced to R.

∴ External ∠APQ = ∠PAQ + ∠AQP = ∠2AQP

Similarly, from ∠BQP, ∠BPR = 2∠BQP

∴ ∠APB = ∠APR + ∠BPR = 2 (∠AQP + ∠BQP)

= ∠AQB Proved

 

Application 1. The circle with centre O passes through three points A, B and C; if ZABO = 35° and ZACO = 45°, let us write by calculat- ing the value of ZBOC.

In ΔOAB, OA = OB (radii of the same circle) OAB= ∴ ∠OBA = 35°

In ΔOAC, OA = OC (radii of the same circle) OAC= ∴ ∠OCA = 45°

 

 

Solution: ∠BAC = 35° + 45° 80°

Maths WBBSE Class 10 Solutions

Application 2. Let us write by calculating the value of x and y from the two figures given below.

 

 

 

Application 3.

Solution: BX BY (Radius of equal circles)

Maths WBBSE Class 10 Solutions

Application 4. ABC is an isosceles triangle with sides AB = AC; we draw ADBC in such a way that ADBC and AABC lie on the same side of BC and ZBAC = 2/BDC. Let us prove that the circle drawn with centre A and with radius AB is passing through point D, i.e., point D lies on the circle.

Solution:

Proof: Let the circle is not passing through D. If cuts BD at D’.

Join C, D’.

Now, BAC is the angle at the centre of the circle with

centre A and BDC is the angle on the circumference on

the same arc.

∴ ∠BAC = 2 ZBDC

But BAC 2 ZBDC (given)

∴ ∠BDC = <BD’C

It is not possible until point D. Coincide with D’ as an external angle of a triangle cannot be equal with its internal opposite angle.

∴Point D is on the circle.

 

 

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

Class 10 WBBSE Math Solution In English Chapter 6 Compound Interest And Uniform Rate Of Increase Or Decrease Exercise 6.1

Application 1. If I take a loan of Rs. 1400 at 5% compound interest per annum for 2 years, let us write by calculating how much compound interest and the total amount I shall pay.

Solution: When Principal = Rs. 1400,

Compound interest =10.25×1400 / 100

= Rs. 143.50

Principal & compound interest are in direct proportion.

Amount = Rs. (1400+ 143.50) Rs. 1543,50.

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Application 2. Let us write by calculating what is the amount of Rs. 1000 for 2 years at the rate of 5% compound interest per annum.

Solution: Amount = Rs. 1000 (1+5/100)²

= Rs. 1000 x (22/20)²

=Rs.1000 x 441/400

=Rs. 2205/2

=Rs. 1102.50

 

Application 3. If Rs. 4000 is invested in a bank at the rate of 5% compound inter- est per annum, then let us write by calculating the amount after 3 years.

Solution: Compound interest = Rs. (4630.50-4000) = Rs. 630.50.

 

Application 4. At 5% compound interest per annum, let us find the compound inter- est on Rs. 10,000 for 3 years.

 

Application 5. If interest is compounded half-yearly, let us write by calculating compound interest and amount of Rs. 8,000 at the rate of 10% compound interest per annum of 1 1/2  years.

Solution: Compound interest for 1 1/2 years = Rs. 1261

Class 10 WBBSE Math Solution In English

Application 6. Let us find compound interest on Rs. 1000 at the rate of 10% compound interest per annum and the interest being compounded at 6 monthly intervals.

 

Application 7. Let us write by calculating compound interest on Rs. 10,000 at the rate of 8% compound interest per annum for 9 months, compounded at an interval of 3 months.

Solution: Compound interest for 9 months = Rs. (1102.50-1000)

= Rs. 102.50.

 

Application 8. If the rate of compound interest for the first year is 4% and for 2nd year is 5%, let us find the compound interest on Rs. 25000 for 2 years.

Class 10 WBBSE Math Solution In English

Application 9. I lend Rs. 10,000 at the rate of 4% compound interest per annum for 2 1/2 years, let us write by calculating how much total money I shall pay.

Solution: Amount for 2 1/2 years = Rs. (10816+216.32) = Rs. 11032.32.

Application 10. Let us find the amount of Rs. 30,000, at the rate of 6% compound interest per annum for 2 1/2 years.

Application 11. Let us write by calculating what sum of money will amount to Rs. 3528 after 2 years at the rate of 5% compound interest per annum.

WBBSE Solutions Guide Class 10

Application 12. Let us see by calculating alternatively that Minatididi deposits Rs. 3,00,000 in the bank.

WBBSE Solutions Guide Class 10

Application 13. The simple interest and compound interest of a certain sum of money for 2 years are Rs. 840 and Rs. 869.40 respectively. Let us calculate that sum of money and the rate of interest.

 

Application 14. Let us calculate at what rate of compound interest, Rs. 5,000 will amount to Rs. 5832 in 2 years.

Application 15. Let us write by calculating in how many years Rs. 5000 will by compound interest at the rate of 10% per annum amount to Rs. 6050.

WBBSE Solutions Guide Class 10

Question 1. I have Rs. 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 it at the end of 3 years.

WBBSE Solutions Guide Class 10

Question 2. Let us calculate the amount of Rs. 5000 at the rate of 8% compound interest per annum for 3 years.

Maths WBBSE Class 10 Solutions

Question 3. Goutam babu borrowed Rs. 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 3 years he will pay.

 

Question 4. Let us write by calculating the amount of Rs. 30,000 at the rate of 9% compound interest per annum for 3 years.

Maths WBBSE Class 10 Solutions
Question 5. Let us write by calculating the amount of Rs. 80,000 for 2 years at the rate of 5% compound interest per annum.

 

Question 6. Chandadavi borrowed some money for 2 years in compound interest at the rate of 8% per annum. Let us calculate, if the compound interest is Rs. 2496, then how much money she had lent?

Maths WBBSE Class 10 Solutions

Question 7. Let us write by calculating the principal which becomes Rs. 2648 after getting 8% compound interest per annum for 3 years.

Ganit Prakash Class 10 Solutions Pdf In English

Question 8. Rahaman chacha deposited some money in a cooperative bank at the rate of 9% compound interest and he received an amount of Rs. 29702.50 after 2 years. Let us calculate how much money Rahaman chacha had deposited in the cooperative bank.

 

Question 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 Rs. 31492.80.

Ganit Prakash Class 10 Solutions Pdf In English

Question 10. Let us calculate the difference between the compound interest and simple in- terest on Rs. 12,000 for 2 years, at 7.5% interest per annum.

Ganit Prakash Class 10 Solutions Pdf In English

Question 11. Let us write by calculating the difference between compound interest and simple interest of Rs. 10,000 for 3 years at 5% per annum.

 

Question 12. Let us write by calculating the sum of money, if the difference between com- pound interest and simple interest for 2 years at the rate of 9% interest per annum is Rs. 129.60.

Ganit Prakash Class 10 Solutions Pdf In English

Question 13. Let us write by calculating the sum of money if the difference between compound interest and simple interest for 3 years becomes Rs. 930 at the rate of 10% interest per annum.

West Bengal Board Class 10 Math Book Solution In English

Question 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 Rs. 6000 for 2 years.

 

Question 15. If the rate of compound interest for the first and second year are 5% and 6% respectively, let us calculate the compound interest on Rs. 5000 for 2 years.

West Bengal Board Class 10 Math Book Solution In English

Question 16. If the simple interest on a certain sum of money for 1 year is Rs. 50 and compound interest for 2 years is Rs. 102, let us write by calculating the sum of money and the rate of interest.

 

Question 17. If simple interest and compound interest of a certain sum of money for two years are Rs. 8400 and Rs. 8652, then let us write by calculating the sum of money and the rate of interest.

Solution: Simple interest for 2 years = Rs. 8400

∴ Simple interest for 1 year = Rs. 4200.

Difference between compound interest & simple interest for 2 years = Rs. (8652-8400) = Rs 252.

∴ Interest on Rs. 4200 for 1 year = Rs. 252

∴ Interest on Rs. 100 for 1 year = Rs. 252 / 4200 x 100 = Rs. 6.

∴ Rate of interest = 6%.

∴ Principal S.I x 100 / Rx time

= 4200 x 100 / 6×1

= Rs. 70,000. Ans.

West Bengal Board Class 10 Math Book Solution In English

Question 18. Let us calculate compound interest on Rs. 6000 for 1 year at the rate of 8% compound interest per annum compounded at the interval of 6 months.

Question 19. Let us write by calculating compound interest on Rs. 6250 at the rate of 10% compound interest per annum, compounded at the interval of 3 months.

Question 20. Let us write by calculating at what rate of interest per annum Rs. 6000 will amount to Rs. 69984 in 2 years.

 

Question 21. Let us calculate in how many years Rs. 4000 will amount to Rs. 46656 at the rate of 8% compound interest per annum.

West Bengal Board Class 10 Math Book Solution In English

Question 22. Let us write by calculating at what rate of compound interest per annum, the amount on Rs. 10,000 for 2 years is Rs. 12100.

 

Question 23. Let us calculate in how many years Rs. 50000 will amount to Rs. 60500 at the rate of 10% compound interest per annum.

 

Question 24. Let us write by calculating in how many years Rs. 30,000 will amount to Rs. 399300 at the rate of 10% compound interest per annum.

 

Question 25. Let us calculate the compound interest and amount on Rs. 1600 for 1 the rate of 10% compound interest per annum, compounded at an interval of 6 months.

 

Application 1. At present, 4000 students have been taking training from this train- ing centre. In the last 2 years, it has been decided that the facility to get a chance for a training programme in this centre will be increased by 5% in comparison to its previous year. Let us see by calculating how many students will get a chance to join this training programme at the end of 2 years.

Solution: After 2 years the total number of candidates

= 4000 x (1+ 5 / 100)

= 4000 x105 /100 x  105 100 

= 4410. Ans.

 

Application 2. The price of a motor car is Rs. 3 lakhs. If the price of the car depreciates at the rate of 30% every year, let us write by calculating the price of the car after 3 years.

Solution: After 3 years, the price of the car

= Rs. 300000 X (1- 30/100)3

= Rs. 30,0000 X (7/10)3

Rs. 300000 X (7x7x7 / 10×10×10)

= Rs. 102900.

 

Application 3. At present the population of a city is 576000; if it is being increased  at the rate of 6 2/3 % every year, let us calculate its population 2 years ago.

 

 

Maths WBBSE Class 10 Solutions Chapter 5 Ration And Proportion Exercise 5.3

 

Question 1. If a: b = c:d, let us show that:

1.  (a² + b²): (a²-b²) = (ac + bd): (ac-bd)

 

2. (a² + ab + b²): (a²- ab+ b²) = (c² + cd + d²): (c² – cd + d²).

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Maths WBBSE Class 10 Solutions

3. √a²+ b²: √b² + d² = (pa + qc): (pb + qd)

 

Maths WBBSE Class 10 Solutions

Question 2. If x: a=y: b=z: c, let us prove that:

Let x/a = y/b = z/c = k (where k ≠ 0)

x=ak, y=bk & z= ck.

1.  x³/a² + y³/b² + z³/c² = (x+y+z)³/(a+b+c)²

 

Class 10 WBBSE Math Solution In English

2. x³+y³+z³ / a³+b³+c³ = xyz/abc

 

 

3. (a²+ b² + c²) (x² + y²+z²) = (ax + by + cz)²

Solution: To prove, (a² + b² + c²) (x² + y²+z²) = (ax + by + cz)²

L.H.S= (a²+ b² + c²) (x² + y² + z²)

= (a²+ b² + c²) (a²k² + b²k² + c²k²) = k²(a² + b² + c²) (a² + b² + c²) = k²(a² + b² + c²)²

= k²(a² + b² + c²)²

R.H.S. = (ax+by+ cz)²

(a.ak + b.bk+c.ck)² = {k(a² + b² + C²)}²

(a² + b²+ c²) (x² + y²+z²) = (ax + by + cz)² Proved.

L.H.S = R.H.S

Class 10 WBBSE Math Solution In English

Question 3. If a: b = c : d = e: f, let us prove that,.

1. Each ratio = 5a-7c-13e / 5b-7d-13f

 

 

2. (a² + c² + e²) (b² + c² + e²) (b² + d² + f²) = (ab + cd + ef)²

 

Class 10 WBBSE Math Solution In English

Question 4. If a: b = b: c, let us prove that

1. (a+b / b+c)² = a²+b² / b²+c²

 

 

2. a²b²c²(1/a³ + 1/b³ + 1/c³) = a³+b³+c³

 

 

3. abc (a+b+c)³ / (ab+bc+ca)³ = 1

 

 

5. If a, b, c, and d are in continued proportion, let us prove that

 1. (a2 + b2 + c2) (b2 + c2+ d2) = (ab + bc + cd)2

 

 

2. (bc)²+(ca)² + (bd)² = (a – d)²

 

 

Question 6:

1. If m/a = n/b, let us show that (m²+n²) (a²+b²) = (am + bn)².

Solution: If m/a = n/b, prove that (m2 + n2) (a2 + b2) = (am + bn)2.

Let, m/a = n/b = k(where k ≠ 0)

m = ak; n=bk

L.H.S = (m²+n²) (a²+b²)

= (a²k² + b²k²) (a²+ b²) = k²(a²+ b²) (a²+ b²) = {k(a²+ b²)}²

R.H.S. = (am bm)²

= (a.ak + b.bk)² = (a²k + b²k)² = {k(a² + b²)}²

  ∴ L.H.S. R.H.S. Proved.

 

2. If a/b = x/y, let us show that (a + b) (a² + b²) x³ = (x + y)(x² + y²) a³.

Solution: If a/b = x/y, prove that (a + b) (a² + b²) x³ = (x + y)(x² + y²) a³.

Let a/b = x/y = k(where k ≠ 0)

∴ a=bk & x=yk

L.H.S = (a + b) (a² + b²) x³

(bk + b) (b²k² + b²) (yk)³ = b(k + 1) b²(k² + 1) y³k³

= b3k3y (k + 1) (k² + 1)

R.H.S.= (x + y) (x²+ y²) a³

=(yk + y) (y²k² + y²) (b³k³) = y(k + 1) y²(k² + 1) 

= b³k³ b³k³y³(k+1) (k² + 1)

∴ L.H.S. = R.H.S. Proved.

 

3. If, x/ Im-n² = y/mn -1² = z = nl – m² , let us show that lx + my + ny = 0.

 

 

4. If x/b+c-a = y/c+a-b = z/ a+b-c let us show that (b-c) x + (c-a)y + (a-b) x=0.

 

WBBSE Solutions Guide Class 10

5. If x/y = a+2 / a-2 , let us show that x²-y² / x²+y² = 4a/a²+4.

 

 

6. If x = 8ab / a+b let us write by calculating the value of (x + 4a /x-4a + x + 4b / x-4b)

 

 

Question 7.

1. If a/3 = b/4 = c/7, let us show that a+b+c/c = 2.

 

WBBSE Solutions Guide Class 10

Question 8.

 

 

3. If x+y/3a-b = y+z/3b-c = z+x / 3c-a, let us show that x+y+z/a+b+c = ax+by+cz/a²+b²+c².

 

 

4. If x/a = y/b = z/c , let us show that x²-yz/a²-bc = y²-zx/b²-ca = z²-xy/c²-ab.

 

WBBSE Solutions Guide Class 10

Question 9.

1. If 3x+4y/3u+4v = 3x+4y/3u-4v , let us show that x/y = u/v.

 

 

2. If (a+b+c+d) : (a+b-c-d) = (a-b+c-d) : (a-b-c+d),let us prove that a:b = c:d.

 

 

Question 10.

1. If a²/b+c = b²/c+a = c²/a+b = 1, let us show that  1/1+a +1/1+b + 1/1+c = 1.

 

 

2. If x² : (by+cz)=y² :(cz + ax) = z² : (ax+by)=1, let us prove that a/a+x + b/b+y+ c/c+z = 1.

 

 

Question 11.

1. If x/xa+yb+zc = y/ya+zb+xc = z/za+xb+yc and x+y+z ≠ 0, let us show that each ratio is equal to 1/a+b+c.

 

WBBSE Solutions Guide Class 10

2. If x²-yz/a = y²-zx/b = z²-xy/c , let us prove that (a+b+c) (x+y+z) = ax +by+cz.

 

 

3. If a/y+z = b/z+x = c/x+y, let us prove that a(b-c)/y²-z² = b(c-a)/z²-x² = c(a-b)/x²-y².

 

 

Chapter 5 Ration And Proportion Exercise 5.3 Multiple Choice Questions

1. The fourth proportion of 3, 4, and 6 are

1. 8
2. 10
3. 12
4. 24

Answer. The 4th proportional of 3,4 & 6 =4×6/3 = 8———-(1)

 

2. The 3rd proportion of 8 and 12 is

1. 12
2. 16
3. 18
4. 20

Answer. The 3rd proportional of 8 &12 = 12 x 12/8 = 18 ———-(3)

 

3. The mean proportion of 16 and 25 is

1. 400
2. 100
3. 20
4. 40

Answer. The mean proportional of 16 & 25 = √16×25 = 4 x 5=20———-(3)

 

6. a is a positive number and if a: 27/64 = 3/4: a, then the value of a is

1. 81/256
2. 9
3. 9/16
4. 16/9

Answer: If a: 27/64 = 3/4: a

∴ a= √27/64 x 3/4 = √81/256 = 9/16—————–(3)

5. If 2a = 3b = 4c, then a:b:c is

1. 3:4:6
2. 4:3:6
3. 3:6:4
4. 6:4:3

Answer: If 2a=3b=4c, find a:b:c

or, 2a/12 = 3b/12 = 4c/12

or a/6 = b/4 = c/3

∴ a:b:c = 6:4:3——–(4)

 

Chapter 5 Ration And Proportion Exercise 5.3 True or False

 

1. Compound ratio of ab:c², bc:a² and ca:b² is 1:1

Answer: ab/c² x bc/a² x ca/b² = 1:1

True

2. x³y, x²y² and xy³ are continued proportional.

Answer: If x³y/x²y² = x²y²/xy³

or, x/y = x/y

 

Chapter 5 Ration And Proportion Exercise 5.3 True Or False

 

1. If the product of three positive consecutive numbers is 64, then their mean proportion is 4

2. If a: 2 = b: 5 = c: 8, then 50% of a = 20% of b = 12.5%  of c.

 

Chapter 5 Ration And Proportion Exercise 5.3 Short Answers

 

1. If a/2 = b/3 = c/a = 2a-3b+4c/p , let us find the value of p.

 

 

 

 

West Bengal Board Class 10 Math Book Solution In English

Chapter 5 Ration And Proportion Exercise 5.2

 

 Question 1. Let us find the value of x for the following proportions :

1.10:35::x:42

Solution: 10:35::x:42

Or, 10/35= X/42

or, 35X = 10 x 42

X=10 x 42 / 35

=12

 

2. X: 50 :: 3:2

Solution: X: 50:: 3:2

or, 2X = 50 x 3

X= 50 x 3/2

=75

Read and Learn More WBBSE Solutions For Class 10 Maths

Question 2. Let us find the fourth proportional of the following:

1. 1/3, 1/4, 1/5

Solution: 1/3, 1/4, 1/5

4th Proportional = \(\frac{\frac{1}{4} \times \frac{1}{5}}{1 / 3}\)

= 3/20

Class 10 WBBSE Math Solution In English

2. 9.6 kg, 7.6 kg, 28.8 kg.

solution: 9.6 kg, 7.6 kg, 28.8 kg.

4th proportional = 7.6kg x 28.8kg / 9.6kg

=22.8

 

 

Question 3. Let us find the 3rd proportional of the following positive numbers

 

Class 10 WBBSE Math Solution In English

Question 4. Let us find the mean proportional of the following positive numbers

1. 5 and 80

Solution: 5 and 80

Required mean proporitnal = √5×80 = √400 = 20.

 

2. 8.1 & 2.5

Solution: 8.1 & 2.5

Mean proportional = √8.1×2.5 = √20.25 = 4.5.

 

3. x³y and xy³

Solution: x³y and xy³

Mean proportional √x³y.xy³

=√x⁴y⁴

=x²y²

 

4. (x − y)², (x + y)²

Solution: (x-y)², (x + y)²

Mean proportional = √(x-y)². (x+y)²

=(x-x) (x + y). 

Class 10 WBBSE Math Solution In English

Question 5. If the two ratios a: b and c : d express mutually opposite relations, let us write what relation will be expressed by their inverse relation.

Answer. Each other are inverse proportion.

 

Question 6. Let us write how many continued proportions can be constructed by three numbers in continued proportions.

Answer. Two.

 

Question 7. If the first and second of the five numbers in continued proportion are 2 and 6 respectively, let us find the fifth number.

 

WBBSE Solutions Guide Class 10

Question 8. Let us write by calculating what should be added to each of 6, 15, 20, and 43 to make the sums proportional.

 

 

Question 9. Let us find what should be subtracted from each of 23, 30, 57, and 78 to make the results proportional.

WBBSE Solutions Guide Class 10

WBBSE Solutions Guide Class 10

Question 10. Let us find what should be subtracted from each of p, q, r, and s to have four numbers in proportion.

WBBSE Solutions Guide Class 10

Question 11. We understand, if 2: 3:: 10: 15 then 2: 10:3:

Solution: If 2:3: :10:15

then 2: 10:: 3:15

 

Question 12. We understand, if 6: 10::9: 15 then 10:6:: :

Solution: 6: 10::9: 15

then 6: 10::9: 15

Question 13. We understand, if 4: 5: 8: 10, then with the help of componendo of proportion, we get (4+5): 5= 10

Solution: If 5:48:10, with the help of componendo, we get (4 + 5):5 = (8+ 10): 10

 

Question 14. We understand, if 5:4 10: 8, then with the help of dividends of proportion we get (5-4):4:8=

Solution: If 5:4 = 10: 8, with the help of components we get (5-4): 4 = (10-8): 8

 

WBBSE Solutions For Class 10 Maths Chapter 5 Ration And Proportion Exercise 5.2 Applications

 

Application 1. If 5: 4: 10: 8, then with the help of componendo and dividendo of proportion, we get, (5+ 4): (5-4)::(10+8):

Solution: (5+ 4) : (5-4) = (10+8): (10-8)

 

Application 2. If a b::c: d, prove that (4a + 7b): (4a7b) :: (4c+ 7d): (4c-7d)

 

Ganit Prakash Class 10 Solutions Pdf In English

Application 3. We can write by applying the addenda property of proportion, 2/3 = 6/9 = 8/12 = 2+6+8 / + + .

Solution: 2/3 = 6/9 = 8/12

= 2+6+8/ 3+9+12

= 16/24

 

Application 4: X/a+b+c   Y/b+c-a  Z/c+a-b , let us prove that each ratio = x+y+z / a+b+c

 

 

Application 5. If (4a+5b) (4c-5d) = (4a-5b) = (4a-5b) (4c + 5d), let us prove that a, b, c, and d are in proportion.

Solution: (4a+5b) (4c-5d) = (4a-5b) = (4a-5b) (4c + 5d)

or, 16ac +20bc20ad25bd = 16ac – 20bc + 20ad – 25bd

or, 20bc+20bc = 20ad + 20ad

or, 40bc = 40ad

or, bc = ad

a/b = c/d

a,b,c & d are in proportion

 

Class 10 Math Solution WBBSE Chapter 5.1

Chapter 5 Ration And Proportion Exercise 5.1

Class 10 Math Solution WBBSE Chapter 5.1

Application 1. Let us see the ratios given below and let us write in the blank places.

Solution:

 

Application 2. Let us write the inverse ratio of the lowest form of x2yp: xy2p

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Class 10 Math Solution WBBSE Chapter 5.1

Application 3. If the ratio of two numbers is 2: 3 and their H.C.F is 7, let us write two numbers.

Solution: Numbers are (2 x 7) = 14 and (3 x 7) = 21.

 

Application 4. Let us write the inverse ratio of the mixed ratio of three ratios p2q: r, qr: p, and r2p: q.

 

 

Application 5. If A: B=3:7 and B: C = 8:5, let us find A: C [Let me do it myself]. Solution: A B = 4:5 and B: C=6:7

 

 

Application 6. If A: B=5:9 and B: C = 4:5, let us write the value of A: B: C .

 

Class 10 Math Solution WBBSE Chapter 5.1

Application 7. If x: y = 7:4, let us show that (5x-6y): (3x+11y) 11: 65 .

 

 

Application 8. If (2x+5y): (5x-7y)= 5: 3, let us find x: y. 

Application 9. If (7x – 5y): (3x+4y) = 7: 11, let us find the value of (5x-3y): (6x +5y).

Class 10 Math Solution WBBSE Chapter 5.1

Application 10. Let us write what should be added to each term of the ratio 5: 3 to make the ratio 7: 6.

 

Question 1. Let us express the following as ratio and let us write by understanding ratio of equality, ratio of leser inequality, ratio of greater inequality in each case :

1. 4 months and 1 year 6 months

Solution: 4 months:

1 year 6 months = 4 months (126) months 

= 4 months: 18 months

 = 2: 9 The ratio is of lesser inequality.

 

2. 75 paise and Re. 1 and 25 paise

Solution: 75p & Re. 1 & 25p

= 75p: 125p

=3:5 This ratio is of lesser inequality.

Maths WBBSE Class 10 Solutions

3. 60 cm and 0.6 meter

Solution: 60 cm. & 0.6 m.

= 60 cm: (0.6 x 100)cm 60 cm: 60 cm

 = 1:1 This is equality.

 

4. 1.2 kg and gram.

Solution: 1.2 kg & 60 gm

= 1.2 kg: 60 gm 

= (1.2 x 1,000) gm: 60 gm

=1200 gm 60 gm

= 20:1 This ratio is of greater inequality.

Maths WBBSE Class 10 Solutions

Question 2.

1. Let us write the ratio of p kg and q gram.

Solution: p kg & q gram

= (px 1000)gm: q gram 

= 1000 p: q

 

2. Let us write when it is possible to find the ratio of x days and z months. 

Solution: It is possible if both quantities are expressed in the same unit.

3. Let us write what type of mixed ratio of a ratio and its inverse ratio. 

Solution: It will be equality.

 

4. Let us find the mixed ratio of   a/b: c, b/c: a c/a: b

5. Let us write by calculating what ratio x²: yz will form with the mixed ratio xy: z².

Maths WBBSE Class 10 Solutions

6. Let us calculate the compound ratio of inverse ratios of x²: yz/x, y²: zx/y, x²:yx/z    

 

Question 3. Let us find the mixed ratio or compound ratio of the following ratios

1. 4:5, 5:7 and 9: 11

Solution: The compound ratio of (4: 5), (5: 7) & (9: 11) is (4 x 5 x 9): (5 x 7 x 11)= 36: 77

 

2. (x+y): (xy), (x² + y²) (x + y)² and (x² y²)²: (x⁴ – y⁴)

Maths WBBSE Class 10 Solutions

Question 4.

1. If A: B 6: 7 and B: C = 8: 7, let us find A: C.

 

2. If A: B=2: 3, B: C=4:5 and C:D = 6:7,let us find A:D

3. If A: B=3: 4 and B: C = 2:3, let us find A: B: C.

WBBSE Solutions Guide Class 10

4. If x y = 2:3 and y: z = 4: 7 let us find x:y: z. 

Solution: x : y = 2:3=8:12

y: z = 4:7 =12:21

x:y: z= 8: 12:21

Question 5.

1. If x:y3: 4, let us find (3y-x): (2x + y).

2. If a:b = 8: 7, let us show that (7a-3b): (11a-9b) = 7:5.

WBBSE Solutions Guide Class 10

3. If p q = 5: 7 and p-q=-4, let us find the value of 3p+ 4q.

 

6.
1. If (5x-3y): (2x+4y) 11: 12, let us find x: y.

WBBSE Solutions Guide Class 10

2. If (3a +7b): (5a-3b) = 5:3, let us find a : b.

 

7.
1. If (7x-5y): (3x+4y) = 7: 11, let us show that (3x-2y): (3x+4y) = 127: 473.

Class 10 WBBSE Math Solution In English

 

2. If (10x+3y): (5x+2y)=9:5, let us show that (2x + y): (x+2y) = 11 : 13.

 

8.
1. Let us calculate what term should be added to both terms of the ratio 2: 5 to make the ratio 6:11.

Class 10 WBBSE Math Solution In English

2. Let us calculate what term should be subtracted from each term of the ratio a: b to make the ratio m : n.

3. What term should be added to the antecedent and subtracted from the consequent of ratio 4: 7 to make a compound ratio of 2: 3 and 5: 4?

 

Application 1. Let us see whether the four numbers 2, 3, 4, and 6 are in proportion or not.

Solution: 2 x 6=3 x 4 and 3 x 4 =2×6

2:3::4:6

Class 10 WBBSE Math Solution In English

Application 2. Let us see whether the four numbers 2.5, -2, -5, and 4 are in proportion or not.

Solution: 2.5 x 4 =(-2) x (-5) and (-2) x (-5)=2.5 x 4

 

Application 3. Let us see whether four the numbers 2, 7, 12, and 42 are in proportion or not.

Solution: 2 x 42 = 84 &.7 x 12 = 84

2:7:: 12:42 (Are in proportion)

 

Application 4. Let us see whether four terms -√2, 6, 1, and √18 are in proportion or not.

Solution: (-√2) * (-√18)

= √36

= 6. & 6 1=6

(√2):6::1: (√18) (Are in proportion).

Application 5. Let us see whether 2a, 3b, 6ac, and 9bc are proportional or not. 

Solution: 2a x 9bc=18abc & 3b x 6ac=18abc

Class 10 WBBSE Math Solution In English

Application 6. Let us see whether 8x, 5yz, 40qx, and 25qyx are proportional or not.

Solution: 8x, 5yz, 40qx & 25qyz

8x x 25qyz = 200qxyz & 5yz x 40qx=200qxyz

8x 5yz: 40qx: 25qyz (Are in proportion).

 

Application 31. If 8: y:: 2:21, let us write by calculating the value of y. 

Solution: If 8 : y:: 2:21, find the value of y.

Ans. 8 x 21 = y x 2 or 2y= 168 

y=168/2

= 84.

 

Application 32. Let us find the fourth proportion of 6, 9, and 12.

Solution: 6x = 9 x 12

x =9×12/6

= 18. Answer

Class 10 WBBSE Math Solution In English

Application 33. Let us write by calculating the fourth proportion of 5, 4, 25.

Solution: Let the 4th proportional = x

5:4:: 25: x

or, 5x = 4 x 25

x = 4×25/5

=20.

Fourth proportional = 20 Answer

 

Application 34. Let us write by calculating how many independent proportions can be obtained from the proportional numbers 5, 6, 10, and 12 and mention those proportions. 

Solution: 5, 6, 10, and 12 are proportional.

5:6::10: 12

1. 5, 10, 6, 12;
2. 6, 5, 12, 10;
3. 10, 5, 12, 6

These 3 are the different forms of proportion.

West Bengal Board Class 10 Math Book Solution In English

Application 36. Let us write by calculating which number is to be added to each of 12, 22, 42, and 72 to make the sums proportional. 

 

 

Application 38. Let us write by calculating which number is to be added to each of 3.6, 7, and 10 to make the sums proportional.

 

 

Application 39. Let us find the 3rd proportion of 9 and 15.

Solution: x = 15×15 / 9

x=25

West Bengal Board Class 10 Math Book Solution In English

Application 40. Let us find 3rd proportional of Rs. 3 and Rs. 12.

 

Application 44. Let us write by calculating the mean proportional of 0.5 and 4.5.

 

West Bengal Board Class 10 Math Book Solution In English

Application 45. If the extreme terms of three positive continued proportional numbers are pqr, pr/q let us find the mean proportional.

Solution: Mean proportional = √0.5×4.5= √2.25 1.5 Ans.

 

Application 46. Let us find the mean proportional of positive numbers xy2 and xz2 [Let me do it myself].

 

 

West Bengal Board Class 10 Math Book Solution In English Chapter 4 Rectangular Parallelopiped or Cuboid Exercise 4.2

Question 1. Let us write the names of 4 cuboids and 4 cube-shaped solid things in our environment.

Wbbse Ccircles class 10 exercise 4.2 Solution:

1. Name of 4 rectangular parallelopipeds:
Brick;
Matchbox;
Mathematics book;
Geometry box.

2. Name of 4 cubes:
Dice of ludo;
Ice cube;
Rubix cube;
Square suitcase.

Read and Learn More WBBSE Solutions For Class 10 Maths

 

 

Question 2. Let us write the names of surfaces, edges, and vertices of the adjoining cuboidal.

Wbbse Ccircles class 10 exercise 4.2 Solution:

Name of the surfaces:
ABCD;
EFGH;
ADEF;
BCGH;
ABEH;
CDFG.

Name of the edges:
AB,
BC,
CD,
DA,
EF,
FG,
EH,
GH,
AE,
BH,
CG & F
G.

Vertexes are:
A,
B,
C,
D,
E,
F,
G
& H.

Question 3. The length, breadth, and height of a cuboidal room are 5 m, 4 m, and 3 m respectively. Let us write the length of the longest rod which can be kept in that room.

Question 4. The area of one surface of a cube is 64 sq.m., let us calculate the volume of the cube.

Wbbse class 10 maths chapter 4.2 Question 5.

In our Bokultala village, a canal is cut whose breadth is 2 m and depth is 8 dcm. If the total quantity of soil extracted is 240 cubic metres, then let us calculate the total surface area of the cube.

Wbbse class 10 maths chapter 4.2 Question 6.

If the length of the diagonal of a cube is 4√3 cm, then let us calculate the total surface area of the cube.

Wbbse class 10 maths chapter 4.2 Question 7.

The sum of the length of the edges of a cube is 60 cm, let us out the volume of the cube.

Wbbse class 10 maths chapter 4.2 Question 8.

If the sum of the areas of 6 surfaces of a cube is 216 sq. cm, then let us calculate the volume of the cube.

 

Wbbse class 10 maths chapter 4.2 Question 9.

The volume of a rectangular parallelopiped is 432 sq. cm. If it is converted into two cubes of equal volumes, then let us calculate the length of each edge of each cube.

Wbbse class 10 maths chapter 4.2 Question 10.

Each side of a cube is decreased by 50%. Let us calculate the ratio of the volumes of the original cube and the changed cube.

 

Wbbse Class 10 Maths Chapter 4 Exercise 4.2 Question 11.

If the ratio of length, breadth and height of a cuboidal box is 3: 2:1 and its volume is 384 cc then let us calculate the total surface area of the box.

 

Wbbse Class 10 Maths Chapter 4 Exercise 4.2 Question 12.

The inner length, breadth and height of a box of tea are 7.5 dcm and 5.4 dcm respectively, if the weight of the box filled with tea is 52 kg 350gm, but in an empty state, its weight is 3.75 kg then let us write by calculating, the weight of 1 cubic dcm. tea.

Wbbse Class 10 Maths Chapter 4 Exercise 4.2 Question 13.

The length, breadth and weight of a brass plate with a squared base are x cm, 1 mm and 4725 gm respectively, if the weight of 1 cubic dcm of brass is 8.4 gm, then let us write by calculating the value of x.

Wbbse Class 10 Maths Chapter 4 Exercise 4.2 Question 14.

The height of Chandamri Road is to be raised. So, 30 cuboidal holes with equal depth and of equal measure are dug out on both sides of the road and with this soil, my road is elevated. If the length and breadth of each hole are 14 m and 8 m respectively and if the total quantity of soil required to make the road is 2520 cubic metres men let us calculate the depth of each hole.

Wbbse Class 10 Maths Chapter 4 Exercise 4.2 Question 15.

If 64 water-filled buckets of equal measure are taken out from a cubical water-filled tank, then 1/3rd of the water remains in the tank. If the length of one edge of the tank is 1.2 m, then let us calculate and write the quantity of water that can be held in each bucket.

 

 

Question 16. If the length, breadth and height of one packet of one gross matchbox are 2.8 dcm and 0.9 dcm respectively, then let us calculate the volume of one match box [one gross 12 dozen]. But if the length and breadth of one matchbox be 5 cm and 3.5 cm, then let us calculate its height of it.

 

 

Question 17. Half of a cuboidal water tank with a length of 2.1 m and breadth of 1.5 m is filled with water. If 630 litres of water is poured more into the tank, then let us calculate and write the depth that will be increased.

 

 

Question 18. The length and breadth of a rectangular field of the village are 20 m and 15 m respectively. For the construction of pillars in the 4 corners of that field 4 cubic holes having lengths of 4 m are dug out and the soils removed are dispersed on the remaining land. Let us calculate and write the height of the surfaces of the field that is increased by.

 

 

Question 19. For elevating 6.5 DCM of low land with a length of 48 m and breadth of 31.5 m, it is decided that the soil will be collected by scooping a hole in a nearby land with a length of 27 m and breadth of 18.2 m, let us calculate the depth of the hole in metre.

 

 

Question 20. There were 800 lit, 725 lit and 575 lit kerosene oil in three kerosene oil drums of the house. The oil of these three drums is poured into a cuboidal pot and for this, the depth of oil in drums becomes 7 dams. If the ratio of the length and breadth of the cuboidal pot is 4: 3, then let us write by calculating the length and breadth of the pot. If the depth of the cuboidal pot would be 5 dcm, then let us calculate whether 1620 lit oil can be kept or not in that pot.

 

 

Question 21. The daily requirements of water for three families in our three-story flat are 1200 lit, 1050 lit and 950 lit respectively. After fulfilling these requirements in order to put up a tank again and to deposit to store 25% of the required water, only land having a length of 2.5 m and breadth of 1.6 m has been produced. Let us calculate the depth of the tank in metres that should be made. If the breadth of the land would be more by 4 dcm, then let us calculate the depth of the tank to be made.

 

 

Question 22. The weight of a wooden box made of wooden planks with a thickness of 5 cm along with its covering is 115.5 kg. But the weight of the box filled with rice is 880.5 kg. The length and breadth of the inner side of the box are 12 dcm and 8.5 dcm respectively and the weight of 1 cubic dcm rice is 1.5 kg. Let us write the inner height of the box after calculation. Let us calculate the total expenditure of colour the outside of the box if the rate is Rs. 1.50 per sq. dcm.

 

 

Question 23. The depth of a cuboidal pond with a length of 20 m and breadth of 18.5 m is 3.2 m, let us write by calculating the time required to irrigate the whole water of the pond with a pump having the capacity to irrigate 160-kilo lit water per hour. If that quantity of water is poured on a paddy field with a ridge having a length of 59.2 m and a breadth of 40 m, then what is the depth of water in that land – let us write by calculating it. [1 cubic metre = 1 kilo litre]

 

 

West Bengal Board Class 10 Math Book Solution In English Chapter 4 Theorems Related To Circle Exercise 4.2 Multiple-Choice Questions

 

Question 1. The inner volume of a cuboidal box is 440 cc. and the area of the inner base is 88 sq. cm, the inner height of the box is

1. 4 cm
2. 5 cm
3. 3 cm
4. 6 cm

Answer. The inner volume of the rectangular box = 440 cu. cm.

& inner area of the bottom of the box = 88 sqm.

Height of the box =440/88=5cm————–(b)

 

Question 2. The length, breadth and height of a cuboidal hole are 40 m, 12 m and 16 m respectively. The number of planks having a height of 5 m breadth of 4 m and thickness of 2 m, can be kept in that hole is

1. 190
2. 192
3. 184
4. 180

 

 

Question 3. The surface area of a cube is 256 sq.m, and the volume of the cube is

1. 64 cubic metre
2. 216 cubic metre
3. 256 cubic metre
4. 512 cubic metre

 

 

Question 4. The ratio of the volumes of the two cubes is 1: 27, and the ratio of the total surface areas of the two cubes is

1. 1:3
2. 1:8
3. 1:9
4. 1: 18

 

 

Question 5. If the total surface area of a cube is s sq. unit and the length of the diagonal is d unit, then the relation between s and d is

 

1. s = 6d²

2. 3s = 7d

3. S³ = d2

4. d² = s/2

 

 

Chapter 4 Theorems Related To Circle Exercise 4.2 True or False

 

1. If the length of each edge of a cube is twice of that 1st cube then the volume of this cube is 4 times more than that of the 1st cube.

False

2. In the rainy season, the height of rainfall in 2 hectares of land is 5 cm, and the volume of rainwater is 1000 cubic metres.

True

 

Chapter 4 Theorems Related To Circle Exercise 4.2 Fill In The Blanks.

 

1.  The number of diagonals of a cuboid is No. of diagonals = 4

2. The length of the diagonal on the surface of a cube =  √2  x the length of one edge.

 

 

West Bengal Board Class 10 Math Book Solution In English Chapter 4 Rectangular Parallelopiped or Cuboid Exercise 4.1

Application 1. I am observing that the length, breadth, and height of the rectangular parallelopiped box, brought by my brother, are 40 cm, 25 cm, and 15 cm respectively.

Solution: Total surface area of a rectangular box

=2 (LxB+ L x H + B x H)

=2 (40 x 25+ 40 x 15+ 25 x 15) sqcm.

=2 (1000+ 600 + 375) sqcm.

= 2 x 1975 sqcm.

= 3950 sqcm. Ans.

Read and Learn More WBBSE Solutions For Class 10 Maths

Application 2. The length, breadth, and height of a rectangular parallelopiped box are 15 cm, 12 cm, and 20 cm respectively. Let me write its total surface area. 

Solution: Total Surface area of a rectangular parallelopiped

=2 (15 x 12 + 15 x 20 + 12 x 20) sqcm.

=2 (180 + 300 +240) sqcm = 2 x (720) sqcm

= 1440 sqcm. Ans.

Application 3. By measuring, I see that the length of one side of the cube, brought by Rajia is 27 cm.

Application 4. Let us write by calculating the number of colored papers required to cover the whole surface of a cube whose length of one side is 12 cm.

Application 5. The length, breadth, and height of our living room are 7 m, 5 m, and 4 m respectively. The shape of the room is [cube/rectangular parallelopiped] Let us calculate the total area to color four walls of the room.

Application 6. Let us write the length of one side of a cube whose total surface area is 150 sq. cm.

Application 7. Let us calculate the length of one side of a cube whose total surface area is 486 sq.m.

 

wbbse class 10 maths chapter 4 exercise 4.1

Application 8. A cuboidal room has its length, breadth, and height as a, b, and c units respectively and if a + b + c = 25 units, ab + bc + ca= 240.5 units, then let us write the length of the longest rod that can be kept in this room.

WBBSE Solutions Guide Class 10 Application 9. I have made a rectangular parallelopiped by joining two cubes side by side made by Mita, whose edge is 8 cm in length. Let us calculate the total surface area and the length of the diagonal of the rectangular parallelopiped which is made in this way.

Application 10. By measuring the cuboidal box, I am observing that the length is 32 cm, breadth is 21 cm, and height is 15 cm.

Solution: Volume of the sand = 32 x 21 x 15 cu. cm = 10080 cu. cm.

Application 11. The length of one side of a cube is 5 cm and its volume is c.c = c.c

wbbse class 10 maths chapter 4 exercise 4.1

Application 12. If the length, breadth, and volume of a cuboidal room are 8 m, 6 m, and 192 cubic m respectively, then let us calculate the height of the room and the area of the four walls of the room.

wbbse class 10 maths chapter 4 exercise 4.1

Application 13. The length and breadth of cuboidal water land in the neighboring villages are 18 m and 11 m respectively. In this water land, water is being irrigated from a nearby pond with a pump. If the pump can irrigate 39,600 lit. water per hour then for how much time is required to raise the height of the water level by 3.5 cm of the water land [1 lit. – 1 cubic DCM.]

Application 14. If the pump can fill 37,400 lit. water in 1 hour, then what time will be required to raise the height of water 17 dcm of a cuboidal water land whose length is 18 m and breadth is 11 m. Let us write by calculating it. 

WBBSE Solutions Guide Class 10 Application 15. From a wooden log with a length of 5 cm., a breadth of 5 dcm. and the thickness of 3 cm, 40 planks are 2 cm in length and 2 cm in breadth are cloven. For cleaving, wood has been destroyed. But still now 108 cubic dcm of wood remains in the log. What is the thickness of each plank that is cloven.

 

Maths WBBSE Class 10 Solutions Chapter 3 Theorems Related To Circle Exercise 3.1

Question 1. Let us see the adjoining figure of the circle with center O and write the rate which is situated in the segment PAQ.

Answer: In the circle with center O, OP, OA, OC, and OQ are the radii in the A segment PAQ. 

             

Question 2. Let us write in the following      by understanding it.

1. In a circle, there is  number of points.

Answer: Infinite.

Read and Learn More WBBSE Solutions For Class 10 Maths

2. The greatest chord of the circle is

Answer: Diameter.

3. The chord divides the circular region into two

Answer:  Sector.

4. All diameters of the circle pass-through 

Answer: Centre.

5. If two segments are equal, then their two arcs are         in length.

Answer: Equal.

6. The sector of the circular region is the region enclosed by the arc and the two 

Answer: Radii.

7. The length of the line segment joining the point outside the circle and the center is    than the length of the radius.

Answer: Greater.

Wbbse Class 10 Maths Chapter 3 Exercise 3.1 Question 3.

With the help of scale and pencil, compass let us draw a circle and indicate centre, chord diameter, radius, major arc, minor arc on it

Answer: Centre O

Chord – CD

Diameter – AB

Radius OE = OA = OB

Minor arc – \(\overline{x y}\)

Major arc – \(\overline{x p y}\)

 

 

Chapter 3 Theorems Related To Circle Exercise 3.1 True or False

 

1. The circle is a plane figure.

Answer:  Circle is a rectilinear figure
True

2. The segment is a plane region.

Answer:  A segment of a circle is a rectilinear figure.
True

3. The sector is a plane region.

Answer:  The sector is rectilinear.
True

4. The chord is a line segment.

Answer: A chord is a line segment.
True

5. The arc is a line segment.

Answer:  Arc is a line segment.
False

6. There are a finite number of chords of the same length in a circle.

Answer:  A circle contains an infinite number of equal chords.
False

7. One and only one circle can be drawn by taking a fixed point as its center.

Answer:  Only one circle can be drawn with a center.
False

8. The lengths of the radii of two congruent circles are equal.

Answer:  The lengths of the radius of two congruent circles are equal.
True

Application 1. I draw a chord PQ of the circle with the Centre, which is not a diameter. I draw a perpendicular  on  from . I prove with reason that  MQ:

 

 

Application 2. I prove the theorem-33 by the proof of congruency of an OBD with the help of the S-A-S axiom of congruency. 

 

 

Wbbse Class 10 Maths Chapter 3 Exercise 3.1

Application 3. The perpendicular distance of a chord from the center of a circle, having a radius of  is in length. Let us write by calculating the length of its chord. 

 

Application 4. In a circle with the radius of in length, the two parallel chords of length and are situated on opposite sides of the center. Let us write by calculating the distance between two chords. 

 

 

 

Application 5. I prove with the reason that two equal chords of any circle are equidistant from its centre.

 

 

 

Application 6. Let us prove that the perpendicular bisector of a chord of a circle passes through its centre. 

 

 

Application 7. Let us prove that a straight line cannot intersect a circle at more than two points. 

 

 

 

 

Maths WBBSE Class 10 Solutions Chapter 3 Theorems Related To Circle Exercise 3.2 Solutions

Question 1. The length of a radius of a circle with its centre O is 5cm and the length of its chord AB is 8cm. Let us write by calculating the distance of chord AB from the centre O.

 

 

 

Question 2. The length the diameter of a circle with its centre at O is 26cm. The distance of the chord PQ from point O is 5. Let us write by calculating the length of the chord PQ.

Read and Learn More WBBSE Solutions For Class 10 Maths

 

 

Question 3. The length of a chord PQ of a circle with its centre O is  4cm and the distance of PQ from the point O  is 2.1 cm. Let us write by calculating, the length of its diameter

Wbbse class 10 maths chapter 3 exercise 3.2 solutions

 

 

Question 4. The lengths of two chords of a circle with its centre at O are  6CM and 8CM. If the distance of the smaller chord from the centre is 4CM, then let us write by calculating, the distance of another chord from the centre.

 

 

 

Question 5. If the length of a chord of a circle is 48CM  and the distance of it from the centre is 7CM, then let us write by calculating the length of the radius of the circle.

Wbbse class 10 maths chapter 3 exercise 3.2 solutions

 

                                                           

 

Question 6. In the circle of the adjoining figure with its centre at O, OP⊥AB;  if  AB = 6CM and PC = 2CM, then let us write by calculating the length of the radius of the circle.

 

 

 

Wbbse class 10 maths chapter 3 exercise 3.2 Question 7.

A straight line intersects one of the two concentric circles at points A and B and the other at points C and D. I prove with the reason that AC=DB.

 

 

Question 8. I prove that the two intersecting chords of any circle cannot bisect each other unless both of them are diameters of the circle.

 

 

Wbbse class 10 maths chapter 3 exercise 3.2 Question 9. The two circles with centres X and Y intersect each other at points  A and B. A is joined with the mid-point ‘ S ‘ of  XY and the perpendicular on SA through the point A is drawn which intersects the two circles at the points P and Q. Let us prove that PA = AQ.

 

 

Question 10. The two parallel chords  AB and CD  with lengths of 10 CM  and 24 CM in a circle are situated on the opposite sides of the centre. If the distance between two chords AB and DC is 17CM, then let us write by calculating, the length of the radius of the circle.

 

 

 

Question 11. The centres of the two circles are P and Q; they intersect at points  A and B. The straight line parallel to the line segment PQ  through point A intersects the two circles at points  C and D.1 prove that CD = 2PQ.

 

Question 12. The two chords AB and AC of a circle are equal. I prove that the bisector of ∠BAC passes through the centre.

 

Wbbse class 10 maths chapter 3 exercise 3.2 Question 13. If the angle-bisector of two intersecting chords of a circle passes through its centre, then let me prove that the two chords are equal.

 

 

Wbbse class 10 maths chapter 3 exercise 3.2 Question 14. I prove that, among two chords of a circle, the length of the chord nearer to the centre is greater than the length of the other.

 

 

 

Wbbse class 10 maths chapter 3 exercise 3.2 Question 15. Let us write by proving the chord with the least length through any point in a circle.

 

 

 

Chapter 3 Theorems Related To Circle Exercise 3.2 Multiple Choice Questions

Question 1. The lengths of two chords of a circle with centre O are equal. If ∠AOB = 60°, then the value of ∠COD is

1. 40°
2. 30°
3. 60°
4. 90°

Answer: AB & CD are two equal chords of the circle with O if LAOB = 60°; then /COD = 60°———-(3)

Question 2. The length of the radius of a circle is 13 cm and the length of a chord of a circle is 10 cm, the distance of the chord from the centre of the circle is

1. 12.5 cm
2. 12 cm
3. √69 cm
4. 24 cm

Answer: Radius = 13 cm. length of chord = 10 cm, Distance from the centre = √13² -5²

= √169-25

= √144

=12-——(2)

Question 3. AB and CD are two equal chords of a circle with its centre O. If the distance of the chord AB from the point O is 4 cm then the distance of the chord from the centre O of the circle is

1. 2 cm
2. 4 cm
3. 6 cm
4. 8 cm

Answer: AB & CD are two equal chords.

If the distance of AB from the centre O = 4 cm.

then the distance of CD from the centre O 4 cm. ——- -(2)

 

Question 4. The length of each of the two parallel chords is 16 cm. If the length of the radius of the circle is 10 cm., then the distance between two chords is

1. 12 cm.
2. 16 cm.
3. 20 cm.
4. 5 cm.

Answer: Two parallel chords of equal length 16 cm.

Length of the radius = 10 cm.

Distance between them = √102-82 + √102-826+6 12 cm. ————(1)

 

Question 5. The centre of two concentric circles is O; a straight line intersects a circle at points A and B and the other circle at the point C and D. If AC = 5 cm, then the length of BD is

1. 2.5 cm
2. 5 cm
3. 10 cm
4. None of these

Answer: When a straight line cuts two concentric circles at A & B and C & D points, AC = BD.

BD 5 cm.———–(2)

 

Chapter 3 Theorems Related To Circle Exercise 3.2 True or False

1. Only one circle can be drawn through three collinear points.

Answer: False

2. The two circles ABCDA and ABCEA are the same circle.

Answer: True

3. If two circles AB and AC of a circle with its centre O are situated on opposite sides of the radius OA, then LOAB = LOAC.

Answer: False

 

Chapter 3 Theorems Related To Circle Exercise 3.2  Fill In The Blanks

Question 1. If the ratio of two chords PQ and RS of a circle with its centre O is 1: 1, then, LPOQ: LROS =—————–

Answer: LPOQ: LROS = 1:1.

Question 2. The perpendicular bisector of any chord of a circle is—————–

Answer: Passing through the centre.

Wbbse class 10 maths chapter 3 exercise 3.2 Solutions

Chapter 3 Theorems Related To Circle Exercise 3.2 Short Answers

Question 1. Two equal circles of radius 10 intersect each other and the length of their common chord is 12 cm. Let us determine the distance between the two centres of the two circles.

 

 

Question 2. AB and AC are two equal chords of a circle having a radius of 5 cm. The centre of the circle is situated at the outside of the triangle ABC. IF AB = AC = 6 cm, then let us calculate the length of the chord BC.

 

 

 

Question 3. The length of two chords AB and CD of a circle with its centre O are equal. If LAOB = 60° and CD = 6 cm, then let us calculate the length of the radius of the circle. Ans. AB = CD = 6 cm.

Wbbse class 10 maths chapter 3 exercise 3.2 Solutions

 

Question 4. P is any point in a circle with its centre O. If the length of the radius is 5 cm and OP = 3 cm, then let us determine the least length of the chord passing through the point P.

 

 

Question 5. The two circles with their centres at P and Q intersect each other at points A and B. Through point A, a straight line parallel to PQ intersects the two circles at points C and D respectively. If PQ = 5 cm, then let us determine the length of the CD.