Children of a society planned to plant trees in their area. -Maths 9th

Children of a society planned to plant trees in their area. -Maths 9th
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Environmental care, beauty, conservation.
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The resident of society decided to paint the hall of cancer detective centre in their premises. -Maths 9th

Description : The resident of society decided to paint the hall of cancer detective centre in their premises. -Maths 9th

Last Answer : Perimeter = 2(l + b) = 260 = l + b = 130 (a) Surface area of four walls = 2h (l + b) = 2 x 6 x 130 = 1560 m2 Cost of painting = 9 x 1560 = ₹14,040 ( b) Amount contributed = 14040/50 = ₹280.8 (c) Cooperation, social cohesion.

The number of members of society A who participated -Maths 9th

Description : The number of members of society A who participated -Maths 9th

Last Answer : The number of participants from society A and C is equal. Things which are double of the same thing are equal to one another. Social service, helpfulness, cooperation, environmental concern.

In a society, the number of persons using CNG -Maths 9th

Description : In a society, the number of persons using CNG -Maths 9th

Last Answer : x + 15 = 25 ⇒ x + 15 - 15 = 25 - 15 (Using Euclid's third axiom) ⇒ x = 10 Environmental care, responsible citizens, futuristic.

A survey was conducted on 50 persons of a society -Maths 9th

Description : A survey was conducted on 50 persons of a society -Maths 9th

Last Answer : Each value with justification is correct. (Write yourself)

The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ? -Maths 9th

Description : The letters of the word ‘SOCIETY’ are placed at random in a row. What is the probability that three vowels come together ? -Maths 9th

Last Answer : There are 7 letters in the word SOCIETY. ∴ Total number of ways of arranging all the 7 letters = n(S) = 7!. When the case of three vowels being together is taken, then the three vowels are considered as one unit, so the ... = 5! 3! ∴ Required probability = \(rac{5! imes3!}{7!}\) = \(rac{1}{7}\)

The number of trees planted on Van Mahotsav -Maths 9th

Description : The number of trees planted on Van Mahotsav -Maths 9th

Last Answer : 103 x 98 = (100 + 3) (100 - 2) = 1002 + (3 - 2) x 100 - 3 x 2 = 10,000 + 100 - 6 = 10,094 Environmental care, social, happiness

What is imagined a completely planned society?

Description : What is imagined a completely planned society?

Last Answer : What is the answer ?

The following table gives the pocket money (in Rs) given to children per day by their parents : Represent the data in the form of a histogram. -Maths 9th

Description : The following table gives the pocket money (in Rs) given to children per day by their parents : Represent the data in the form of a histogram. -Maths 9th

Last Answer : The required histogram is as below :

The following table gives the pocket money (in Rs) given to children per day by their parents : Represent the data in the form of a histogram. -Maths 9th

Description : The following table gives the pocket money (in Rs) given to children per day by their parents : Represent the data in the form of a histogram. -Maths 9th

Last Answer : The required histogram is as below :

What Can A Teacher Do To Develop Integrated Personalities Of His Pupil? Options: A) Children Should Be Encouraged And Helped To Set Themselves Well-Defined Goals And Objectives Which They Can Hope To Achieve. B) They Should Be Helped To Build Their Self-Esteem To Develop A Sense Of Worthiness And Positive Ego Image. C) Life In Schools Should Be So Planned That Children Are Able To Express Themselves Fully. D) All Of These

Description : What Can A Teacher Do To Develop Integrated Personalities Of His Pupil? Options: A) Children Should Be Encouraged And Helped To Set Themselves Well-Defined Goals And Objectives Which They Can Hope To ... Should Be So Planned That Children Are Able To Express Themselves Fully. D) All Of These

Last Answer : D) All Of These 

An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Description : An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Last Answer : (i) From figure , it is clear that angle between the lines joining stalls C,D and stalls C, B is ∠BCD . Given, ∠DBC = 60° and ∠BAC = 40° CD is a chord of circle. Here, ∠CBD and ... for society. (iii) Other social issues for which such compaigns are required, are old age home, orphanages, etc.

Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Description : Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Last Answer : (i) Frequency distribution table (ii) From the above frequency distribution table, we observe that number of children in the class - interval 15 - 20 is 2. So, 2 children view television for 15 hours or more than 15 hours a week .

In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. -Maths 9th

Description : In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. -Maths 9th

Last Answer : Total children = 364 Number of children like potato chips = 91 Number of children do not like potato chips = 364 – 91 = 273 273 Required probability = 273 / 364 =0.75

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Description : An amusement fair was organised in a circular park for the children of slum clusters. Free food was supplied to them at 4 stalls situated at A, B, C and D as shown in the given figure. -Maths 9th

Last Answer : (i) From figure , it is clear that angle between the lines joining stalls C,D and stalls C, B is ∠BCD . Given, ∠DBC = 60° and ∠BAC = 40° CD is a chord of circle. Here, ∠CBD and ... for society. (iii) Other social issues for which such compaigns are required, are old age home, orphanages, etc.

Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Description : Thirty children were asked about the number of hours they watched TV programmes in the previous week. The results were found as follows : -Maths 9th

Last Answer : (i) Frequency distribution table (ii) From the above frequency distribution table, we observe that number of children in the class - interval 15 - 20 is 2. So, 2 children view television for 15 hours or more than 15 hours a week .

In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. -Maths 9th

Description : In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. -Maths 9th

Last Answer : Total children = 364 Number of children like potato chips = 91 Number of children do not like potato chips = 364 – 91 = 273 273 Required probability = 273 / 364 =0.75

750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Description : 750 families with 3 children were selected randomly and the following data recorded If a family member is chosen at random, compute the probability that it has : -Maths 9th

Last Answer : (i) P(no boy child) =100 / 750 = 2 / 15 (ii) P (no girl child) = 120 /750 =4 / 25

In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. -Maths 9th

Description : In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. -Maths 9th

Last Answer : NEED ANSWER

In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. -Maths 9th

Description : In a survey of 364 children aged 19-36 months, it was found that 91 liked to eat potato chips. -Maths 9th

Last Answer : (c) Total number of survey children's age from 19-36 months, n(S) = 364 In those of them 91 out of them liked to eat potato chips. ∴ Number of children who do not like to eat potato chips, n(E) = ... S) = 273/364 = 0.75 Hence, the probability that he/she does not like to eat potato chips is 0.75.

30 children were asked about the number of hours -Maths 9th

Description : 30 children were asked about the number of hours -Maths 9th

Last Answer : No, the number of children who watched TV for 10 or more hours a week is 4 + 2, i.e., 6 .

Thirty children were asked about the number -Maths 9th

Description : Thirty children were asked about the number -Maths 9th

Last Answer : The frequency distribution of above data in tabular form is given as: (ii) Two children watched television for 15 or more hours a week.

In a survey of 364 children aged 19-36 months, -Maths 9th

Description : In a survey of 364 children aged 19-36 months, -Maths 9th

Last Answer : Children who do not like potato chips = 364 - 91 = 273 P (a child does not like potato chips) = 273/364 = 0.75

In a survey of 364 children aged 20-40 months, -Maths 9th

Description : In a survey of 364 children aged 20-40 months, -Maths 9th

Last Answer : Number of children = 364 Number of children not like to eat potato chips = 364 - 91 = 273. The required probability = 273/364 = 0.75

1500 families with 2 children were selected -Maths 9th

Description : 1500 families with 2 children were selected -Maths 9th

Last Answer : (i) P (a family having 2 girls) = Number of families having 2 girls/Total number of families = 475/1500 = 19/60 (ii) P (a family having 1 girl) = Number of families having 1 girl/Total number of ... families = 211/1500 Sum of probabilities = 475/1500 + 814/1500 + 211/1500 = 1500/1500 = 1

In a school, 5 out of every 7 children participated in 'Save Wild Life' -Maths 9th

Description : In a school, 5 out of every 7 children participated in 'Save Wild Life' -Maths 9th

Last Answer : 5/7 = 0.714285(recurring), Non - terminating repeating decimal. Caring, social, helpful, environmental concern.

A group of children prepared some decorative pieces -Maths 9th

Description : A group of children prepared some decorative pieces -Maths 9th

Last Answer : In △AEC, ∠A + ∠E + ∠C = 180° ...(i) (Angle sum property of a triangle) Similarly, in △BDF, ∠B + ∠D + ∠F = 180° ...(ii) Adding (i) and (ii), we get ∠A + ∠B + ∠C + ∠D + ∠E + ∠F = 360° Social, caring, cooperative, hardworking.

The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Description : The whole surface area of a rectangular block is 1300 cm2. Find its volume, if their dimensions are in the ratio of 4 : 3 : 2. -Maths 9th

Last Answer : Let the length, breadth and height of the rectangular box be 4x, 3x and 2x, respectively. ∵ Total surface area = 1300 cm2 2(4x × 3x + 3x × 2x + 4x × 2x) = 1300 52x2 =1300x2 = 25x = 5 ∴ Volume of rectangular box = 4x × 3x × 2x = 24(5)2 = 3000 cm3

Three circles of radius a, b, c touch each other externally. The area of the triangle formed by joining their centres is: -Maths 9th

Description : Three circles of radius a, b, c touch each other externally. The area of the triangle formed by joining their centres is: -Maths 9th

Last Answer : (a) \(\sqrt{(a+b+c).a.b.c}\)As shown in the figure, AB = a + b, BC = b + c, CA = a + c∴ Area of ΔABC = \(\sqrt{s(s-AB)(s-BC)(s-CA)}\)where, s = \(rac{1}{2}\) (AB + BC + CA)= \(rac{a+b+b+c+c+a}{2}\) = a + b + ... (\sqrt{(a+b+c)[(a+b+c)-(a+b)][(a+b+c)-(b+c)][(a+b+c)-(c+a)]}\)= \(\sqrt{(a+b+c).a.b.c}\)

Were your children planned or not planned?

Description : Were your children planned or not planned?

Last Answer : answer:Even when you plan ‘em they are really unplanned! The first one I asked if she was sure it was mine! this did not go over well, as you might imagine!

the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Description : the curved surface area of a cylinder is 154 cm. the total surface area of the cylinder is three times its curved surface area. find the volume of the cylinder. -Maths 9th

Last Answer : T.S.A = 3*154 = 462 cm² C.S.A = 154 cm² C.S.A = 2πrh T.S.A = 2πr(r+h) Now, In T.S.A = 2πrr + 2πrh 462 = 2πrr + 2πrh 462 = 2*22/7*r*r + 154 462 - 154 = 2*22/7*r*r 308*7/2*22 = r*r 49 = r*r R = 7 cm ... 7*h 154/44 = h 7/2 =h H = 3.5 cm or 7/2 cm Now volume = πrrh = 22/7 * 7* 7 *7/2 = 11*49 = 539 cm³

case study questions class 9 maths surface area and volume -Maths 9th

Description : case study questions class 9 maths surface area and volume -Maths 9th

Last Answer : Q. Read the source or text given below and answer the following questions: A conical circus tent has to be made with a cloth that is 5m wide, whose height is 24m, and the radius of the base is 7m. ... 4. Find the Curved Surface Area Answers: 1. Curved Surface Area 2. 550 m 3. Rs.3850 4.110m²

A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes. -Maths 9th

Description : A 4 cm cube is cut into 1 cm cubes. Calculate the total surface area of all the small cubes. -Maths 9th

Last Answer : Side of cube = 4 cm But cutting into 1 cm cubes, we get = 4 x 4 x 4 = 64 Now surface area of one cube = 6 x (1)² = 6 x 1=6 cm² and surface area of 64 cubes = 6 x 64 cm² = 384 cm²

Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Description : Three equal cubes are placed adjacently in a row. Find the ratio of total surface area of the new cuboid to that of the sum of the surface areas of the three cubes. -Maths 9th

Last Answer : Let each side of a cube = a cm Then surface area = 6a² cm² and surface area of 3 such cubes = 3 x 6a² = 18a² cm² By placing three cubes side by side we get a cuboid whose ... + 3a²] = 14 a² ∴ Ratio between their surface areas = 14a² : 18a² = 7 : 9

Find the ratio of the total surface area and lateral surface area of a cube. -Maths 9th

Description : Find the ratio of the total surface area and lateral surface area of a cube. -Maths 9th

Last Answer : Let a be the edge of the cube, then Total surface area = 6a2² and lateral surface area = 4a² Now ratio between total surface area and lateral surface area = 6a² : 4a² = 3 : 2

Find the lateral surface area and total surface area of a cube of edge 10 cm. -Maths 9th

Description : Find the lateral surface area and total surface area of a cube of edge 10 cm. -Maths 9th

Last Answer : Edge of cube (a) = 10 cm (i) ∴ Lateral surface area = 4a² = 4 x (10)² = 4 x 100 cm²= 400 cm² (ii) Total surface area = 6a² = 6 x(10)² cm² = 6 x 100 = 600 cm²

Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. -Maths 9th

Description : Find the lateral surface area and total surface area of a cuboid of length 80 cm, breadth 40 cm and height 20 cm. -Maths 9th

Last Answer : Length of cuboid (l) = 80 cm Breadth (b) = 40 cm Height (h) = 20 cm (i) ∴ Lateral surface area = 2h(l + b) = 2 x 20(80 + 40) cm² = 40 x 120 = 4800 cm² (ii) Total surface area = 2(lb ... x 40 + 40 x 20 + 20 x 80) cm² = 2(3200 + 800 + 1600) cm² = 5600 x 2 = 11200 cm²

The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2. -Maths 9th

Description : The diameter of a roller is 84 cm and its length is 120 cm. It takes 500 complete revolutions to move once over to level a playground. Find the area of the playground in m2. -Maths 9th

Last Answer : The roller is in the form of a cylinder of diameter = 84 cm ⇒ Radius of the roller(r) = 842 cm = 42 cm Length of the roller (h) = 120 cm Curved surface area of the ... roller = 31680 cm2 = 3168010000m2 ∴ Area of the playground levelled in 500 revolutions = 500 x 3168010000m2 = 1584m2

A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Description : A field is in the shape of a trapezium whose parallel sides are 25 m and 10 m. The non-parallel sides are 14 m and 13 m. Find the area of the field. -Maths 9th

Last Answer : Let the given field is in the form of a trapezium ABCD such that parallel sides are AB = 10 m and DC = 25 m Non-parallel sides are AD = 13 m and BC = 14 m. We draw BE || AD, such that BE = 13 m. ... = 112 m2 So, area of the field = area of ∆BCE + area of parallelogram ABED = 84 m2 + 112 m2 = 196 m2

A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Description : A rhombus shaped field has green grass for 18 cows to graze. If each side of the rhombus is 30 m and its longer diagonal is 48 m, how much area of grass field will each cow be getting? -Maths 9th

Last Answer : Here, each side of the rhombus = 30 m. Let ABCD be the given rhombus and the diagonal, BD = 48 m Sides ∆ABC are a = AB = 30m, b = AD = 30m, c = BD = 48m Since, a diagonal divides the rhombus into ... Area of grass for 18 cows to graze = 864 m2 ⇒ Area of grass for 1 cow to graze = 86418 m2 = 48 m2

A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Description : A triangle and a parallelogram have the same base and the same area. If the sides of the triangle are 26 cm, 28 cm and 30 cm, -Maths 9th

Last Answer : For the given triangle, we have a = 28 cm, b = 30 cm, c = 26 cm Area of the given parallelogram = Area of the given triangle ∴ Area of the parallelogram = 336 cm2 ⇒ base x height = 336 ⇒ ... be the height of the parallelogram. ⇒ h = 33628 = 12 Thus, the required height of the parallelogram = 12 cm

Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used. -Maths 9th

Description : Radha made a picture of an aeroplane with coloured paper as shown in figure. Find the total area of the paper used. -Maths 9th

Last Answer : For surface I: It is an isosceles triangle whose sides are a = 5 cm, b = 5 cm, c = 1 cm = (0.75 x 3.3) cm2 = 2.475 cm2 (approx.) For surface II: It is a rectangle with length 6.5 cm and breadth 1 cm. ∴ Area of ... surface V) = [2.475 + 6.5 + 1.3 + 4.5 + 4.5] cm2 = 19.275 cm2 = 19.3 cm2 (approx.)

Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Description : Find the area of a quadrilateral ABCD in which AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. -Maths 9th

Last Answer : Given a quadrilateral ABCD with AB = 3 cm, BC = 4 cm, CD = 4 cm, DA = 5 cm and AC = 5 cm. For ∆ABC, a = AB = 3 cm, b = BC = 4 cm and c = AC = 5 cm Now, area of quadrilateral ABCD = area of ∆ABC + area of ∆ACD = 6 cm2 + 9.2 cm2 = 15.2 cm2 (approx.)

Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Description : Sides of a triangle are in the ratio of 12 : 17 : 25 and its perimeter is 540 cm. Find its area. -Maths 9th

Last Answer : Let the sides of the triangle be a = 12x cm, b = 17x cm, c = 25x cm Perimeter of the triangle = 540 cm Now, 12x + 17x + 25x = 540 ⇒ 54x = 54 ⇒ x = 10 ∴ a = (12 x10)cm = 120cm, b = (17 x 10) cm = 170 cm and c = (25 x 10)cm = 250 cm Now, semi-perimeter, s = 5402cm = 270 cm

A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula. -Maths 9th

Description : A traffic signal board, indicating ‘SCHOOL AHEAD’, is an equilateral triangle with side a. Find the area of the signal board, using Heron’s formula. -Maths 9th

Last Answer : Let each side of the equilateral triangle be a. Semi-perimeter of the triangle,

A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Description : A right circular cylinder just encloses a sphere of radius r (see fig. 13.22). Find (i) surface area of the sphere, (ii) curved surface area of the cylinder -Maths 9th

Last Answer : Surface area of sphere = 4πr2, where r is the radius of sphere (ii) Height of cylinder, h = r+r =2r Radius of cylinder = r CSA of cylinder formula = 2πrh = 2πr(2r) (using value of h) = 4πr2 (iii) Ratio ... sphere)/CSA of Cylinder) = 4r2/4r2 = 1/1 Ratio of the areas obtained in (i) and (ii) is 1:1.

Find the total surface area of a hemisphere of radius 10 cm -Maths 9th

Description : Find the total surface area of a hemisphere of radius 10 cm -Maths 9th

Last Answer : Radius of hemisphere, r = 10cm Formula: Total surface area of hemisphere = 3πr2 = 3×3.14×102 = 942 The total surface area of given hemisphere is 942 cm2.

Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Description : Find the surface area of a sphere of radius: (i) 10.5cm (ii) 5.6cm (iii) 14cm -Maths 9th

Last Answer : Formula: Surface area of sphere (SA) = 4πr2 (i) Radius of sphere, r = 10.5 cm SA = 4 (22/7) 10.52 = 1386 Surface area of sphere is 1386 cm2 (ii) Radius of sphere, r = 5.6cm Using formula, SA = 4 (22 ... 75 cm Surface area of sphere = 4πr2 = 4 (22/7) 1.752 = 38.5 Surface area of a sphere is 38.5 cm2

A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Description : A joker’s cap is in the form of right circular cone of base radius 7 cm and height 24cm. Find the area of the sheet required to make 10 such caps. -Maths 9th

Last Answer : Radius of conical cap, r = 7 cm Height of conical cap, h = 24cm Slant height, l2 = (r2+h2) = (72+242) = (49+576) = (625) Or l = 25 cm CSA of 1 conical cap = πrl = (22/7)×7×24 = 550 CSA of 10 caps = (10×550) cm2 = 5500 cm2 Therefore, the area of the sheet required to make 10 such caps is 5500 cm2.