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

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

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

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 .

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.

The students of a Vidyalaya were asked to participate in a -Maths 9th

Description : The students of a Vidyalaya were asked to participate in a -Maths 9th

Last Answer : Radius of penholder = r = 3 cm Height of penholder = h = 10.5 cm Total surface area of penholder = πr2 + 2 πrh = πr(r + 2h) = 22/7 x 3(3 + 2 x 10.5) = 66/7 x 24 cm2 Cardboard required for 35 competitors = 35 x 66/7 x 24 = 7920 cm2

For annual day, Sakshi and Nidhi were asked to make one rangoli -Maths 9th

Description : For annual day, Sakshi and Nidhi were asked to make one rangoli -Maths 9th

Last Answer : In △ABC and △PQR, BC = QR (Given) ⇒ 1/2BC = 1/2QR ⇒ BM = QN In triangles ABM and PQN, we have AB = PQ (Given) BM = QN (Proved above) AM = PN ... BC = QR (Given) ∴ ΔΑΒC ≅ ΔPOR (SAS congruence criterion) Participation, beauty, hardworking.

While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Description : While discussing the properties of a parallelogram teacher asked about the relation between two angles x and y of a parallelogram as shown ... -Maths 9th

Last Answer : (a) Yes , x < y is correct (b) Ð ADB =Ð DBC = y (alternate int. angles) since BC < CD (angle opp. to smaller side is smaller) there for, x < y (c) Truth value

Teacher asked the students 'Can we write 0.47(recurring) -Maths 9th

Description : Teacher asked the students 'Can we write 0.47(recurring) -Maths 9th

Last Answer : Yes, Let x = 0.477777.... ....(i) 10x = 4.77777... .....(ii) Subtracting (i) from (ii), we get 9x = 4.3 or x = 43/90 Scientific temper, knowledge, curosity.

For her records, a teacher asked the students about their heights. -Maths 9th

Description : For her records, a teacher asked the students about their heights. -Maths 9th

Last Answer : Yes, Things which are equal to the same thing are equal to one another. Knowledge, curiosity, truthfulness

In the linear equation y = 4x + 13, if x is the number of hours a labourer is on work and y are his wages in rupees then draw the graph. Also find the wages when work is done for 6 hours. -Maths 9th

Description : In the linear equation y = 4x + 13, if x is the number of hours a labourer is on work and y are his wages in rupees then draw the graph. Also find the wages when work is done for 6 hours. -Maths 9th

Last Answer : when the work is done for 6 hours x=6 y=4(6)+13 y=24+13 y=37 the labourer gets Rs.37 if he works for 6hrs

A boat covers 14 kms in upstream and 20 kms downstream in 7 hours. Also it covers 22 kms upstream and 34 kms downstream in 10 hours. Find the speed of the boat in still water and of that the stream. -Maths 9th

Description : A boat covers 14 kms in upstream and 20 kms downstream in 7 hours. Also it covers 22 kms upstream and 34 kms downstream in 10 hours. Find the speed of the boat in still water and of that the stream. -Maths 9th

Last Answer : Given, The boat covers 14 km upstream and 20km downstream . at time 7 hours also cover 22km ups. and 34km dwn in10 hours total speed = total distance/total time :.total distance = 14+20+22+34=90km and total time=7+10= ... =90/17 => 5.294km/h => 5.294km/h the speed of boat in still water is 5.29 km/h

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

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

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

Three coins were tossed 30 times simultaneously. -Maths 9th

Description : Three coins were tossed 30 times simultaneously. -Maths 9th

Last Answer : Frequency disribution of above data in tabular form is given as:

Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Description : Construct a histogram for the marks of the student given below : - Marks 0-10,10-30,30-45,45-50,50-60 and number of students 8,32,18,10,6 -Maths 9th

Last Answer : see in book okay!!!

In the class intervals 10-20, 20-30, the number 20 is included in -Maths 9th

Description : In the class intervals 10-20, 20-30, the number 20 is included in -Maths 9th

Last Answer : NEED ANSWER

The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

Description : The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

Last Answer : NEED ANSWER

In the class intervals 10-20, 20-30, the number 20 is included in -Maths 9th

Description : In the class intervals 10-20, 20-30, the number 20 is included in -Maths 9th

Last Answer : B is the correct answer

The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

Description : The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is -Maths 9th

Last Answer : Excluded number is

On planet A , each alien has 3 eyes and on planet B , each alien has 4 eyes. A group of aliens from planets A and B landed on earth. They had 30 eyes in all. Assuming number of aliens from planets A and B respectively , express the situation in the form of linear equation in two variables. -Maths 9th

Description : On planet A , each alien has 3 eyes and on planet B , each alien has 4 eyes. A group of aliens from planets A and B landed on earth. They had 30 eyes in all. Assuming number of ... planets A and B respectively , express the situation in the form of linear equation in two variables. -Maths 9th

Last Answer : answer:

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.

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 :

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

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.

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 :

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

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.

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

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.

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

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

Last Answer : Environmental care, beauty, conservation.

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.

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

Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Description : Find the cost of digging a cuboidal pit 8m long, 6m broad and 3m deep at the rate of Rs 30 per m3 -Maths 9th

Last Answer : The given pit has its length(l) as 8m, width (b)as 6m and depth (h)as 3 m. Volume of cuboidal pit = l×b×h = (8×6×3) = 144 (using formula) Required Volume is 144 m3 Now, Cost of digging per m3 volume = Rs 30 Cost of digging 144 m3 volume = Rs (144×30) = Rs 4320

he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th

Description : he frame has a base diameter of 20 cm and height of 30 cm. A margin of 2.5 cm is to be given for folding it over the top and bottom of the frame. -Maths 9th

Last Answer : Say h = height of the frame of lampshade, looks like cylindrical shape r = radius Total height is h = (2.5+30+2.5) cm = 35cm and r = (20/2) cm = 10cm Use curved surface area formula to find the ... 2πrh = (2 (22/7) 10 35) cm2 = 2200 cm2 Hence, 2200 cm2 cloth is required for covering the lampshade.

The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Description : The perimeter of a right triangle is 30 cm. If its hypotenuse is 13 cm, then what are two sides? -Maths 9th

Last Answer : The other two sides of the triangle are 12 cm and 5 cm Explanation: Let the other two sides of triangle be x and y It's hypotenuse is 13 cm Perimeter of triangle = Sum of all sides ... When y = 12 x=17-y = 17-12 =5 So, the other two sides of the triangle are 12 cm and 5 cm

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Description : 4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.30). Show that F is the mid-point of BC. -Maths 9th

Last Answer : . Solution: Given that, ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. To prove, F is the mid-point of BC. Proof, BD intersected EF at G. In ΔBAD, E is the ... point of BD and also GF || AB || DC. Thus, F is the mid point of BC (Converse of mid point theorem)

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Description : Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Last Answer : Here, the arranged data is 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 Total number of observations = 10 Here, 10 is an even number , therefore median will be the mean of (10 / 2)th and (10 / 2 + 1)th observation. ... ⇒ 3x + 12 / 2 = 24 ⇒ 3x - 12 = 48 ⇒ 3x = 60 ⇒ x = 20 ∴ The value of x = 20

The following data given the weight (in grams) of 30 oranges picked from a basket: -Maths 9th

Description : The following data given the weight (in grams) of 30 oranges picked from a basket: -Maths 9th

Last Answer : Here , class width = 20 class mark = 70 Half of the class width =20 /2 =10 Upper limit of first class interval = 70 + 10 = 80 Lower limit of first class interval = 70 - 10 = 60 Thus, class interval becomes 60 ... than 180 g = 1 + 1 = 2 (b) Number of oranges weights less than 100 g = 3 + 10 = 13

Factorise : 2x3 -3x2 -17x + 30 -Maths 9th

Description : Factorise : 2x3 -3x2 -17x + 30 -Maths 9th

Last Answer : Factorisation of following

ABCD is trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. -Maths 9th

Description : ABCD is trapezium in which AB || DC, DC = 30 cm and AB = 50 cm. -Maths 9th

Last Answer : According to question prove that ar (DCYX) = 7/9 ar (XYBA).

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the centre of the circle is -Maths 9th

Last Answer : (d) Given, AD = 34 cm and AB = 30 cm In figure, draw OL ⊥ AB. Since, the perpendicular from the centre of a circle to a chord bisects the chord.

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Description : AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, then find the distance of AB from the centre of the circle. -Maths 9th

Last Answer : ∵ The perpendicular drawn from the centre to the chord bisects it. ∴ AM = 1/2 AB = 1/2 × 30 cm = 15 cm Also, OA = 1/2 AD = 1/2 × 34 cm = 17 cm In rt. △OAM, we have OA2 = OM2 + AM2 172 = OM2 + 152 ⇒ 289 = OM2 + 225 ⇒ OM2 = 289 - 225 ⇒ OM2 = 64 ⇒ OM = √64 = 8 cm

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Description : Ten observations 6, 14, 15, 17, x + 1, 2x – 13, 30, 32, 34, 43 are written in ascending order. -Maths 9th

Last Answer : Here, the arranged data is 6, 14, 15, 17, x + 1, 2x - 13, 30, 32, 34, 43 Total number of observations = 10 Here, 10 is an even number , therefore median will be the mean of (10 / 2)th and (10 / 2 + 1)th observation. ... ⇒ 3x + 12 / 2 = 24 ⇒ 3x - 12 = 48 ⇒ 3x = 60 ⇒ x = 20 ∴ The value of x = 20