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

1 Answer

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

Related questions

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.

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

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 .

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 .

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 .

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.

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.

Description : Three students were made to stand on the points P, Q and S -Maths 9th

Last Answer : Coordinates of R are (6, 6). Reasoning, enjoyment, physical fitness.

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

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!!!

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

Description : The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Last Answer : x̅ = 1 / n (Σxi) ⇒ 55 = x1 + x2 + .......+ ⇒ x7 / 7 ⇒ x1 + x2 + ...... + x7 = 55 × 7 = 385 x1 + x2 + ...... + x6 = 52 + 54 + 55 + 53 + 56 + 54 = 324 ∴ x7 = 385 - 324 = 61 kg ∴ weight of the seventh student is 61 kg.

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88

Description : In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Last Answer : For Mean : As we know that ⇒ x̅ = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60 / 15 x̅ = 822 / 15 = 54.8 ⇒ x̅ = 54.8 For Median : First of ... 52 For Mode : Make a frequency table for given data : Here, the marks 52 has the maximum frequency '3'. ∴ Mode = 52

Description : In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Last Answer : Here, class size = 135 - 305 = 10 ∴ Lower limit of first class interval is 305 - 10 / 2 = 300 Upper limit of first class interval is 305 + 10 / 2 = 310 Thus, first class interval is 300 - 310 Required histogram and frequency polygon is given on the graph paper .

Description : Following is the frequency distribution of total marks obtained by the students of different section of class-IX. -Maths 9th

Last Answer : Since class intervals of the given frequency distribution are not of equal width. We would make modifications in the lengths of the rectangles in the histogram, so that the areas of rectangles are proportional ... lengths as given in the last column .The histogram of the data is given below.

Description : Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. -Maths 9th

Last Answer : Clearly, the mean score of two sections A and B is same

Description : In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

Description : The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

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!!!

Description : A school provides milk to the students daily in cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Diameter d = 7 cm Radius r = 7 / 2 cm and h = 12 cm ∴ V = πr2h = 22 / 7 × 7 / 2 × 7 / 2 × 12 = 462 Total milk for 1600 students = 462 × 1600 = 739200 cm3 = 739200 / 1000 litres = 739.2 litres .

Description : The mean weight per student in a group of 7 students is 55 kg. -Maths 9th

Last Answer : x̅ = 1 / n (Σxi) ⇒ 55 = x1 + x2 + .......+ ⇒ x7 / 7 ⇒ x1 + x2 + ...... + x7 = 55 × 7 = 385 x1 + x2 + ...... + x6 = 52 + 54 + 55 + 53 + 56 + 54 = 324 ∴ x7 = 385 - 324 = 61 kg ∴ weight of the seventh student is 61 kg.

Description : Given are the scores (out of 25) of 9 students in a Monday test : 14, 25, 17, 22, 20, 19, 10, 8 and 23 -Maths 9th

Last Answer : Ascending orders of scores is : 8, 10, 14, 17, 19, 20, 22, 23, 25 Now, new score = 8 + 10 + 14 + 17 + 19 + 20 + 22 + 23 + 25 / 9 = 158 / 9 = 17.5 marks Median = (n + 1 / 2)th observation because n is odd = (9 + 1 / 2)th observation = 5th observation = 19 marks.

Description : The scores of an English test out of 100 of 20 students are given below : 75, 69, 88, 55, 95, 88, 73, 64, 75, 98, 88, 95, 90, 95, 88, 44, 59, 67, 88, 99. -Maths 9th

Last Answer : Median=n=even =n/2=20/2=10th observation =98 Mode =88

Description : In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

Last Answer : For Mean : As we know that ⇒ x̅ = 41 + 39 + 48 + 52 + 46 + 62 + 54 + 40 + 96 + 52 + 98 + 40 + 42 + 52 + 60 / 15 x̅ = 822 / 15 = 54.8 ⇒ x̅ = 54.8 For Median : First of ... 52 For Mode : Make a frequency table for given data : Here, the marks 52 has the maximum frequency '3'. ∴ Mode = 52

Description : In a school marks obtained by 80 students are given in the table. Draw a histogram. Also, make frequency polygon. -Maths 9th

Last Answer : Here, class size = 135 - 305 = 10 ∴ Lower limit of first class interval is 305 - 10 / 2 = 300 Upper limit of first class interval is 305 + 10 / 2 = 310 Thus, first class interval is 300 - 310 Required histogram and frequency polygon is given on the graph paper .

Description : Following is the frequency distribution of total marks obtained by the students of different section of class-IX. -Maths 9th

Last Answer : Since class intervals of the given frequency distribution are not of equal width. We would make modifications in the lengths of the rectangles in the histogram, so that the areas of rectangles are proportional ... lengths as given in the last column .The histogram of the data is given below.

Description : Following table gives the distribution of students of sections A and B of a class according to the marks obtained by them. -Maths 9th

Last Answer : Clearly, the mean score of two sections A and B is same

Description : In a medical examination of students of a class, the following blood groups are recorded : -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 Number of students having blood group ‘B’ = 12 Required probability =12 / 40 = 3 / 10

Description : The given table shows the month of birth of 40 students of class IX of a particular section in a school. -Maths 9th

Last Answer : (a) P (later half of the year) = 23 / 40 (b) P (month having 31 days) = 26 / 40 = 13 / 20 (c) P(month having 30 days) = 10 / 40 = 1 / 4

Description : A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : Volume of milk in 1 glass =πr2h =π×(3.5)2×12=461.58cm2​for 1600 students milk needed is =1600×461.58=738258litre​

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : NEED ANSWER

Description : A school provides milk to the students daily in a cylindrical glasses of diameter 7 cm. -Maths 9th

Last Answer : According to question find the litres of milk is needed to serve 1600 students.

Description : In a diagnostic test in mathematics given to students, the following marks (out of 100) are recorded 46, 52, 48, 11, 41, 62, 54, 53, 96, 40, 98 and 44. -Maths 9th

Last Answer : Median will be a good representative of the data, because 1.each value occurs once. 2.the data is influenced by extreme values.

Description : The ratio of girls and boys in a class is 1: 3. Set up an equation between the students of a class and boys and then draw its graph. Also find the number of boys in a class of 40 students from the graph. -Maths 9th

Last Answer : Total number of boys and girl = 40, Ratio = 1 : 3 Number of girls be A and Number of boys be B. Ratio of number of girls and boys is 1 : 3, so Therefore 3A=B To find number of boys we ... the number 30 represents the number of girls. 40 as total on the line A = 10, which is the common equation.

Description : The blood groups of 30 students of class VIII are recorded as follows: -Maths 9th

Last Answer : Frequency Distribution Table Blood group O is most common as it has highest frequency, i.e., 12. Blood group AB is rarest as it has lowest frequency, i.e., 3.

Description : The heights of 50 students, measured to the nearest centimetre, -Maths 9th

Last Answer : Frequency distribution of above data in tabular form is given as: (ii) One conclusion we can draw from the above table is that more than 50% of students are shorter than 165 cm.

Description : A class consists of 50 students out of which 30 are girls. -Maths 9th

Last Answer : Mean marks scored by girls ( x̅1 ) = 73 Number of girls (n1) = 30 Mean marks scored by boys ( x̅2 ) = 71 Number of boys (n2) = 50 - 30 = 20 Mean score of the whole class ( x̅12 ) = n1 x̅1 + n2 x̅2 /n1 + n2 = 30 x 73 + 20 x 71/30 + 20 = 2190 + 1420/50 = 3610/50 x̅ 2 = 72.2

Description : The marks obtained (out of 100) by a class of 80 students are given below: -Maths 9th

Last Answer : In the given frequency distribution, the class intervals are not of equal width. Therefore, we would make modification in the lengths of the rectangle in the histogram so that the areas of rectangle ... draw rectangles with lengths as given in the last column. The histogram of data is given below:

Description : In a medical examination of students of a class, -Maths 9th

Last Answer : Total number of students = 10 + 13 + 12 + 5 = 40 P (a student has blood group B) = 12/40 = 3/10

Description : To know the opinion of the students about the -Maths 9th

Last Answer : (i) P (a student likes statistics) = Numbers of students who like statistics/Total number of students = 135/200 = 0.675 (ii) P (a students does not like statistics) = Numbers of students who dislikes statistics/Total number of students = 65/200 = 0.325

Description : Some students of a school started a campaign against -Maths 9th

Last Answer : Area of rectangle ABCD = 5 x 3 = 15 cm2 Cost incurred on one banner = ₹3 x 15 = ₹45 Cost incurred on 5 banners = 5 x ₹45 = ₹225 Social awareness, caring, cooperative.

Description : 95 students each from 102 schools participated -Maths 9th

Last Answer : 95 x 102 = (100 - 5)(100 + 2) 1002 + (- 5 + 2) x 100 + (- 5 x 2) = 10,000 - 300 – 10 = 9,690 Fraternity, patriotism.

Description : The number of sincere students (x) in a class -Maths 9th

Last Answer : x = 2 + 2y Sincere students always progress in life as they value time and channelise their talent in productive activities while a careless student always wastes his talent and time.

Description : A class consists of 80 students, 25 of them are girls and 55 boys. 10 of them are rich and 20 are fair complexioned. -Maths 9th

Last Answer : Let P (A) = Probability of selecting a fair complexioned person. ThenP(A) = \(rac{20}{80}\) = \(rac{1}{4}\)Let P(B) = Probability of selecting a rich person. Then P(B) = \(rac{10}{80}\) = \(rac{1}{8}\)Let P (C) = ... ) = \(rac{1}{4}\)x \(rac{1}{8}\)x \(rac{5}{16}\) = \(rac{5}{512}\) = 0.009.

Description : A problem in mathematics is given to four students A, B, C and D. Their chances of solving the problem -Maths 9th

Last Answer : Given, P(A) = \(rac{1}{2}\) ⇒ P(\(\bar{A}\)) = P(A not solving the problem) = 1 - \(rac{1}{2}\) = \(rac{1}{2}\)P(B) = \(rac{1}{3}\) ⇒ P(\(\bar{B}\)) = 1 - \(rac{1}{3}\) = \(rac{2}{3}\)P(C) = \ ... 3}{4}\)x \(rac{4}{5}\) = \(rac{1}{5}\)∴ P(problem will be solved) = 1 - \(rac{1}{5}\) = \(rac{4}{5}\).

Description : There are 5 professors and 6 students out of whom a committee of 2 professors and 3 students is to be formed such that a -Maths 9th

Last Answer : answer:

Description : Two students A and B solve an equation of the form x^2 + px + q = 0. A starts with a wrong value of p and obtains the roots as 2 and 6. -Maths 9th

Last Answer : Let αα and ββ be the roots of the quadratic equation x2+px+q=0x2+px+q=0 Given that, A starts with a wrong value of p and obtains the roots as 2 and 6. But this time q is correct. i.e., a product of roots ... 1 Now, from Eqs. (ii) and (iii), we get α=−3 and β=−4α=−3 and β=−4 which are correct roots.

Description : How do you refinance mortgage w/ bad credit now after finances change do to lost of job 2 years ago. found knew job about 3 month after but pays less.Asked about modification w/mortgage company said they don't participate. ?

Last Answer : Just because your mortgage company say they don't refinance does not mean you should stop. Make a list of the places that do participate and apply there to see if you can get the help you need.