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

1 Answer

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

Related questions

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 : 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 : 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 : 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 : 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 : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

Description : In a class, there are x girls and y boys, a student is selected at random, then find the probability of selecting a boy. -Maths 9th

Last Answer : Number of boys = y Total students = (x + y) Thus,P(a boy)= y/(x+y)

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 : The average age of students of a college is 31.6 yrs. The average age of boys in the class is 32.8 yrs and that of girls is 30.8. The ratio of number of boys to the number of girls in the class. A) 2:3 B) 3:1 C) 1:3 D) 3:2

Last Answer : A) Let the ratio be D:1.then, (D * 32.8) + ( 1* 30.8)= (D+1) * 31.6 32.8 D +30.8 = 31.6 D + 31.6 32.8 D – 31.6D = 31.6 – 30.8 1.2D = 0.8 D = 0.8/1/2 = 0.4/0.6 D = 2/3 Required ratio = 2/3 :1 = 2:3.

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 : 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 : 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 : 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 : 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 : 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).

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).

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

Last Answer : NEED ANSWER

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : NEED ANSWER

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

Last Answer : B is the correct answer

Description : A grouped frequency distribution table with classes of equal sizes using 63-72 (72 included) as one of the class is constructed for the following data 30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88, -Maths 9th

Last Answer : (b) We arrange the given data into groups like 13-22,23-32 103-112. (since, our data is from 14 to 112). The class width in this case is 9. Now, the given data can be arranged in tabular form as follows. Hence, the number of classes in distribution will be 10.

Description : An article manufactured by a company consists of two parts X and Y. In the process of manufacture of the part X, 9 out of 100 parts may be defective. -Maths 9th

Last Answer : (b) \(rac{1}{4}\)Since A and B are mutually exclusive and exhaustive events, therefore, P(A ∩ B) = 0, P(A ∪ B) = 1 We know that, P(A ∪ B) = P(A) + P(B) - P(A ∩ B) ⇒ 1 = P(A) + 3 P(A) (Given P(B) ... ∵ P(A) + P(B) = 1)⇒ P(\(\bar{B}\)) = 1 - P(B) = 1 - \(rac{3}{4}\) = \(rac{1}{4}\).

Description : Three girls Reshma, Salma and Mandeep are playing a game by standing on a circle of radius 5 m drawn in a park. -Maths 9th

Last Answer : Solution :- Let R, S and M represent the position of Reshma, Salma and Mandeep respectively. Clearly △RSM is an isosceles triangle as RS = SM = 6m Join OS which intersect RM at A. In △ROS and △MOS OR = OM ( ... . ∴ RM = 2RA RM = 2 x 4.8 = 9.6m Hence, distance between Reshma and Mandeep is 9.6m.

Description : The following data on the number of girls to -Maths 9th

Last Answer : Solution :- (i) (ii) Number of girls to the nearest ten per thousand boys are maximum in scheduled tribes whereas they are minimun in urban areas.

Description : The following data on the number of girls to the nearest -Maths 9th

Last Answer : (i) (ii) The gender equity exists most in scheduled tribe and least in urban areas. Yes, gender equity leads to economic growth which, in turn, helps in the development of a society.

Description : Four boys and three girls stand in a queue for an interview. What is the probability that they will be in alternate positions ? -Maths 9th

Last Answer : Total number of ways of arranging 4 boys and 3 girls, i.e., 7 people in a queue (row) = n(S) = 7! Let A : Event in which the 4 boys and 3 girls occupy alternate position. This is possible when the ... {4 imes3 imes2 imes1 imes3 imes2 imes1}{7 imes6 imes5 imes4 imes3 imes2 imes1}\) = \(rac{1}{35}.\)

Description : 6 boys and 6 girls are sitting in a row randomly. The probability that boys and girls sit alternately is -Maths 9th

Last Answer : (b) \(rac{1}{462}\)Let S be the sample space. Then, n(S) = Number of ways in which 6 boys and 6 girls can sit in a row = 12! Let E : Event of 6 girls and 6 boys sitting alternately. Then, the ... )= \(rac{2 imes6 imes5 imes4 imes3 imes2 imes1}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{462}\).

Description : 6 boys and 6 girls are seated in a row. Probability that all the boys sit together is -Maths 9th

Last Answer : (d) \(rac{1}{132}\)Let S be the sample space for arranging 6 boys and 6 girls in a row. Then, n(S) = 12! If all 6 boys are to sit together, then consider the 6 boys as one entity. Now the ... ) = \(rac{7 imes6 imes5 imes4 imes3 imes2}{12 imes11 imes10 imes9 imes8 imes7}\) = \(rac{1}{132}\).

Description : A group of 2n boys and 2n girls is divided at random into two equal batches. -Maths 9th

Last Answer : (c) \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)Total number of boys and girls = 2n + 2n = 4n Since, there are two equal batches, each batch has 2n members ∴ Let S (Sample space) : Selecting one batch out of 2 ⇒ S : ... )2∴ Required probability = \(rac{n(E)}{n(S)}\) = \(rac{(^{2n}C_n)^2}{^{4n}C_{2n}}\)

Description : There are 5 boys and 3 girls. In how many ways can they stand in a row so that no two girls are together? -Maths 9th

Last Answer : Have the 55 boys stand in a line. This can be done in 5!5! ways. For the moment, add a boy at each end, who will be removed when we're done. Now send the 33 girls, one at a time, to ... that each girl can either wedge herself between two boys or else stand next to a boy at one end or the other.

Description : There are 6 numbered chairs placed around a circular table. 3 boys and 3 girls want to sit on them such that neither of two boys nor two girls -Maths 9th

Last Answer : Since the chairs are numbered, so they are distinguishable. Therefore 3 boys can be arranged on 3 alternate chairs in 3! ways. 3 girls can be arrenged in 3! ways Also, the girls can be seated before the boys. Total number of required ways = 3! × 3! + 3! × 3! = 2 × (3!)2

Description : A storage tank consists of a circular cylinder with a hemisphere adjoined on either end. If the external diameter of the cylinder be 1.4 m and its length be 8 m, find the cost of painting it on the outside at the rate of Rs. 10 per m -Maths 9th

Last Answer : Answer We have, r=0.7m, h=8m ∴ Total surface area = 2πr2+2πrh=2πr(r+h)=2×722​×0.7×8.7m2 Required cost = Rs. {2×722​×0.7×8.7×10}=Rs.382.80

Description : A lead pencil consists of a cylinder of wood -Maths 9th

Last Answer : Radius of solid graphite cylinder (r) = 1/2 mm = 1/2 x 1/10 = 1/20 cm Length of graphite cylinder (h) = 14 cm Volume of graphite cylinder (h) = πr2h = 22/7 x 1/20 x 1/20 x 14 = 0.11 cm3 Radius ... Volume of the wood = Volume of the pencil - Volume of the graphite = 5.39 cm3 - 0 .11 cm3 = 5.28 cm3

Description : A machine manufactured by a firm consists of two parts A and B. But of 100 A’s manufactured, -Maths 9th

Last Answer : Let E = {part A is defective}, F = {part B is defective}.Then, P (E) = \(rac{9}{100}\), P(F) = \(rac{5}{100}\)∴ P(\(\bar{E}\)) = 1 - \(rac{9}{100}\) = \(rac{91}{100}\) and P(\(\bar{F}\)) = 1 - \(rac{5 ... . P(\(\bar{F}\)) = \(rac{91}{100}\) x \(rac{95}{100}\)= 0.91 x 0.95 = 0.8645 = 0.86 to 2 d.p.

Description : In a class there are 16 girls out of 40 students in total. Girls as a percentage of total students are `"_____"`

Last Answer : In a class there are 16 girls out of 40 students in total. Girls as a percentage of total students are `"_____"`

Description : there are 24 students in a class 8 of the student are boys a) write the amount of boys in the class as a fraction in its simplest form b) write the amount of girls in the class as a fraction in its simplest form?

Last Answer : two thirds

Description : In a class there are 15 boys and 10 girls. Three students are selected at random. The difference between the probability that 2 boys and 1 girl are selected compared to 1 boy and 2 girls are selected is: a) 23/78 b) 19/88 c) 15/92 d) 4/23 e) 7/46

Last Answer : 2 boys, 1 gi(Prl = (15c2×10c1) / 25c3 = 1050/2300 1 boy, 2 girls = (15c1×10c2) / 25c3 = 675/2300 Difference = (1050 - 675)/2300 = 375/2300 = 15/92 Answer: c)

Description : The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ratio of the number of boys to the number of girls in the class is: A. 1 : 4 B. 2 : 3 C. 3 : 4 D. 4 : 2 E. None of these

Last Answer : Answer – B (2 : 3) Explanation – Let the ratio be k:1 Then, k x 16.4 + 1 x 15.4 = (k + 1) x 15.8 (16.4 – 15.8) k = (15.8 – 15.4) k=2/3

Description : The male students of your class are annoyed with you' on the pretext you have been easily approached the girl students and do a favor to them (girls). In such an embarrassing situation how would you like ... it is difficult for you refuse the request made by female students. (d) None of the above.

Last Answer : (a) You will tell the male students that girls have no option except to depend on the school teacher for their academic assistance.

Description : The male students in your class are annoyed with you on the pretext you that have a favour to the female students. In such an embarrassing situation how would you like to control them in class? ... students that it is difficult for you refuse the request made by female students D) None of these

Last Answer : B) You will justify that most of the female students are more sincere towards their studies than male students.

Description : Number of students in 4th and 5th class is in the ratio 6 : 11. 40% in class 4 are girls and 48% in class 5 are girls. What percentage of students in both the classes are boys? A) 62.5% B) 54.8% C) 52.6% D) 55.8% E) 53.5%

Last Answer : Answer: B Total students in both = 6x+11x = 17x Boys in class 4 = (60/100)*6x = 360x/100 Boys in class 5 = (52/100)*11x = 572x/100 So total boys = 360x/100 + 572x/100 = 932x/100 = 9.32x % of boys = [9.32x/17x] * 100=54.8%

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 : 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.