There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

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Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be – 3.5. Find the mean of the given numbers. -Maths 9th

Last Answer : Let x be the mean of 50 numbers. ∴ sum of 50 numbers = 50x Since each number is subtracted from 53. According to question, we have 53 × 50 - 50x / 50 = - 3.5 ⇒ 2650 - 50x = -175 ⇒ 50x = 2825 ⇒ x = 2825 / 50 = 56.5

Description : There are 50 numbers. Each number is subtracted from 53 and the mean of the number so obtained is found to be – 3.5. -Maths 9th

Last Answer : Find the mean of the given number is

Description : What should be subtracted from 27x^3 – 9x^2 – 6x – 5 to make it exactly divisible by (3x – 1) -Maths 9th

Last Answer : answer:

Description : Two numbers are in the ration 3:5. If 9 is subtracted from the numbers, the ratio becomes 12:23. The numbers are (A) 30, 50 (B) 36, 60 (C) 33, 55 (D) 42, 70

Last Answer : Answer: C

Description : What number must be subtracted from each of the numbers 32, 38, 17 and 20, so that the terms formed will be in proportion ?

Last Answer : What number must be subtracted from each of the numbers 32, 38, 17 and 20, so that the terms formed will be in proportion ?

Description : The number of positive integral values of m satisfying the inequalities 8m + 35 > 75 and 5m + 18 < 53 is -Maths 9th

Last Answer : answer:

Description : The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

Last Answer : Here the marks are out of 50 , so we find its percentage (i.e. out of 100)

Description : The table shows the marks obtained by a student in unit tests out of 50 : -Maths 9th

Last Answer : Here the marks are out of 50 , so we find its percentage (i.e. out of 100)

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 : 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 : In Fig. 8.53,ABCD is a parallelogram and E is the mid - point of AD. A line through D, drawn parallel to EB, meets AB produced at F and BC at L.Prove that (i) AF = 2DC (ii) DF = 2DL -Maths 9th

Last Answer : Given, E is mid point of AD Also EB∥DF ⇒ B is mid point of AF [mid--point theorem] so, AF=2AB (1) Since, ABCD is a parallelogram, CD=AB ⇒AF=2CD AD∥BC⇒LB∥AD In ΔFDA ⇒LB∥AD ⇒LDLF​=ABFB​=1 from (1) ⇒LF=LD so, DF=2DL

Description : What is the probability of getting 53 Sundays or 53 Tuesdays or 53 Thursdays in a non–leap year ? -Maths 9th

Last Answer : A non-leap year consists of 365 days. Therefore in a non-leap year there are 52 complete weeks and 1 day over which can be one of the seven days of the week. Possible outcomes n(S) = 7 = {Sunday, Monday, Tuesday, Wednesday, Thursday, ... ⇒ n(A) = 3∴ P(A) = \(rac{n(A)}{n(S)}\) = \(rac{3}{7}.\)

Description : Find four numbers, the sum of which is 45, so that if 2 is added to the first number, 2 is subtracted from the second number, the third number is multiplied by 2 and the fourth number is divided by 2, then the four numbers so produced are all the same. What are the four numbers? -Riddles

Last Answer : 8 + 2 = 10 12 - 2 = 10 5 x 2 = 10 20 ÷ 2 = 10 45

Description : If 6 years are subtracted from the present age of Anuj and the remainder is divided by 18, then the present age of his grandson Gopal is obtained.If Gopal is 2 years younger to Mohan whose age is 5 years, then what is the age of Anuj? A.44 B.60 C.80 D.92

Last Answer : Answer : B (60) Explanation - Let Anuj's age be X Gopal is 2 years younger than Mohan, so Gopal is 3 years (i.e 5 - 2 = 3) If Arun had born 6 years before, his age would had been X - 6. As per the ... as that of Gokul's age. i.e. (X - 6) /18= 3 X-6 =3 x 18 x = 60

Description : How many small cubes each of 96 cm^2 surface area can be formed from the material obtained by melting a larger cube of 384 cm^2 surface area ? -Maths 9th

Last Answer : answer:

Description : Mean of 50 observations was found to be 80.4. -Maths 9th

Last Answer : Here, n = 50, x̅ = 80.4 So, x̅ = ∑ xi/n ⇒ 80.4 = ∑ xi/50 ⇒ ∑ xi = 80.4 x 50 = 4020 Correct value of ∑ xi = 4020 - 69 + 96 = 4047 Correct mean = Correct value of ∑ xi/n = 4047/50 = 80.94

Description : Plot the points A (5, 5) and B (–5, 5) in cartesian plane. Join AB, OA and OB. Name the type of triangle so obtained. -Maths 9th

Last Answer : Solution :- The obtained triangle is an isosceles triangle.

Description : A right DABC with sides 5 cm, 12 cm and 13 cm is revolved about the side 12 cm. What is the volume of the solid so obtained ? -Maths 9th

Last Answer : From the figure it is clear that a cone is formed. Here, h = 12 cm, r = 5 cm Volume of cone = = 314 cm3

Description : If x is subtracted from the numbers 7, 31 and 199, then the remainders will be in continued proportion. What is the value of x? a) 5 b) 3 c) 8 d) 4 e) None of these

Last Answer : Answer: B (7-x)/(31-x)= (31-x)/(199-x) gives x=3

Description : What two square numbers when subtracted make 9?

Last Answer : They are 25 minus 16 = 9

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

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

Description : A box contains 50 bolts and 150 nuts. On checking the box, it was found that half of the bolts and half of the nuts are rusted. -Maths 9th

Last Answer : Total number of nuts and bolts in the box = 150 + 50 = 200 Number of nuts and bolts rusted = 1 / 2 × 200 = 100 P(a rusted nut or bolt) = 100 / 200 = 1 / 2

Description : A box contains 50 bolts and 150 nuts. On checking the box, it was found that half of the bolts and half of the nuts are rusted. -Maths 9th

Last Answer : Total number of nuts and bolts in the box = 150 + 50 = 200 Number of nuts and bolts rusted = 1 / 2 × 200 = 100 P(a rusted nut or bolt) = 100 / 200 = 1 / 2

Description : At a petrol pump, it was found that out of 50 vehicles -Maths 9th

Last Answer : (a) 11/25 (b) By using more of public transport wherever possible and using substitutes of petrol such as diesel and CNG.

Description : An umbrella is made by stitching 10 triangular pieces of cloth of two different colours (see figure), each piece measuring 20 cm, 50 cm and 50 cm. -Maths 9th

Last Answer : Let the sides of each triangular piece be a = 20 cm, b = 50 cm, c = 50 cm

Description : A bus stop is barricaded from the remaining part of the road, by using 50 hollow cones made of recycled cardboard. Each cone has a base diameter of 40 cm and height 1 m. -Maths 9th

Last Answer : Given: Radius of cone, r = diameter/2 = 40/2 cm = 20cm = 0.2 m Height of cone, h = 1m Slant height of cone is l, and l2 = (r2+h2) Using given values, l2 = (0.22+12) = (1.04) Or l ... (32.028 12) = Rs.384.336 = Rs.384.34 (approximately) Therefore, the cost of painting all these cones is Rs. 384.34.

Description : 50 circular plates each of radius .... Find its total surface area -Maths 9th

Last Answer : 50 circular plate diameter of plate =14cm Radius of plate = d/2=214​=7cm thickness =0.5cm → Height of the cylinder = No. of plates × thickness of plate =50×0.5 =25cm → total surface Area = curved surface + 2 × area of base =2πr.h+2×πr2 =2πr(h+r) =2π×7(25+7) =2×722​×7×32 =1408cm2

Description : Plot the points a(5,5) and b(-5,5) in the cartesian plane .join OA AB and OB name the figure obtained and find its area -Maths 9th

Last Answer : This answer was deleted by our moderators...

Description : Find the polynomial of least degree which should be subtracted from the polynomial x4 + 2x3 – 4x2 + 6x – 3 so that it is exactly divisible by x2 – x + 1. -Maths 10th

Last Answer : Here, p(x) = x4 + 2x3 - 4x2 + 6x - 3, g(x) = x2 - x +1 On dividing p(x) by g(x) Therefore (x-1) must be subtracted from the polynomial p(x) to make it divisible by g(x).

Description : show the following numbers on the number line. (a) 0.2 (b) 1.9 (c) 1.1 (d) 2.5 -Maths 9th

Last Answer : (a) 0.2 lies between the points 0 and 1 on the number line. The space between 0 and 1 is divided into 10 equal parts. Therefore each equal part will be equal to one-tenth. 0.2 is the second point ... parts. Therefore each equal part will be equal to one-tenth. 2.5 is the fifth point between 2 and 3

Description : Determine the mean of first 10 natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 / 10 = 55 / 10 = 5.5

Description : Determine the mean of first 10 natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 / 10 = 55 / 10 = 5.5

Description : Find the arithmetic mean of first-five natural numbers. -Maths 9th

Last Answer : Mean = 1 + 2 + 3 + 4 + 5/5 = 15/5 = 3

Description : The mean of ten numbers is 55. -Maths 9th

Last Answer : Solution :-

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 : Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

Description : The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Last Answer : According to question the mid-points of the sides of a rhombus, taken in order.

Description : The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

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 : Plot the following points and write the name of the figure obtained by joining, them in order -Maths 9th

Last Answer : Let X’ OX and Y’ OY be the coordinate axes and mark point on it. Here, point P(-3,2) lies in II quadrant, Q(-7,-3) lies in III quadrant, R(6, -3) lies in IV quadrant and S(2,2) lies in I quadrant. Plotting the points on the graph paper, the figure obtained is trapezium PQRS.

Description : The figure obtained by joining the mid-points of the sides of a rhombus, taken in order, is -Maths 9th

Last Answer : According to question the mid-points of the sides of a rhombus, taken in order.

Description : The figure obtained by joining the mid-points of the adjacent sides of a rectangle of sides 8 cm and 6 cm, is -Maths 9th

Last Answer : Here, length of rectangle ABCD = 8 cm and breadth of rectangle ABCD = 6.cm Let E, F, G and H are the mid-points of the sides of rectangle ABCD, then EFGH is a rhombus.

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 percentage of marks obtained by a student in monthly unit tests are given below. -Maths 9th

Last Answer : (i) Number of tests in which the student scored more than 70% marks = 3 ∴ P(more than 70% marks) = 3/6 = 1/2 (ii) Number of tests in which the student scored less than 70% marks = 3 ∴ P(less ... ) Number of tests in which the student scored at least 60% marks = 5 ∴ P(at least 60% marks) = 5/6