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

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
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(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.
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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

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

A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data -Maths 9th

Description : A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data -Maths 9th

Last Answer : NEED ANSWER

A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data -Maths 9th

Description : A grouped frequency table with class intervals of equal sizes using 250-270 (270 not included in this interval) as one of the class interval is constructed for the following data -Maths 9th

Last Answer : (c) Here, we arrange the given data into groups like 210-230, 230-250 390-410. (since, our data is from 210 to 406). The class width in this case is 20. Now, the given data can be arrange in tabular form as follows

Nesha scored 120 out of 150 in English, 120 out of 180 in mathematics and 160 out of 200 in Science. Find Nesha’s score as percentage: (i) in Mathematics (ii) in all the three subjects (on the whole). a) 66 2/3, 78 22/51 b) 71 23/5, 45 7/31 c) 54 13/33, 82 3/4 d) 60 3/4, 76 32/41

Description : Nesha scored 120 out of 150 in English, 120 out of 180 in mathematics and 160 out of 200 in Science. Find Nesha’s score as percentage: (i) in Mathematics (ii) in all the three subjects (on the whole). a) 66 2/3, 78 22/51 b) 71 23/5, 45 7/31 c) 54 13/33, 82 3/4 d) 60 3/4, 76 32/41 

Last Answer :  Answer: A (i) Percentage scored in Mathematics = 120/180 100  = 12000/180  =1200/18  = 66 2 /3 % (ii) Total maximum of all the three subjects = 150 + 160 + 200 = 510 and Total score in the ... percentage on the whole = (400/510 100)  = (40000/510)  = 4000/51  = 78 22/51 %

A group of 210 students appeared in some test. The mean of of students is found to be 60. The mean of the remaining students is found to be 78. The mean of the whole group will be: (A) 80 (B) 76 (C) 74 (D) 72

Description : A group of 210 students appeared in some test. The mean of of students is found to be 60. The mean of the remaining students is found to be 78. The mean of the whole group will be: (A) 80 (B) 76 (C) 74 (D) 72

Last Answer : Answer: D

A dumper is normally filled in 18 hours but takes 6 hours longer to fill because of a leak in the bottom of the dumper. If the dumper is full the leak will empty it in how many hours? A) 76 hours B) 78 hours C) 72 hours D) 74 hours

Description : A dumper is normally filled in 18 hours but takes 6 hours longer to fill because of a leak in the bottom of the dumper. If the dumper is full the leak will empty it in how many hours? A) 76 hours B) 78 hours C) 72 hours D) 74 hours

Last Answer : C Work done by leak in 1 hr=(1/18-1/24) =>4-3/72 =>1/72 Leak will empty the dumper in 72 hours

Harish buys petrol at rs 21, Rs.24 and Rs.27 per litre for 3 successive years. What approximately is the average cost per litre of petrol if he spends Rs.12000 each year? A) 23.74 B) 54.76 C) 24.66 D) 28.99

Description : Harish buys petrol at rs 21, Rs.24 and Rs.27 per litre for 3 successive years. What approximately is the average cost per litre of petrol if he spends Rs.12000 each year? A) 23.74 B) 54.76 C) 24.66 D) 28.99

Last Answer : A)  Total quantity of petrol consumed in 3 yr = (12000/21 +12000/24+12000/27) =12000(1/2 + 1/24 + 1/27) =12000(72+63+56/1512) =12000(191/1512) =(2292000/1512)litres Total amount spent = rs(3 *12000) = Rs.36000 Average cost = Rs(36000*1512/2292000) =Rs.23.74

The energy consumption and production patterns in a chemical plant over a 9 month period is provided in the table below; Month 1 2 3 4 5 6 7 8 9 Production in Tonnes / month 493 297 381 479 585 440 234 239 239 Energy Consumption MWh /month 78.2 75.7 76.3 76.1 78.1 70.7 73.7 64.4 72.1 Estimate the cumulative energy savings at end of the 9th month and give your inference on the result ? ( consider 9 month data for evaluation for predicted energy consumption)

Description : The energy consumption and production patterns in a chemical plant over a 9 month period is provided in  the table below; Month 1 2 3 4 5 6 7 8 9 Production in Tonnes / month 493 ... and give your inference on the  result ? ( consider 9 month data for evaluation for predicted energy consumption)

Last Answer : It is required to use the equations Y= mX + C and  nC + mΣX = ΣY  cΣX + mΣX2 = ΣXY Month X =  Production in  Tonnes /  month Y =Energy  Consumption MWh  /month  X 2 XY 1 493 78.2 ... 73.7 54756 17240.63 8 239 64.4 57121 15402.96 9 239 72.1 57121 17228.98 3387 665.3 1410683 253221

The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Last Answer : NEED ANSWER

The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Description : The median of the data 78, 56, 22, 34, 45, 54, 39, 68, 54 and 84 is -Maths 9th

Last Answer : According to question find the median of the data

Prepare a continuous grouped frequency distribution from the following data: -Maths 9th

Description : Prepare a continuous grouped frequency distribution from the following data: -Maths 9th

Last Answer : If m is mid-point of a class and h is the class size, lower and upper limits of the class intervals are m - h/2 and m + h/2 respectively. Class size (h) = 15 - 5 = 10 So, the class interval formed ... 2) - (5 + 10/2) i.e., 0 - 10 Continuing in the same manner, the continuous classes formed are:

What is the spectrum of stereophonic FM signal? A. 19 to 38 kHz B. 30 to 53 kHz C. 59.5 to 74.5 kHz D. 88 to 108 MHz

Description : What is the spectrum of stereophonic FM signal? A. 19 to 38 kHz B. 30 to 53 kHz C. 59.5 to 74.5 kHz D. 88 to 108 MHz

Last Answer : B. 30 to 53 kHz

There are two groups of a class consisting of 72 and 88 students respectively. If the average weight of 1st group is 80kg and the weight of 2nd group is 70kg. Find the average weight of whole class? A) 64kg B) 74.5kg C) 84.5kg D) 54kg

Description : There are two groups of a class consisting of 72 and 88 students respectively. If the average weight of 1st group is 80kg and the weight of 2nd group is 70kg. Find the average weight of whole class? A) 64kg B) 74.5kg C) 84.5kg D) 54kg

Last Answer :  B)  Total weight of (72 + 88) = (72*80) + (88 * 70)kg = (5760 + 6160)kg  = 11,920kg  Average weight of the whole class = 11920 / 160  = 74.5 kg

The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Description : The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Last Answer : NEED ANSWER

The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Description : The width of each of five continuous classes in a frequency distribution is 5 and the lower class limit of the lowest class is 10. -Maths 9th

Last Answer : (c) Let x and y be the upper and lower class limit of frequency distribution. Given, width of the class = 5 ⇒ x-y= 5 (i) Also, given lower class (y) = 10 On putting y = 10 in Eq. (i), we get ... 20-25, 25-30 and 30-35. Thus, the highest class is 30-35, Hence, the upper limit of this class is 35.

P,Q, R enter into a partnership.P initially invests Rs 54 lakh and withdraws Rs 18 lakhs after 4 years. Q initially Rs 72 lakh ans adds another Rs 18 lakhs after 6 years and R invests Rs 108 lakh and adds another Rs 18 lakh few years. At the end of 10 years profit Of R is equal to sum of profit of P and Q. Then for how many years did R invests Rs 126 lakh per annum A) 5 B) 7 C) 8 D) 11

Description : P,Q, R enter into a partnership.P initially invests Rs 54 lakh and withdraws Rs 18 lakhs after 4 years. Q initially Rs 72 lakh ans adds another Rs 18 lakhs after 6 years and R invests Rs 108 lakh and adds another ... Then for how many years did R invests Rs 126 lakh per annum A) 5 B) 7 C) 8 D) 11 

Last Answer : Answer: C) Ratio of their profit P, Q, R is 54*4+36*6:72*6+90*4:108*x+126*(10-x) =12:22:35 - 0.5x Profit of R=profit of P +profit of Q 35-0.5x=12+22 X=2 year The number of years for R invests Rs 126 lakh =10-2=8years

67,74,78,?,92,94 a) 86 b) 85 c) 83 d) 84 e) 88

Description : 67,74,78,?,92,94 a) 86 b) 85 c) 83 d) 84 e) 88

Last Answer : The next term is obtained by adding the unit’s digit to the number. Hence the missing number =78+8=86 Answer: a)

A moist soil sample of volume 60 cc. weighs 108 g and its dried weight is 86.4 g. If its absolute density is 2.52, the degree of saturation is (A) 54 % (B) 64 % (C) 74 % (D) 84 %

Description : A moist soil sample of volume 60 cc. weighs 108 g and its dried weight is 86.4 g. If its absolute density is 2.52, the degree of saturation is (A) 54 % (B) 64 % (C) 74 % (D) 84 %

Last Answer : (D) 84 %

Assume postal rates for 'light letters' are: $0.25 up to 10 grams; $0.35 up to 50 grams; $0.45 up to 75 grams; $0.55 up to 100 grams. Which test inputs (in grams) would be selected using boundary value analysis? A. 0, 9,19, 49, 50, 74, 75, 99,100 B. 10, 50, 75,100, 250, 1000 C. 0, 1,10,11, 50, 51, 75, 76,100,101 D. 25, 26, 35, 36, 45, 46, 55, 56

Description : Assume postal rates for 'light letters' are: $0.25 up to 10 grams; $0.35 up to 50 grams; $0.45 up to 75 grams; $0.55 up to 100 grams. Which test inputs (in grams) would be selected using boundary value analysis? A. 0, 9,19, ... C. 0, 1,10,11, 50, 51, 75, 76,100,101 D. 25, 26, 35, 36, 45, 46, 55, 56

Last Answer : C. 0, 1,10,11, 50, 51, 75, 76,100,101

The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Description : The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Last Answer : NEED ANSWER

The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Description : The class marks of a continuous distribution are 1.04, 1.14, 1.24, 1.34, 1.44,1.54 and 1.64. -Maths 9th

Last Answer : It is not correct. Because the difference between two consecutive class marks should be equal to the class size. Here, difference between two consecutive marks is 0.1 and class size of 1.55-1.73 is 0.18, which are not equal.

Find the orbital dipole moment in a field of dipoles of radius 20cm and angular velocity of 2m/s(in 10 -22 order) a) 64 b) 76 c) 54 d) 78

Description : Find the orbital dipole moment in a field of dipoles of radius 20cm and angular velocity of 2m/s(in 10 -22 order) a) 64 b) 76 c) 54 d) 78

Last Answer : a) 64

19/14 x 12/72 x 18/29 x 767 = ? (Take 2 digits after the decimal point) a) 110.69 b) 118.78 c) 111.56 d) None e) 108.98

Description : 19/14 x 12/72 x 18/29 x 767 = ? (Take 2 digits after the decimal point) a) 110.69 b) 118.78 c) 111.56 d) None e) 108.98

Last Answer : 19/14 x 12/72 x 18/29 x 767 =? Sol: 1.35 x 0.16 x 0.62 x 767 = ? 102.71 = ? Answer: d)

A, B and C play cricket. A's runs are to B's runs and B's runs are to C's as 3:2 , 3:2. They get altogether 342 runs. How many runs did A make? A) 162 B) 108 C) 72 D) 78 E) None

Description : A, B and C play cricket. A's runs are to B's runs and B's runs are to C's as 3:2 , 3:2. They get altogether 342 runs. How many runs did A make? A) 162 B) 108 C) 72 D) 78 E) None

Last Answer : Answer: A A:B = 3:2 = 9:6; B:C = 3:2 = 6:4 (making B equal) Therefore, A:B:C = 9:6:4 Therefore, the runs made by A = (9/19) X 342 = 162.

A sample of natural gas containing 80% methane (CH4 ) and rest nitrogen (N2 ) is burnt with 20% excess air. With 80% of the combustibles producing CO2 and the remainder going to CO, the Orsat analysis in volume percent is (A) CO2 : 6.26, CO : 1.56, O2 : 3.91, H2O :15.66, N2 : 72.60 (B) CO2 : 7.42, CO : 1.86, O2 : 4.64, N2 :86.02 (C) CO2 : 6.39, CO : 1.60, O2 : 3.99, H2O:25.96, N2 :72.06 (D) CO2 : 7.60, CO : 1.90, O2 : 4.75, N2 : 85.74

Description : A sample of natural gas containing 80% methane (CH4 ) and rest nitrogen (N2 ) is burnt with 20% excess air. With 80% of the combustibles producing CO2 and the remainder going to CO, the Orsat analysis in volume percent is (A) CO2 : 6 ... 96, N2 :72.06 (D) CO2 : 7.60, CO : 1.90, O2 : 4.75, N2 : 85.74

Last Answer : (B) CO2 : 7.42, CO : 1.86, O2 : 4.64, N2 :86.02

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : NEED ANSWER

For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Description : For drawing a frequency polygon of a continuous frequency distribution, we plot the points whose ordinates are the frequencies of the respective classes and abscissae are, respectively -Maths 9th

Last Answer : (c) Class marks i.e., the mid-point of the classes are abscissa of the points, which we plot for frequency polygon.

A batsman average for 20 innings is 25 runs. His highest score exceeds his lowest score by 64 runs. If these 2 innings are excluded, the average of the remaining 18 innings is 24 runs. The highest score of the player is. A) 66 B) 72 C) 77 D) 88

Description : A batsman average for 20 innings is 25 runs. His highest score exceeds his lowest score by 64 runs. If these 2 innings are excluded, the average of the remaining 18 innings is 24 runs. The highest score of the player is. A) 66 B) 72 C) 77 D) 88

Last Answer : A) Let the highest score be ‘x’ Then lowest score = x-64 Then (25 *20)- 18 * 24 = 432 X+X-64=68 X=66

The average age of 160 boys in a class is 58 yrs. The average group of 30 boys in the class is 42 yrs and the average of another group of 50 boys in the class is 36 years. What is the average age of the remaining boys? A) 72.58 B) 74.25 C) 77.75 D) 75.68

Description : The average age of 160 boys in a class is 58 yrs. The average group of 30 boys in the class is 42 yrs and the average of another group of 50 boys in the class is 36 years. What is the average age of the remaining boys? A) 72.58 B) 74.25 C) 77.75 D) 75.68 

Last Answer : C) Total age of 160 boys = 160* 58= 9280 total age of 30 boys = 30 * 42= 1260 total age of next 50boys = 50 * 36= 1800 average of the remaining boys = [(9280-{1260+1800})/[160 - (30 + 50)] =>9280-3060/80 =>6220/80 =77.75yrs

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!!!

what is the next number in this pattern0.54, 0.6, 0.66, 0.72?

Description : what is the next number in this pattern0.54, 0.6, 0.66, 0.72?

Last Answer : i think it is 0.8 what about you

You are a project manager for Move It Now trucking company. Your company specializes in moving household goods across the city or across the country. Your A. 54 B. 66 C. 60 D. 30project involves upgrading the nationwide computer network for the company. Your lead engineer has given you the following estimates for a critical path activity: 60 days most likely, 72 days pessimistic, 48 days optimistic. What is the weighted average or expected value?

Description : You are a project manager for Move It Now trucking company. Your company specializes in moving household goods across the city or across the country. Your project involves upgrading the nationwide computer network for the ... the weighted average or expected value? A. 54 B. 66 C. 60 D. 30

Last Answer : C. 60

In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Description : In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Last Answer : Frequency distribution table

In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Description : In figure, there is a histogram depicting daily wages of workers in d factory. Construct the frequency distribution table. -Maths 9th

Last Answer : Frequency distribution table

The average age of 150 students in a class is 40% of the number of students in the class and the average age of a group of 50 students present in the class is 32yrs and the average age of another 50 students in the class is 36yrs. What is the average age of the remaining students in the class? A) 102 B) 118 C) 112 D) 108

Description : The average age of 150 students in a class is 40% of the number of students in the class and the average age of a group of 50 students present in the class is 32yrs and the average age of another 50 students ... is the average age of the remaining students in the class? A) 102 B) 118 C) 112 D) 108

Last Answer : C) 150 students average 150×40/100=60 years According to the question, 150×60=50×32+50×36+50×X 50×X=9000-1600-1800 50x=5600 X=112

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Description : The median of the data 26,56,32,33,60,17,34,29,45 is 33. If 26 is replaced by 62, then find the new median. -Maths 9th

Last Answer : Here, the given data in ascending order is 17, 29, 32, 33, 34, 45, 56, 60, 62 Now median is (9 + 1 / 2)th term i.e. , 5th term Hence, new median is 34.

What is the mean median mode and range of 78 25 57 98 20 48 45 37 88 24 19?

Description : What is the mean median mode and range of 78 25 57 98 20 48 45 37 88 24 19?

Last Answer : 19, 20, 24, 25, 37, 45, 48, 57, 78, 88, 98Mean: 49Median: 45Range: 79There is no mode.

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

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

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

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

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

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 class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

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

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.

The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Description : The class marks of a frequency distribution are 104, 114, 124, 134, 144, 154, 164. -Maths 9th

Last Answer : Since the class marks are equally spaced. ∴ Class size = 114 - 104 = 10 If a is a class mark and h is size of class interval, then lower limit and upper limit of the class interval area a - h / 2 and a + h / ... are 99 - 109, 109 - 119, 119 - 129, 129 - 139, 139 - 149, 149 - 159, 159 - 169.

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

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.

In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is -Maths 9th

Description : In a frequency distribution, the mid value of a class is 10 and the width of the class is 6. The lower limit of the class is -Maths 9th

Last Answer : NEED ANSWER

If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Description : If m is the mid-point and l is the upper class limit of a class in a continuous frequency distribution, then lower class limit of the class is -Maths 9th

Last Answer : NEED ANSWER

The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Description : The class marks of a frequency distribution are given as follows 15, 20, 25, …. The class corresponding to the class mark 20 is -Maths 9th

Last Answer : NEED ANSWER