If 80 is 72 What is full marks?

If 80 is 72 What is full marks?
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If 80 is 70% then full marks is 111.11... (repeating). Asomewhat unlikely value.
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A students maks were wrongly entered as 72 instead of 52. Due to that the average marks for the class got increased by half. The number of students in the class is. A) 50 B) 56 C) 46 D) 40

Description : A students maks were wrongly entered as 72 instead of 52. Due to that the average marks for the class got increased by half. The number of students in the class is. A) 50 B) 56 C) 46 D) 40

Last Answer : D)  Let there be y students in the class Total increase in marks = y *1/2 = y/2 Therefore y/2 = (72 – 52) y/2 = 20 y = 40

In an competitive examination, the average was found to be 72 marks. After detecting the errors the marks of 94 candidates had to be charged from 90 to 66 each, and the average is reduced to 64 marks. Find the total number of candidates who took the exam. A) 282 B) 382 C) 828 D) 200

Description : In an competitive examination, the average was found to be 72 marks. After detecting the errors the marks of 94 candidates had to be charged from 90 to 66 each, and the average is reduced to 64 marks. Find the total number of candidates who took the exam. A) 282 B) 382 C) 828 D) 200 

Last Answer :  A)  let the number of candidates be A total marks of candidates = 72A after detecting error the change of marks for one candidate = 90 - 66 = 24 marks change of marks for 94 candidates = 94*24 = ... after detecting error of N candidates = 64A then, 72A- 2256= 64A , 8A=2256 A=282

Sanjay gets 72 % marks in examinations. If these are 864 marks, find the maximum marks. a) 1050 b) 860 c) 1225 d) 1200 e) 1500

Description : Sanjay gets 72 % marks in examinations. If these are 864 marks, find the maximum marks. a) 1050 b) 860 c) 1225 d) 1200 e) 1500

Last Answer : Answer: D  Let the maximum marks be s  Then 72 % of s = 864  72/100 × s = 864  s = (846 × 100)/72  s = 86400/92  s = 1200 Therefore, maximum marks in the examinations are 1200.

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

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 .

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

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 .

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

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:

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Description : What is the average adult pulse rate? A. 140-150 B. 60-80 C. 72-80 D. 115-125

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According to India's Report of 2001 percentage of population living in rural areas is – (1) 70 (2) 72 (3) 74 (4) 80

Description : According to India's Report of 2001 percentage of population living in rural areas is – (1) 70 (2) 72 (3) 74 (4) 80

Last Answer : (2) 72 Explanation: As per the 2001 Census of India, in a total population of 1027 million in India, about 742 million (or 72.2%) lived in rural areas and 285 million (or 27.8%) lived in ... during 1991-2001. According to 2011 census, population of rural India is 31.1% and urban India is 68.9%.

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Last Answer : C. 72-80 (Answer)

An electric field is given as E = 6y 2 z i + 12xyz j + 6xy 2 k. An incremental path is given by dl = -3 i + 5 j – 2 k mm. The work done in moving a 2mC charge along the path if the location of the path is at p(0,2,5) is (in Joule) a) 0.64 b) 0.72 c) 0.78 d) 0.80

Description : An electric field is given as E = 6y 2 z i + 12xyz j + 6xy 2 k. An incremental path is given by dl = -3 i + 5 j – 2 k mm. The work done in moving a 2mC charge along the path if the location of the path is at p(0,2,5) is (in Joule) a) 0.64 b) 0.72 c) 0.78 d) 0.80

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?% of 725 + 14% of 2100 = 700 a) 64 b) 80 c) 72 d) 56 e) 48

Description : ?% of 725 + 14% of 2100 = 700 a) 64 b) 80 c) 72 d) 56 e) 48

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The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C,D and E becomes 79 kg. The weight of A is : a) 70 kg b) 72 kg c) 75 kg d) 80 kg e) 78 kg

Description : The average of weight of three men A,B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg.more than that of D, replaces A, then the average weight of B,C ... 79 kg. The weight of A is : a) 70 kg b) 72 kg c) 75 kg d) 80 kg e) 78 kg

Last Answer : A+B+C = 3×84=252 A+B+C+D= (4×80)=320 D = (320-252)=68 and E = (68+3)=71 Now, B+C+D+E = (4×79)=316 B+C+D=(316-71)=245 kg So, A = (320-245)=75 kg Answer: c)

Milling a hundred pounds of wheat should yield ____ percent straight flour. a. 72 to 75 c. 88 to 90 b. 75 to 80 d. 68 to 70

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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

The average age of Anu, banu and tonu is 74 yrs. 10years hence the average age of anu and tonu is 86 yrs.6yrs ago the average age of aarthi and banu was 72 yrs. Find the present age of aarthi. A) 80 B) 86 C) 84 D) 83

Description : The average age of Anu, banu and tonu is 74 yrs. 10years hence the average age of anu and tonu is 86 yrs.6yrs ago the average age of aarthi and banu was 72 yrs. Find the present age of aarthi. A) 80 B) 86 C) 84 D) 83

Last Answer : B) anu+ banu+ tonu = 74 * 3 = 222yrs 10yrs hence, anu + 10 + tonu +10 = 86 * 2 anu + tonu= 152 banu = 222-152 = 70 yrs aarthi -6 + banu - 6= 72 * 2 aarthi + banu = 144 + 12 = 156 aarthi age = 156-70 aarthi age = 86yrs

Use capital letters, full stops, question marks, commas and inverted commas wherever necessary in the following para-graph. -English 9th

Description : Use capital letters, full stops, question marks, commas and inverted commas wherever necessary in the following para-graph. -English 9th

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Nusselt number for full developed, laminar, constant property flow in a pipe at uniform heat flux is (A) 0.72 (B) 4.364 (C) 18 (D) 83

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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

3 faucet A,B,C can fill a cistern from empty to full in 45mins, 35mins, and 25 mins respectively. When the cistern is empty all the 3 faucets are opened. A,B,c discharge chemical solutions A,B,C respectively. What is the proportion of solution C in the liquid in the tank after 5 mins? A) 63/143 B) 72/151 c) 67/173 D) 81/167

Description : 3 faucet A,B,C can fill a cistern from empty to full in 45mins, 35mins, and 25 mins respectively. When the cistern is empty all the 3 faucets are opened. A,B,c discharge chemical solutions A,B,C respectively. What is ... the liquid in the tank after 5 mins? A) 63/143 B) 72/151 c) 67/173 D) 81/167

Last Answer : A Part filled by (A+B+C) in 5 mins = 5(1/45 +1/35 + 1/25) = 5 (35+45+63/1575) = 143/315 Part filled by c in 5 mins = 5/25 = 1/5 Required ratio = 1/5 * 315/143 =63/143.

A tap can fill a cistern in 11hours, but due to a leakage it took 13hours to fill the cistern. If the cistern is full, in what time will the cistern become empty due to leakage ? A) 73.30hrs B) 77.50hrs C) 71.5 hrs D) 72.30hrs E) None of these

Description : A tap can fill a cistern in 11hours, but due to a leakage it took 13hours to fill the cistern. If the cistern is full, in what time will the cistern become empty due to leakage ? A) 73.30hrs B) 77.50hrs C) 71.5 hrs D) 72.30hrs E) None of these

Last Answer : C 1/11-1/13 = 13-11/143 =>2/143 =>1/71.5 =>71.5hours