The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, 48, 29, 40, 45, 48, 42, 23, 35. Find the average mark

The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, 48, 29, 40, 45, 48, 42, 23, 35. Find the average mark
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The marks obtained by 15 students in an examination are given below : 40, 20, 24, 19,20, 35, 12, ... , 23, 35. Find the average mark of the students.
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1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Description : 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50'. What number is missing? -Riddles

Last Answer : 22

Two students appeared at an examination. One of them secured 18marks more than the other and his marks was 72% of the sum of their marks. What are the marks obtained by them? a) 12.5,23.3 b) 26.7,16.0 c) 13.3,14.2 d) 11.45, 29.45 e) 29.8,15.4

Description : Two students appeared at an examination. One of them secured 18marks more than the other and his marks was 72% of the sum of their marks. What are the marks obtained by them? a) 12.5,23.3 b) 26.7,16.0 c) 13.3,14.2 d) 11.45, 29.45 e) 29.8,15.4 

Last Answer : Answer: D  Let the marks secured by them be x and (x + 18)  Then sum of their marks = x + (x + 18) = 2x + 18  Given that (x + 18) was 72% of the sum of their marks  =>(x+18) = 72/100(2x+18)  => ... 11x = 126 x = 11.45  Then (x + 18) = 11.45 + 18 = 29.45  Hence their marks are 11.45 and 29.45

In a monthly test, 10 students were awarded marks in a Mathematics examination as follows: 23, 25, 15, 20, 17, 10, 24, 15, 19 If a student is selected

Description : In a monthly test, 10 students were awarded marks in a Mathematics examination as follows: 23, 25, 15, 20, 17, 10, 24, 15, 19 If a student is selected

Last Answer : In a monthly test, 10 students were awarded marks in a Mathematics examination as follows: 23, 25 ... the probability that he gets more than 18 marks?

In a Examination a candidate has to score minimum of 24 marks inorder to clear the exam. The maximum that he can score is 40 marks. Identify the Valid Equivalance values if the student clears the exam. A. 22,23,26 B. 21,39,40 C. 29,30,31 D. 0,15,22

Description : In a Examination a candidate has to score minimum of 24 marks inorder to clear the exam. The maximum that he can score is 40 marks. Identify the Valid Equivalance values if the student clears the exam. A. 22,23,26 B. 21,39,40 C. 29,30,31 D. 0,15,22

Last Answer : C. 29,30,31

The Average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of failed candidates is 15, what is the number of candidates who passed the examination? A.90 B.100 C.108 D.115 E.None of these

Description : The Average of marks obtained by 120 candidates in a certain examination is 35. If the average marks of passed candidates is 39 and that of failed candidates is 15, what is the number of candidates who passed the examination? A.90 B.100 C.108 D.115 E.None of these

Last Answer : Answer – B (100) Explanation – Let the number of passed candidates be a Then total marks =>120 x 35 = 39 a + (120 – a) x 15 4200 = 39 a + 1800 – 15 a a = 100

The average marks of a class of 90 students is 126. Out Of them, 4 scores zero, first 60 students scored an average of 116, next 24 scored an average of 118. What is the mark obtained by the remaining student in the class? A) 750 B) 862 C) 774 D) 875

Description : The average marks of a class of 90 students is 126. Out Of them, 4 scores zero, first 60 students scored an average of 116, next 24 scored an average of 118. What is the mark obtained by the remaining student in the class? A) 750 B) 862 C) 774 D) 875

Last Answer : C) According to the question, 90*126=(4*0)+(60*116)+(24*118)+2y 11340=6960+2832+2y 2y=1548 y=774

In an examination mahi scores 64% of marks, nitesh scores 52% of marks and ritesh scores 48% of marks. The maximum mark of the exam is a three digit number, whose sum is 10 and the middle digit is equal to the sum of the other two digits. The number will decreased by 297, if its digit are reversed. approximately what is the average mark obtained by mahi, nitesh and ritesh? A) 247 B) 248 C) 264 D) 284

Description : In an examination mahi scores 64% of marks, nitesh scores 52% of marks and ritesh scores 48% of marks. The maximum mark of the exam is a three digit number, whose sum is 10 and the middle digit is equal to ... what is the average mark obtained by mahi, nitesh and ritesh? A) 247 B) 248 C) 264 D) 284

Last Answer : A) Maximum mark consist of a three digit number let's consider Unit digit place is Z,ten's place digit is Y and hundred's place digit is X. According to the question, Y=x+z---------1 X+Y ... Total number of Mahi, Nitesh and Ritesh =164/100 451=739.61 Required average =739.64/3=246.54~247.

In a post graduate examination the marks obtained by a student is 75 per paper. If he had obtained 33 marks more in Evs paper & 27 more marks in science paper, then his average per paper is increased by 3 marks. Then how many papers were there in exam? A) 10 B) 12 C) 14 D) 20

Description : In a post graduate examination the marks obtained by a student is 75 per paper. If he had obtained 33 marks more in Evs paper & 27 more marks in science paper, then his average per paper is increased by 3 marks. Then how many papers were there in exam? A) 10 B) 12 C) 14 D) 20

Last Answer : Answer: D)  Let the number of paper be A. Then total marks earned him = 75A from questions, 75A + 33+ 27= 78A 3A = 60=> A=20 = number of subjects

The number of students in four classes A, B, C, D and their respective mean marks obtained by each of the class are given below : Image The combined mean of the marks of four classes together will be : (A) 32 (B) 50 (C) 20 (D) 15

Description : The number of students in four classes A, B, C, D and their respective mean marks obtained by each of the class are given below :   The combined mean of the marks of four classes together will be : (A) 32 (B) 50 (C) 20 (D) 15 

Last Answer : Answer: A

A national random sample of 20 ACT scores from 2010 is listed below. Calculate the sample mean and standard deviation. 29, 26, 13, 23, 23, 25, 17, 22, 17, 19, 12, 26, 30, 30, 18, 14, 12, 26, 17, 18 a. 20.50, 5.79 b. 20.50, 5.94 c. 20.85, 5.79 d. 20.85, 5.94

Description : A national random sample of 20 ACT scores from 2010 is listed below. Calculate the sample mean and standard deviation. 29, 26, 13, 23, 23, 25, 17, 22, 17, 19, 12, 26, 30, 30, 18, 14, 12, 26, 17, 18 a. 20.50, 5.79 b. 20.50, 5.94 c. 20.85, 5.79 d. 20.85, 5.94

Last Answer : Katie earned 84,92,84,75 and 70 on her first 5 tests. What is the minimum grade Katie needs to earn on the next test to have a mean of 84?

The average of marks obtained by 60 students in a computer examination is 18. If the average marks of passed students is 20and that of the failed students is 8, what is the number of students who passed the examination? A) 100 B) 75 C) 50 D) 85

Description : The average of marks obtained by 60 students in a computer examination is 18. If the average marks of passed students is 20and that of the failed students is 8, what is the number of students who passed the examination? A) 100 B) 75 C) 50 D) 85 

Last Answer : C Let the number of passed students be x. Then total marks = 60 × 18= 20x + (60– x) × 8 1080= 20x + 480– 8x 12x = 600 ∴ x = 50 ∴ number of passed students = 50

The average score in a bank examination of 13 students of a class is 50. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark was 35 and the maximum mark was 100. It is also known that none of the students failed. If each of the top five scorers had distinct integral scores, the maximum possible score of the topper is. A) 85 B) 90 C) 95 D) 100

Description : The average score in a bank examination of 13 students of a class is 50. If the scores of the top five students are not considered, the average score of the remaining students falls by 5. The pass mark ... integral scores, the maximum possible score of the topper is. A) 85 B) 90 C) 95 D) 100

Last Answer : D Average = Sum of observations/Number of observations Given, average score in a bank examination of 13 students of a class is 50. Sum of total scores = 13 50= 650 Given, if the scores of the top five students are not ... b + c + d + e = 290 ⇒46+ 47 + 48+ 49 +e = 290 ⇒e = 100

Students of two university appeared for a common test of maximum 60 marks. The average of their marks for university 1 & university 2 are 39 & 42 respectively. If the no of students of university 1 is twice the no of students of university 2. Then what is the average marks of all students of both the university? A) 40 B) 42 C) 26 D) 36

Description : Students of two university appeared for a common test of maximum 60 marks. The average of their marks for university 1 & university 2 are 39 & 42 respectively. If the no of students of university 1 is twice the no ... the average marks of all students of both the university? A) 40 B) 42 C) 26 D) 36

Last Answer : Answer: A) Let number of students of university2 be N and the no of students of university 1 be 2N the average of university1 and university 2 is 39 and 42 total marks of university 1 students and university 2 ... 78N and N*42 =42N average of both university,  =(78N+42N) /(N+2N) = 40 marks

In an exam the average marks obtained by a candidate is 82 per paper. If he had obtained 32 marks more in science paper & 28more marks in social paper, then his average per paper is increased by 15 marks. Then how many papers when there in examination? A) 10 B) 6 C) 4 D) 8

Description : In an exam the average marks obtained by a candidate is 82 per paper. If he had obtained 32 marks more in science paper & 28more marks in social paper, then his average per paper is increased by 15 marks. Then how many papers when there in examination? A) 10 B) 6 C) 4 D) 8 

Last Answer : C)  let the number of paper be x total mark earned him = 82x then,, 82x+32+28= 97x 15x=60 x= 4 =number of subjects

Mark the correct option relating to age limit a) For WLI & EA policies the insurant should not be less than 19 year an not more than 55 year on the date of next birthday b) For CWLA the upper age limit is 50 year c) The max age limit for 15 and 20 year AEA is 45 and 40 respectively d) Under Ugal suraksha policy age at entry of both the spouses should not be less than 21 year and not more than 45 year. e) All the above

Description : Mark the correct option relating to age limit a) For WLI & EA policies the insurant should not be less than 19 year an not more than 55 year on the date of next birthday b) For CWLA the upper age ... of both the spouses should not be less than 21 year and not more than 45 year. e) All the above

Last Answer : e) All the above

The mean marks obtained by a class of 40 students is 65. The mean marks of half of the students is found to be 45. The mean marks of the remaining students is (A) 85 (B) 60 (C) 70 (D) 65

Description : The mean marks obtained by a class of 40 students is 65. The mean marks of half of the students is found to be 45. The mean marks of the remaining students is (A) 85 (B) 60 (C) 70 (D) 65

Last Answer : Answer: A

What are the characteristics of Continuous and Comprehensive Evaluation ? (a) It increases the workload on students by taking multiple tests. (b) It replaces marks with grades. (c) It evaluates every aspect of the student. (d) It helps in reducing examination phobia. Select the correct answer from the codes given below: (A) (a), (b), (c) and (d) (B) (b) and (d) (C) (a), (b) and (c) (D) (b), (c) and (d)

Description : What are the characteristics of Continuous and Comprehensive Evaluation ? (a) It increases the workload on students by taking multiple tests. (b) It replaces marks with grades. (c) It evaluates every aspect of the student. (d) It helps in ... ) and (d) (C) (a), (b) and (c) (D) (b), (c) and (d)

Last Answer : Answer: D

Adam and Smith are working on a project. Adam takes 18 hrs to type 108 pages on a computer, while Smith takes 15 hrs to type 120 pages. How much time will they take, working together on two different computers to type a project of 360 pages? a) 23 hrs 24 min b) 45 hrs 42 min c) 25 hrs 42 min d) 42 hrs 45 min

Description : Adam and Smith are working on a project. Adam takes 18 hrs to type 108 pages on a computer, while Smith takes 15 hrs to type 120 pages. How much time will they take, working together on two different computers to type a ... ? a) 23 hrs 24 min b) 45 hrs 42 min c) 25 hrs 42 min d) 42 hrs 45 min

Last Answer : C Number of pages typed by Adam=108 Number of pages typed by Adam in 1 hour = 108/18=6 Number of pages typed by Smith =120 Number of pages typed by Smith in 1 hour = 120/15 =8 Number of pages typed by both in ... (6 + 8) = 14 Time taken by both to type 360 pages, = (360 * 1/14) =25 hrs 42 min

Gugan obtained a total of 1313 marks out of 1400 in an examination. What is his approximate percentage in the examination? a) 25 b) 60 c) 45 d) None e) 98

Description : Gugan obtained a total of 1313 marks out of 1400 in an examination. What is his approximate percentage in the examination? a) 25 b) 60 c) 45 d) None e) 98

Last Answer : Answer: D Required percentage = 1313/ 1400 × 100 =131300/1400 =93.7

The smallest number which when divided by 20, 25, 35 and 40 leaves a remainder of 14, 19 , 29 and 34 respectively is (a) 1994 (b) 1494 (c) 1394 (d) 1496

Description : The smallest number which when divided by 20, 25, 35 and 40 leaves a remainder of 14, 19 , 29 and 34 respectively is (a) 1994 (b) 1494 (c) 1394 (d) 1496

Last Answer : 1394 Hint: Note that 20 -4 = 6; 25 -19 = 6;35-29= 6;40- 34 = 6 Required number = L.C.M. of ( 20, 25, 35 and 40)-6

A candidate scoring 50% in an examination fails by 60 marks , while another candidate scores 75 % mark, gets 40 marks more than the minimum pass marks . Find the minimum pass mark. a) 125 b) 220 c) 140 d) 260 e) 298

Description :  A candidate scoring 50% in an examination fails by 60 marks , while another candidate scores 75 % mark, gets 40 marks more than the minimum pass marks . Find the minimum pass mark. a) 125 b) 220 c) 140 d) 260 e) 298

Last Answer : Answer: B  Let x be the maximum marks,  Then (50% of x)+60 = (75% of x)-40  x/2 +60 = 3x/4 -20  60+20 = 3x/4 – x/ 2 X=320  Hence maximum marks = 320  Minimum pass marks = 320/2 + 60 = 220

The average marks scored by two Class A1 and A2 students are 120 and 130 respectively. If 8 students are moved from Class A2 to Class A1 and the average marks of the two Class get interchanged. Find the total number of students in two Class put together, if the average marks scored by the 8 students who moved is 150 A) 25 B) 30 C) 35 D) 40

Description : The average marks scored by two Class A1 and A2 students are 120 and 130 respectively. If 8 students are moved from Class A2 to Class A1 and the average marks of the two Class get interchanged. Find the total number ... average marks scored by the 8 students who moved is 150 A) 25 B) 30 C) 35 D) 40

Last Answer :  D)  let the number of students in Class A1 be x and class A2 be y. Total marks scored by the students will be 120x and 130y, the average gets interchanged after moving student from y Thus we get, 130y- ... (x+8) 120x+1200=130x+1040 10x=160 X=16 Thus the total number of students =24+16=40

P and Q can do a piece of job in 24 days; Q and R can do it in 30 days while R and P can finish it in 40 days. If P, Q, Rworks together, in how many days will they finish the job? In how many days will each one of them finish it, working alone? A) 15,25,35,45 B) 20,30,40,120 C) 20,30,40,50 D) 20,25,30,60

Description : P and Q can do a piece of job in 24 days; Q and R can do it in 30 days while R and P can finish it in 40 days. If P, Q, Rworks together, in how many days will they finish the job? In how many days will each ... finish it, working alone? A) 15,25,35,45 B) 20,30,40,120 C) 20,30,40,50 D) 20,25,30,60 

Last Answer : B Time taken by (P + Q) to finish the job = 24days. (P + Q)'s 1 day's job = 1/24 Time taken by (Q +R) to finish the job = 30days. (Q + R)'s 1 day's job = 1/30 Time taken by (R + P) to finish the ... - {(P + Q)'s 1 day's job } = (1/20 -1/24)=1/120 Hence, R alone can finish the job in 120 days.

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

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

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

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.

The mean of 20, 40, 35, 42, and 45 is ____.

Description : The mean of 20, 40, 35, 42, and 45 is ____.

Last Answer : The mean of 20, 40, 35, 42, and 45 is ____.

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.

In a class of 55 students, the number of students studying different subjects are 23 in Mathematics, 24 in Physics, 19 in Chemistry, 12 in Mathematics

Description : In a class of 55 students, the number of students studying different subjects are 23 in Mathematics, 24 in Physics, 19 in Chemistry, 12 in Mathematics

Last Answer : In a class of 55 students, the number of students studying different subjects are 23 in Mathematics, 24 in Physics, 19 ... ` B. `30` C. `22` D. `27`

A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

Description : A problem Drum ( 3 ft. diameter ; 6 ft. height ) is field with a fluid whose density is 50 lb/ft^3. Determine the total volume of the fluid.  A. 42.41 ft^3  B.44.35 ft^3  C.45.63 ft^3  D.41.23 ft^3 Formula: Vf = (pi d^2 h) / 4

Last Answer : 42.41 ft^3

The average age of the whole class is 12.05 years. The average age of all girls is 12.5 years and the average age of boys is 11.75 years. If the total number of boys is 45 then what is the total number of girls in the class? a) 20 b) 25 c) 30 d) 35 e) 40

Description : The average age of the whole class is 12.05 years. The average age of all girls is 12.5 years and the average age of boys is 11.75 years. If the total number of boys is 45 then what is the total number of girls in the class? a) 20 b) 25 c) 30 d) 35 e) 40

Last Answer : Let the number of girls be x. The number of boys is 45. :. Total age of boys and girls in the class = (x+45) × 12.05 or, (x+45) × 12.05 = 45 × 11.75 + x × 12.5 or, 12.05x + 542.25 = 528.75 + 12.5x or, 13.5 = 0.45x or, x = 13.5/0.45 = 30 Answer is: c)

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!!!

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:

What percent of the total number of girls in the college know Bangla? (1) 45 (2) 40 (3) 48 (4) 42 (5) 50

Description : What percent of the total number of girls in the college know Bangla? (1) 45 (2) 40 (3) 48 (4) 42 (5) 50

Last Answer : 5) 50

By selling 45 lemons for Rs 40, a man loses 20%.How many should he sell for Rs 24 to gain 20% in the transaction? a) 19 b) 22 c) 20 d) 18 e) 14

Description : By selling 45 lemons for Rs 40, a man loses 20%.How many should he sell for Rs 24 to gain 20% in the transaction? a) 19 b) 22 c) 20 d) 18 e) 14

Last Answer : Selling price = Rs. 40 Loss = 20% Therefore, cost price = 40×100 /(100 - 20) = Rs. 50 Therefore, cost price of each lemon = Rs. 50/45 Now, selling price = Rs. 24 Profit = 20% Cost price = 24×100/120 = Rs. 20 Number of lemons for Rs. 20 = 20/(50/45) = 900/50 = 18 Answer: d)

In a post-office, stamps of three different denominations of Rs 7, Rs 8, Rs 10 are available. The exact amount for which one cannot buy stamps is (A) 19 (B) 20 (C) 23 (D) 29

Description : In a post-office, stamps of three different denominations of Rs 7, Rs 8, Rs 10 are available. The exact amount for which one cannot buy stamps is (A) 19 (B) 20 (C) 23 (D) 29

Last Answer : Answer: A

Insert the missing number in the following: 3, 8, 18, 23, 33, ?, 48 (A) 37 (B) 40 (C) 38 (D) 45

Description : Insert the missing number in the following: 3, 8, 18, 23, 33, ?, 48 (A) 37 (B) 40 (C) 38 (D) 45

Last Answer : Answer: C The Difference between 2 number in series is 5 then 10, 5 then 10. Therefore after 33 + 5 = 38.

The average marks obtained by a student in Tamil, english, maths, science, and social together is 65% above the average mark obtained in maths and science together. How many more marks exceeded by the average of maths and science together than the sum of the tamil and english together? A) 50 B) 65 C) Data insufficient D) None of these

Description : The average marks obtained by a student in Tamil, english, maths, science, and social together is 65% above the average mark obtained in maths and science together. How many more marks exceeded by the average of ... of the tamil and english together? A) 50 B) 65 C) Data insufficient D) None of these

Last Answer : C) According to the question  There is no other information about the marks in any one of the subject. The data gives is insufficient to answer the question.

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

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.

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

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.

The pressure at a point 4 m below the free surface of water is (A) 19.24 kPa (B) 29.24 kPa (C) 39.24 kPa (D) 49.24 kPa

Description : The pressure at a point 4 m below the free surface of water is (A) 19.24 kPa (B) 29.24 kPa (C) 39.24 kPa (D) 49.24 kPa

Last Answer : Answer: Option C

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

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 : (b) Since, the difference between mid value is 5. So, the corresponding class to the class mark 20 must have difference 5. Therefore, option (c) and (d) are wrong. Since, the mid value is 20 which can get only, if we take option (b)

On what dates of April 1994 did SUNDAYfall? (A) 2,9,16,23,30 (B) 3,10,17,24 (C) 4,11,18,25 (D) 1,8,15,22,29

Description : On what dates of April 1994 did SUNDAYfall? (A) 2,9,16,23,30 (B) 3,10,17,24 (C) 4,11,18,25 (D) 1,8,15,22,29

Last Answer : Answer: B Solution:. Let us find the day on 1st April, 1994. 2000 years have 0 odd day. We have to calculate day on 1st April, 1994. 2000-1994 = 6 Years. 1996 is leap year and have 4 odd days Ordinary years ... was Friday Thus, on 3 rd , 10 th , 17 th , 24 th will be Sunday fall on April 1994.

Find the greatest unit of time with which 5 hours 15 minutes and 8 hours 24 minutes can both be represented as integers. (a) 70 min. (b) 63 min. (c) 48 min. (d) 42 min.

Description : Find the greatest unit of time with which 5 hours 15 minutes and 8 hours 24 minutes can both be represented as integers. (a) 70 min. (b) 63 min. (c) 48 min. (d) 42 min.

Last Answer : 63 min. 

The average age of A and B is 20 years. If C were to replace A, the average would be 19 and id C were to replace B,the average would be 21 . The ages of A, B and C are (in years) (a) 22, 17,16 (b) 22, 18, 20 (c) 30, 18, 15 (d) 23, 17 ,15

Description : The average age of A and B is 20 years. If C were to replace A, the average would be 19 and id C were to replace B,the average would be 21 . The ages of A, B and C are (in years) (a) 22, 17,16 (b) 22, 18, 20 (c) 30, 18, 15 (d) 23, 17 ,15

Last Answer : 22, 18, 20 Hint : A + B = 2 x 20 C + B = 2 x 19 A + C = 2 x 21

The ratio of the ages of Bala and Babu is 13: 11. The total of their ages is 3.6decades. The proportion of their ages after 0.95decades will be [1 Decade = 10 years] a) 24:23 b) 12:11 c) 27:24 d) 29:26

Description : The ratio of the ages of Bala and Babu is 13: 11. The total of their ages is 3.6decades. The proportion of their ages after 0.95decades will be [1 Decade = 10 years] a) 24:23 b) 12:11 c) 27:24 d) 29:26

Last Answer : D Let, Bala’s age = 13A and babu’s age = 11A Then 13A + 11A = 36 A = 1.5 Bala’s age = 19.5years and Babu’s age = 16.5 years Proportion of their ages after 9.5 is = (19.5+9.5) : (16.5 + 9.5) = 29 : 26

5, 8, 12, 18, 24, 30, 36, 42, ____ a) 52 b) 46 c) 48 d) 54 e) 56

Description : 5, 8, 12, 18, 24, 30, 36, 42, ____ a) 52 b) 46 c) 48 d) 54 e) 56

Last Answer : Each term in the series is the addition of successive prime numbers. Like Prime numbers are2, 3, 5, 7, 11, 13, 17, 19, 23, 29 and so on. So first term is addition of (2 + 3). Ie 5, then second term is (3 + 5) ie ... +17)ie. 30, then (17+19)ie.36, then (19+23)ie., 42, then (23+29)i.e. 52. Answer: a)

A sum of amount of money is adequate to pay peter’s salary for 42 days and Harish’s salary for 56 days. The same money is adequate to pay the salary of both for. a) 12 days b) 24 days c) 36 days d) 48 days

Description : A sum of amount of money is adequate to pay peter’s salary for 42 days and Harish’s salary for 56 days. The same money is adequate to pay the salary of both for. a) 12 days b) 24 days c) 36 days d) 48 days

Last Answer : B Let the total money be Rs.x Peter’s 1 day salary = Rs. x/42 Harish 1 day salary = Rs. x/56 (peter+harish)’s 1 day salary = Rs(x/42+x/56) =(4x+3x)/168 =7x/168 =x/24 Money is adequate to pay the salary of both for 24 days.