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

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
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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?
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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

Description : 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

Last Answer : 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.

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

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.

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

The percentage of marks obtained by a student in a monthly test is as follows: What is the probability that the student gets more than 75% marks in a

Description : The percentage of marks obtained by a student in a monthly test is as follows: What is the probability that the student gets more than 75% marks in a

Last Answer : The percentage of marks obtained by a student in a monthly test is as follows: What is the ... student gets more than 75% marks in a test?

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`

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 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

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)

The ratio of the number of boys to that of girls in a school is 3:2. If 30% of boys and 70% of girls appeared in an examination, the ratio of students appeared in the examinations to that not appeared in the examination is A) 23:27 B) 22:25 C) 21:17 D) 18:17 E) None of these

Description : The ratio of the number of boys to that of girls in a school is 3:2. If 30% of boys and 70% of girls appeared in an examination, the ratio of students appeared in the examinations to that not appeared in the examination is A) 23:27 B) 22:25 C) 21:17 D) 18:17 E) None of these 

Last Answer :  Answer: A 3:2 = 30:20 30% boys = (30/100) × 30= 9 70% girls = (70/100) × 20 = 14 (9+14) : (50-23) = 23 : 27

In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

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

In a mathematics test given to 15 students, the following marks (out of 100) are recorded : -Maths 9th

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

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

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

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)

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 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

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.

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?

f the temperature falls below 18 degrees, the heating is switched on. When the temperature reaches 21 degrees, the heating is switched off. What is the minimum set of test input values to cover all valid equivalence partitions? A. 15, 19 and 25 degrees B. 17, 18, 20 and 21 degrees C. 18, 20 and 22 degrees D. 16 and 26 degrees

Description : f the temperature falls below 18 degrees, the heating is switched on. When the temperature reaches 21 degrees, the heating is switched off. What is the minimum set of test input values to cover all valid equivalence ... 17, 18, 20 and 21 degrees C. 18, 20 and 22 degrees D. 16 and 26 degrees

Last Answer : A. 15, 19 and 25 degrees

Normal haemoglobin level for male is _______ g/L a. 13.8 to 17 b. 12.8 to 15 c. 20 to 23 d. 16 to 19

Description : Normal haemoglobin level for male is _______ g/L a. 13.8 to 17 b. 12.8 to 15 c. 20 to 23 d. 16 to 19

Last Answer : a. 13.8 to 17

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

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Description : Find the range of the given data : 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, 20 -Maths 9th

Last Answer : Here, the minimum and maximum values of given data are 6 and 32 respectively. Range = 32 – 6 = 26

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : In conclusion, the range of data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11, and 20 is 26.

The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Description : The range of the data 25, 18, 20, 22, 16, 6, 17, 15, 12, 30, 32, 10, 19, 8, 11 and 20 is -Maths 9th

Last Answer : (d) In a given data, maximum value = 32 and minimum value = 6 We know, range of the data = maximum value – minimum value = 32 – 6 = 26 Hence, the range of the given data is 26.

Consider the following database Student:Roll No. First Name Last Name Class Marks Obtained(%) Scholarship Awarded -Technology

Description : Consider the following database Student:Roll No. First Name Last Name Class Marks Obtained(%) Scholarship Awarded -Technology

Last Answer : 1. 95.42. Mohan Garg and scholarship awarded is 50000.3. 54. 6

Suresh and Ramesh are twins. In a form by mistake, Suresh reverses the last 2 digits of his year of birth, and this makes him 9 years older than Ramesh. With this mistake, the sum of their ages in 2015 becomes 43. How old will Ramesh be 2021? a) 17 b) 19 c) 21 d) 23 e) 25

Description : Suresh and Ramesh are twins. In a form by mistake, Suresh reverses the last 2 digits of his year of birth, and this makes him 9 years older than Ramesh. With this mistake, the sum of their ages in 2015 becomes 43. How old will Ramesh be 2021? a) 17 b) 19 c) 21 d) 23 e) 25

Last Answer : Let the correct ages of Ramesh and Suresh be X in 2015 (since they are twins, they will have the same age) Given: 2x + 9 = 43 Or x = 17 years in 2015. Age in 2021 = 17 + 6 = 23 Note: year of birth = 2015 – 17 = 1998. Reversing, Suresh = 1989 or 9 year older. Answer: d)

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

In an examination it is required to get 672 aggregate marks to pass. A student gets 70% marks and is declared failed by 126 marks. What are the maximum aggregate marks a student can get? a) 780 b) 840 c) 741 d) 805 e) 983

Description : In an examination it is required to get 672 aggregate marks to pass. A student gets 70% marks and is declared failed by 126 marks. What are the maximum aggregate marks a student can get? a) 780 b) 840 c) 741 d) 805 e) 983

Last Answer : Answer: A Difference = 672-126 = 546 According to the question, 70% of total aggregate = 546 Total aggregate marks = 546 × 100 /70  = 54600/70  = 780

Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? -Maths 10th

Description : Had Ravita scored 10 more marks in her Mathematics test out of 30 marks, 9 times these marks would have been the square of her actual marks. How many marks did she get in the test? -Maths 10th

Last Answer : answer:

A thermometer measures temperature in whole degrees only. If the temperature falls below 18 degrees, the heating is switched off. It is switched on again when the temperature reaches 21 degrees. What are the best values in degrees to cover all equivalence partitions? A. 15,19 and 25. B. 17,18 and19. C. 18, 20 and22. D. 16, 26 and 32.

Description : A thermometer measures temperature in whole degrees only. If the temperature falls below 18 degrees, the heating is switched off. It is switched on again when the temperature reaches 21 degrees. What are the best values in degrees to ... and 25. B. 17,18 and19. C. 18, 20 and22. D. 16, 26 and 32.

Last Answer : A. 15,19 and 25.

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

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

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

How do you write an algorithm which takes students marks and grades them as follows Mark greater than 60 Pass?

Description : How do you write an algorithm which takes students marks and grades them as follows Mark greater than 60 Pass?

Last Answer : Need answer

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

The ages of Gaurav and Goutham will be in the ratio 5 : 7 after ten years from now and will be in the ratio 13 : 18 after twelve years from now. Find the ratio of the sum of their ages 10 years hence to the sum of their ages 12 years hence. 1 : 25 : 27 2 : 13 : 15 3 : 14 : 17 4 : 18 : 23 5 : 30 : 31

Description : The ages of Gaurav and Goutham will be in the ratio 5 : 7 after ten years from now and will be in the ratio 13 : 18 after twelve years from now. Find the ratio of the sum of their ages 10 years hence to the sum of their ages 12 ... hence. 1 : 25 : 27 2 : 13 : 15 3 : 14 : 17 4 : 18 : 23 5 : 30 : 31

Last Answer : 5 : 30 : 31

What is the probability that a two digit number selected at random will be a multiple of '3' and not a multiple of '7'? e) 13/45 f) 14/45 g) 23/90 h) 19/45

Description : What is the probability that a two digit number selected at random will be a multiple of '3' and not a multiple of '7'? e) 13/45 f) 14/45 g) 23/90 h) 19/45

Last Answer : Answer: a

According to Indian Standards Institute, the actual size of modular bricks is (A) 23 cm × 11.5 cm × 7.5 cm (B) 25 cm × 13 cm × 7.5 cm (C) 19 cm × 9 cm × 9 cm (D) 20 cm × 10 cm × 10 cm

Description : According to Indian Standards Institute, the actual size of modular bricks is (A) 23 cm × 11.5 cm × 7.5 cm (B) 25 cm × 13 cm × 7.5 cm (C) 19 cm × 9 cm × 9 cm (D) 20 cm × 10 cm × 10 cm

Last Answer : (C) 19 cm × 9 cm × 9 cm

In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Description : In a class test, the sum of marks obtained by P in Mathematics and Science is 28. Had he got 3 more marks in Mathematics and 4 marks less in Science, the product of marks obtained in the two subjects would have been 180? Find the marks obtained in two subjects separately. -Maths 10th

Last Answer : Let P has obtained x in mathematics and y in science. Then by the problem, x+y=28.......(1). If P would have got 3 marks more in mathematics, then P would have got (x+3) in mathematics ... P obtained 12 marks in mathematics and 16 in science. or, P obtained 9 marks in mathematics and 19 in science.

One of the fields on a form contains a text box which accepts numeric values in the range of 18 to 25. Identify the invalid Equivalance class A. 17 B. 19 C. 24 D. 21

Description : One of the fields on a form contains a text box which accepts numeric values in the range of 18 to 25. Identify the invalid Equivalance class A. 17 B. 19 C. 24 D. 21

Last Answer : A. 17

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!!!

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 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

Saravanan has scored an total of 90% in an examination with five subjects in the ratio 18:14:15:21:22. If 85% is the marks required to get an ‘A’ grade in each subject, then find in how many subjects did he get an ‘A’ grade. Given that the maximum marks in each subject is 120. A) 2 B) 3 C) 4 D) 5

Description : Saravanan has scored an total of 90% in an examination with five subjects in the ratio 18:14:15:21:22. If 85% is the marks required to get an ‘A’ grade in each subject, then find in how many subjects did he get an ‘A’ grade. Given that the maximum marks in each subject is 120. A) 2 B) 3 C) 4 D) 5 

Last Answer : Answer: B) Maximum total marks = 5 120 = 600 Marks scored by Saravanan = 90/100*600=540 Let the marks scored by him in the 5 subjects be 18x,14x,15x,21x and 22x respectively. 18x + 14x + 15x + 21x + ... A' grade i.e. 0.85 120 = 102 Therefore, he has got an A' grade in 3 subjects.

In an examination, 900 students appeared. Out of these students; 56 % got first division, 27 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed. a) 225 b) 153 c) 245 d) 148 e) 298

Description : In an examination, 900 students appeared. Out of these students; 56 % got first division, 27 % got second division and the remaining just passed. Assuming that no student failed; find the number of students who just passed. a) 225 b) 153 c) 245 d) 148 e) 298

Last Answer : Answer: B The number of students with first division = 56 % of 900  = 56/100 × 900 = 50400/100  = 504 And, the number of students with second division = 27 % of 900  = 27/100 × 900  =24300/100  = 243 Therefore, the number of students who just passed = 900 – (504 + 243)  = 900- 747  = 153