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
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
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
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
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.
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
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
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.
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
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
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.
Description : The batting average for 20 innings of a cricketer is 25 runs. His highest score exceeds his lowest score by 86 runs. If these two innings are excluded, the average of the remaining 18 innings is 22runs. Find out the highest score of the player. A) 85 B) 90 C) 95 D) 100
Last Answer : C) Total runs scored by the player in 20 innings = 20*25 Total runs scored by the player in 18 innings after excluding two innings = 18 22 Sum of the scores of the excluded innings = 20 25 - 18 22 = 104 Given that ... Now x + 86 + x = 104 => 2x = 18 then x = 9 Highest score = 9+ 86 = 95
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
Description : A batch of 50 people has the oldest person with 160 years of age. The average of the batch is reduced by 2, if the oldest person is replaced by someone new, Find the age of the new person. A) 45 B) 30 C) 60 D) 75
Last Answer : Answer: C) Average=Sum of observations/Number of observations Given, a batch of 50 people has the oldest person with 160 years of age. Let the average of the group be a'. Sum total of ages = 50 a Given, average of the ... person be b' . 50a - 160 + b = 50(a - 2) b= 160-100 b = 60 years
Description : Students of two colleges appeared for Talent test carrying 250marks as maximum. The average of their marks for college1 & college 2 are 160 & 180respectively. If the number of students of college 1 is the half of the ... marks of all students of both the college? A) 170 B) 110 C) 173.33 D) 177.33
Last Answer : C) let the number of students of college 2 be '2N' then the number of students of college1 is 'N' the average marks for college1 is 160 the average marks for college 2 is 180 total marks of ... = 360N average marks of all students of both the colleges = (160N + 360N)/N+2N = 173.33marks
Description : on analyzing the result of an competitive exam the teacher found that the average for the entire the class was 69 marks. If we say that average of 10 % of the students scored 77 marks and average of 28 % of the ... marks of the remaining students of the class A) 67.54 B) 68.26 C) 66.91 D) 69.06
Last Answer : D) average of entire class = 69marks average of 28 % of the students = 66 marks average of 10% of the students = 77 marks then % of remaining students = (100- 10 - 28) = 62% let the average of 62% of the ... (62*x)+770+1848= 6900 62 * x = 6900-770-1848 62*x = 4282 x = 69.06
Description : A group of students of arithmetic mean of the marks in a test was 63. The brightest 30% of them secured a mean score of 70 and the dullest 15% a mean score of 41. The mean score of remaining 55 % is A) 63.675 B) 61.785 C) 65.181 D) 66.67
Last Answer : C) Let the requird mean score be ‘x’ Then (30*70)+(15*41)+55x = 63*100 2100+615+55x=6300 2715+55x=6300 55x=3585 X=65.181
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
Description : A certain company employed 600 men and 400 women and the average wage was 2.55 per hour.If a women got 50 paise less than a man, what were their wages per hour ? (a) Man rs 3.00,Woman Rs 2.50 (b) Man Rs 3.50, Woman Rs 3.00 (c) Man Rs 2.75, Woman Rs 2.25 (d) Man Rs 3.25, Woman Rs 2.75
Last Answer : Man = Rs 2.75, woman = Rs 2.25
Description : The average monthly expenditure of Mr. Abi family for the first 4 months is Rs.4210, next 4 month expenditure Rs.4450 for the last 4 months Rs. 4360 . If his family saves Rs. 6800 for 12 months, find the average monthly ... the 12 months? A) Rs. 4707.25 B) Rs.4564.75 C) Rs. 4906.66 D) Rs. 4806.50
Last Answer : C) Mr Abi family first 4 month expenditure=4210*4=16840 Mr Abi family next 4 month expenditure=4450*4=Rs17800 Mr Abi family last 4 month expenditure=4360*4=Rs.17440 Mr Abi family total ... 17440+6800=Rs.58880. Mr Abi family average expenditure in 12 months =58880/12=Rs.4906.66
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
Description : The average age of a family of 24 members is 132 years. If the age of the youngest member is 22 years, the average age of the family at the birth of the youngest member was? A) 75 B) 110 C) 100 D) 95
Last Answer : Answer: B) Total age of the family of 24 members = 132* 24 = 3168 yrs Total age of the family members 22 yrs ago = 3168 - (24*22) = 2640 at the time total members in a family = 24 average age of the family at the birth of the youngest member = 2640/24 = 110yrs.
Description : The average height of 50 students in a class is 182 cm. 40 students whose average height is 182.5 cm left the class and 50 students whose average height is 180.5 cm joined the class. Find the average height of the present? A) 180.41 B) 175.5 C) 180.55 D) 185.5
Last Answer : A) The average of the students leaving the class as well as joining the class to be 182 so that the average remains the same. But it is given that the average of the 40 students leaving the class is 182. ... strength of the class. Hence the average of the present class = 182 - 95/60 = 180.5 cm
Description : There are three different categories of jobs A, B and C. The average salary of the students who got the job of A and B categories is Rs.32 lakh per annum. The average salary of the student who got the job of B ... 30 & 44 B) lies between 48 & 56 C) lies between 38 & 45 D) lies between 49 & 50
Last Answer : Answer: C) Let the number of students who got the jobs of A, B and C categories is a, b and c respectively, ∴Total salary of Aand B = 32 (a +b) ∴Total salary of B and C = 54 (b+ ... the minimum salary must be Rs.38 lakh and the maximum salary cannot exceed 45, which is the highest of the three..
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
Description : Find the average of first 85 natural numbers? A) 43 B) 41 C) 45 D) 47
Last Answer : A) Sum of 1st n natural numbers = n(n+1)/2 So, sum of 1st 85 natural numbers = 85(85+1)/2 =85(86)/2 =7310/2 =3655 Required average = 3655/85 =43.
Description : The average age of students of a college is 31.6 yrs. The average age of boys in the class is 32.8 yrs and that of girls is 30.8. The ratio of number of boys to the number of girls in the class. A) 2:3 B) 3:1 C) 1:3 D) 3:2
Last Answer : A) Let the ratio be D:1.then, (D * 32.8) + ( 1* 30.8)= (D+1) * 31.6 32.8 D +30.8 = 31.6 D + 31.6 32.8 D – 31.6D = 31.6 – 30.8 1.2D = 0.8 D = 0.8/1/2 = 0.4/0.6 D = 2/3 Required ratio = 2/3 :1 = 2:3.
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.
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
Description : In an examination, 35%of the total students failed in Hindi, 45%failed in English and 20%in both. The percentage of those who passed in both subjects is (A) 10 (B) 20 (C) 30 (D) 40
Last Answer : Answer: D
Description : Of the 5 numbers, the first is twice the second, the second is one – third of the third, the third is 5 times of the fourth, and the fourth is three – seventh of fifth. The average of the numbers is 35. The largest of these number is. A) 30.63 B) 65.625 C) 43.75 D) 75.356
Last Answer : B) Let the fifth number be x'. 4th number = 3/7 x 3rd number = 5 (3/7 x) = 15/7x 2nd number = 1/3 (15/7 x) = 15 /21 x 1st number = 2 (15/21 x) = 30/21 x X + 3/7 x + 15 /7 ... 120x = 3675 X = 30.625 So the numbers are 30.62, 13.125, 65.625, 21.875, 43.75. Therefore largest number is 65.625.
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
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!!!
Description : The average age of the men in a team is 76 years and that of the women is 75 years. The average age for the whole team is. A) 75 yrs B) 75.5 yrs C) Cannot be computed with the given information D) None of the above
Last Answer : C) Clearly to find the average, we ought to number of men, women or total persons in the class, neither of which has been given. So the data provided is inadequate.
Description : The average age of three -seventh of class is 49, what should be the average of remaining four-seventh students so that the average of the entire class is 63? A) 24.5 B) 86.5 C) 73.5 D) 25.5
Last Answer : C) According to the question, 63=3/7×49+4/7×y 63=147/7+4y/7 441=147+4y 4y=294 Y=73.5
Description : The average age of 78 students of a batch is 30 years. If the age of the head master be included, then the average increases by 6 months. Find the age of head master? A) 56yrs 5 months B) 66yrs 6 months C) 56yrs 5 months D) 69yrs 5 months
Last Answer : D) Total age of 78 students = (780 * 30)yrs = 2340 yrs Average of 79 persons = 30 yrs 6 months = 6 ½ yrs Total age of 79 persons = 61/2 * 79 = 2409.5 Age of headmaster = (2409.5 – 2340)yrs = 69.5 yrs = 69 yrs 5 months.
Description : When a girl weighing 90 kgs left a class, the average weight of the remaining 119 students increased by 400 g. What is the average weight of the remaining 119 students? A) 124 B) 138 C) 145 D) 116
Last Answer : B Let the average weight of the 119 students be X. Therefore, the total weight of the 119 of them will be 119X. The questions states that when the weight of this student who left is added, the total ... , the average weight decreases by 0.4 kgs. (119X+90)/120=X−0.4 ⇒119X+90=120X−48 ⇒X=138
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
Description : The average height of a batch is 344 cm. 16more students with an average height of 320 cm joined the batch therefore decreasing the average height of the batch by 12 cm. Find the total strength of the batch? A) 12 B) 11 C) 14 D) 16
Last Answer : Answer: D) let the initial strength of the batch be X total height of the batch initially = 344X total height of 16 new students = 320 * 16 = 5120 cm Then, the average height of X+ 16students = 332 332 = [5120 + 344 X]/ [X + 16] X= 16 students
Description : The average height of the students in a group was 360cm. When 10 students whose height is 292.8cm are newly admitted. The average height of the group was reduced by 24cm.How many students are present in the group? A) 29 B) 25 C) 28 D) 14
Last Answer : Answer: C) let the number of students initially in the class be A, then, total height = A * 360------ 1 again the no of students increased = A + 10 then, total height A + 10 student = (A +10) * 336 ... 3 (A+10) *336-2928 =A*360 336A+3360-2928=360A A=18 Number of students in class = 18+10=28
Description : The average scores of a batch of students in a test is 84. The 36% of students's average score is 57 and 42% of students's score is 84. Then find the average score of remaining 22% of students approximately, A) 128 B) 131 C) 119 D) 106
Last Answer : Answer: A) let the number of students be 100, then, 100*84 = 36*57 + 42*84 + 22*x 22x = 8400-2052-3528 x = 128.18
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
Description : The Cricketer average score in 40 over is 21, if the man scores 23runs in 8over, 26 runs in 10 over and 18runs in 14 over. What is the average score in remaining over? A) 18 B) 19 C) 16 D) 20
Last Answer : A) According to the question 40*21=23*8+26*10+18*14+8×X 840=184+260+252+8x 8x=144 X=144/8=18 runs
Description : At the average age of siva and sasi is equal to the average age of somu and ramu. if the ratio of age of siva and sasi 1:3 and somu age is equal to the three by two of siva age. What is the present age of somu if ramu age is 25years? A) 12 B) 15 C) 18 D) 21
Last Answer : B) siva:sasi=x:3x siva+sasi/2=somu + ramu/2 X+3X=3/2×X+25 5X=50 X=10 Siva’s age is 10 then somu's age = 3/2×10=15 years
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
Description : 18 friends went to a restaurant for taking their breakfast. 17 of them spent RS.24 each on their breakfast and the last one spent RS.16 more then the average expenditure of all the 18. What was the total money spent by them? A) 228 B) 448.9 C) 458.6 D) 428.0
Last Answer : B) Let the average expenditure of all the 18 be Rs ‘x’. Then (24 * 17) + (x + 16) = 18x 424 + x = 18x x = 24.9 therefore total money spent 18x = 18 * 24.9 = Rs.448.9
Description : The average height of 60 girls was calculated to be 300 cm. It was detected later that one value of 330 cm was wrongly copied as 270 cm for the computation of the mean. Find the correct mean. A) 301cm B) 300.5cm C) 307cm D) 311cm
Last Answer : A) Calculated average height of 60 girls = 300cm. Incorrect sum of the heights of 60 girls = (300 60)cm = 18000 cm. Correct sum of the heights of 60 girls = (incorrect sum) - (wrongly copied item) ... 330) cm = 18060cm. Correct mean = correct sum/number of girls = (18060/60) cm = 301 cm.
Description : There are 50 compartments in a Chennai express carrying an average of 70 passengers per compartments. At least 24 passengers were sitting in each compartment, not any compartment has equal number of passengers, ... can be accommodated in 50th compartment? A. 748 B. 705 C.739 D.cannot be determined
Last Answer : C) Total number of passengers in Chennai express = 50*70 = 3500 Total number of candidates from 1 to 49 compartments = 24+25+....+70 = (70*71)/2 +(23*24)/2 = 2761 number of passengers in 50th compartment = 3500 - 2761= 739
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