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

1 Answer

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

Related questions

Description :  The average age of K and L is 48 years. If M replaced K, the average age would be 40 and if M replaced L, the average would be 44. What are the ages of K,L,M repectively? A) 58, 46, 22 B) 32, 44, 36 C) 52,44,36 D) 56, 22,38

Last Answer : Answer: C)  Total age of K& L = 48*2 = 96 yrs----->1 Total age of L& M = 40*2 = 80yrs----->2 Total age of K& M = 44*2 = 88 yrs----->3 from 1, 2, 3 K+L+M =132---->4 from 1 & 4, M= 36 from 2 & 4, K=52 from 3 & 4, L=44 Hence the required answer is 52,44,36 

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 : 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 : 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 : 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 : 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

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 :  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 : 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 : 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 : 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 age of 14 mens is increased by 1 year when one of them whose age is 44 years is replaced by a lady. What is the age of the lady? A) 54 B) 58 C) 50 D) 56 

Last Answer : B) suppose the average age of 14 men is x years and the age of lady is y years. the total age of 14 mens will be = 14x years the average age of 14 mens is increased by 1 years who of them whose age is 44 years is replaced by ... 14x - 44 + y = 14* (x+1) 14x - 44 + y = 14x + 14 y = 14+ 44 y=58

Description : The mean of 8 article was found to be 15. On rechecking, it was found that two article were wrongly taken as 11 and 9 instead of 16 and 14 respectively. Find the correct mean. A) 17.25 B) 13.65 C) 16.54 D) 16.25

Last Answer : D) Calculated mean of 8 articles = 15 Incorrect sum of these 8 articles = (15*8) = 120. Correct sum of these 8 articles = (incorrect sum) - (sum of incorrect articles) + (sum of actual articles) = [120 ... (30)] = 130 Therefore, correct mean = 130/8 = 16.25 Hence, the correct mean is 16.25.

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 : 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 : 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 : 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 : 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 : When the average age of a father, mother and their son was 90 years, the son got married and a child was born just 6 year after the marriage when child turned 14 years the average age of the family is 80yrs. Find the age of daughter - in - law at present? A) 55 B) 56 C) 57 D) 58

Last Answer : B) total age of father, mother and son at the time of son's marriage = 90*3 =270 present age of family father, mother, son, daughter-in-law, child = (father, mother, son age at the time of marriage ... 60)+ daughter-in-law present age + 14 = 80*5 =400 daughter-in-law present age= 400-344 = 56yrs

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 : 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 : The average age of a morning class is 108, if the average age of 72 ladies in the class is as same as the total average and the number of gents in the class is 2 more than the ladies in the class, what is the average age of gents in the class? A) 152 B) 108 C) 156 D) 154 

Last Answer : B) According to question, (72+74)X=72×108+74X 146x-74x=7776 72x=7776 X=7776/72= 108years

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

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.

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 average age of Mr and Mrs Rahim at the time of their marriage in 1981 was 56 yrs. On the occasion of their anniversary in 1986, they observed that the average age of their family had come down by 15 yrs compared to ... varshini was born. What was the age of varshini in 1990? A) 6 B) 5 C) 4 D) 1

Last Answer : B) Sum of the ages of Mr and Mrs Rahim=56×2=112years Sum of the ages in 1986=41×3=123 years Sum of the ages of Mr and Mrs Rahim =112+10=122 Daughters age in 1986=123-122=1 years  Daughters age in 1990 =1+4=5 years.

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 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 : A group of 4 persons joins in javelin throw competition. The best player scored 42.5 points. If he had scored 46 points, the average score for the team would have been 42. The number of points the team scored. A) 164.5 B) 166.5 C) 169.7 D) 162.5

Last Answer : A) Let the total score be ‘x’. (X+46 – 42.5)/4 = 42 X + 3.5 = 42 * 4 X + 3.5 = 168 X = 168 – 3.5 X = 164.5

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 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 : 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 weight of 45 passenger on board an aircraft is 50 kg. If the weight of 5 members of the crew is added, the average is reduced by half kilogram . What is the average weight of the crew members?

Last Answer : Total weight of 45 passenger = 45 x 50 = 2250 kg Total weight of 45 passenger and 5 crews = 50 x 49.5 = 2475 kg Total weight of 5 crews = 2475-2250 = 225kg Average weight of 5 crews = 225/5 = 45kg.

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

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 : There were 32 boys in a hostel. If the number of boys be increased by 15, then the expenditure on food increases by Rs. 43 per day while the average of expenditure of boys is reduced by Rs. 2. What was the initial expenditure on food per day? A) Rs.108.2 B) Rs.125.6 C) Rs.135.8 D) Rs.144.5

Last Answer : Answer: A)  Let initial expense of 32 boys = Rs. x average expense of 32 boys = x/32 average expense of 47 boys = (x + 43)/ 47 then, (x/32) - ((x+43)/47) = 2  x= Rs.108.28

Description : Mr. Sebastine, a famous author, recently got his new novel released. To his utter dismay, he found that for the 1974 pages on an average there were 3 mistakes in every page. While, in the first 1024 pages there ... the average number of mistakes per page for the remaining pages. A) 5 B) 2 C) 3 D) 4

Last Answer : D According to the question X is the remaining pages average. 2122+(1974-1024)×x/1974=3 950x=5922-2122=3800 X=3800/950=4

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 :  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 :  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 : In a famous restaurant rooms were numbered from 401 to 430, each room gives an earning of Rs. 4675/ day for the first 15 days a month and for the later half, Rs. 5370/ day per room. Find the average income room per day over the month of 30 days? A) 5700.5 B) 5872.7 C) 5900.5 D) 5022.5  

Last Answer : Answer: D) Total number of rooms = 30 earning in 1st 15 days for 30 rooms  = 15 * 4675* 30 = Rs. 2103750 earning in 2nd 15 days for 30 rooms  = 15 * 5370 * 30 = Rs. 2416500 Average income per ... day over the month of 30 days  = [ 2103750 + 2416500] / [ 30 * 30]  = Rs. 5022.5

Description : The average age of the assembalage of people was 114 yrs. When the 2 people of the assembalage went out with age of age of 84 yrs and 108yrs, then the average of the assembalage increased by 12yrs. How many people are there initially? A) 3 B) 4 C) 5 D) 6 

Last Answer : Answer: C)  let the number of people initially be N total age of N people = 114N total age of N-2 people = 114N - ( 84+108) = 114N -192 average age of N-2 people = 126 = [ 114N - 192] / [ N - 2] N=5 Hence th required answer is 5

Description : An industry employed 1200 men and 800 women and the average salary was Rs 51 per day. If a woman got Rs 10 less than a man , then what are their daily salary?(mens and women respectively) A) men Rs 25 and women Rs ... Rs 45 and women Rs 55 C) men Rs 35 and women Rs 25 D) men Rs 55 and women Rs 45

Last Answer : D) Let the daily salary of man be Rs’x’ Then the daily salary of a women = Rs (x-100 Now 1200x + 800(x-10) = 51 *(1200+800) 1200x+800x -8000 = 51*2000 2000x = 102000+8000 =110000 X=55 Therefore mens daily salary Rs 55 and womens daily salary Rs 45

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 : 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 : If the average of two numbers are 138& product is 2241. Then find the difference of both numbers? A) 225.40 B) 259.25 C) 267.81 D) 294.57

Last Answer : B) let the two number be a & b sum of two numbers, a+b = 138*2 =276---1 product of two numbers, a*b=2241 (a+b)2 = a2 + 2ab + b2 ------2 (a-b)2 = a2 -2ab +b2 ------3 solving 2 & 3, 76176 - (a-b)2 = 4*2241 a-b = 259.25

Description : 1/2of a certain travel is covered at the rate of 30km/hr, one-third at the rate of 40 km/hr and the rest at 35km/hr. Find the average speed for the whole travel. A) 33 1/3 B) 33 3/5 C) 34 3/7 D) None of these

Last Answer : B) let the total travel be X km.  Then X/2 km at the speed of 30 km/hr and X/3 km at 40 km/hr and the rest distance( X - X/2 - X/3) =1/6 X at the speed of 35km/hr. Total time taken during the travel ... 2*30 hrs+ X/3*40hrs+ X/6*35hrs = 5X/168 hrs Average speed =X/(5x/168) = 168/5 =33 3/5km hr

Description : The average of 7 consecutive odd numbers is 41. Find the largest of these numbers A) 40 B) 57 C) 47 D) 43

Last Answer : C) Let the numbers be x , x+2,x+4,x+6, x+8, x+10 and x+12 Then x+x+2+x+4+x+6+x+8+x+10+x+12/7 7x+42=287 7x=245 X=35 Largest number = x+12 = 35+12 =47