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 : 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 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 weight of a group of 22students is 34kg. When the weight of the staff is also included, the average weight increases by 2kg. What is the weight of the staff? A. 80 kgs B. 54 kgs C. 47 kgs D. 33kgs
Last Answer : A) The average weight of a group of 22students = 34 kgs. Therefore, the total weight of the group = 22*34 = 748 kg When the weight of the staff is included, there are 23 individuals. The average ... , the total weight of the 22 students plus the staff = 23* 36= 828 828-748=80 kg.
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 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 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 : The average weight of M,N and O is 168 kg. If P joins the group, the average weight of the group becomes 160 kg. If another man Q who weighs 6kg more than P replaces M, then the average of N,O,P and Qbecomes158 kg. What is the weight of M? A) 175 B) 140 C) 145 D) 150
Last Answer : D) M+N+O =168 * 3 =504 M+N+O+P=160*4=640……(1) P=136 kg and Q =142 kg N+O+P+Q =158*4 =632 N+O+P =632-142 =490…..(2) from 1 & 2, M=640-490 =150kg
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 : 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 : 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 : The average state of the city in the first 4 days of a month was 116 degrees. The average for the second, third, fourth and fifth days was 120 degrees. If the state of the first and fifth days ... :8 then what is the state on the fifth day. A) 112degrees B) 128degrees C) 132degrees D) 130degrees
Last Answer : B) Sum of states on 1st, 2nd, 3rd, 4th days = (116 *4) = 464 =======>A Sum of states on 2nd, 3rd, 4th and 5th days = (120 * 4)= 480 ======>b From a and b Subtracting each other ... be 7x and 8x degrees respectively Then 8x - 7x = 16 x = 16 therefore state on the 5th day = 8x = 128 degrees.
Description : The average age of 11 daughters of a family is 12yrs. Average age of the daughters together with their parents is 27yrs. If mother is 14yrs younger than father. Then find the father's age A) 127 yrs B) 116.5 yrs C) 168yrs D) 170yrs
Last Answer : Answer: B) Total age of 11 daughters of a family = 11*12 = 132 father age = 14 +mother age total age of the family (11 daughters + father + mother) is, 132+mother +14+ mother = 27 * 13 mother = 102.5 father age = 14+102.5 =116.5years
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 : 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 : Distance between two stations P & Q is 1556km. A passenger train covers the journey from P to Q at 168km per hr and return back to P with a uniform speed of 112km/hr. Find the Average speed of the train during the whole journey? A) 124.4 km/hr B) 130.4km/hr C) 134.4 km/hr D) 130.0 km/hr
Last Answer : C) Given x= 168 , y = 112 Required average speed = (2xy / x + y) km/hr = 2 * 168 *112 / 168 + 112 = 37632 / 280 =134.4 km/hr
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 : 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 : 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 : 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 : 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 : The average of 12 numbers is 7.9. The average of 5 of them is 6.8, while the average of other five is 7.7. what is the average of the remaining 2 numbers? A) 10.25 B) 11.15 C) 12.65 D) 13.25
Last Answer : B) Sum of the remaining 2 numbers = [(12 * 7.9) – (586.8) – (5*7.7)] = 94.8 – 34 – 38.5 = 22.3 Required average = 22.3 / 2 = 11.15
Description : The average price of 12 note books is Rs.15 while the average price of 10 of these note books is Rs.13. Of the remaining 2 note books, if the price of one note book is 50% more than the price of the other, what is the price of ... Rs. 30, Rs.20 B) Rs. 8, Rs. 30 C) Rs. 10, Rs. 16 D) Rs. 12, Rs. 20
Last Answer : Answer: A) Total price of 12 note books = Rs. 180 Total price of 10 note books= Rs. 130 ⇒ The price of 2 note books = Rs. 50 Let the price of each book be x and y. ⇒ x + y = 50 ----- ... note book is 50% more than the other price 150y/100+y=50 y=20 Substituting y value in (1) we get, x=30
Description : If the arithmetic mean of 150 numbers is calculated 70. If each number is increased by 10, then mean of new number is. A) 70 B) 80 C) 90 D) 60
Last Answer : B) Arithmetic mean of numbers = 70 Sum of 150 numbers = (70 * 150) =10500 Total increase = (150 * 10) = 1500 Increased sum = 10500 + 1500 = 12000 Increased average = 12000/150 =80
Description : What is the median of 112 135 124 115 119 90 112 134 113 114 115.?
Last Answer : The median is 115.
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 : 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 : 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 : Rohit married 21 years ago at the age of 90 years. His wife's present age 99years. If 24 years later the average age of rohit, his wife and their son was 90 years, then what is son's present age? A) 15yr B) 14yr C) 13yr D) 12yr
Last Answer : Answer: D) Rohit present age = 111years total age of family after 24 yrs = 90*3= 270 present age of rohit & his wife = 111+99 = 210yrs present age of rohit, his wife & his son= 270- 24*3 =198 present age of son= 210-198=12yrs
Description : 11 girls went to a canteen. 10 of them spent 10 each and the 11th girl spent 25 more than the average expenditure of all. Find the total money spent by them? A) 172 .5 B) 178 .5 C) 137.5 D) 165.5
Last Answer : C) Let the average money spent by the 11girls = x Money spent by the 11th girl = (x + 25) Money spent by the other 10 girls = (10*10) = 100 Total money spent by the 11 girls = (100 + x + 25) =(x + ... = (x + 125)/ 11 =>10x=125 x =12.5 Total money spent by the 11 girls = 11*12.5 = 137.5.
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 : 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 : 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 : 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 : 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 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 : 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 : 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 : 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 : 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 : 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 : 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 weight of x,y,z is 90kg. If the average weight of x and y be 80kg and that of y and z be 86kg. Find the weight of y A) 66kg B) 63kg C) 62kg D) 70kg
Last Answer : C) Let x ,y ams z represent their individual weights. Then, X + y + z = ( 90 * 3) = 270kg X+ y = 80 *2 = 160kg Y + z = 86 * 2 = 172kg Therefore y = (x + y) + ( y+z) – (x+y+z) = 160 + 172 – 270 Y= 62kg.
Description : In shewak's opinion, his weight is greater than 76 kg but less than 83 kg. His sister doest not agree with shewak and she thinks that shewak's weight is greater than 71 kg but less than 81 kg. His ... , what is the average of different probable weights of shewak? A) 67kg B) 68kg C) 72kg D) 78kg
Last Answer : Answer: D) Let shewak's weight be X kg. According to shewak , 76 < X < 83 According to shewak 's sister, 71 < X < 81. According to shewak 's father, X ≤ 79 The values satisfying all the above conditions are 77, 78and 79. Required average=[ (77+78+79)/3] = 201/ 3 = 78kg.
Description : 119 + 784 ÷ 32 x 116 = ? a) 2961 b) 3121 c) 2564 d) 2861 e) 2941
Last Answer : 119 + 784 ÷ 32 x 116 = ? Sol: 119 + 24.5 x 116 =? 119 + 2842 = ? ? = 2961 Answer: a)
Description : What is the free space loss, in dB, between two microwave parabolic antennas 38.0 kilometer apart operating at 7.0 GHz? A. 145.6 dB B. 138.5 dB C. 135.5 dB D. 140.89 dB
Last Answer : D. 140.89 dB