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 : 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 : An Aided school has only three classes which contain 70, 36 and 40students respectively. The pass percentage of these classes are30,25 and 20 respectively. The pass percentage of the aided school is A) 23 % B) 15% C) 26 % D) 30%
Last Answer : C) Total number of students passed in class1 = 30/100 *70 = 21 Total number of students passed in class 2= 25/100 *36 = 9 Total number of students passed in class 3= 20/100 *40 = 8 Total number of students passed in the school = 38 Thus, pass percentage of the school = 38 /146*100 = 26 %
Description : BDL Ltd. is currently preparing its cash budget for the year to 31 March 2014. An extract from its sales budget for the same year shows the following sales values. Rs March 60,000 April 70,000 May 55,000 June 65,000 40% ... 2013 is (a) Rs 60,532 (b) Rs 61,120 (c) Rs 66,532 (d) Rs 86,620
Last Answer : (a) Rs 60,532
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 : 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 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 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 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 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 : Nandita scores 60% marks in five subjects together, viz., Hindi, Science, Mathematics, English and Sanskrit, where in the maximum marks of each subject were 105. How many marks did Nandita score in Science, if she scored 69 ... and 51 marks in English? a) 66 b) 68 c) 55 d) 65 e) None of these
Last Answer : Answer: D Total of maximum marks of all subjects=105×5=525 75% of 525=525 ×60/100= 315 Obtained marks of foru subjects (Hindi, Sanskrit, mathematics and English) =69+62+68+51=250 So, the obtained marks in Science=315-250=65
Description : In a batch, 40% of the students offered Maths, 30% offered science and 15% offered both. If a student is selected at random, what is the probability that they has offered science or maths? A) 0.55 B) 0.65 C) 0.45 D) 0.75
Last Answer : Answer: A) P(M) = 0.40 P(S) =0.30 and P(M∩S) = 0.15 P(M∪S) = P(M) + P(S) - P(M∩S) = 0.55
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 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 : 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 : 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 : 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 : 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 : 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 : 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 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 : 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 : 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 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 : 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 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 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 : 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 : Percentage of water in buffalo milk is:- a. 65-67 % b. 70-75 % c. 80-85 % d. 87-90 %
Last Answer : d. 87-90 %
Description : Average age of A, B, C is 90yrs, however when D joins them, then the average comes down to 80 yrs. Now a new person E, whose age is six-fifth of the age of D, replaces A and the new average is 78. What is age of A? A) 64 B) 68 C) 62 D) 66
Last Answer : B) according to the question A+B+C/3=90 => A+B+C =270…………..(1) A+B+C +D/4=80 => A+B+C +D=320………(2) From the equation 1 and 2 D=50 E =6/5×50=60 E+B+C +D/4=78 => E+B+C +D =312 B+C +D =312-60=252……………..(3) From 2and 3 The age of A=320-252=68
Description : The average of 3,8,7 and a is 6 and the average of 19,2,7, a and b is 11. What is the value of b? A) 11 B) 21 C) 31 D) 41
Last Answer : B) We have (3+8+7+a+6/5)=6 24 =a = 30 A=6 Also (19+2+7+a+b/5)=11 19+2+7+6+b=55 34+b=55 b=21
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
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 : In the series 1, 5, 13, 25, 41, ……., the next term is (A) 59 (B) 63 (C) 61 (D) 68
Last Answer : Answer: C
Description : Ronit spends his money from his saving in different way that is he spends 25% on travel,15% on food,20% on wages and 17% on shopping a clothes and after that all expenditure he saved 6900.Find the how much he spent on clothes. A) 4400 B) 5100 C) 4000 D) 6000
Last Answer : B) Let the total income of Ronit x then total expenditure from income X x (25%+15%+20%+17%)=X*77% X x 77% =Total savings= X x 23% X = 6900 x 100 / 23 = 30000 expenditure on clothes = 17% so,30000 x 17 / 100 = 5100
Description : An electric motor drives a windless which hauls up a crate A of weight W = 500 kg up an inclined plane through a height of 10 m in 2 minutes. The torque acting on the windless is 25 kgm and the drum of the ... The efficiency of the inclined plane would be a.0.41 b.0.61 c.0.265 d.0.5305 e.0.66
Last Answer : d. 0.5305
Description : Charge report prepared in a) ACG 67 b) ACG 63 c) ACG 61 d) ACG 40
Last Answer : c) ACG 61
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 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 : 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 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 : 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 : Harish buys petrol at rs 21, Rs.24 and Rs.27 per litre for 3 successive years. What approximately is the average cost per litre of petrol if he spends Rs.12000 each year? A) 23.74 B) 54.76 C) 24.66 D) 28.99
Last Answer : A) Total quantity of petrol consumed in 3 yr = (12000/21 +12000/24+12000/27) =12000(1/2 + 1/24 + 1/27) =12000(72+63+56/1512) =12000(191/1512) =(2292000/1512)litres Total amount spent = rs(3 *12000) = Rs.36000 Average cost = Rs(36000*1512/2292000) =Rs.23.74
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 : 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 : 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 : 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 : 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 : A cricket player had a certain average of runs for his 43 innings. In his 44th innings, he is bowled out for no score on his part. This brings down his average by 4runs. His new average of runs is ? A) 172 B) 182 C) 166 D) 162
Last Answer : A) Let the cricket player’s average of runs for his 43 innings be X runs. Total number of runs in 43innings=43x According to the question, [43x + 0]/ 44= x-4 43x = 44x – 176 x = 176 new average of runs = 176-4 =>172
Description : A cricketer played 3 matches in tournament. The respective ratio between the scores of first and second matches was 3:7 and that between the scores of second and third matches between was 7 :2. the difference ... the cricketer average score in all the matches together? A) 336 B) 146 C) 168 D) 189
Last Answer : A) The ratio between first and second matches equal to 3 :7 The ratio between second and third matches=7:2 The difference between first and third matches=3x-2x=84 runs = 84 Required average=1008/3=336 .