Easy Way of Calculating Averages

Average

Q. 1. The average of five positive numbers is 213. The average of the first two numbers is 233.5 and the average of last two numbers is 271. What is the third number?

(a) 64   (b) 56  (c) 106  (d) Cannot be determined  (e) None of these

Ans: (b)56

Explanation:

The sum of the five numbers = 5 × 213 =1065

The sum of the first two numbers = 2 × 233.5 = 467

The sum of  the last two numbers = 542

Then the sum of the four numbers = 467 + 542 =1009

So, the third number will be = 1065 – 1009 = 56. Ans.

Q. 2. The average age of a woman and her daughter is 16 years. The ratio of their ages is 7:1 respectively. What is the woman’s age?

(a) 4 years  (b) 28 years  (c) 32 years  (d) 6 years   (e) None of these

Ans: (b) 28 years

Explanation:

Sum of their ages = 2 × 16 = 32

Let 7x and x be their respective ages,  then, 8x = 32  and x = 4

So, the age of the woman = 7x = 7 × 4 = 28 years.Ans.

Q. 3. Find the average of the following set of scores:

        568, 460, 349, 987, 105, 178, 426

(a)442   (b)441  (c)440  (d)439  (e)None of these

Ans: (d) 439

Explanation:

Average = sum of the scores/No. of scores

              = 3073/7 = 439.Ans.

Q. 4. The average age of 54 girls in a class was calculated as 14 years. It was later realized that the actual age of one of the girls in the class was calculated as 13 years. What is the actual average age of the girls in the class? (Rounded off to two digits after decimal).

(a) 10.50 years  (b) 12.50 years  (c) 12 years  (d) 13.95 years  (e) None of these

Ans: (d) 13.95

Explanation:

The sum of the ages of the 54 girls entered as error = 54 × 14 = 756.

 Now, deduct the error i.e 13 – 10.5 = 2.5

Then the actual sum of the ages = 756 – 2.5 = 753.5

So, the actual average = 753.5/54 = 13.95 years. Ans.

Q. 5.  The average age of a man and his son is 40.5 years. The ratio of their ages is 2 : 1 respectively. What is the man’s age?

(a) 56 years  (b) 52 years  (c) 54 years   (d) Cannot be determined (e) None of these

Ans: (c) 54 years.

Explanation:

Let the age of the son = x

Then, 2x + 1x        i.e. 3x = 40.5 × 2 = 81

    Therefore, x = 27   and so the man’s age = 2x = 2 × 27 = 54 years.  Ans.

Q. 6. A car covers a distance from town A to town B at the speed of 58 kmph and covers the distance from town B to town A at the speed of 52 kmph. What is the approximate average speed of the car?

(a ) 55 kmph  (b) 52 kmph  (c) 48 kmph   (d) 50 kmph   (e) 60 kmph

Ans: (a) 55 kmph

Explanation:

Average speed = sum of the two speeds/2  = 58 + 52 =110/2 = 55 kmph.Ans.

Note:

Otherwise, If  two equal distances are covered at two different speeds of x kmph and y kmph, then, Average speed = 2xy/x + y kmph.

Q. 7. If 36a + 36b = 576, then what is the average of a and b?

(a) 16  (b) 8  (c) 12  (d) 6 (e) None of these

Ans: (b) 8

Explanation:

36 (a+b) = 576,    given,  then, a + b = 576/36 = 16

So, the average of  a and b = a + b/2   = 16/2 = 8. Ans.

Q. 8. The average marks of 65 students in a class was calculated as 150. It was later realized that the marks of one of the students was calculated as 142, whereas his actual average marks were 152.  What is the actual average marks of the group of 65 students? (Rounded off to two digits after decimal)

(a) 151.25   (b) 150.15   (c) 151.10   (d) 150.19    (e) None of these

Ans: (b) 150.15

Explanation:

Increase in total marks = 152 – 142 = 10

Therefore the New average = 150 + 10/65 = 150.15.Ans.

OR

Sum of the   total average marks of 65 students = 65 × 150 = 9750

Here, add the difference of 10 marks = 9760.

Therefore the New average = 9760/65 = 150.15. Ans.

Q. 9. In a class there are 32 boys and 28 girls. The average age of the boys in the class is 14 years and the average age of the girls in the class is 13 years. What is the average age of the whole class?  (Rounded off to two digits after decimal)

(a) 13.50   (b) 13.53  (c) 12.51   (d) 13.42  (e) None of these

Ans: (b) 13.53

Explanation:

The sum of the ages of  32 boys = 32 × 14 = 448

The sum of the ages of 28 girls = 28 × 13 = 364

Therefore, the sum of the ages of the whole class of  60 students =812

The average age of the whole class of 60 students = 812/60 = 13.53. Ans.

Q. 10. The average of 5 consecutive even numbers A, B, C, D and E respectively is 74. What is the product of C and E?

(a) 5928  (b)5616  (c) 5538   (d) 5772   (e) None of these

Ans: (d) 5772

Explanation:

Let the sum of the 5 consecutive even numbers = x + x +2 + x + 4 + x + 6 + x + 8 = 5 × 74 = 370

5x + 20 = 370;   5x = 370 – 50 = 350

Therefore, x = 350/5 =70,   then the  nos. A,B, C, D and E are 70, 72, 74, 76 and 78 respectively.

Product of  C and E = 74 × 78 = 5772. Ans.

Q. 11. Average of four consecutive odd numbers is 106. What is the third number in ascending order?

(a) 107   (b) 111  (c) 113     (d) Cannot be determined  (e) None of these

Ans: (a) 107

Explanation:

Let the sum of the four consecutive odd nos. = x + x +2 + x + 4 + x + 6

Then, 4x + 12 = 4 × 106 = 424

    Therefore, x = 103,  The nos. in ascending order is 103, 105, 107 and 109 and the third number is = 107. Ans.

Q. 12. Of  the three numbers, the average of the first and the second is greater than the average of the second and the third by 15. What is the difference between the first and the third of the three numbers?

(a) 15    (b) 45  (c) 60   (d) Data inadequate  (e) None of these

Ans: (e) None of these

Explanation:

Let x,  y, z be the three numbers,

Then x + y/2 – y + z/2 = 15

x+y – (y+z)/2 = x + y – y – z/2 = 15

So, x – z = 2 × 15 = 30.Ans.

Q. 13. The ratio of roses and lilies in a garden is 3:2 respectively. The average number of roses and lillies is 180. What is the number of lilies in the garden?

(a) 144   (b) 182   (c)216  (d)360  (e) None of these

Ans: (a) 144

Explanation:

Let the nos. of roses and lilies be 3x and 2x respesctively.

Then, 3x + 2x/2 = 180; 5x = 360  and so, x = 72

Therefore, the number of lilies = 2x = 2 × 72 = 144. Ans.

Q. 14. The average monthly income of a family of four earning members was Rs.15130. One of  the daughters in the family got married and left home, so the average monthly income of the family came down to Rs.14660. What is the monthly income of the married daughter?

(a) Rs.15350   (b)Rs.12000   (c)Rs.16540   (d) Cannot be determined   (e) None of these

Ans: (c) Rs.16540

Explanation:

The sum of the incomes o f the four members = 4 × 15130 = Rs.60520

The sum of the incomes of the family except the married daughter = 3 × 14660 = 43980

Therefore, the income of the married daughter = Rs.60520 – Rs.43980 = Rs.16540. Ans.

Q. 15.  The average temperature of Monday, Tuesday, Wednesday and Thursday was 36.5º C and for Tuesday, Wednesday, Thursday and Friday was 34.5º C. If the temperature on Monday was 38º C, find the temperature on Friday.

(a) 34ºC   (b)  36º C  (c) 37.4º C   (d) 32º C  (e) 30º C

Ans: (e) 30º

Explanation:

The sum of temperatures on Monday, Tuesday, Wednesday and Thursday = 36.5 × 4 = 146º C

The sum of the temperatures on Tuesday, Wednesday and Thursday  = 146 – Temperature on Monday    i.e. = 146- 38 = 108º C.

Sum of the temperatures on Tuesday, Wednesday , Thursday and Friday = 4 × 34.5 = 138º C

Therefore, the temperature on Friday = 138 – 108 = 30º C. Ans.

Q.16. The average age of a man and his two sons born on the same day is 30 years. The ratio of the ages of father and one of his sons is 5 : 2 respectively. What is the father’s age?

(a) 50 years   (b) 30 years  (c) 45 years   (d) 20 years  (e)None of these

Ans: (a) 50 years

Explanation:

Sum of their ages = 3 × 30 =90

i.e. 5x + 2x+ 2x = 90        => 9x  = 90  and x = 10

Then, father’s age = 5x = 5 × 10 = 50 years. Ans.

Q. 17. A cricketer has an average score of 49 runs in 24 innings. How many runs must he score in the 25^th innings to make his average 50?

(a) 94   (b) 84  (c) 74  (d) 76    (e)None of these

Ans: (c) 74

Explanation:

His total score in 24 innings = 24 × 49 = 1176 runs.

To get an average of 50 in 25^th innings his score should be = 25 × 50 =1250

Therefore, the score required to obtain in his 25^th innings = 1250 – 1176 = 74. Ans.

Q. 18. The average age of 14 boys in a class is 13 years. If the class teacher’s age is included, the average age is increased by one year. What is the class teacher’s age?

(a) 31 years   (b) 27.5 years    (c) 24 years   (d) 28 years   (e) None of these

Ans:  (d) 28 years

Explanation:

The teacher’s age = New average + Old No. of  persons × difference in average

i.e.   14 + 14 × 1  =  14 + 14 = 28 years.  Ans.

Q. 19. The average of four positive integers is 73.5. The highest integer is 108 and the lowest integer  is 29. The difference between the remaining two integers is 15. Which of the following is the smaller of the remaining two integers?

(a) 80   (b) 86  (c) 73  (d) Cannot be determined  (e)None of these

Ans: (e) None of these

Explanation:

The sum of the four Integers   =  73.5 × 4 = 294

Sum of the highest and the lowest integer = 108 + 29 = 137

And the sum of the remaining two integers = 294 – 137 = 157

Subtract the difference between them i.e. 157 – 15 = 142

Here, 142 is the double of the smallest integer, therefore, the smallest integer = 142/2 = 71. Ans.

Q. 20. The average age of a woman and her daughter is 46 years. The ratio of their ages is 15 : 8 respectively. What will be the respective ratio of their ages after 8 years?

(a) 8 : 5  (b) 10 : 17   (c) 17 : 10    (d) 5 : 8   (e) None of these

Ans: (c) 17 : 10

Explanation:

The sum of their ages, 15x + 8x = 2 × 46 = 92

23x = 92 :.    x= 92/23 = 4

Therefore  their respective ages are 60 and 32 .

After 8 years their ages will be 68 and 40 respectively

So, the required ratio = 68 : 40  = 17 :  10 Ans.

Q. 21. In a class of 75 students, the average age is 23 years. The average age of male students is 25 years and that of female students is 20 years. Then the ratio of male to female students is

(a) 3:2  (b)7:3  (c) 8:7  (d)6:9  (e) 9:6

Ans: (a) 3 : 2

Explanation:

Total age of 75 students = 75 × 23 = 1725

Let x be the no. of boys and y be the no. of girls.

Then, 25x + 20y = 1725   ----- (i)

 And  x + y = 75   -------(ii) × 20 gives, 20x + 20y   = 1500 ---- (iii)

Subtracting (iii) from (ii) we get, x = 45 and y = 30

Required ratio = 45 : 30

                       = 3 : 2.Ans.

Q. 22. The average age of 3 friends is 32 years. If the age of the fourth friend is added, their average age comes to 31 years. What is the age of the fourth friend?

(a) 32 years  (b) 28 years (c) 24 years   (d) 26 years   (e) None of these

Ans: (b) 28 years

Explanation:

The age of the fourth friend = new average + old no. of persons × difference in average

                                            = 31  + 3 × -1 = 31 -3 = 28 years. Ans.

Q. 23. Average salary of 19 workers in an industry Rs.2500. The salary of supervisor is Rs.5550. Find the average salary of all the 20 employees.

(a)Rs.3355.5  (b) Rs.4500.00  (c) Rs.4642.5    (d) Rs.2652.5   (e) None of these

Ans: (d) Rs.2652.5

Explanation:

The total salary of 19 workers = 19 × 2500 = Rs.47500

Total salary of 20 workers including the supervisor = 47500 + 5550 = Rs.53050

Therefore the required average = 53050/20 = Rs.2652.5. Ans.

Q. 24. The average of three even consecutive numbers is 24. What is the summation of the three numbers?

(a) 24  (b) 72   (c) 26   (d) 80    (e) None of these

Ans: (b) 72

Explanation:

Let the three consecutive even numbers be x, x+2 and x + 4

Then, x + x +2 + x+4 /3  =24

i.e.   3x + 6 = 3 × 24

3x = 72-6      => x  = 66/3 = 22

Then , their summation = 22 + 24 + 26 = 72.Ans.

Q. 25. The sum of seven consecutive even numbers of a set is 532. What is the average of first four consecutive even numbers of the same set?

(a) 76   (b) 75  (c) 74  (d) 73   (e) None of these

Ans:  (d) 73

Explanation:

See that, their summation as, x + x +2 + x +4 + x +6 +x + 8 + x + 10 + x + 12 = 532

7x + 42 = 532       and so, x = 70   (first number)

Therefore, the seven consecutive even numbers are 70, 72, 74, 76, 78, 80, 82

So, the required average = 70 + 72 + 74 + 76/4 = 292/4 = 73.Ans.

Q. 26. There are 50 boys in a class. One boy weighing 40 kg goes away and at the same time another boy joins the class. If the average weight of the class is thus decreased by 100 g, find  the weight of the new boy.

(a) 35kg  (b) 43kg  (c) 36kg  (d) 30 kg  (e) None of these

Ans: (a) 35 kg

Explanation:

The weight of the new boy = weight of the boy who left + Number of boys × difference in average.

i.e.  weight of the new boy = 40kg + 50 × - .1kg = 40kg – 5kg = 35 kg. Ans.

Q. 27. Kamlesh bought 65 books for Rs.1050 from one shop and 50 books for Rs.1020 from another. What is the average price he paid per book?

(a) Rs. 36.40   (b) Rs.18.20   (c) Rs. 24   (d)Rs.18  (e) None of these

Ans: (d) Rs.18

Explanation:

The total price = Rs.1050 + Rs.1020 = Rs.2070

Total number of books = 65 + 50 = 115

So, the average price per book = 2070/115 = Rs.18. Ans.

Q. 28. A car covers the first 39 kms. of it’s journey in  45 minutes and covers the remaining 25 kms. in 35 minutes. What is the average speed of the car?

(a)  40 kms/hr   (b) 64 kms./hr  (c) 49 kms./hr  (d) 48 kms./hr   (e) none of these

Ans: (d) 48 kms/hr

Explanation:

Total speed = 39 + 25 kms. = 64 kms.

Total time taken = 45 + 35 = 80 mins.   = 80/60 hrs.

So, the average speed of the car = 64 ÷ 80/60  = 64 × 60/80 = 48 km/hr. Ans.

Q. 29. The sum of five numbers 260. The average of the first two numbers is 30 and the average of the last two numbers is 70. What is the third number?

(a) 33  (b)60  (c)75  (d) Cannot be determined  (e) None of these

Ans: (b) 60

Explanation:

The sum of  the first two numbers = 30 × 2 = 60

The sum of the last two numbers = 70 × 2 = 140

Therefore, the third number = 260 – (60 + 140) = 260 – 200 = 60. Ans.

Q. 30. The average weight of 12 boys is 35 kgs. If the weight of an adult is added, the average becomes 37 kgm. What is the weight of the adult?

(a) 65kgs   (b) 68 kgs  (c) 62 kgs  (d) 63 kgs  (e) None of these

Ans: (e) None of these

Explanation:

The weight of the adult = New average + No. of boys × difference in average

                                     = 37 + 12 × 2  = 37 + 24 = 61 kgs.Ans.

Q. 31. The average marks in Science subject of a class of 20 students is 68. If the marks of two students were misread as 48 and 65 of the actual marks 72 and 61  respectively, then what would be the correct average?

(a)  68.5   (b)69   (c)69.5   (d)70  (e)66

Ans: (b) 69

Explanation:

The total number of marks = 20 × 68 = 1360

Add difference of marks misread against the actual marks i.e. 20 = 1360 + 20 = 1380

Now the correct average = 1380/20 = 69 Ans.

Q. 32.  The average speed of a car is 75 kms/hr. What will be the average speed of the car if the driver decreases the average speed of the car by 40 percent?

(a) 50 kms/hr  (b) 45 kms/hr  (c) 40 kms/hr   (d) 55 kms/hr   (e)None of these

Ans: (b) 45 kms/hr

Explanation:

40% of 75 kms/hr = 30 kms/hr

Required average = 75 – 30 kms/hr = 45 kms/hr. Ans.

Q. 33. The average marks of a student in seven subjects is 41. After re-evaluation in one subject the marks were changed to 42 from 14 and in remaining subjects the marks remained unchanged. What are the new average marks?

(a) 45  (b) 44   (c)46   (d) 47   (e) None of these

Ans: (a) 45

Explanation:

Present total marks = 41 × 7 = 287

Add the change in marks = 28

Now,  the total marks = 287 + 28 = 315; 

So, the new average marks = 315/7 = 45. Ans.

Q. 34. The body weight of six boys is recorded as 42 kgs.; 72 kgs; 85 kgs; 64 kgs; 54 kgs and 73 kgs.  What is the average body weight of all the six boys?

(a) 64 kgs.   (b) 67 kgs.   (c) 62 kgs.  (d) 65 kgs.   (e) None of these

Ans: (d) 65 kgs.

Explanation:

Required average = 42 + 72 + 85 + 64 + 54 + 73 /6  = 390/6 = 65. Ans.

Q. 35. The rainfall in a city in 5 consecutive years was recorded as 20.54 inches, 33.10 inches, 11.62 inches, 19.20 inches and 21.74 inches. What was the average annual rainfall?

(a) 21.44 inches  (b) 21.24 inches  (c) 20.24 inches  (d) 17.94 inches  (e) None of these

Ans: (b) 21.24 inches

Explanation:

The average annual rainfall = 20.54 + 33.10  + 11.62  + 19.20  + 21.74/5

 = 106.20/5 = 21.24 inches. Ans.

Q. 36. The average score of a cricketer in two matches is 27 and in three other matches is 32. Then find the average score in all the five matches?

(a)  25    (b) 20  (c) 30  (d) 35   (e) None of these

    Ans: (c) 30

Explanation:

Sum of the scores in two matches = 2 × 27 = 54

And sum of the scores in three matches = 3 × 32 = 96

Total score in five matches = 54 + 96 =150

So, the average scores in five matches = 150/5 = 30 Ans.

Q. 37. The average marks of nine students in a group is 63. Three of them scored 78, 69 and 48 marks. What are the average marks of remaining six students?

(a) 63.5   (b) 64  (c) 63   (d) 62.5   (e) None of these

Ans: (e) None of these

Explanation:

The total marks of nine students = 9 × 63 = 567

Sum of the marks of three students = 78 + 69 + 48 = 195

Therefore, the sum of marks of the remaining six students= 567 – 195 = 372

Average marks of remaining six students = 372/6 = 62. Ans.

Q. 38. Out of the three given numbers, the first number is twice the second and thrice the third. If the average of the three numbers is 154. What is the difference between the first and the third number?

(a)126  (b)42  (c) 166 (d)52  (e) None of these

Ans: (e) None of these

Explanation:

Let x be the  first number, then, the second number is x/2 and the third number is x/3

Then , the sum of numbers i.e.  x + x/2 + x/3 = 3 × 154 = 462

ð  11x/6 = 462  and so,   x = 252

ð  Then third number = 252/3 = 84

ð  So, the difference between first and the third number = 252 – 84 = 168 Ans.

Q. 39. The average speed of a bus is three-fifth the average speed of a car which covers 3250 kms. in 65 hours. What is the average speed of the bus?

(a) 30 Kms/hr.  (b) 20 Kms/hr  (c) 35 Kms/hr  (d) 36 Kms/hr  (e) None of these.

Ans: (a) 30 Kms./hr

Explanation:

The average speed of the car = 3250/65 = 50km/hr

Therefore, the average speed of the bus = 3/5^th of the average speed of the car = 3 × 50/5

       = 30km/hr.Ans.

Q. 40. The average speed of a car is twice the average speed of a truck. The truck covers 648 kms. in 24 hours. How much distance will the car cover in 15 hours?

(a) 820 kms.  (b) 1014 kms.  (c) 810 kms.  (d) 980 kms  (e) None of these

Ans: (c) 810 kms.

Explanation:

The average speed of the truck = 648/24 = 27 kms /hr

Therefore, the average speed of the car = 2 × 27 = 54 kms/hr

The  distance covered by the car in 15 hours = 54 × 15 = 810 kms.Ans.

Q.41. An aeroplane flies with an average speed of 756 km/hr. A helicopter takes 48 hours to cover twice the distance covered by aeroplane in 9  hours. How much distance will the helicopter cover in 18 hours.?  ( assuming that flights are non-stop and moving with uniform speed)

(a) 5014 km    (b)5140 km   (c) 5130km  (d) 5103 km  (e) none of these

Ans: (d) 5103 km.

Explanation:

The distance covered by the aeroplane in 9 hours = 9 ×756 = 6804 kms.

The distance covered by the helicopter in 48 hours = 2 × 6804 kms.

The distance covered by the helicopter in 1 hour = 2 × 6804/48 kms.

Therefore, the distance covered by the helicopter in 18 hours = 2 × 6804 × 18/48

                                                                                                  = 5103 kms. Ans.

Q. 42. The average age of A,B and C is 26 years. If the average age of A and C is 29 years. What is the age of B?

(a) 26   (b) 20  (c) 29  (d)23  (e) None of these

Ans: (b) 20

Explanation:

The sum of the ages of A, B and C = 3 × 26 = 78

The sum of the ages of  A and C = 2 × 29 = 58

So, age of B =  78 – 58 = 20. Ans.

Q. 43. The average of four positive integers is 124.5. The highest integer is 251 and the lowest integer is 65. The difference between the remaining two integers is 26. Which of the following integers is higher of the remaining two integers?

(a) 78   (b) 102  (c) 100  (d) Cannot be determined  (e) None of these

Ans: (e) None of these

Explanation:

The sum of the four positive integers = 4 × 124.5 = 498

The sum of the highest and lowest integer = 251 + 65 = 316

The sum of the remaining two integers = 498 – 316 = 182

The difference between  these two integers = 26

Subtract 26 from 182 and divide the result by 2, then we get the lowest one of the remaining two integers.

i.e. 182 – 26 = 156;   156/2 = 78   and the higher one of the remaining = 78 + 26 = 104. Ans.

Q. 44. The average height of 21 girls was recorded as 148 cms. If the teacher’s height was added, the average increased by one. What was the teacher’s height?

(a) 156 cms.  (b) 168 cms.  (c) 170 cms.  (d) 162 cms.  (e) none of these

Ans: (c) 170 cms.

Explanation:

The height of the teacher = New average + No. of girls × increase in average

                                         = 149 + 21 × 1 = 149 + 21 =170 cms.Ans.

 Q. 45. The average speed of a tractor is two-fifths the average speed of a car. The car covers 450    km in 6 hours. How much distance will the tractor cover in 8 hours?

a) 210 km b) 240 km   c) 420 km  d) 480 km  e) None of these

Ans: (b) 240 km

Explanation:

The average speed of the car = 450/6 = 75 km/hr

Therefore, the average speed of the tractor = 75 × 2/5 = 30 km/hr

So, the distance covered by the tractor in 8 hours = 30 × 8 = 240 km. Ans.

Q.46. Find the average of the following set of numbers.

          354, 281, 623, 518, 447, 702, 876

a)      538 b) 555 c) 568 d) 513 e) None of these

    Ans: (e) None of these 

  Explanation:

 Required average = 354 + 281 + 623 + 518 + 447 + 702 + 876 = 3801/7 = 543.Ans.

Q. 47.  If  average of 11 consecutive odd numbers is 17, what is the difference between the smallest and the largest number?

(a) 18  (b)  20  (c) 22  (d) 24   (e) None of these

Ans:  (b) 20

 Explanation:

  Let the numbers be x, x + 2, x+ 4, x +6, x+8, x + 10, x+12, x+14, x + 16, x + 18 and x + 20;

 Here, the smallest number = x  and   the largest number = x + 20

Therefore, their difference = x + 20 – x = 20.Ans.

Q. 48. The average of five consecutive odd numbers is 95. What is the fourth number in the descending order?

 (a) 91  (b) 95  (c) 99 (d) 97   (e)None of these

Ans: (e) None of these

Explanation:

Let the sum of the numbers be x + x +2 +  x +4 + x + 6 + x + 8 = 5 × 95 = 475

 5x + 20 = 475        :. x= 91

The fourth number in descending order = x + 2  = 91 + 2 =93. Ans.

Q. 49.  The average of marks obtained in 120 students was 35. If the average of passed candidates was 39 and that of failed candidates is 15, the number of candidates who  passed the examination is:

(a) 100   (b) 110   (c) 120   (d) 80

Ans: (a) 100

Explanation:

The sum of the marks obtained by 120 students = 120 × 35 = 4200

Let ‘x’ be the number of passed candidates

Then, sum of the marks obtained by the passed candidates = x × 39 = 39x

So, sum of the marks obtained by the failed candidates = (120 – x) 15

i.e.  39x + (120 – x ) 15 = 4200

       39x + 1800 – 15x = 4200

      39x -15x = 4200 – 1800 = 2400

      24x = 2400       and  so,   x = 100. Ans.

Q. 50.  The average marks in Hindi subject  of a class of 54 students is 76. If the marks of two students are misread as 60 and 77 of the actual marks 36 and 47 respectively, then what would be the correct average?

 (a) 75.5   (b) 77  (c) 75  (d) 76.5  (e) None of these

 Ans: (c) 75

Explanation:

The sum of the marks of 54 students = 54 × 76 = 4104

Difference of marks misread = 54

Subtracting this difference = 4104 – 54 = 4050

Now, the correct average = 4050/54 = 75. Ans.

Q. 51. In the Annual Examination Ramit scored 64 percent marks and Sangeet scored 634 marks. The maximum marks of the examination are 850. What are the average marks scored by Ramit and Sangeet  together?

(a) 544  (b) 567  (c) 589  (d) 591  (e) None of these

Ans: (c) 589

 Explanation:

The marks scored by Ramit = 64% of 850 = 544

The sum of  the marks of Ramit and Sangeet = 544 + 634 = 1178

So, the required average = 1178/2  = 589. Ans.

Q. 52. The average of 8 numbers is 14. The average of 6 of these numbers is 16. What is the average of the remaining 2 numbers?

(a)16   (b) 4   (c) 8  (d) Data inadequate  (e) None of these

Ans: (c) 8

Explanation:

The sum of 8 numbers = 8 × 14 = 112

The sum of 6 numbers = 6 × 16 = 96

Therefore, the sum of the last 2 numbers = 112 – 96 = 16

And the required average = 16/2 = 8. Ans.

Q. 53. Find the missing number if the average of all the eight numbers is 474.

           533, 128,  429,  225,  ______, 305, 601, 804

(a)    767    (b) 781   (c) 776   (d) 758   (e) None of these

Ans: (a) 767

Explanation:

The sum of the 8 numbers = 8 × 474 = 3792

The sum of the 7 numbers except the missing number = 533 + 128 + 429 + 225 + 305 + 601 + 804 = 3025

Therefore, the missing number = 3792 – 3025 = 767. Ans.

Q. 54. What is the average age of a family of five members, whose ages are 42, 49, 56, 63 and 35 years respectively?

(a) 60 years   (b) 49 years   (c) 45 years   (d)58 years  (e) None of these

Ans: (b) 49 years

Explanation:

Sum of their ages = 42 + 49 + 56 + 63 + 35 = 245

Their average age = 245/5 = 49 Years. Ans.

Q. 55. Average score of Rahul, Manish and Suresh is 63. Rahul’s score is 15 less than Manish. If Ajay scored 30 marks more than the average score of Rahul, Manish and Suresh, what is the sum of Manish’s and Suresh’s scores?

(a) 120   (b) 111   (c) 117   (d) Cannot be determined  (e) None of these

Ans: (b)  111

Explanation:

Sum of the marks of Rahul, Manish and Suresh = 63 × 3 = 189

Ajay’s score = 63 + 30 = 93

Rahul’s score = 93 – 15 = 78

Manish’s score = 78 – 10 = 68

Therefore, the sum of Manish’s and Suresh’s score = 189 – 78 = 111. Ans.

Q. 56. The average marks of a student in seven subjects is 41. After re-evaluation in one subject the marks were changed to 42 from 14 and in remaining subjects the marks remained unchanged. What are the new average marks?

(a) 45   (b) 44   (c) 46   (d) 47   (e) None of these

Ans: (a) 45

Explanation:

The old sum of marks = 7 × 41 = 287

New sum = 287 + ( 42-14)  = 287 + 28 = 315

So, the new average of his marks = 315/7 = 45. Ans.

Q. 57. Last month the number of employees in Bank A was 3/4^th of the number of employees in Bank B and the average number of employees of both the banks was 3108. Two employees resigned from Bank B this month. What is the total number of employees in Bank B at present?

(a) 3550  (b) 2662  (c) 3560  (d) 2642   (e) 3650

Ans: (a) 3550

Explanation:

Let the number of employees in Bank B be ‘x’’

Then, the number of employees in Bank A = 3x/4

Then,  x + 3x/4 = 3108 × 2 = 6216,       i.e.  x = 3552

Present number of employees in Bank B = 3552 – 2 = 3550. Ans.

Q. 58. The average speed of a train is 3 1/3 times the average speed of a tractor. The tractor covers 495 km. in 15 hours. How much distance will the train cover in 7 hours?

(a) 670 km.  (b) 770 km.   (c) 660 km.   (d) 760 km.   (e) None of these

Ans: (b) 770 km.

Explanation:

The average speed of the tractor = 495/15 = 33 kmph

Therefore, the average speed of the train = 33 × 10/3 = 110 kmph

So, the distance covered by the train in 7 hours = 110 × 7 = 770 km. Ans.

Q. 59. The average of 4 consecutive even numbers is 103. What is the product of smallest and the largest number?

(a) 10400    (b) 10504    (c) 10605    (d) 10600   (e) None of these

Ans: (d) 10600

Explanation:

Let the sum of 4 consecutive even numbers = x + x +2  + x + 4 + x + 6 = 103 × 4

4x + 12  = 412  and  x = 100 is the smallest number

 And x + 6 = 100 + 6 = 106 is the largest number

So, the required product = 100 × 106 = 10600. Ans.

Q. 60. The sum of five numbers is 150. The average of the first two numbers is 41 and average of the last two numbers is 23. What is the third number?

(a) 18    (b) 23  (c) 22  (d) 31  (e ) None of these

Ans: (c)22

Explanation:

The third number = 150 – ( Sum of   first two numbers + Sum of the last two numbers)

                            = 150 – (82 + 46)  = 150 – 128 = 22.Ans.

Q. 61. A, B, C, D are four consecutive odd numbers and their average is 42. What is the product of A and B?

(a) 1599    (b) 1505    (c) 1608     (d) 1635   (e ) None of these

Ans: (a) 1599

Explanation:

Let A = x;  B = x +2;  C = x + 4;  D = x + 6

Their sum = x + x+2 + x+ 4+ x + 6 = 42 × 4 = 168

4x + 12 = 168      ;     x = 156/4 =39= A

The product of A and B = x × (x+2) = 39 × 41 = 1599. Ans.

 Q. 62. There are six numbers, 30 , 72, 53, 68, ‘x’ and 87, out of which ‘x’ is unknown. The average of the numbers is 60.  What is the value of ‘x’?

(a) 40   (b ) 60   (c) 70     (d) 30   (e) None of these

Ans: (e) None of these

Explanation:

The sum of six numbers = 6 × 60 =360

The sum of five numbers except ‘x’ = 30 + 72 + 53  + 68  + 87 = 310

Therefore, x = 360 – 310 = 50.Ans.

Q. 63. If  the average of the 1^st  and 2^nd  of three numbers is 9 more than the average of the 2^nd and 3^rd number, what is the difference between the 1^st and the 3^rd number?

(a) 12   (b) 14    (c) 16    (d) 18    (e) 20

Ans: (d) 18

 Explanation:

Let a, b and c be the three numbers.

Then, given that a + b /2 = b + c/2 + 9

i.e. a + b/2   = b + c +  18/2  =>  a + b = b + c + 18

i.e. a – c = 18. Ans.

Q. 64. The average of the ages of Sumit, Krishna and Rishabh is 43 and the average of the ages of Sumit, Rishabh and Rohit is 49. If Rohit is 54 years old, what is Krishna’s age?

(a) 45 years   (b) 24 years   (c) 36 years   (d) Cannot be determined  (e) None of these

Ans: (c) 36 years

Explanation:

Sum of the ages of Sumit, Krishna and Rishabh = 43 × 3 = 129

Sum of the ages of Sumit , Rohit and Rishabh = 49 × 3= 147

Given that, the age of Rohit = 54 years

Therefore, Sum of the ages of Sumit and Rishabh = 147 – 54 =93

Then, Krishnas’ age = 129 – 93 = 36 years. Ans.

Q. 65. The average age of 8 men is increased by 2 years when two of them whose ages are 21 and 23 years are replaced by two new men. The average age of the two new men is

(a) 22 years  (b) 24 years  (c) 28 years   (d) 30 years   (e) None of these

Ans: (d) 30 years

Explanation:

The increase in total age when two men are replaced by two men = 2 × 8 = 16 years

Sum of the ages of the two men left = 21 + 23 = 44 years

Therefore, the sum of the ages of the two new men = 44 + 16 = 60 years

So, the required average = 60/2 = 30 years. Ans.

Q. 66. The average of fifty numbers is 28. If two numbers namely 25 and 35 are discarded, then the average of the remaining numbers is nearly

 (a) 29.27   (b) 27.92  (c) 27.29    (d) 29.72   (e) None of these

Ans: (b) 27.92

Explanation:

The sum of fifty numbers = 50 × 48 = 1400

Sum of the remaining 46 numbers =  1400  - (25+ 35)  = 1400 – 60 = 1340

Then, the required average = 1340/46= 27.916 = 27.92 nearly. Ans.

Q. 67. The average of three numbers is 77. The first number is twice the second and the second number is twice the third. Find the first number?

(a) 33   (b) 66   (c) 77   (d) 132   (e) None of these

Ans: (d) 132

Explanation:

Let ‘x’ be the third number, then second number = 2x   and the first number = 4x

4x + 2x + x = 3 × 77 = 231

7x = 231   and  ‘x’ the third number = 33

Therefore, the first number i.e. 4x  = 4 × 33 = 132. Ans.

Q. 68.  The average weight of 21 girls was recorded as 58 kgs. If the teacher’s weight was added, the average increased by one. What was the teacher’s weight?

(a) 96 kgs.    (b) 78 Kgs.   (c) 80 kgs.   (d) 62 kgs.    (e) None of these

Ans: (c) 80 kgs.

Explanation:

The teacher’s weight = New average + number of girls × difference in average

                                  = 59 + 21 × 1   = 59 + 21 = 80 Ans.

Q.69. A, B, C, D and E are five consecutive even numbers. Average of A and E is 46. What is the largest number?

(a) 52   (b) 42   (c) 50   (d) 48    (e) None of these

Ans: (c) 50

Explanation:

Let A = x,  B = x + 2,   C = x +4,   D = x + 6,   E = x + 8

Then, x + x + 8 = 2 × 46 = 92  i.e.. 2x + 8 =92

 And x= 42,

So, the largest number i.e x + 8 = 42 + 8 =50. Ans.

Q. 70. If 56a + 56b = 1008, what is the average of a and b?

(a) 6   (b) 9    (c) 12   (d) 18    (e) None of these

Ans: (b) 9

Explanation:

56(a+b) = 1008

Then, a + b = 1008/56 = 18

So. The required average = 18 /2 = 9 Ans.

Q. 71. The average of 5 consecutive odd numbers A,  B, C, D and E is 45. What is the product of B & D?

(a) 2107   (b) 2205   (c) 1935    (d) 2021     (e) None of these

Ans: (d) 2021

Explanataion:

The average of odd number of odd numbers will always be the middle number.

So, here 45 will be the middle i.e. the third number.

Therefore, the numbers are A = 41,  B = 43,  C = 45, D = 47 and E = 49

So, the required product = B × D = 43 × 47 = 2021.Ans.

Q. 72. The average of runs of a cricketer of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4?

(a) 76  (b) 70  (c) 4   (d) 2   (e) None of these

Ans: (a) 76

Explanation:

The score of the cricketer after 10 innings = 10 × 32 = 320

The score after 11 innings to make average 36 = 11 × 36 = 396

Therefore, the runs required in his 11^th innings to make his average 36 = 396 – 320 =76.Ans.

OR

The required runs = old average + ( total innings × Difference in average)

                             = 32 + ( 11 × 4) = 32 + 44 = 76.Ans.

Q. 73. The average of five consecutive odd numbers A, B, C, D and E is 25. What percentage of E is C?

(a) 86.2   (b) 87.5   (c) 56.8   (d) 88.9  (e) 78.5

Ans: (a) 86.2

Explanation:

Because the average of odd number of odd numbers is the middle number,  here, C = 25,

So, E = 29

Therefore, the required percentage = 25 × 100/29 = 86.2.Ans.

Q. 74. The sum of five consecutive even numbers of set A is 220. What is the sum of a different set of five consecutive numbers whose second lowest number is 37 less than double of the lowest number of set A?

(a) 223   (b) 225  (c) 235   (4) 243   (e) None of these

Ans: (e) None of these

Explanation:

Let the sum of first set A = x + x +2 + x + 4 +x+ 6 +x+8 = 220

5x + 20 = 220 and so, x = 40

So, the consecutive even numbers in set A = 40, 42, 44, 46 and 48

So, the second lowest number in another set = 80 – 37 = 43

And so, numbers in that set = 42, 43, 44, 45, 46

And the required sum = 42 + 43+ 44+ 45 +46 = 220.Ans.

Q. 75. A, B, C, D and E are five consecutive odd numbers. The average of these five numbers is 35. What is the product of C and E?

(a) 1245  (b) 1365   (c) 1380   (d) 1401   (e) 1425

Ans: (b) 1365

Explanation:

Because, average of odd number of odd numbers is the middle number, so, C = 35,

So, E = 39  and the product  of C and E = 35 × 39 = 1365.Ans.

Q. 76. The average of five positive integers is 385. The average of the first two integers is 568.5. The average of the fourth and fifth integers is 187.5. What is the third integer?

(a) 420  (b) 382   (c) 415   (d) Cannot be determined (e) None of these

Ans: (e) None of these

Explanation:

The sum of the five integers = 5 × 385 = 1925

The sum of the first two integers = 2 × 568.5 = 1137

The sum of the last two integers = 2 × 187.5 = 375

Therefore, the third number = 1925 – ( 1137 + 375) = 1925 – 1512 = 413.Ans.

Q. 77. The average of four consecutive odd numbers A, B, C and D respectively is 94. What is the product of A & C?

(a) 8835   (b) 8463   (c) 8827   (d) 8645   (e) None of these

Ans: (d) 8645

Explanation:

Let the sum x + x + 2 + x +4 + x + 6 = 94 × 4 =376

4x + 12 = 76    and so, x = 91    i.e.   A= 91, B = 93,  C = 95  and  D = 97

Therefore, the required product, A × C = 91 × 95 = 8645.Ans.

Q. 78. Find the average of the following set of scores:

288, 420, 166, 80, 24, 108, 340, 222

(a)    202  (b) 218  (c) 224   (d) 206   (e) None of these

Ans: (d) 206

Explanation:

The Sum of the scores = 288 + 420 + 166 + 80 + 24 + 108 + 340 + 222 = 1648

Therefore, the required average = 1648/8 = 206.Ans.

Q. 79. The average age of 8 persons is increased by 2 years, when one of them, whose age is 24 years is replaced by a new person. The age of the new person is

(a) 42 years  (b) 40 years  (c) 38 years  (d) 45 years  (e) None of these

Ans: (b) 40 years

Explanation:

The age of the new person = the age of the person who left + (No. of persons × Difference in average)

 i.e. = 24 + (8 × 2) = 24 + 16 = 40 years. Ans.

Q. 80. Out of three numbers, the first is twice the second and is half of the third. If the average of the three numbers is 56, then difference of the first and third number is

(a) 12  (b) 20  (c) 24   (d) 48   (e) None of these

Ans: (d) 48

Explanation:

Let the second no. be ‘x’

Then, the first no. = 2x  and the third no. = 2 × 2x = 4x.

The sum, 2x + x + 4x = 3 × 56

   7x = 3 × 56,    So, x = 56 × 3/7 = 8 × 3 = 24, is the second number,

  First number = 2 × 24 = 48

And the third number = 4 × 24 = 96

Then, the difference between the first and the third number = 96 – 48 = 48. Ans.

Q. 81. The average of four positive integers is 73.5. The highest integer  is 108 and the lowest integer is 29. The difference between the remaining two integers is 15. Which of the following is the smaller of the remaining two integers?

(a) 80   (b) 86  (c) 73   (d) Cannot be determined  (e) None of these

Ans: (e) None of these

Explanation:

The sum of  the four positive integers = 4 × 73.5 = 294

The sum of the remaining two integers = 294 – ( 108 + 29) = 157

Less the their difference = 157 – 15 = 142  and  then, the smaller of the remaining two numbers = 142/2 = 71.Ans.

Q. 82. Average of five consecutive odd numbers is 95. What is the fourth number in descending order?

 (a) 91   (b) 95  (c) 99  (d) 97   (e) None of these

Ans: (e) None of these

Explanation:

Because the middle number is the average of odd number of odd numbers = 95

So, the numbers are 91, 93, 95, 97, 99

The numbers in descending order is 99, 97,  95, 93, 91

So, the required number = 93. Ans.

Q. 83. The average age of A and B is 20 years, that of B and C is 19 years and that of A and C is 21 years. What is the age (in years) of B?

(a) 39     (  b) 21    (c) 20     (d) 18     (e) None of these

Ans: (d) 18 years

Explanations:

The sum of ages, A + B = 40,  B + C = 38,   A + C = 42

Then, A + B + B + C = 40 + 38 = 78   i.e. A + C + 2B = 78, (where, A + C = 42)

2B = 78 – 42 = 36,    So, B = 36/2 = 18 years. Ans.

Q. 84. In an experiment, the average of 11 observations was 90. If the average of the first five observations is 87 and that of the last five is 84, then the sixth observation is

(a) 135  (b) 145   (c) 150   (d) 165  (e) None of these

Ans: (a) 135

Explanations:

The sum of the 11 observations = 11 × 90 = 990

The sum of the first five observations = 5 ×  87 = 435

The sum of the last five observations = 5 × 84 = 420

So, the sum of the 10 observations = 435 + 420 = 855

Therefore, the sixth observation = 990 – 855 = 135.Ans.

Q. 85. If the average of four numbers a, b, c and d is A, then the average of a, b, c, d and 3 A/2 is

(a) 11 A/10   (b) A/2    (c)5 A/2    (d) 2A  (e) None of these

Ans: (a) 11 A/10

Explanations:

The sum of a, b, c, and d = 4 A

Sum of 4A + 3 A/2 = 11 A/2, Therefore, the required average = 8 A + 3 A/2 = 11 A/10. Ans.

Q. 86. The average age of a man and her daughter is 40 years. The ratio of their ages is 6 : 2. Then what is the daughter’s age (in years)?

 (a) 20  (b) 18   (c) 22  (d) 16  (5) 60

Ans: (a) 20 years

Explanations:

The sum of their ages = 2 × 40 =80

6x + 2x = 80

8x = 80 ,    So,   x = 10 The daughter’s age = 2x = 2 × 10 = 20 years. Ans.

Q. 87. The average of x and y is 18. If z is equal to 9, the average of x, y and z is

(a) 3   (b) 9   (c) 12     (d) 15   (e) None of these

Ans: (d) 15

Explanation:

The sum of x and y = 36

The sum of x + y + z = 36 + 9 =45

The required average = 45/3 = 15. Ans.

Q. 88. The average of 11 numbers is 10.9. If the average of the first six numbers is 10.5 and that of the last six numbers is 11.4, then the middle (6^th)  number is

(a) 11.5  (b) 11   (c) 11.3   (d) 11.0    (e) None of these

Ans: (a) 11.5

Explanation:

Sum of the first 6 numbers = 6 × 10.5 = 63

Sum of the last 6 numbers = 6 × 11.4 = 68.4

Sum of first 6 and last 6 numbers = 63 + 68.4 = 131.4

Sum of 11 numbers = 11 × 10.9 = 119.9

Then the middle number = 131.4 – 119.9 = 11.5. Ans.

Q. 89. There are 30 students in a class. The average age of the first 10 students is 12.5 years. The average age of the next 20 students is 13.1 years. The average age ( in years) of the whole class is

(a) 12.5  (b) 12.7   (c) 12.8   (d) 12.9   (e) None of these

Ans: (d) 12.9

Explanation:

The sum of the ages of 10 students = 10 × 12.5 = 125

Sum of the ages of the remaining 20 students = 20 × 13.1 = 262

Therefore, the sum of the ages of 30 students = 125 + 262 = 387

So, the required average = 387/30 = 12.9. Ans.

Q. 90. The average of 50 numbers is 38. If two numbers 45 and 55 are left, then the average of the remaining numbers is

(a) 50   (b) 37.5     (c) 38.5    (d) 37.2    (e) None of these

Ans: (b) 37.5

Explanation:

The sum of the 50 numbers = 50 × 38 = 1900

Less sum of 45 & 55  i.e.   Sum of the remaining 48 numbers = 1900 – 100 = 1800

Therefore the average of the remaining 48 numbers = 1800/48 = 37.5.   Ans.

Q. 91. The mean of 50 observations was 36. It was found later that an observation 48 was  wrongly taken as 23. The correct (new) mean is

(a) 35.2   (b) 36.1   (c) 36.5   (d) 39.1   (e) None of these

Ans: (c) 36.5

Explanation:

Sum of the 50 observations = 50 × 36 = 1800

Add the difference error = 48 -23 = 25

Then, the correct sum = 1825

The correct (new ) average = 1825/50 = 36.5. Ans.

Q. 92. The sum of three consecutive even numbers is 44 more than the average of these numbers. The product of the smallest and the largest numbers is equal to

(a) 385   (b) 422  (c)  480    (d) 504     (e) None of these

Ans: (c) 480

Explanation:

The sum of the numbers i.e. x + x+2 + x+ 4   = 3x + 6 

3x + 6 = 44 + 3x + 6/3

3 ( 3x + 6) = 132 + 3x + 6

 9x + 18 = 132 + 3x + 6 = 138 + 3x

6x = 120   and   x = 20, the smallest number

The largest number = x + 4 = 20 + 4 = 24

Therefore, the required product = 20 × 24 = 480. Ans.

Q. 93. The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is

(a) 50 kg    (b) 55 kg  (c) 45 kg    (d) 53 kg   (e) None of these

Ans: (c) 45 kg

Explanation:

The weight of the teacher = New average + (No. of students × difference in average)

                                          = 35.4   + 24 × .4 = 35.4 + 9.6 = 45 kg. Ans.

Q. 94. Of three numbers, the first is twice the second and the second 3 times the third. If their averages is 100, the largest of the three numbers is

(a) 120    (b) 150    (c) 180    (d) 300    (e) None of these

Ans: (c) 180

Explanation:

Let the third number be ‘x’

Then, the second number will be = 3x

 The first number = 6x

Their average,   10x/3 = 100,   i.e. x = 30   the third number.

The largest number = 6x = 6 × 30 =180. Ans.

Q. 95. A grocer has a sale of Rs.6435,  Rs.6927, Rs.6855, Rs.7230 and Rs.6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs.6500?

(a) Rs.4991   (b) Rs.5991   (c) Rs.6991    (d) Rs.6001    (e) None of these

Ans: (a) Rs.4991

Explanation:

The sum of the sales of the 5 consecutive months = Rs.6435 + 6927 + 6855 +7230 + 6562

                                                                                = Rs.34009

To get an average of Rs.6500 in 6 months, the sum of sales should be = 6 × 6500 = Rs.39000

Therefore the sixth month sale should be = 39000 – 34009 = Rs.4991. Ans.

Q. 96. Of three numbers, the second is twice the first and it is also thrice the third. If the average of three numbers is 44, the difference of the first number and the third number is

(a) 24    (b) 18   (c) 12   (d) 6    (e) None of these

Ans: (c) 12

Explanation:

Let the 3^rd number be ‘x’, then the second number = 3x  and  the first number = 3x/2

Their sum = 3x/2 + 3x + x  (multiplying by 2), we get, 3x + 6x + 2x = 44 × 3

11x = 44 × 3,  and  x = 12

Now, the first number = 3x = 3× 12 = 36

And the third number = 2x = 2 × 12 = 24

Therefore, the required difference = 36 – 24 = 12. Ans.

Q. 97. The average of Babu’s marks in 7 subjects is 75. His average in 6 subjects excluding Science is 72. How many marks did he get in Science?

(a) 72   (b) 90   (c) 93   (d) cannot be determined   (e) None of these

Ans: (c) 93

Explanation:

Sum of his marks in 7 subjects = 7 × 75 = 525

Sum of the marks in 6 subjects excluding Science = 6 × 72 = 432

Therefore,  his marks in Science = 525 – 432 = 93. Ans.

Q. 98. The average of 15 numbers is 7. If the average of the first 8 numbers be 6.5 and the average of the last 8 numbers be 9.5, then the middle number is

(a) 20   (b) 21   (c) 23    (d) 18    (e) None of these

Ans: (c) 23

Explanation:

The sum of 15 numbers = 15 × 7 = 105

The sum of the first 8 numbers = 8× 6.5 = 52

The sum of the last 8 numbers = 8 × 9.5 = 76

Sum of the first 8 and last 8 numbers = 52 + 76 = 128

Therefore, the middle number = 128 – 105 = 23. Ans.

Q. 99. The average of 5 consecutive even numbers is 28. What is the sum of the largest number and the square of the smallest number?

(a) 603   (b) 612   (c) 608    (d) 605    (e) None of these

Ans: (c)608

Explanation:

Because, the average of odd number of even numbers is the middle number, here the middle number = 28 and the 5 consecutive even numbers are 24, 26, 28, 30 ,32

The required sum = 32 + 24²  = 32 + 576 = 608.Ans.

Q. 100. The average of five numbers is 371.8. The average of the first and second number is 256.5 and the average of the fourth and fifth number is 508. Which of the following is the  third number?

(a) 360  (b) 310   (c)430   (d) 380   (e) 330

Ans: (e) 330

Explanation:

The sum of the five numbers = 5 × 371.8 = 1859

The sum of the first two numbers = 2 × 256.5 = 513

The sum of the last two numbers = 2 × 508 = 1016

Therefore, the sum of the four numbers = 513 + 1016 = 1529

So, the third number = 1859 – 1529 = 330.Ans.