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 25th
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 25th
innings his score should be = 25 × 50 =1250
Therefore, the score required to
obtain in his 25th 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
11x/6
= 462 and so, x = 252
Then
third number = 252/3 = 84
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 = 3/5th 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 3/4th 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 1st
and 2nd of three
numbers is 9 more than the average of the 2nd and 3rd
number, what is the difference between the 1st and the 3rd
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
11th 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 (6th) 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 3rd 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.
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