1. A cricketer scored some runs in his 21st innings, as a result, his average runs increased by 3. If the present average run is 40, how many runs he scored in the final innings?
a) 85
b) 103
c) 82
d) 100
Discussion
Explanation: Let he scored x runs in final innings
Now,
37 × 20 + x = 40 × 21
x = 840 - 740
x = 100
2. Average of 80 numbers are 42. When 5 more numbers are included, the average of 85 numbers become 45. Find the average of 5 numbers.
a) 82
b) 89
c) 93
d) 98
Discussion
Explanation: Total of 80 numbers
= 80 × 42 = 3360
Now, total of 85 numbers
= 85 × 45 = 3825
Hence, sum of 5 numbers
= 3825 - 3360 = 465
Average of five numbers
= $$\frac{{465}}{5}$$ = 93
3. The average age of A, B, C, D and E is 40 years. The average age of A and B is 35 years and the average of C and D is 42 years. Age of E is :
a) 46 years
b) 45 years
c) 48 years
d) 42 years
Discussion
Explanation: A + B + C + D + E = 40 × 5 = 200
A + B = 35 × 2 = 70
C + D = 42 × 2 = 84
Therefore,
E = (A + B + C + D + E) - (A + B + C + D)
E = 200 - 70 - 84
E = 46 years
4. A man travels equal distances of his journey at 40, 30 and 15 km/h. respectively. Find his average speed for whole journey.
a) 24
b) 25
c) 27
d) 28
Discussion
Explanation: Required average speed,
$$ = {\frac{{\left( {3 \times 40 \times 30 \times 15} \right)}}{{ {\left( {40 \times 30} \right) + \left( {40 \times 15} \right) + \left( {30 \times 15} \right)} }}} $$
$$ = 24\,{\text{km/hr}}$$
5. A batsman makes a score of 270 runs in the 87th inning and thus increase his average by a certain number of runs that is a whole number. Find the possible values of the new average.
a) 12
b) 98
c) 184
d) All of these
Discussion
Explanation: Part of the runs scored in the 87th innings will go towards increasing the average of the first 86 innings to the new average and remaining part of the runs will go towards maintaining the new average for the 87th innings. The only constraint in this problem is that there is increase in average by a whole number of runs. This is possible for all three options.
6. A school has only four classes that contain 10, 20, 30 and 40 students respectively. The pass percentage of these classes are 20%, 30%, 60% and 100% respectively. Find the pass % of the entire school.
a) 76%
b) 66%
c) 56%
d) 34%
Discussion
Explanation: The number of pass candidates are 2 + 6 + 18 + 40 = 66 out of total 100.
Hence, Pass percentage = 66%
7. One-fourth of certain journey is covered at the rate of 25 km/h, one-third at the rate of 30 km/h and the rest at 50 km/h. Find the average speed for the whole journey.
a) $$\frac{{600}}{{53}}$$ km/h
b) $$\frac{{1200}}{{53}}$$ km/h
c) $$\frac{{1800}}{{53}}$$ km/h
d) $$\frac{{1600}}{{53}}$$ km/h
Discussion
Explanation: Let distance be 120 km
Hence 30 km is covered by 25 kmph and 40 km covered by 30 kmph and rest 50 km has been covered 50 kmph
$$\eqalign{ & {\text{average}} = {\frac{{120}}{{{\text{total}}\,{\text{time}}\,{\text{taken}}}}} \cr & = \frac{{120}}{{\frac{{30}}{{25}} + \frac{{40}}{{30}} + \frac{{50}}{{50}}}} \cr & = \frac{{3600}}{{106}} \cr & = \frac{{1800}}{{53}}\,{\text{km/h}} \cr} $$
8. A man started his journey from Lucknow to Kolkata, which is 200 km, at the speed of 40 kmph then he went to Banglore which is 300 km, at the speed of 20 kmph. Further he went to Ahmedabad which is 500 km, at the speed of 10 kmph. The average speed of the man is :
a) 15.6 kmph
b) 16.1 kmph
c) 14$${\frac{5}{7}}$$ Kmph
d) 14$${\frac{2}{7}}$$ kmph
Discussion
Explanation:
$$\eqalign{ & {\text{Average}}\,{\text{speed}}, \cr & = {\frac{{{\text{Total}}\,{\text{Distance}}}}{{{\text{Total}}\,{\text{time}}}}} \cr & = {\frac{{ {200 + 300 + 500} }}{{ { {\frac{{200}}{{40}}} + {\frac{{300}}{{20}}} + {\frac{{500}}{{10}}} } }}} \cr & = \frac{{1000}}{{70}} \cr & = 14{\frac{2}{7}}\, \text{kmph} \cr} $$
9. Five years ago, the average age of A, B, C and D was 45 yr. with E joining them now, the average of all the five is 49 yr. How old is E?
a) 45 years
b) 25 years
c) 64 years
d) 40 years
Discussion
Explanation: Total present age of A, B, C and D
= (45 × 4) + (4 × 5)
= 200 years
Total age present age of A, B, C, D and E
= 49 × 5
= 245 years
So, Age of E = 45 years
10. There are five boxes in cargo hold. The weight of the first box is 200 kg and the weight of the second box is 20% higher than the weight of the third box, whose weight is 25% higher than the first box's weight. The fourth box at 350 kg is 30% lighter than the fifth box. Find the difference in the average weight of the four heaviest boxes and the four lightest boxes.
a) 51.5 kg
b) 75 kg
c) 37.5 kg
d) 112.5 kg
Discussion
Explanation: The weight of boxes is :
1st box = 200 kg
2nd box = 300 kg
3rd box = 250 kg
4th box = 350 kg
5th box = 500 kg
difference between heavier 4 and lighter 4 is 300
difference in average is 75
11. The mean weight of 100 students in a class is 46 kg. The mean weight of boys is 50 and of girls is 40 kg. Therefore, the number of boys is:
a) 50
b) 60
c) 70
d) 65
Discussion
Explanation: Let number of boys are x and then number of girls
= (100 - x)
50x + (100 - x) × 40 = 46 × 100
x = 60
Number of boys = 60
12. The average income of A, B and C is Rs. 12,000 per month and average income of B, C and D is Rs. 15,000 per month. If the average salary of D be twice that of A, then the average salary of B and C is in Rs. :
a) 13,500
b) 9,000
c) 8,000
d) 18,000
Discussion
Explanation:
$$\eqalign{ & A + B + C = 12000 \times 3 \cr & B + C + D = 15000 \times 3 \cr & \to D - A = 3000 \times 3 = 9000 \cr & \,D = 2A \cr & \,D = 18000\,{\text{and}}\,A = 9000 \cr & {\text{Therefore}}, \cr & {\text{Average}}\,{\text{salary}}\,{\text{of}}\,B\,{\text{and}}\,C, \cr & = \frac{{ {45000 - 18000} }}{2} \cr & = 13500 \cr} $$
13. The average marks of four subjects is 120. If 33 was misread as 13 during the calculation, what will be the correct average?
a) 122
b) 120
c) 125
d) 121
Discussion
Explanation:
$$\eqalign{ & {\text{Correct average}} \cr & = 120 + \left( {\frac{{33 - 13}}{4}} \right) \cr & = 120 + 5 \cr & = 125 \cr} $$
Average given is 120.
Difference of 33 and 13 is 20.
That means 20 must be added to total.
Then average of is and so must be added to average, i.e.
Correct average = 120 + 5 = 125
14. Find the average increase rate, if increase in the population in the first year is 30% and that in the second year is 40%.
a) 41%
b) 56%
c) 40%
d) 38%
Discussion
Explanation: Let 100 be the original population.
1st year's population increased = 30%
So, Population after first year
= (100 + 30% of 100)
= 130
Population in second year increases by 40%,
Then Population
= (130 + 40% of 130)
= 182
The final population become 182 which was originally at 100.
It means there is 82% increment in the population in 2 years
So, Average increment = $$\frac{{82}}{2}$$ = 41%
15. The average weight of 47 balls is 4 g. if the weight of the bag (in which the balls are kept) be included; the calculated average weight per ball increases by 0.3 g. What is the weight of the bag?
a) 14.8 g
b) 14.4 g
c) 15 g
d) 14.1 g
Discussion
Explanation: Total increased weight
= 0.3 × 47
= 14.1 g
16. The average of 20 students is 12 years, if the teacher's age is included, average increases by one. The age of the teacher is:
a) 28 years
b) 30 years
c) 33 years
d) 35 years
Discussion
Explanation: Average of 20 students = 12 years
Total age of 20 students
= 20 × 12
= 240 years
When teacher included average become 13 years
Now, total age 20 students and teacher
= 13 × 21 = 273 years
∴ Age of teacher
= 273 - 240
= 33 years
17. The average monthly salary of 660 workers in a factory is Rs. 380. The average monthly salary of officers is Rs. 2100 and the average monthly salary of the other workers is Rs. 340. Find the number of other workers.
a) 645
b) 650
c) 640
d) 642
Discussion
Explanation:
$$\eqalign{ & {\text{Total}}\,{\text{salary}}\,{\text{of}}\,660\,{\text{workers}} \cr & = 660 \times 380 \cr & = Rs.\,250800 \cr & {\text{If}}\,{\text{other}}\,{\text{workers}}\,{\text{be}}\,x;\,{\text{then}}, \cr & = \left[ {\left( {660 - x} \right) \times 2100} \right] + 340x \cr & = 250800 \cr & \,1386000 - 2100x + 340x = 250800 \cr & 1760x = 1135200 \cr & \,x = \frac{{1135200}}{{1760}} = 645 \cr & {\text{Number}}\,{\text{of}}\,{\text{other}}\,{\text{workers}} = 645 \cr} $$
18. 19 people went to a hotel for combine dinner party 13 of them spent Rs. 79 each on their dinner and rest spent 4 more than the average expenditure of all the 19. What was the total money spent by them.
a) 1628.4
b) 1534
c) 1496
d) None of these
Discussion
Explanation: Let average expenditure of 19 people be x
19x = 13 × 79 + 6 × (x + 4)
19x = 13 × 79 + 6x + 24
x = 80.84
So, total money spent
= 80.84 × 19
= Rs. 1536.07
19. Students of three different classes appeared in common examination. Pass average of 10 students of first class was 70%, pass average of 15 students of second class was 60% and pass average of 25 students of third class was 80% then what will be the pass average of all students of three classes?
a) 69%
b) 72%
c) 74%
d) 75%
Discussion
Explanation:
$$\eqalign{ & {\text{Sum}}\,{\text{of}}\,{\text{pass}}\,{\text{student}}\,{\text{}}\,{\text{first,}}\,{\text{second}}\,{\text{and}}\,{\text{third}}\,{\text{class}}, \cr & = \left( {70\% \,{\text{of}}\,10} \right) + \left( {60\% \,{\text{of}}\,15} \right) + \left( {80\% \,{\text{of}}\,25} \right) \cr & = 7 + 9 + 20 = 36 \cr & {\text{Total}}\,{\text{students}}\,{\text{appeared}}, \cr & = 10 + 15 + 25 = 50 \cr & {\text{Pass}}\,{\text{average}}, \cr & = 36 \times \frac{{100}}{{50}} = 72\% \cr} $$
20. A passenger travels from Delhi to Merut at a speed of 30 kmph and return with a speed of 60 kmph. What is the average speed?
a) 45
b) 48
c) 40
d) 50
Discussion
Explanation: Average Speed
$$\eqalign{ & = \frac{{2 \times 60 \times 30}}{{60 + 30}} \cr & = \frac{{3600}}{{90}} \cr & = 40\,{\text{kmph}} \cr} $$
21. My Scooty gives an average of 40 kmpl of petrol. But after recent filling at the new petrol pump, its average dropped to 38 kmpl. I investigated and found out that it was due to adulterated petrol. Petrol pimps add kerosene, which is $$\frac{2}{3}$$ cheaper than petrol, to increase their profits. Kerosene generates excessive smoke and knocking and gives an average of 18 km per 900 ml. If I paid Rs. 30 for a litre of petrol, What was the additional amount the pump-owner was making?
a) Rs. 1.75
b) Rs. 1.80
c) Rs. 2
d) Rs. 2.30
Discussion
Explanation: Let x ml of kerosene be there in 1 litre mixture.
Then, quantity of petrol in 1 litre mixture = (1000 - x) ml
$$ \frac{{40}}{{1000}}\left( {1000 - x} \right)$$ $$ + \frac{{18}}{{900}}x$$ = 38
$$\eqalign{ & \frac{x}{{25}} - \frac{x}{{50}} = 2 \cr & \frac{x}{{50}} = 2 \cr & x = 100 \cr} $$
So, 1 litre mixture has 900 ml petrol and 100 ml kerosene.
Cost of 1 litre petrol = Rs. 30
Cost of 1 litre kerosene
= Rs. $$\left[ {\left( {1 - \frac{2}{3}} \right) \times 30} \right]$$
= Rs. 10
Coast of 1 litre mixture
= Rs. $$\left( {\frac{{30}}{{1000}} \times 900 + \frac{{10}}{{1000}} \times 100} \right)$$
= Rs. 28
Additional amount earned by pump-owner
= Rs. (30 - 28)
= Rs. 2
22. The body weight of seven students of a class is recorded as 54 kg, 78 kg, 43 kg, 82 kg, 67 kg, 42 kg and 75 kg. What is the average body weight of call the seven students?
a) 63 kg
b) 69 kg
c) 741 kg
d) 73 kg
Discussion
Explanation: Average body weight
$$\eqalign{ & = {\frac{{54 + 78 + 43 + 82 + 67 + 42 + 75}}{7}} {\text{ kg}} \cr & = {\frac{{441}}{7}} {\text{ kg}} = 63{\text{ kg}} \cr} $$
23. There are five boxes in a cargo hold. The weight of the first box is 200 kg and the weight of the second box is 20% more than the weight of third box, whose weight is 25% more than the first box’s weight. The fourth box at 350 kg is 30% lighter than the fifth box. The difference in the average weight of the four heaviest boxes and the four lightest boxes is
a) 37.5 kg
b) 51.5 kg
c) 75 kg
d) 112.5 kg
Discussion
Explanation: Weight of first box = 200 kg
Weight of third box
= 125 % of 200 kg
= 250 kg
Weight of second box
= 120% of 250 kg
= 300 kg
Weight of fourth box = 350 kg
Let the weight of fifth box be x kg
70% of x = 350 kg
$$\eqalign{ & x = \left( {\frac{{350 \times 100}}{{70}}} \right) \cr & x = 500{\text{ kg}} \cr} $$.
Average weight of four heaviest boxes
$$\eqalign{ & {\text{ = }}\left( {\frac{{500 + 350 + 300 + 250}}{4}} \right){\text{kg}} \cr & {\text{ = 350 kg}} \cr} $$
Average weight of four lightest boxes
$$\eqalign{ & = \left( {\frac{{200 + 250 + 300 + 350}}{4}} \right){\text{kg}} \cr & = 275{\text{ kg}} \cr} $$
Required difference
= (350 - 275)
= 75 kg
24. The average of 4 positive integers is 59. The highest integer is 83 and the lowest integer is 29. The difference between the remaining two integers is 28. Which of the following integers is higher of the remaining two integers ?.
a) 39
b) 48
c) 76
d) Cannot be determined
Discussion
Explanation: Sum of four integers = 59 × 4 = 236
Let the required integers be x and x -28
x + (x - 28) = 236 - (83 + 29)
2x - 28 = 124
2x = 152
x = 76
Required integer = 76
25. A family consists of grandparents, parents and three grandchildren. The average age of the grandparents is 67 years that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?
a) $$28\frac{4}{7}{\text{ years}}$$
b) $$31\frac{5}{7}{\text{ years}}$$
c) $$32\frac{1}{7}{\text{ years}}$$
d) None of these
Discussion
Explanation: Required average
$$\eqalign{ & {\text{ = }} {\frac{{67 \times 2 + 35 \times 2 + 6 \times 3}}{{2 + 2 + 3}}} \cr & {\text{ = }} {\frac{{134 + 70 + 18}}{7}} \cr & = \frac{{222}}{7} \cr & = 31\frac{5}{7}{\text{ years}} \cr} $$
26. The average age of seven boys sitting in a row facing North is 26 years. If the average age of the first three boys is 19 years and the average age of the last three boys is 32 years, what is the age of the boy who is sitting in the middle of the row?
a) 24 years
b) 28 years
c) 29 years
d) 31 years
Discussion
Explanation: Age of the boy sitting in the middle
= [26 × 7 - (19 × 3 + 32 × 3)]
= (180 + 153) years
= 29 years
27. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:
a) 17 kg
b) 20 kg
c) 26 kg
d) 31 kg
Discussion
Explanation: Let A, B, C represent their respective weights.
A + B + C = (45 × 3) = 135.....(i)
A + B = (40 × 2) = 80.....(ii)
B + C = (43 × 2) = 86.....(iii)
Adding (ii) and (iii),
A + 2B + C = 166.....(iv)
Subtracting (i) from (iv), B = 31
B's weight = 31 kg
28. The average weight of 45 students in a class is 52 kg. Five of them whose average weight is 48 kg leave the class and other 5 students whose average weight is 54 kg join the class. What is the new average weight (in kg) of the class?
a) $$52\frac{1}{3}$$
b) $$52\frac{1}{2}$$
c) $$52\frac{2}{3}$$
d) None of these
Discussion
Explanation: Sum of the weights of the students after replacement
= [(52 × 45) - (48 × 5) + (54 × 5)] kg
= 2370 kg
New average
= $$\left( {\frac{{2370}}{{45}}} \right)$$ kg
= $$52\frac{2}{3}$$ kg
29. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average of the whole team. What is the average age of the team?
a) 23 years
b) 24 years
c) 25 years
d) None of these
Discussion
Explanation: Let the average age of the whole team be x years.
11x - (26 + 29) = 9 (x - 1)
11x - 9x = 46
2x = 46
x = 23
30. A motorist travels to a place 150 km away at an average speed of 50 km/hr and returns at 30 km/ hr. His average speed for the whole journey in km/hr is-
a) 35 km/hr
b) 37 km/hr
c) 37.5 km/hr
d) 40 km/hr
Discussion
Explanation: Average speed
$$\eqalign{ & = \left( {\frac{{2xy}}{{x + y}}} \right){\text{ km/hr}} \cr & = \left( {\frac{{2 \times 50 \times 30}}{{50 + 30}}} \right){\text{ km/hr}} \cr & = 37.5{\text{ km/hr}} \cr} $$
31. 3 years ago the average of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is:
a) 1 years
b) $$\frac{3}{2}$$ years
c) 2 years
d) 3 years
Discussion
Explanation: Let age of the baby is x.
3 years ago total age of the family = 5 × 17 = 85 years
Total age of the 5 member at present time = 85 + 3*5 = 100 years
Total age of the family at present time including baby = 100 + X
The average of the family including baby at present time = 17 years
$$\frac{{100 + {\text{x}}}}{6} = 17$$
100 + X = 102
X = 102 - 100 = 2 years
32. The average salary of all the workers in a workshop is Rs. 8,000. The average salary of 7 technicians is Rs. 12,000 and the average salary of the rest is Rs. 6,000. The total number of workers in the workshop is:
a) 20
b) 21
c) 22
d) 23
Discussion
Explanation: Let the rest workers = x
(7 + x) × 8000 = 12000 × 7 + 6000x
56000 + 8000x = 84000 + 6000x
2000x = 28000
x = 14
Total number of worker = 14 + 7 = 21
33. In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach the target of 282 runs?
a) 6.25
b) 6.5
c) 6.75
d) 7
Discussion
Explanation:
$$\eqalign{ & {\text{Required run rate}} \cr & = {\frac{{282 - \left( {3.2 \times 10} \right)}}{{40}}} \cr & = \frac{{250}}{{40}} \cr & = 6.25 \cr} $$
34. A family consists of two grandparents, two parents and three grandchildren. The average age of the grandparents is 67 years, that of the parents is 35 years and that of the grandchildren is 6 years. What is the average age of the family?
a) $$28\frac{4}{7}$$ years
b) $$31\frac{5}{7}$$ years
c) $$32\frac{1}{7}$$ years
d) None of these
Discussion
Explanation:
$$\eqalign{ & {\text{Required average}} \cr & {\text{ = }} {\frac{{67 \times 2 + 35 \times 2 + 6 \times 3}}{{2 + 2 + 3}}} \cr & = {\frac{{134 + 70 + 18}}{7}} \cr & = \frac{{222}}{7} \cr & = 31\frac{5}{7}{\text{years}} \cr} $$
35. 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. 6001
d) Rs. 6991
Discussion
Explanation: Total sale for 5 months
= Rs. (6435 + 6927 + 6855 + 7230 + 6562)
= Rs. 34009
Required sale = Rs. [(6500 × 6) - 34009]
= Rs. (39000 - 34009)
= Rs. 4991
36. Mr. Joe’s family consists of six people-himself, his wife and their four children. It is known that the average age of the family immediately after the birth of the first, second, third and fourth child was 16, 15, 16 and 15 years respectively. Find the age of Mr. Joe’s eldest son if the present average age of the entire family is 16 years
a) 8 years
b) 12 years
c) 15 years
d) 16 years
Discussion
Explanation: When the first child was born, the total age of all the family members = (16 × 3) years
= 48 years
When the second child was born, the total age of all the family members = (15 × 4) years
= 60 years
By the time the second child was born, each one of the 3 family members had grown by
$$\eqalign{ & = \left( {\frac{{60 - 48}}{3}} \right) \cr & = \frac{{12}}{3} \cr} $$
= 4 years
Hence, the age of eldest son when the second child was born = 4 years
When the third child was born, the total age of all the family members = (16 × 5) years
= 80 years
By the time, the third child was born, each one of the four family members had grown by
= $$\left( {\frac{{80 - 60}}{4}} \right)$$
= 5 years
The age of the eldest son when the third child was born = (4 + 5) years
= 9 years
When the fourth child was born, the total age of all the family members = (15 × 6) years
= 90 years
By the time, the fourth child was born, each of the five family members had grown by
= $$\left( {\frac{{90 - 80}}{5}} \right)$$
= 2 years
So, the age of the eldest son when the fourth child was born = (9 + 2) years
= 11 years
At present, the total age of all the 6 family members = (16 × 6) years
= 96 years
By now, each one of the 6 members have grown by
= $$\left( {\frac{{96 - 90}}{6}} \right)$$ years
= 1 year
The present age of the eldest son
= (11 + 1) years
= 12 years
37. Out of 10 teachers of a school, on teacher retires and in place of him a new teacher 25 years old joins. As a result of it average age of the teachers reduces by 3 years. Age of the retired teacher ( in years) is :
a) 55
b) 60
c) 58
d) 56
Discussion
Explanation: Total number of teachers = 10
Age of new teacher = 25 years
Age of the retire teacher
= (25 + 3 × 10) years
= 55 years
38. The average salary of all the workers in a workshop is Rs. 8000. The average salary of 7 technicians is Rs. 12000 and the average salary of the rest is Rs. 6000. The total number of workers in the workshop is-
a) 20
b) 21
c) 22
d) 23
Discussion
Explanation: Let the total number of workers be x.
8000x = (12000 × 7) + 6000(x - 7)
2000x = 42000
x = 21
39. The average age of a husband and his wife was 23 years at the times of their marriage. After five years they have a one-year old child. The average age of the family now is
a) 19 years
b) 23 years
c) 28.5 years
d) 29.3 years
Discussion
Explanation: Sum of the present ages of husband, wife and child
= (23 × 2 + 5 × 2) + 1
= 57 years
Required average = $$\left( {\frac{{57}}{3}} \right)$$ = 19 years
40. In a class with a certain number of students, if one student weighting 50 kg is added then the average weight of the class increased by 1 kg. If one more student weighting 50 kg is added, then the average weight of the class increased by 1.5 kg over the original average. What is the original average weight (in kg) of the class?
a) 2
b) 4
c) 46
d) 47
Discussion
Explanation: Let the original average weight of the class be x kg and let there be n students.
Sum of weights of n students = (nx) kg
$$\eqalign{ & \frac{{nx + 50}}{{n + 1}} = x + 1 \cr & nx + 50 = \left( {n + 1} \right)\left( {x + 1} \right) \cr & nx + 50 = nx + x + n + 1 \cr & x + n = 49 \cr & 2x + 2n = 98.....(i) \cr & {\text{And,}} \cr & \frac{{nx + 100}}{{n + 2}} = x + 1.5 \cr & nx + 100 = \left( {n + 2} \right)\left( {x + 1.5} \right) \cr & nx + 100 = nx + 1.5n + 2x + 3 \cr & 2x + 1.5n = 97.....(ii) \cr} $$
Subtracting (ii) from (i), we get: 0.5n = 1 or n = 2
Putting n = 2 in (i), we get: x = 47
41. The average height of 25 boys is 1.4 m. When 5 boys leave the group, then the average height increased by 0.15 m. What is the average height of the 5 boys who leave ?
a) 0.8 m
b) 0.9 m
c) 0.95 m
d) 1.05 m
Discussion
Explanation: Sum of height of the 5 boys
= (25 × 1.4 - 20 × 1.55) m
= 4 m
Required average = $$\left( {\frac{4}{5}} \right)$$ = 0.8 m
42. 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. 12000
b) Rs. 15350
c) Rs. 16540
d) Cannot be determined
Discussion
Explanation: Monthly income of the married daughter
= Rs. (15130 × 4 - 14660 × 3)
= Rs. (60520 - 43980)
= Rs. 16540
43. The average marks obtained by 22 candidates in an examination are 45. The average marks of the first ten are 55 and that of the last eleven are 40. The number of marks obtained by the 11th candidates is
a) 0
b) 45
c) 47.5
d) 50
Discussion
Explanation: Mark obtained by the 11th candidate
= [(45 × 22) - (55 × 10 + 40 × 11)]
= (990 - 990) = 0
44. The average of 10 numbers is 40.2. Later it is found that two numbers have been wrongly added. The first is 18 greater than the actual number and the second number added is 13 instead of 33. Find the correct average.
a) 40.2
b) 40.4
c) 40.6
d) 40.8
Discussion
Explanation: Correct sum
= (40.2 × 10 - 18 + 33 - 13)
= 404
Correct average = $$\left( {\frac{{404}}{{10}}} \right)$$ = 40.4
45. The average height of 35 girls in a class was calculated as 160 cm. It was later found that the height of one of the girls in the class was wrongly written as 144 cm, whereas her actual height was 104 cm. What is the actual average height of the girls in the class? ( rounded off to 2 digits after decimal)
a) 158.54 cm
b) 158.74 cm
c) 159.56 cm
d) None of these
Discussion
Explanation: Correct sum
= (160 × 35 + 104 - 144) cm
= 5560 cm
Actual average height
= $$\left( {\frac{{5560}}{{35}}} \right)$$ cm
= 158.857 cm $$ \approx $$ 158.86 cm
46. The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?
a) 0
b) 1
c) 10
d) 19
Discussion
Explanation: Average of 20 numbers = 0
Sum of 20 numbers (0 x 20) = 0
It is quite possible that 19 of these numbers may be positive and if their sum is a then 20th number is (-a)
47. The average weight of 8 person's increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?
a) 76 kg
b) 76.5 kg
c) 85 kg
d) Data inadequate
Discussion
Explanation: Total weight increased = (8 x 2.5) kg = 20 kg.
Weight of new person = (65 + 20) kg = 85 kg.
48. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?
a) 23 years
b) 24 years
c) 25 years
d) None of these
Discussion
Explanation: Let the average age of the whole team by x years
11x - (26 + 29) = 9(x - 1)
11x - 9x = 46
2x = 46
x = 23
49. The average monthly income of P and Q is Rs. 5050. The average monthly income of Q and R is Rs. 6250 and the average monthly income of P and R is Rs. 5200. The monthly income of P is:
a) 3500
b) 4000
c) 4050
d) 5000
Discussion
Explanation: Let P, Q and R represent their respective monthly incomes.
P + Q = (5050 x 2) = 10100 .... (i)
Q + R = (6250 x 2) = 12500 .... (ii)
P + R = (5200 x 2) = 10400 .... (iii)
Adding (i), (ii) and (iii), we get: 2(P + Q + R) = 33000 or P + Q + R = 16500 .... (iv)
Subtracting (ii) from (iv), we get P = 4000
P's monthly income = Rs. 4000
50. The average age of husband, wife and their child 3 years ago was 27 years and that of wife and the child 5 years ago was 20 years. The present age of the husband is:
a) 35 years
b) 40 years
c) 50 years
d) None of these
Discussion
Explanation: Sum of the present ages of husband, wife and child = (27 x 3 + 3 x 3) years
= 90 years
Sum of the present ages of wife and child = (20 x 2 + 5 x 2) years
= 50 years
Husband's present age = (90 - 50) years
= 40 years
51. A person purchase 1 kg of tomatoes from each of the 4 places at the rate of 1 kg, 2 kg, 3 kg, 4 kg per rupee respectively. On an average, he has purchased x kg of tomatoes per rupee. Then the value of x is
a) Rs. 1.92
b) Rs. 2
c) Rs. 2.50
d) None of these
Discussion
Explanation: Total quantity purchased = 4 kg
Total Money paid
$$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right) \cr & = {\text{Rs}}{\text{. }}\frac{{25}}{{12}} \cr} $$
Required average
$$\eqalign{ & = \left( {4 \times \frac{{12}}{{25}}} \right){\text{ kg/rupee}} = \left( {\frac{{48}}{{25}}} \right){\text{ kg/rupee}} \cr & = 1.92{\text{ kg/rupee}} \cr} $$
52. Out of 9 persons, 8 persons spent Rs. 30 each for their meals. The ninth one spent Rs. 20 more than the average expenditure of all the nine. The total money spent by all of them was
a) Rs. 260
b) Rs. 290
c) Rs. 292.50
d) Rs. 400.50
Discussion
Explanation: Let the average expenditure be Rs. x
9x = 8 × 30 + (x + 20)
9x = x + 260
8x = 260
x = 32.50
Total money spent = Rs. 9x = Rs. (9 × 32.50)
= Rs. 292.50
53. 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) None of these
Discussion
Explanation: Let the numbers be x, y and z.
$$\eqalign{ & \left( {\frac{{x + y}}{2}} \right) - \left( {\frac{{y + z}}{2}} \right) = 15 \cr & \left( {x + y} \right) - \left( {y + z} \right) = 30 \cr & x - z = 30 \cr} $$
54. The average of runs of a cricket player 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) 2
b) 4
c) 70
d) 76
Discussion
Explanation: Average after 11 innings = 36
Required number of runs = (36 × 11) - (32 × 10) = 396 - 320
= 76
55. There are 3 groups of students, each containing 25, 50 and 25 students respectively. The mean marks obtained by the first two groups are 60 and 55. The combined mean of all the three groups is 58. What is the mean of the marks scored by the third group ?
a) 52
b) 57
c) 58
d) 62
Discussion
Explanation: Let the mean marks of the third group be x.
$$\eqalign{ & \frac{{25 \times 60 + 50 \times 55 + 25 \times x}}{{25 + 50 + 25}} = 58 \cr & 1500 + 2750 + 25x = 5800 \cr & 25x = 1550 \cr & x = 62 \cr} $$
56. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is:
a) 53.33
b) 54.68
c) 55
d) None of these
Discussion
Explanation:
$$\eqalign{ & {\text{Required}}\,{\text{average}} \cr & = {\frac{{55 \times 50 + 60 \times 55 + 45 \times 60}}{{55 + 60 + 45}}} \cr & = {\frac{{2750 + 3300 + 2700}}{{160}}} \cr & = \frac{{8750}}{{160}} \cr & = 54.68 \cr} $$
57. A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half $$\frac{1}{2}$$ . The number of pupils in the class is:
a) 10
b) 20
c) 40
d) 73
Discussion
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{there}}\,{\text{be}}\,x\,{\text{pupils}}\,{\text{in}}\,{\text{the}}\,{\text{class}}{\text{.}} \cr & {\text{Total}}\,{\text{increase}}\,{\text{in}}\,{\text{marks}} = {x \times \frac{1}{2}} = \frac{x}{2} \cr & \frac{x}{2} = \left( {83 - 63} \right) \cr & \frac{x}{2} = 20 \cr & x = 40 \cr} $$
58. The marks of six boys in a group are 48, 59, 87, 37, 78 and 57. What are the average marks of all six boys?
a) 61
b) 65
c) 69
d) None of these
Discussion
Explanation: Total marks of six boys
= 48 + 59 + 87 + 37 + 78 + 57
= 366
Required Average = $$\frac{{366}}{6}$$ = 61
59. Six numbers are arranged in decreasing order. The average of the first five numbers is 30 and the average of the last five numbers is 25. The difference of the first and the last numbers is
a) 20
b) 25
c) 5
d) 30
Discussion
Explanation: Numbers are
x > y > z > p > q > r
Average of first five numbers = 30
Sum of first five numbers
= a + y + z + p + q = 5 × 30 = 150.....(i)
Average of last five numbers
= y + z + p + q + r = 5 × 25 = 125.....(ii)
By equation (i) and (ii)
a - r = 150 - 125 = 25
60. The average of 12 numbers is 15 and the average of the first two is 14. What is the average of the rest?
a) 15
b) $$15\frac{1}{5}$$
c) 14
d) $$14\frac{1}{5}$$
Discussion
Explanation: Average of 12 numbers = 15
Total of 12 numbers = 15 × 12 = 180
Average of first two number = 14
Total of first two number = 14 × 2 = 28
Total of remaining ten numbers = 180 - 28 = 152
Required average of remaining ten number
$$\eqalign{ & = \frac{{152}}{{10}} \cr & = \frac{{76}}{5} \cr & = 15\frac{1}{5} \cr} $$
61. A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?
a) Rs. 7.98
b) Rs. 8
c) Rs. 8.50
d) Rs. 9
Discussion
Explanation: Total quantity of petrol consumed in 3 years
$$\eqalign{ & = \left( {\frac{{4000}}{{7.50}} + \frac{{4000}}{8} + \frac{{4000}}{{8.50}}} \right){\text{litres}} \cr & = 4000\left( {\frac{2}{{15}} + \frac{1}{8} + \frac{2}{{17}}} \right){\text{litres}} \cr & = {\frac{{76700}}{{51}}} {\text{ litres}} \cr & {\text{Total}}\,{\text{amount}}\,{\text{spent}} = Rs.\,\left( {3 \times 4000} \right) \cr & = Rs.\,12000 \cr & {\text{Average}}\,{\text{Cost}} = Rs.\, {\frac{{12000 \times 51}}{{76700}}} \cr & = Rs.\,\frac{{6120}}{{767}} \cr & = Rs.\,7.98 \cr} $$
62. In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
a) 67 kg
b) 68 kg
c) 69 kg
d) Data inadequate
Discussion
Explanation: Let Arun's weight by X kg.
According to Arun, 65 < X < 72
According to Arun's brother, 60 < X < 70
According to Arun's mother, X < 68
The values satisfying all the above conditions are 66, 67 and 68
$$\eqalign{ & {\text{Required}}\,{\text{average}} = {\frac{{66 + 67 + 68}}{3}} \cr & = {\frac{{201}}{3}} \cr & = 67\,kg \cr} $$
63. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:
a) 17 kg
b) 20 kg
c) 26 kg
d) 31 kg
Discussion
Explanation: Let A, B, C represent their respective weights.
A + B + C = (45 x 3) = 135 .... (i)
A + B = (40 x 2) = 80 .... (ii)
B + C = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31
B's weight = 31 kg.
64. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.
a) 47.55 kg.
b) 48 kg.
c) 48.55 kg.
d) 49.25 kg.
Discussion
Explanation:
$$\eqalign{ & {\text{Required}}\,{\text{average}} \cr & = {\frac{{50.25 \times 16 + 45.15 \times 8}}{{16 + 8}}} \cr & = {\frac{{804 + 361.20}}{{24}}} \cr & = \frac{{1165.20}}{{24}} \cr & = 48.55 {\text{ kg.}} \cr} $$
65. A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
a) 250
b) 276
c) 280
d) 285
Discussion
Explanation: Since the month begins with a Sunday, to there will be five Sundays in the month.
$$\eqalign{ & {\text{Required}}\,{\text{average}} \cr & = {\frac{{510 \times 5 + 240 \times 25}}{{30}}} \cr & = \frac{{8550}}{{30}} \cr & = 285 \cr} $$
66. There were 35 students in a hostel. If the number of the students is increased by 7, then the expenses of the mess increased by Rs. 42 per day, while the average expenditure per head diminishes by Rs. 1. The original expenditure of the mess per day was
a) Rs. 400
b) Rs. 420
c) Rs. 432
d) Rs. 442
Discussion
Explanation: Let the original expenditure of the mess per day be Rs. x
Then, new expenditure
= Rs. (x + 42)
$$\eqalign{ & \therefore \frac{x}{{35}} - \frac{{\left( {x + 42} \right)}}{{42}} = 1 \cr & 6x - 5\left( {x + 42} \right) = 210 \cr & x - 210 = 210 \cr & x = 420 \cr} $$
67. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
a) 25
b) 26
c) 27
d) 28
Discussion
Explanation: Excluded number = (18 × 5) - (16 × 4)
= 90 - 64
= 26
68. The average of the reciprocals of x and y is-
a) $$\frac{(x + y)}{(x - y)}$$
b) $$\frac{(x + y)}{2xy}$$
c) $$\frac{2(x + y)}{xy}$$
d) $$\frac{2xy}{(x + y)}$$
Discussion
Explanation: Required average
$$\eqalign{ & = \frac{{\left( {\frac{1}{x} + \frac{1}{y}} \right)}}{2} \cr & = \frac{{x + y}}{{2xy}} \cr} $$
69. The arithmetic mean of 15 numbers is 41.4. Then the sum of these numbers is
a) 414
b) 420
c) 620
d) 621
Discussion
Explanation: Sum of numbers = (41.4 × 15)
= 621
70. If the average of m numbers is n2 and that of n numbers is m2 , then the average of (m + n) numbers is
a) m - n
b) mn
c) m + n
d) $$\frac{m}{n}$$
Discussion
Explanation: Sum of m numbers = mn2
Sum of n numbers = nm2
Average of (m + n) numbers
= $$\frac{mn(m + n)}{(m + n)}$$
= mn
71. The average wages of a worker during a fortnight comprising 15 consecutive working days was Rs.90 per day. During the first 7 days, his average wages was Rs.87/day and the average wages during the last 7 days was Rs.92 /day. What was his wage on the 8th day?
a) 83
b) 92
c) 90
d) 97
Discussion
Explanation: The total wages earned during the 15 days that the worker worked = 15 × 90
= Rs. 1350
The total wages earned during the first 7 days = 7 × 87
= Rs. 609
The total wages earned during the last 7 days = 7 × 92
= Rs. 644
Total wages earned during the 15 days,
= wages during first 7 days + wage on 8th day + wages during the last 7 days.
1350 = 609 + wage on 8th day + 644
wage on 8th day = 1350 - 609 - 644 = Rs. 97
72. The average temperature on Wednesday, Thursday and Friday was 25°. The average temperature on Thursday, Friday and Saturday was 24°. If the temperature on Saturday was 27°, what was the temperature on Wednesday?
a) 24°
b) 21°
c) 27°
d) 30°
Discussion
Explanation: Total temperature on Wednesday, Thursday and Friday was 25 × 3 = 75°
Total temperature on Thursday, Friday and Saturday was 24 × 3 = 72°
Hence, difference between the temperature on Wednesday and Saturday = 3°
If Saturday temperature =27°, then
Wednesday's temperature = 27 + 3 = 30°
73. When a student weighing 45 kgs left a class, the average weight of the remaining 59 students increased by 200g. What is the average weight of the remaining 59 students?
a) 57
b) 56.8
c) 58.2
d) 52.2
Discussion
Explanation: Let the average weight of the 59 students be X. Therefore, the total weight of the 59 of them will be 59X
The questions states that when the weight of this student who left is added, the total weight of the class = 59X + 45
When this student is also included, the average weight decreases by 0.2 kgs
59X + $$\frac{{45}}{{60}}$$ = X - 0.2
59X + 45 = 60X - 12
45 + 12 = 60X - 59X
X = 57
74. The difference between two angles of a triangle is 24°. The average of the same two angles is 54°. Which one of the following is the value of the greatest angle of the triangle?
a) 45°
b) 60°
c) 66°
d) 72°
Discussion
Explanation: Let a and b be the two angles in the question, with a > b. We are given that the difference between the angles is 24°.
a – b = 24
Since the average of the two angles is 54°, we have $$\frac{{{\text{a}} + {\text{b}}}}{2}$$ = 54
Solving for b in the first equation yields b = a – 24, and substituting this into the second equation yields,
$$ {\frac{{\left\{ {{\text{a}} + \left( {{\text{a}} - 24} \right)} \right\}}}{2}} = 54$$
2a − 24 = 54 × 2
2a − 24 = 108
2a = 108 + 24
2a = 132
a = 66
b = a − 24 = 66 − 24 = 42
Now, let c be the third angle of the triangle. Since the sum of the angles in the triangle is
180°, a + b + c = 180°
Putting the previous results into the equation yields 66 + 42 + c = 180°
Solving for c yields c = 72°
The greatest of the three angles a, b and c is c, which equal.
75. The average age of a family of 5 members is 20 years. If the age of the youngest member be 10 years then what was the average age of the family at the time of the birth of the youngest member?
a) 13.5
b) 14
c) 15
d) 12.5
Discussion
Explanation: At present the total age of the family = 5 × 20 =100
The total age of the family at the time of the birth of the youngest member,
= 100 - 10 - (10 × 4)
= 50
Average age of the family at the time of birth of the youngest member,
$$ = \frac{{50}}{4} = 12.5$$
76. The average of five different positive numbers is 25. x is the decrease in the average when the smallest number among them is replaced by 0. What can be said about x?
a) x is less than 5
b) x is greater than 5
c) x is equal to 5
d) Cannot be determined
Discussion
Explanation: Let a, b, c, d, and e be the five positive numbers in the decreasing order of size such that e is the smallest number.
We are given that the average of the five numbers is 25. Hence, we have the equation
$$\frac{{{\text{a}} + {\text{b}} + {\text{c}} + {\text{d}} + {\text{e}}}}{5} = 25$$
a + b + c + d + e = 125 ----- (1) by multiplying by 5.
The smallest number in a set is at least less than the average of the numbers in the set if at least one number is different.
For example, the average of 1, 2, and 3 is 2, and the smallest number in the set 1 is less than the average 2. Hence, we have the inequality
0 < e < 25
0 > -e > -25 by multiplying both sides of the inequality by -1 and flipping the directions of the inequalities. Adding this inequality to equation (1) yields
0 + 125 > (a + b + c + d + e) + (-e) > 125 - 25
125 > (a + b + c + d) > 100
125 > (a + b + c + d + 0) > 100 by adding by 0
25 > $$\frac{{{\text{a}} + {\text{b}} + {\text{c}} + {\text{d}} + 0}}{5}$$ ⇒ 20 by dividing the inequality by 5
25 > The average of numbers a, b, c, d and 0 > 20
Hence, x equals
(Average of the numbers a, b, c, d and e) – (Average of the numbers a, b, c, and d)
= 25 − (A number between 20 and 25)
⇒ A number less than 5
Hence, x is less than 5
77. In 2011, the arithmetic mean of the annual incomes of Ramesh and Suresh was Rs. 3800. The arithmetic mean of the annual incomes of Suresh and Pratap was Rs. 4800, and the arithmetic mean of the annual incomes of Pratap and Ramesh was Rs. 5800. What is the arithmetic mean of the incomes of the three?
a) Rs. 4000
b) Rs. 4200
c) Rs. 4400
d) Rs. 4800
Discussion
Explanation: Let a, b, and c be the annual incomes of Ramesh, Suresh, and Pratap, respectively.
The arithmetic mean of the annual incomes of Ramesh and Suresh was Rs. 3800.
$$\frac{{{\text{a}} + {\text{b}}}}{2}$$ = 3800
a + b = 2 × 3800 = 7600
The arithmetic mean of the annual incomes of Suresh and Pratap was Rs. 4800.
$$\frac{{{\text{b}} + {\text{c}}}}{2}$$ = 4800
b + c = 2 × 4800 = 9600
The arithmetic mean of the annual incomes of Pratap and Ramesh was Rs. 5800.
$$\frac{{{\text{c}} + {\text{a}}}}{2}$$ = 5800
c + a = 2 × 5800 = 11,600
Adding these three equations yields:
(a + b) + (b + c) + (c + a) = 7600 + 9600 + 11,600
2a + 2b + 2c = 28,800
a + b + c = 14,400
The average of the incomes of the three equals the sum of the incomes divided by 3,
$$\eqalign{ & \frac{{{\text{a}} + {\text{b}} + {\text{c}}}}{3} \cr & = \frac{{14,400}}{3} \cr & = {\text{Rs}}{\text{.}}\,4800 \cr} $$
78. In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother does not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
a) 67 kg
b) 68 kg
c) 69 kg
d) Data inadequate
Discussion
Explanation: Let Arun's weight by X kg.
According to Arun,
65 < X < 72
According to Arun's brother,
60 < X < 70
According to Arun's mother,
X $$ \leqslant $$ 68
The values satisfying all the above conditions are 66, 67 and 68.
Required average,
$$\eqalign{ & = \frac{{66 + 67 + 68}}{3} = \frac{{201}}{3} = 67\,{\text{kg}}{\text{.}} \cr} $$
79. A student finds the average of 10 positive integers. Each integer contains two digits. By mistake, the boy interchanges the digits of one number say ba for ab. Due to this, the average becomes 1.8 less than the previous one. What was the difference of the two digits a and b?
a) 8
b) 6
c) 2
d) 4
Discussion
Explanation: Let the original number be (10a + b)
After interchanging the digits, the new number becomes (10b + a)
The question states that the average of 10 numbers has become 1.8 less than the original average. Therefore, the sum of the original 10 numbers will be (10 × 1.8 = 18 more than the sum of the 10 numbers with the digits interchanged.
10a + b = 10b + a + 18
9a - 9b = 18
a - b = 2
80. The average of first five multiples of 3 is:
a) 9
b) 10
c) 8
d) 11
Discussion
Explanation: First five multiples of three are:
3, 6, 9, 12, 15.
$$\eqalign{ & {\text{Average}} = \frac{{ {3 + 6 + 9 + 12 + 15} }}{5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{{45}}{5} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 9 \cr} $$
81. Ajit has a certain average for 9 innings. In the tenth innings, he scores 100 runs thereby increasing his average by 8 runs. His new average is:
a) 20
b) 21
c) 28
d) 32
Discussion
Explanation: Let Ajit's average be x for 9 innings. So, Ajit scored 9x run in 9 innings.
In the 10th inning, he scored 100 runs then average became (x+8). And he scored (x + 8) × 10 runs in 10 innings.
$$\eqalign{ & 9x + 100 = 10 \times \left( {x + 8} \right) \cr & 9x + 100 = 10x + 80 \cr & x = 100 - 80 \cr & x = 20 \cr & {\text{New}}\,{\text{average}} = \left( {x + 8} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 28\,{\text{runs}} \cr} $$
82. The average temperature for Wednesday, Thursday and Friday was 40°C. The average for Thursday, Friday and Saturday was 41° C. If temperature on Saturday was 42° C, what was the temperature on Wednesday?
a) 39° C
b) 44° C
c) 38° C
d) 41° C
Discussion
Explanation: Average temperature for Wednesday, Thursday and Friday = 40° C
Total temperature = 3 × 40 = 120° C
Average temperature for Thursday, Friday and Saturday = 41° C
Total temperature = 41 × 3 = 123° C
Temperature on Saturday = 42° C
(Thursday + Friday + Saturday) - (Wednesday + Thursday + Friday) = 123 - 120;
Saturday - Wednesday = 3
Wednesday = 42 - 3 = 39° C
83. The average of the first five multiples of 9 is:
a) 20
b) 27
c) 28
d) 30
Discussion
Explanation:
$$\eqalign{ & {\text{Required}}\,{\text{average}} \cr & = {\frac{{{\text{total}}\,{\text{sum}}\,{\text{of}}\,{\text{multiple}}\,{\text{of}}\,9}}{5}} \cr & = {\frac{{9 + 18 + 27 + 36 + 45}}{5}} \cr & = 27 \cr} $$
Average of 9 and 45 is also 27.
And average of 18 and 36 is also 27.
84. The speed of the train going from Nagpur to Allahabad is 100 km/h while when coming back from Allahabad to Nagpur, its speed is 150 km/h. find the average speed during whole journey.
a) 125 km/hr
b) 75 km/hr
c) 135 km/hr
d) 120 km/hr
Discussion
Explanation:
$$\eqalign{ & {\text{Average speed}}, \cr & = \frac{{ {2 \times x \times y} }}{{ {x + y} }} \cr & = \frac{{ {2 \times 100 \times 150} }}{{ {100 + 150} }} \cr & = \frac{{ {200 \times 150} }}{{250}} \cr & = 120\,{\text{km/hr}} \cr} $$
85. Find the average of first 97 natural numbers
a) 47
b) 37
c) 48
d) 49
Discussion
Explanation: Average of 1st n natural number is given by
$$ = \frac{{ {\frac{{ {{\text{n}} \times \left( {{\text{n}} + 1} \right)} }}{2}} }}{{\text{n}}}$$
Average of 1st 97 natural number is given by
$$\eqalign{ & {\frac{{ {\frac{{ {97 \times \left( {97 + 1} \right)} }}{2}} }}{{97}}} \cr & = 49 \cr} $$
86. There are two sections A and B of a class, consisting of 36 and 44 students' respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class.
a) 30 kg
b) 35 kg
c) 42.5 kg
d) 37.25 kg
Discussion
Explanation: The total weight of (36 + 44) Students of A and B,
= (36 × 40 + 44 × 35) kg = 2980 kg.
Average weight of the whole class = $$\frac{{2980}}{{80}}$$ kg
Average weight = 37.25 kg
87. Distance between two stations A and B is 778km. A train covers the journey from A to B at 84 km per hour and returns back to A with a uniform speed of 56 km per hour. Find the average speed of train during the whole journey
a) 60 km/hr
b) 30.5 km/hr
c) 57 km/hr
d) 67.2 km/hr
Discussion
Explanation:
$$\eqalign{ & {\text{Average speed}} \cr & = {\frac{{2xy}}{{ {x + y} }}} \,\,{\text{km/hr}} \cr & x = 84{\kern 1pt} \,{\text{kmph}} \cr & y = 56\,{\kern 1pt} {\text{kmph}} \cr & {\text{Average speed}} \cr & = {\frac{{ {2 \times 84 \times 56} }}{{ {84 + 56} }}} \cr & = 67.2\,{\kern 1pt} {\text{kmph}} \cr} $$
88. The average of 50 numbers is 30. If two numbers, 35 and 40 are discarded, then the average of the remaining numbers is nearly
a) 28.32
b) 29.68
c) 28.78
d) 29.27
Discussion
Explanation:
Total sum of 48 numbers
= (50 × 30) – (35 +40)
= 1500 – 75
= 1425
Average = $$\frac{{1425}}{{48}}$$ = 29.68
89. The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, then find the average for the last four matches.
a) 33.25
b) 33.5
c) 34.25
d) 35
Discussion
Explanation: Total sum of last 4 matches
= (10 × 38.9) - (6 × 42)
= 389 - 252 = 137
Average = $$\frac{{137}}{4}$$ = 34.25
90. A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Find his average after 17th inning.:
a) 40
b) 39
c) 52
d) 55
Discussion
Explanation: Let the average after 17th innings = x
Then average after 16th innings = (x - 3)
16 × (x - 3) + 87 = 17x
x = 39
91. There were 35 students in a hostel. Due to the admission of 7 new students the expenses of the mess were increased by Rs.42 per day while the average expenditure per head diminished by Re 1. What was the original expenditure of the mess?
a) Rs. 450
b) Rs. 320
c) Rs. 550
d) Rs. 420
Discussion
Explanation: Let the original average expenditure be Rs. x
42(x - 1) - 35x = 42
7x = 84
x = 12
Original expenditure,
= Rs. (35 × 12)
= Rs. 420
92. Nine persons went to a hotel for taking their meals. Eight of them spent Rs. 12 each on their meals and the ninth spent Rs. 8 more than the average expenditure of all the nine. What was the total money spent by them.
a) Rs. 115
b) Rs. 116
c) Rs. 117
d) Rs. 118
Discussion
Explanation: Let the average expenditure of all the nine be x
12 × 8 + (x + 8) = 9x
x = 13
Total money spent = 9x
= Rs. (9 × 13)
= Rs. 117
93. David obtained 76, 65, 82, 67 and 85 marks (out of 100) in English, mathematics, physics, chemistry and biology. What are his average marks?
a) 65
b) 69
c) 75
d) None of these
Discussion
Explanation:
$$\eqalign{ & {\text{Average}} \cr & = \frac{{ {76 + 65 + 82 + 67 + 85} }}{5} \cr & = \frac{{375}}{5} \cr & = 75 \cr} $$
94. The average of runs of a cricket player of 10 innings was 32. How many runes must be made in his next innings so as to increase his average of runs by 4?
a) 72
b) 74
c) 70
d) 76
Discussion
Explanation: Average after 11 innings = 36
Required number of runs = ( 36 × 11) - (32 × 10)
= 396 - 320
= 76
95. There are two sections A and B of a class, consisting of 36 and 44 students respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class
a) 30 kg.
b) 35 kg.
c) 42.5 kg
d) 37.25 kg.
Discussion
Explanation: Total weight of (36 + 44) Students,
= (36 × 40 + 44 × 35) kg
= 2980 kg
Average weight of the whole class,
= $$ {\frac{{2980}}{{80}}} $$ kg.
Average weight = 37.25 kg.
96. The average age of three boys is 15 years. If their ages are in ratio 3 : 5 : 7, the age of the youngest boy is
a) 21 years
b) 18 years
c) 15 years
d) 9 years
Discussion
Explanation: Sum of ages of three boys = 45 years
3x + 5x + 7x = 45
15x = 45
x = 3
Age of youngest boy = 3x
= 3 × 3 = 9 years
97. The average of a group of men is increased by 5 years when a person aged of 18 years is replaced by a new person of aged 38 years. How many men are there in the group?
a) 3
b) 4
c) 5
d) 6
Discussion
Explanation: Let N be the no. of persons in the group.
Required number of person is given by;
Member in group × aged increased = difference of replacement
N × 5 = 38 - 18
5N = 20
N = 4
98. In a boat there are 8 men whose average weight is increased by 1 kg when 1 man of 60 kg is replaced by a new man. What is weight of new comer?
a) 70 kg
b) 66 kg
c) 68 kg
d) 69 kg
Discussion
Explanation: Member in group × age increased = difference of replacement
8 × 1 = new comer - man going out
new comer = 8 + 60
new comer = 68 kg.
99. The average of 25 results is 18. The average of first 12 of those is 14 and the average of last 12 is 17. What is the 13th result?
a) 74
b) 75
c) 69
d) 78
Discussion
Explanation: Sum of 1st 12 results = 12 × 14
Sum of last 12 results = 12 × 17
13th result = x (let)
12 × 14 + 12 × 17 + x = 25 × 18
x = 78
100. A train covers the first 16 km at a speed of 20 km per hour another 20 km at 40 km per hour and the last 10 km at 15 km per hour. Find the average speed for the entire journey.
a) 24 km
b) 26 km
c) 21 km
d) \[23\frac{{23}}{{59}}\] km
Discussion
Explanation:
$$\eqalign{ & {\text{Average}}\,{\text{speed}} \cr & = \frac{{{\text{total}}\,{\text{distance}}\,{\text{covered}}}}{{{\text{total}}\,{\text{time}}}} \cr & {\text{Total}}\,{\text{Distance}} = {16 + 20 + 10} \cr & = 46\,{\text{km}} \cr & {\text{Time}}\,{\text{taken}} = {\frac{{16}}{{20}}} + {\frac{{20}}{{40}}} + {\frac{{10}}{{15}}} \cr & = {\frac{{4}}{{5}}} + {\frac{{1}}{{2}}} + {\frac{{2}}{{3}}} \cr & = {\frac{{24 + 15 + 20}}{{30}}} \cr & = \frac{{59}}{{30}} \cr & {\text{Average}}\,{\text{speed}} \cr & = \frac{{46 \times 30}}{{59}} \cr & = 23\frac{{23}}{{59}}\,{\text{km/hr}} \cr} $$