1. Which one of the following numbers can be removed from the set S = {0, 2, 4, 5, 9} without changing the average of set S?
a) 0
b) 2
c) 4
d) 5
Explanation: The average of the elements in the original set S is:
$$\eqalign{ & \frac{{0 + 2 + 4 + 5 + 9}}{5} \cr & = \frac{{20}}{5} \cr & = 4 \cr} $$
If we remove an element that equals the average, then the average of the new set will remain unchanged. The new set after removing 4 is {0, 2, 5, 9}.
Average of the elements,
$$\eqalign{ & \frac{{0 + 2 + 5 + 9}}{4} \cr & = \frac{{16}}{4} \cr & = 4 \cr} $$
2. Average cost of 5 apples and 4 mangoes is Rs. 36. The average cost of 7 apples and 8 mangoes is Rs. 48. Find the total cost of 24 apples and 24 mangoes
a) 1044
b) 2088
c) 720
d) 3240
Explanation: Average cost of 5 apples and 4 mangoes = Rs. 36
Total cost = 36 × 9 = 324
Average cost of 7 apples and 8 mangoes = 48
Total cost = 48 × 15 = 720
Total cost of 12 apples and 12 mangoes = 324 + 720 = 1044
Cost of 24 apples and 24 mangoes = 1044 × 2 = 2088
3. The average weight of three boys A, B and C is $$\frac{{163}}{3}$$ kg, while the average weight of three boys B, D and E is 53 kg. What is the average weight of A, B, C, D and E?
a) 52.4 kg
b) 53.2 kg
c) 53.8 kg
d) Data inadequate
Explanation: In this question, sum of numbers is provided, net required sum (i.e. A + B+ C+ D + E) cannot be calculated by the given data.
Therefore, Data inadequate.
4. Average of ten positive numbers is x. If each number is increased by 10%, then x :
a) remains unchanged
b) may decrease
c) may increase
d) is increased by 10%
Explanation: Let 10 numbers be x1, x2, x3, . . . . . . . x10
Average of these 10 numbers is 10
$$ \frac{{ {x1 + x2 + x3 + .... + x10} }}{{10}} = x$$
Now if each number is increased by 10%,
then new average, say y.
$$y = \frac{{ {1.1x1 + 1.1x2 + 1.1x3 + .... + 1.1x10} }}{{10}}$$
$${\kern 1pt} y = 1.1 \times {\frac{{ {x1 + x2 + x3 + .... + x10} }}{{10}}} $$
$$\eqalign{ & y = 1.1x \cr & y\,{\text{is }}10\% \,{\text{increased}} \cr} $$
5. The average price of 10 books is Rs.12 while the average price of 8 of these books is Rs. 11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what is the price of each of these two books?
a) Rs. 5, Rs.7.50
b) Rs. 8, Rs. 12
c) Rs. 10, Rs. 16
d) Rs. 12, Rs. 14
Explanation: Total cost of 10 books = Rs. 120
Total cost of 8 books = Rs. 94
The cost of 2 books = Rs. 26
Let the price of each book be x and y.
x + y = 26 - - - -(1)
Given that the price of 1 book is 60% more than the other price
$$\eqalign{ & \left( {\frac{{160}}{{100}}} \right)y + y = 26 \cr & y\left( {\frac{{160}}{{100}} + 1} \right) = 26 \cr & y\left( {\frac{{160 + 100}}{{100}}} \right) = 26 \cr & y = \frac{{\left( {26 \times 100} \right)}}{{260}} \cr & y = 10 \cr & {\text{Substituting}}\,\,y = 10\,\,{\text{in }}{\kern 1pt} \left(1 \right), \cr & x + 10 = 26 \cr & x = 16 \cr} $$
6. Distance between two stations A and B is 778km. A train covers the journey from A to B at 84km per hour and returns back to A with a uniform speed of 56km 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
Explanation:
$$\eqalign{ & {\text{Average speed}} = {\frac{{2xy}}{{ {x + y} }}} {\kern 1pt} {\kern 1pt} {\text{km/hr}} \cr & = {\frac{{ {2 \times 84 \times 56} }}{{ {84 + 56} }}} {\kern 1pt} {\kern 1pt} {\text{km/hr}} \cr & = {\frac{{ {2 \times 84 \times 56} }}{{140}}} {\kern 1pt} {\kern 1pt} {\text{km/hr}} \cr & = 67.2{\kern 1pt} {\kern 1pt} {\text{km/hr}} \cr} $$
7. 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
Explanation: Total sum of 48 numbers =
(50 × 30) - (35 + 40)
1500 - 75 = 1425
Average = $$\frac{{1425}}{{48}}$$ = 29.68
8. 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
Explanation: Total sum of last 4 matches,
= (10 × 38.9) - (6 × 42)
= 389 - 252 = 137
Average = $$\frac{{137}}{4}$$ = 34.25
9. 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
Explanation: Let the average expenditure of all the nine be Rs. X
12 × 8 + (X + 8) = 9X
X = 13
Total money spent = 9X
= Rs. (9 × 13)
= Rs. 117
10. 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
Explanation: Average after 11 innings = 36
Required number of runs
= ( 36 × 11) - (32 × 10)
= 396 - 320
= 76
11. 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 13 years. What is the average age of the whole class (rounded to two digits after decimal) ?
a) 12.51
b) 13.42
c) 13.50
d) 13.53
Explanation: Required average
$$\eqalign{ & = \left( {\frac{{32 \times 14 + 28 \times 13}}{{32 + 28}}} \right) \cr & = \left( {\frac{{448 + 364}}{{60}}} \right) \cr & = \left( {\frac{{812}}{{60}}} \right) \cr & = 13.53 \cr} $$
12. The average annual income (in Rs.) of certain agricultural workers is S and that of other workers is T. The number of agricultural workers is 11 times that of other workers. Then the average monthly income (in Rs.) of all the workers is-
a) $$\frac{S + T}{2}$$
b) $$\frac{S + 11T}{2}$$
c) $$\frac{1}{{11S}} + 1$$
d) $$\frac{11S + T}{12}$$
Explanation: Let the number of other workers be x
Then, number of agricultural workers = 11x
Total number of workers = 12x
∴ Average monthly income
$$\eqalign{ & = \frac{{S \times 11x + T \times x}}{{12x}} \cr & = \frac{{11S + T}}{{12}} \cr} $$
13. The average mark of student in 4 subjects is 75. If the student obtained 80 marks in the fifth subject, then the new average is
a) 72.5
b) 76
c) 77
d) 77.5
Explanation: Sum of marks in 4 subjects
= 75 × 4
= 300
Sum of marks in 5 subjects
= 300 + 80
= 380
∴ New average
= $$\frac{380}{5}$$
= 76
14. The average of 7 consecutive numbers is 20. The largest of these numbers is-
a) 20
b) 22
c) 23
d) 24
Explanation: Let the number be x, x + 1, x + 2, x + 4, x + 5 and x + 6
Then,
$$ \Rightarrow \frac{{x + \left( {x + 1} \right) + \left( {x + 2} \right) + \left( {x + 3} \right) + \left( {x + 4} \right) + \left( {x + 5} \right) + \left( {x + 6} \right)}}{7} = 20$$
$$\eqalign{ & \Rightarrow 7x + 21 = 140 \cr & \Rightarrow 7x = 119 \cr & \Rightarrow x = 17 \cr} $$
∴ Largest number
= x + 6 = 17 + 6
= 23
15. Of four numbers whose average is 60, the first is one-fourth of the sum of the last three. The first number is -
a) 15
b) 42
c) 45
d) 48
Explanation: Let the four numbers be a, b, c and d respectively
Then,
⇒ a = $$\frac{1}{4}$$ (b + c + d)
⇒ b + c + d = 4a
Also,
a + b + c + d = 60 × 4 = 240
⇒ a + 4a = 240
⇒ 5a = 240
⇒ a = 48
Hence, first number = 48
16. The average of five consecutive numbers is x. If the next two numbers are included, how shall the average vary?
a) It shall increase by 1
b) It shall remain the same
c) It shall increase by 1.4
d) It shall increase by 2
Explanation: Let the five consecutive numbers be z, z + 1, z + 3 and z + 4
Then,
$$ \Rightarrow \frac{{z + \left( {z + 1} \right) + \left( {z + 2} \right) + \left( {z + 3} \right) + \left( {z + 4} \right)}}{5} = x$$
$$\eqalign{ & \Rightarrow 5z + 10 = 5x \cr & \Rightarrow z = \frac{{5x - 10}}{5} \cr & \Rightarrow z = x - 2 \cr} $$
So, the numbers are x - 2, x - 1, x, x + 1, x + 2
∴ Required mean
$$ = \frac{{\left( {x - 2} \right) + \left( {x - 1} \right) + x + \left( {x + 1} \right) + \left( {x + 2} \right) + \left( {x + 3} \right) + \left( {x + 4} \right)}}{7}$$
$$\eqalign{ & = \frac{{7x + 7}}{7} \cr & = x + 1 \cr} $$
17. In a certain factory there are five workers A, B, C, D and E. A can complete a work in 4 minutes, B in 5 minutes, C in 6 minutes, D in 10 minutes and E in 12 minutes. The average number of units of work completed per worker per minute will be-
a) 0.16
b) 0.172
c) 0.80
d) 0.87
Explanation: Number of units of work completed by the five workers in 1 minute:
$$\eqalign{ & A \to \frac{1}{4}, \cr & B \to \frac{1}{5}, \cr & C \to \frac{1}{6}, \cr & D \to \frac{1}{{10}}, \cr & E \to \frac{1}{{12}} \cr} $$
∴ Required average
$$\eqalign{ & = \frac{{\left( {\frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{{10}} + \frac{1}{{12}}} \right)}}{5} \cr & = \left( {\frac{4}{5} \times \frac{1}{5}} \right) \cr & = \frac{4}{{25}} \cr & = 0.16 \cr} $$
18. The average expenditure of a man for the first five months of a year is Rs. 5000 and for the next seven months it is Rs. 5400. He saves Rs. 2300 during the year. His average monthly income is-
a) Rs. 5425
b) Rs. 5446
c) Rs. 5500
d) Rs. 5600
Explanation:Total yearly income
= Rs. (5000 × 5 + 5400 × 7 + 2300)
= Rs. (25000 + 37800 + 2300)
= Rs. 65100
∴ Average monthly income
= Rs. $$\left( {\frac{{65100}}{{12}}} \right)$$
= Rs. 5425
19. The average weight of 16 boys in a class is 50.25 kgs and that of the remaining 8 boys is 45.15 kgs. Find the average weight of all the boys in the class.
a) 47.55 kgs
b) 48 kgs
c) 48.55 kgs
d) 49.25 kgs
Explanation: Required mean
$$\eqalign{ & = \left( {\frac{{50.25 \times 16 + 45.15 \times 8}}{{16 + 8}}} \right) \cr & = \left( {\frac{{804 + 361.20}}{{24}}} \right) \cr & = \frac{{1165.20}}{{24}} \cr & = 48.55 \cr} $$
20. 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
Explanation:Since the month begins with a Sunday, so there will be five Sundays in the month.
∴ Required average
$$\eqalign{ & = \left( {\frac{{510 \times 5 + 240 \times 25}}{{30}}} \right) \cr & = \frac{{8550}}{{30}} \cr & = 285 \cr} $$
21. The sum of the three consecutive even numbers is 44 more than the average of these numbers. Which of the following is the third largest of these numbers?
a) 16
b) 18
c) 24
d) Cannot be determined
Explanation: Let the numbers be x, x + 2 and x + 4
Then,
⇒ (x + x + 2 + x + 4) - $$\frac{(x + x + 2 + x + 4)}{3}$$ = 44
⇒ (3x + 6) - $$\frac{(3x + 6)}{3}$$ = 44
⇒ 2 (3x + 6) = 132
⇒ 6x = 120
⇒ x = 20
∴ Largest number = x + 4 = 24
22. The average of five consecutive odd numbers is 95. What is the fourth number in the descending order?
a) 91
b) 95
c) 97
d) None of these
Explanation: Let the numbers be x, x + 2, x + 4, x + 6 and x + 8
Then,
$$ \Rightarrow \frac{{x + \left( {x + 2} \right) + \left( {x + 4} \right) + \left( {x + 6} \right) + \left( {x + 8} \right)}}{5} $$ $$= 95$$
$$\eqalign{ & \Rightarrow 5x + 20 = 475 \cr & \Rightarrow 5x = 455 \cr & \Rightarrow x = 91 \cr} $$
So, the numbers are 91, 93, 95, 97 and 99
Clearly, the fourth number in the descending order is 93
23. The average of two numbers is 6.5 and square root of their product is 6. What are the numbers?
a) 11 and 2
b) 8 and 5
c) 9 and 4
d) 10 and 3
Explanation: Let the two numbers be x and y
Then,
x + y = 6.5 × 2 = 13 and
$$\sqrt {{\text{xy}}} $$ = 6 or xy = 36
⇒ (x - y)2 = (x + y)2 - 4xy
⇒ (x - y)2 = (13)2 - 4 × 36
⇒ (x - y)2 = 169 - 144
⇒ (x - y)2 = 25
⇒ (x - y) = 5
Solving x + y = 13 and x - y = 5
We get : x = 9 , y = 4
24. A, B, C and D are four consecutive even numbers respectively and their average is 65. What is the product of A and D?
a) 3968
b) 4092
c) 4216
d) 4352
Explanation: Let x, x + 2, x + 4 and x + 6 represent numbers A, B, C and D respectively.
Then,
$$\eqalign{ & \Rightarrow \frac{{x + \left( {x + 2} \right) + \left( {x + 4} \right) + \left( {x + 6} \right)}}{4} = 65 \cr & \Rightarrow 4x + 12 = 260 \cr & \Rightarrow 4x = 248 \cr & \Rightarrow x = 62 \cr} $$
So, A = 62, B = 64, C = 66, D = 68
∴ A × D = 62 × 68 = 4216
25. The arithmetic mean of the series 1, 2, 4, 8, 16, . . . . . . , 2n is -
a) $$\frac{{{2^n} - 1}}{{n + 1}}$$
b) $$\frac{{{2^n} + 1}}{{n}}$$
c) $$\frac{{{2^n} - 1}}{{n}}$$
d) $$\frac{{{2^{n + 1}} - 1}}{{n + 1}}$$
Explanation: he given series is a G.P. with first term, a = 1 and common ratio, r = 2,
It has (n + 1) terms.
∴ Sum of the terms of the series
= $$\frac{{{2^{n + 1}} - 1}}{{2 - 1}}$$
= 2n + 1 $$ - $$ 1
Arithmetic mean = $$\frac{{{2^{n + 1}} - 1}}{{n + 1}}$$
26. The average of odd numbers up to 100 is
a) 49
b) 50.5
c) 50
d) 100
Explanation: Sum of odd numbers upto 100 = 1 + 3 + 5 +.....+ 99
= $$\frac{50}{2}$$ [2 + (50 - 1) × 2]
= 2500
[$$\because $$ Sum of n terms of an A.P. with first term a and common difference]
$$\left[ {d = \frac{n}{2}\{ 2n + \left( {n - 1} \right)d\} } \right]$$
∴ Required average
= $$\frac{2500}{50}$$
= 50
27. In a group of 120 people, one-fifth are men, one-fourth are women and the rest children. The average age of women is five-sixth of the average age of men. Average age of children is one-fourth of the average age of men. If average age of men is 60 years, what is the average age of the group?
a) 32.75 yeras
b) 38.45 years
c) 45.25 years
d) 50.5 years
Explanation: Number of men
= $$\frac{1}{5}$$ × 120
= 24
Number of women
= $$\frac{1}{4}$$ × 120
= 30
Number of children
= 120 - (24 + 30)
= 66
Average age of men = 60 years
Average age of children
= $$\frac{1}{4}$$ × 60
= 15 years
Average age of women
= $$\frac{5}{6}$$ × 60
= 50 years
∴ Avarage age of the group
$$\eqalign{ & = \left( {\frac{{60 \times 24 + 50 \times 30 + 15 \times 66}}{{120}}} \right){\text{ years}} \cr & = \left( {\frac{{3930}}{{120}}} \right){\text{ years}} \cr & = 32.75{\text{ years}} \cr} $$
28. The average of ten numbers is 7. If each number is multiplied by 12, then the average of the new set of numbers is :
a) 7
b) 19
c) 82
d) 84
Explanation: Average of 10 numbers = 7
Sum of these 10 numbers = (10 × 7) = 70
$$\eqalign{ & \therefore {x_1} + {x_2} + ..... + {x_{10}} = 70 \cr & \Rightarrow 12{x_1} + 12{x_2} + ..... + 12{x_{10}} = 840 \cr & \Rightarrow \frac{{12{x_1} + 12{x_2} + ..... + 12{x_{10}}}}{{10}} = 84 \cr} $$
⇒ Average of new numbers is 84
29. The mean temperature of Monday to Wednesday was 37°C and of and of Tuesday to Thursday was 34°C. If the temperature on Thursday was $$\frac{4}{5}$$ that of Monday, the temperature of Thursday was-
a) 34°C
b) 35.5°C
c) 36°C
d) 36.5°C
Explanation: M + T + W = (37 × 3)°C = 111°C.....(i)
T + W + Th = (34 × 3)°C = 102°C.....(ii)
Subtracting (ii) from (i), we get:
⇒ M - Th = 9°C
⇒ M - $$\frac{4}{5}$$M = 9
⇒ $$\frac{1}{5}$$ M = 9
⇒ M = 45
∴ Temperature on Thursday
= $$ \left( {\frac{{4}}{{5}}} \times 45 \right) $$ °C
= 36°C
30. The average weight of three boys A, B and C is $$54\frac{1}{3}$$ kg, while the average weight of B, D and E is 53 kg. What is the average weight of A, B, C, D and E ?
a) 52.4 kg
b) 53.2 kg
c) 53.8 kg
d) Data inadequate
Explanation: Total weight of (A + B + C)
=( $$54\frac{1}{3}$$ × 3 ) kg
= 163 kg
Total weight of (B + D + E)
= (53 × 3) kg
= 159 kg
Adding both, we get:
= A + 2B + C + D + E
= (163 + 159) kg
= 322 kg
So, to find average weight of A, B, C, D and E, we ought to know B's weight, which is not given.
So, the data is inadequate.
31. The average price of 10 books is Rs. 12 while the average price of 8 of these books is Rs. 11.75. Of the remaining two books, if the price of one book is 60% more than the price of the other, what is the price of each of these two books ?
a) Rs. 5, Rs. 7.50
b) Rs. 8, Rs. 12
c) Rs. 10, Rs. 16
d) Rs. 12, Rs. 14
Explanation: Total price of the two books
= Rs. [(12 × 10) - (11.75 × 8)]
= Rs. (120 - 94)
= Rs. 26
Let the price of one book be Rs. x
Then, the price of other book
= Rs. (x + 60 % of x)
= Rs. $$\left( {x + \frac{3}{5}x} \right)$$
= Rs. $$\left( { \frac{8x}{5}} \right)$$
So, $${x + \frac{8x}{5}}$$ = 26
⇒ 13x = 130
⇒ x = 10
32. 16 children are to be divided into two groups A and B of 10 and 6 children. The average percent marks obtained by the children of group A is 75 and the average percent marks of all the 16 children is 76. What is the average percent marks of children of group B ?
a) $$77\frac{1}{3}$$
b) $$77\frac{2}{3}$$
c) $$78\frac{1}{3}$$
d) $$78\frac{2}{3}$$
Explanation: Required average
$$\eqalign{ & = \frac{{\left( {76 \times 16} \right) - \left( {75 \times 10} \right)}}{6} \cr & = \left( {\frac{{1216 - 750}}{6}} \right) \cr & = \frac{{466}}{6} \cr & = \frac{{233}}{3} \cr & = 77\frac{2}{3} \cr} $$
33. The average age of 30 students in a class is 15 years. If 6 students of this class have the average age of 16 years, then the average age of the remaining 24 students would be-
a) 14 yrs
b) 14 yrs 3 mths
c) 14 yrs 6 mths
d) 14 yrs 9 mths
Explanation: Sum of the ages of 24 students
= (15 × 30) - (16 × 6)
= 450 - 96
= 354
∴ Required average
= $$\left( {\frac{{354}}{{24}}} \right)$$
= $$14\frac{3}{4}$$ yrs
= 14 yrs 9 mths
34. Of the four numbers, the first is twice the second, the second is one-third of the third and the third is 5 times the fourth. The average of the numbers is 24.75. The largest of these numbers is
a) 9
b) 25
c) 30
d) None of these
Explanation: Let the fourth number be x
Then, third number = 5x
Second number = $$\frac{5x}{3}$$ and
First number = $$\frac{10x}{3}$$
x + 5x + $$\frac{5x}{3}$$ + $$\frac{10x}{3}$$ = (24.75 × 4)
⇒ 11x = 99
⇒ x = 9
So, the numbers are 9, 45, 15 and 30
∴ Largest number = 45
35. A car owner buys petrol at Rs. 17, Rs. 19 and Rs. 20 per litre for three consecutive years. Compute the average cost per litre, if he spends Rs 6460 per year.
a) Rs. 18.49
b) Rs. 18.58
c) Rs. 19.20
d) Rs. 21.66
Explanation: Total quantity of petrol consumed in 3 years
= $$\left( {\frac{{6460}}{{17}} + \frac{{6460}}{{19}} + \frac{{6460}}{{20}}} \right)$$ litres
= (380 + 340 + 323) litres
= 1043 litres
Total amount spent
= Rs. (3 × 6460)
= Rs. 19380
∴ Average cost
= Rs. $$\left( {\frac{{19380}}{{1043}}} \right)$$
= Rs. 18.58
36. The average of the two-digit numbers, which remain the same when the digits interchange their positions, is
a) 33
b) 44
c) 55
d) 66
Explanation: Average
$$ = \frac{{11 + 22 + 33 + 44 + 55 + 66 + 77 + 88 + 99}}{9}$$
$$ = \frac{{\left( {11 + 99} \right) + \left( {22 + 88} \right) + \left( {33 + 77} \right) + \left( {44 + 66} \right) + 55}}{9}$$
$$\eqalign{ & = \frac{{4 \times 110 + 55}}{9} \cr & = \frac{{495}}{9} \cr & = 55 \cr} $$
37. The average of a non-zero number and its square is 5 times the number. The number is
a) 9
b) 17
c) 29
d) 295
Explanation: Let the number be x
Then,
$$\eqalign{ & \Rightarrow \frac{{x + {x^2}}}{2} = 5x \cr & \Rightarrow {x^2} - 9x = 0 \cr & \Rightarrow x\left( {x - 9} \right) = 0 \cr & \Rightarrow x = 0{\text{ or }}x = 9 \cr} $$
So, the number is 9
38. The average age of a cricket team of 11 players is the same as it was 3 years back because 3 of the players whose current average age of 33 years are replaced by 3 youngsters. The average age of the new comers is :
a) 23 years
b) 21 years
c) 22 years
d) 20 years
Explanation: According to the question,
Increased age of 11 players
= 11 × 3
= 33 years
Current age of 3 players who are replaced
= 3 × 33
= 99 years
∴ Age of 3 newcomers
= 99 - 33
= 66 years
∴ Average age = $$\frac{66}{3}$$ = 22 years
39. The average of five numbers is 27. If one number is excluded, the average becomes 25. The excluded number is ?
a) 25
b) 27
c) 30
d) 35
Explanation: Sum of five number
= 27 × 5 = 135
Sum of four number
= 25 × 4 = 100
Excluded number
= 135 - 100 = 35
40. If the average weight of 6 students is 50 kg. If two student of average weight of 51 kg are added and two other students of average weight of 55 kg are also added then the average weight of all the students is :
a) 61 kg
b) 51.5 kg
c) 51 kg
d) 51.2 kg
Explanation: According to the question,
Required Average
$$\eqalign{ & = \frac{{6 \times 50 + 51 \times 2 + 55 \times 2}}{{10}} \cr & = \frac{{300 + 212}}{{10}} \cr & = \frac{{512}}{{10}} \cr & = 51.2{\text{ Kg}} \cr} $$
41.The average of the squares of first ten natural numbers is
a) 35.5
b) 36
c) 37.5
d) 38.5
Explanation:As we know that average of square of "n" natural number is
$$\eqalign{ & = \frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{{6n}} \cr & = \frac{{\left( {n + 1} \right)\left( {2n + 1} \right)}}{6} \cr} $$
According to the question,
Average of square of first ten natural number is
$$\eqalign{ & = \frac{{\left( {10 + 1} \right)\left( {20 + 1} \right)}}{6} \cr & = \frac{{11 \times 21}}{6} \cr & = 38.5 \cr} $$
42. The average of 11 results is 50. If the average of the first six results is 49 and that of the last six is 52, the sixth number is
a) 48
b) 50
c) 52
d) 56
Explanation: According to the question,
Average of 11 numbers is = 50
sum of 11 numbers is = 50 × 11 = 550
∴ VI number
= 312 + 294 - 550
= 56
43. A man bought 13 articles at Rs. 70 each, 15 at Rs. 60 each and 12 at Rs. 65 each. The average price per article is
a) Rs. 60.25
b) Rs. 64.75
c) Rs. 65.75
d) Rs. 62.25
Explanation: According to the question,
$$\eqalign{ & = \frac{{13 \times 70 + 15 \times 60 + 12 \times 65}}{{40}} \cr & = \frac{{910 + 900 + 780}}{{40}} \cr & = \frac{{2590}}{{40}} \cr & = 64.75 \cr} $$
44. The average salary, per head, of all the workers of an institution is Rs. 60. The average salary of 12 officers is Rs. 400; the average salary, per head, of the rest is Rs. 56. The total number of workers in the institution is
a) 1030
b) 1035
c) 1032
d) 1020
Explanation: Let the total number of worker = x
According to the question,
⇒ 12 × 400 + (x - 12) × 56 = 60x
⇒ 4800 + 56x - 672 = 60x
⇒ 4128 = 4x
⇒ x = $$\frac{4128}{4}$$
= x = 1032
45. The average of the three numbers x, y and z is 45. x is greater than the average of y and z by 9. The average of y and z is greater than y by 2. Then the difference of x and z is :
a) 3
b) 5
c) 7
d) 8
Explanation: According to the question,
$$\eqalign{ & \Rightarrow \frac{{x + y + z}}{3} = 45 \cr & \Rightarrow x + y + z = 135.....(i) \cr & \Rightarrow x = \frac{{y + z}}{2} + 9 \cr & \Rightarrow 2x - y - z = 18.....(ii) \cr & x + y + z = 135 \cr & \underline {2x - y - z = 18} \cr & 3x = 153 \cr & x = 51 \cr} $$
From (i)
y + z = 135 - 51 = 84.....(iii)
Also,
$$\eqalign{ & \Rightarrow \frac{{y + z}}{2} = y + 2 \cr & \Rightarrow y + z = 2y + 4 \cr & \Rightarrow z - y = 4 \cr & + y + z = 84 \cr & \underline { - y + z = 4} \cr & \,\,\,\,\,\,\,\,\,2z = 88 \cr & \,\,\,\,\,\,\,\,\,\,\,\,z = 44 \cr} $$
Required difference
= 51 - 44 = 7
46. The total marks obtained by a student in Physics, Chemistry and Mathematics together is 120 more than the marks obtained by him in Chemistry. What is the average marks obtained by him in Physics and Mathematics together?
a) 40
b) 60
c) 120
d) Cannot be determined
Explanation: P + C + M = C + 120
⇒ P + M = 120
∴ Required average
= $$\frac{P + M}{2}$$
= $$\frac{120}{2}$$
= 60
47. A student was asked to find the arithmetic mean of the numbers 3, 11, 7, 9, 15, 13, 8, 19, 17, 21, 14 and x. He found mean to be 12. What should be the number in place of x?
a) 3
b) 7
c) 17
d) 31
Explanation: Clearly, we have
$$ \Rightarrow \frac{{3 + 11 + 7 + 9 + 15 + 13 + 8 + 19 + 17 + 21 + 14 + x}}{{12}} = 12$$
$$\eqalign{ & \Rightarrow 137 + x = 144 \cr & \Rightarrow x = 144 - 137 \cr & \Rightarrow x = 7 \cr} $$
48. If the mean of a, b, c, is M and ab + bc + ca = 0, then the mean of a2, b2, c2 is
a) M2
b) 3M2
c) 6M2
d) 9M2
Explanation: We have :
$$\frac{a + b + c}{3}$$ = M
or (a + b + c) = 3M
Now, (a + b + c)2 = (3M)2 = 9M2
⇔ a2 + b2 + c2 + 2 (ab + bc + ca) = 9M2
⇔ a2 + b2 + c2 = 9M2
[$$\because $$ (ab + bc + ca) = 0]
∴ Required mean
$$\eqalign{ & = \left( {\frac{{{a^2} + {b^2} + {c^2}}}{3}} \right) \cr & = \frac{{9{M^2}}}{3} \cr & = 3{M^2} \cr} $$
49. The average of 2, 7, 6 and x is 5 and the average of 18, 1, 6, x and y is 10. What is the value of y?
a) 5
b) 10
c) 20
d) 30
Explanation: We have:
$$\eqalign{ & \Rightarrow \left( {\frac{{2 + 7 + 6 + x}}{5}} \right) = 5 \cr & \Rightarrow 15 + x = 20 \cr & \Rightarrow x = 5 \cr} $$
Also,
$$\eqalign{ & \Rightarrow \left( {\frac{{18 + 1 + 6 + x + y}}{5}} \right) = 10 \cr & \Rightarrow 25 + 5 + y = 50 \cr & \Rightarrow y = 20 \cr} $$
50. The average of x1, x2, x3 and x4 is 16. Half the sum of x2, x3, x4 is 23. What is the value of x1?
a) 17
b) 18
c) 19
d) 20
Explanation: x1 + x2 + x3 + x4 = 16 × 4 = 64
⇒ $$\frac{1}{2}$$ (x2 + x3 + x4) = 23
⇒ x2 + x3 + x4 = 46
∴ x1 = 64 - 46
= 18
51. The average of 50 numbers is 38. If the numbers 45 and 55 are discarded, then the average of the remaining numbers is
a) 36.5
b) 37
c) 37.5
d) 37.52
Explanation: Sum of 50 numbers = 38 × 50 = 1900
Sum of remaining 48 numbers
= 1900 - (45 + 55)
= 1800
∴ Required average
= $$\left( {\frac{{1800}}{48}} \right)$$
= 37.5
52. 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 first and third numbers is
a) 12
b) 20
c) 24
d) 48
Explanation:Let the second number be x
Then, first number = 2x, 3rd number = 4x
∴ 2x + x + 4x = 56 × 3
⇒ 7x = 168
⇒ x = 24
∴ Required difference
= 4x - 2x
= 2x
= 2 × 24
= 48
53. The average price of three items of furniture is Rs. 15000. If their prices are in the ratio 3 : 5 : 7, the price of the cheapest item is
a) Rs. 9000
b) Rs. 15000
c) Rs. 18000
d) Rs. 21000
Explanation: Let their prices be 3x, 5x and 7x respectively
Then,
⇒ 3x + 5x + 7x = 15000 × 3
⇒ 15x = 45000
⇒ x = 3000
∴ Cost of cheapest item
= Rs. (3000 × 3)
= Rs. 9000
54. A school has 4 sections of chemistry in Class X having 40, 35, 45 and 42 students. The mean marks obtained in Chemistry test are 50, 60, 55 and 45 respectively for the 4 sections. Determine the overall average of marks per student
a) 50.25
b) 51.25
c) 52.25
d) 53.25
Explanation: Average marks
$$\eqalign{ & = \left( {\frac{{50 \times 40 + 60 \times 35 + 55 \times 45 + 45 \times 42}}{{40 + 35 + 45 + 42}}} \right) \cr & = \left( {\frac{{2000 + 2100 + 2475 + 1890}}{{162}}} \right) \cr & = \left( {\frac{{8465}}{{162}}} \right) \cr & = 52.25 \cr} $$
55. The mean of 5 observations is 60, the mean of 10 observations is 30 and the mean of 15 observations is 20. The mean of all the 30 observations is
a) 20
b) 25
c) 30
d) 40
Explanation: Required mean
$$\eqalign{ & = \left( {\frac{{60 \times 5 + 30 \times 10 + 20 \times 15}}{{5 + 10 + 15}}} \right) \cr & = \left( {\frac{{300 + 300 + 300}}{{30}}} \right) \cr & = \left( {\frac{{900}}{{30}}} \right) \cr & = 30 \cr} $$
56. While calculating the average of a batsman as 36 in 100 matches that he played, one of the scores 90 was incorrectly noted as 40. The percentage error is
a) 0.5%
b) 1.21%
c) 1.34%
d) 1.36%
Explanation: Correct sum
= 36 × 100 + 90 - 40
= 3650
Correct average
= $$\frac{3650}{100}$$
= 36.5
Error = (36.5 - 36) = 0.5
∴ Error %
= ( $$\frac{0.5}{36.5}$$ × 100 )%
= $$\frac{100}{73}$$%
= 1.36%
57. The average of 8 numbers is 20. The average of first two numbers is $$15\frac{1}{2}$$ and that of the next three is $$21\frac{1}{3}$$. If the sixth number be less than the seventh and eighth numbers by 4 and 7 respectively, then the eight number is
a) 18
b) 22
c) 25
d) 27
Explanation: Let the eight number be x
Then, sixth number = (x - 7)
Seventh number = (x - 7) + 4 = (x - 3)
So,
$$ \Leftrightarrow \left( {2 \times 15\frac{1}{2}} \right) + \left( {3 \times 21\frac{1}{2}} \right)$$ $$ + \left( {x - 7} \right)$$ $$ + \left( {x - 3} \right)$$ $$ + \,x = 8 \times 20$$
$$\eqalign{ & \Leftrightarrow 31 + 64 + \left( {3x - 10} \right) = 160 \cr & \Leftrightarrow 3x = 75 \cr & \Leftrightarrow x = 25 \cr} $$
58. In an examination, a pupil’s average mark was 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination?
a) 8
b) 9
c) 10
d) 11
Explanation: Let the number of paper be x
Then, pupil's total score = 63x
∴ $$\frac{63x + 20 + 2}{x}$$ = 65
⇒ 2x = 22
⇒ x = 11
59. In a one-day cricket match the captain of one of the teams scored 30 runs more than the average runs scored by the remaining six batsman of that team who batted in the match. If the total runs scored by all the batsmen of that team were 310, how many runs did the captain score?
a) 50
b) 60
c) 70
d) Cannot be determined
Explanation:Let the average score of the remaining 6 batsmen be x runs.
Then, sum of their scroe = 6x
Captain's score = (x + 30)
∴ 6x + (x + 30) = 310
⇒ 7x = 280
⇒ x = 40
Hence, captain's score = x + 30 = 70
60. 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
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
61. 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) 24 years
b) 36 years
c) 45 years
d) Cannot be determined
Explanation: Sumit + Krishna + Rishabh = 43 × 3 = 129.....(i)
Sumit + Rishabh + Rohit = 49 × 3 = 147.....(ii)
Subtracting (i) from (ii), we get:
Rohit - Krishna = 18
⇒ Krishna = 54 - 18 = 36
62. A library has an average of 510 visitors on Sundays and 240 on other days. What is the average number of visitors per day in the month of June beginning with a Sunday?
a) 276
b) 280
c) 285
d) 250
Explanation:If a month beings with Sunday then there are 5 Sundays in that month.
Total number of visitors comes on Sunday
= 510 × 5 = 2550
Total number of visitors come on other days
= 240 × 25 = 6000
∴ Average number of visitor per day
$$\eqalign{ & = \frac{{2550 + 6000}}{{30}} \cr & = \frac{{8550}}{{30}} \cr & = 285 \cr} $$
63. Average of a, b, and c is 11; average of c, d and e is 17; average of e and f is 22 and average of e and c is 17. Find out the average of a, b, c, d, e and f.
a) $$15\frac{2}{3}$$
b) $$18\frac{1}{2}$$
c) $$16\frac{1}{2}$$
d) None of these
Explanation:Average of a, b and c = 11
Total of a, b and c = 33.....(i)
Similarly, average of c, d and e = 17
Sum of c + d + e = 3 × 17 = 51.....(ii)
Average of e and f is 22
Sum of e + f = 2 × 22 = 44.....(iii)
Average of e and c is 17
Sum of e + c = 2 × 17 = 34.....(iv)
By equations (i) + (ii) + (iii) - (iv)
a + b + c + c + d + e + e + f - e - c
= 33 + 51 + 44 - 34
= 128 - 34
= 94
∴ Required average
= $$\frac{94}{6}$$
= $$\frac{47}{3}$$
= $$15\frac{2}{3}$$
64. The average of 6 numbers is 7. The average of three numbers of them is 5. What will be the average of remaining numbers?
a) 15
b) 30
c) 9
d) 42
Explanation:Average of 6 numbers = 7
Sum of 6 numbers = 6 × 7 = 42
Average of three numbers = 5
Sum of three numbers = 5 × 3 = 15
∴ Sum of the remaining three numbers
= 42 - 15 = 27
∴ Required average
= $$\frac{27}{3}$$
= 9
65. Find the average of 205, 302, 108, 403, and 202
a) 450
b) 1125
c) 244
d) 1220
Explanation: Sum of numbers
= 205 + 302 + 108 + 403 + 202
= 1220
∴ Required average
= $$\frac{1220}{5}$$
= 244
66. The average weight of 3 men A, B, and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg more than that of D, replaces A, then the average weight of B, C, D and E becomes 79 kg. The weight if A is
a) 70 kg
b) 72 kg
c) 75 kg
d) 80 kg
Explanation: Let A, B, C, D and E represent their respective weights.
Then,
A + B + C = (84 × 3) = 252 kg
A + B + C + D = (80 × 4) = 320 kg
∴ D = (320 - 252) kg = 68 kg
E = (68 + 3) kg = 71 kg
B + C + D + E = (79 × 4) = 316 kg
Now,
= (A + B + C + D) - (B + C + D + E)
= (320 - 316) kg
= 4 kg
∴ A - E = 4
⇒ A = (4 + E)
⇒ A = 75 kg
67. A team of 8 persons joins in a shooting competition. The best marksman scored 85 points. If had scored 92 points, the average score for the team would have been 84. The number of points, the team scored was
a) 588
b) 645
c) 665
d) 672
Explanation:Let the total score be x
$$\eqalign{ & \therefore \frac{{x + 92 - 85}}{8} = 84 \cr & \Rightarrow x + 7 = 672 \cr & \Rightarrow x = 665 \cr} $$
68. A batsman makes a score of 84 runs in the 21st inning and thus increases his average by 2 runs. His average after 21st inning is
a) 24
b) 34
c) 44
d) 54
Explanation: Let the average for 20 innings be x
$$\eqalign{ & \therefore \frac{{20x + 84}}{{21}} = x + 2 \cr & \Rightarrow 20x + 84 = 21x + 42 \cr & \Rightarrow x = 42 \cr} $$
∴ Average after 21st innings
= 42 + 2
= 44
69. When 15 is included in a list of natural numbers, their mean is increased by 2. When 1 is included in this new list, the mean of the numbers in the new list is decreased by 1. How many numbers were there in the original list ?
a) 4
b) 5
c) 6
d) 7
Explanation:Let there be n numbers in the original list and let their mean be x.
Then, sum of n numbers = nx
$$\eqalign{ & \therefore \frac{{nx + 15}}{{n + 1}} = x + 2 \cr & \Rightarrow nx + 15 = \left( {n + 1} \right)\left( {x + 2} \right) \cr & \Rightarrow nx + 15 = nx + 2n + x + 2 \cr & \Rightarrow 2n + x = 13.....(i) \cr} $$
And,
$$\eqalign{ & \therefore \frac{{nx + 16}}{{n + 2}} = \left( {x + 2} \right) - 1 \cr & \Rightarrow nx + 16 = \left( {n + 2} \right)\left( {x + 1} \right) \cr & \Rightarrow nx + 16 = nx + n + 2x + 2 \cr & \Rightarrow n + 2x = 14.....(ii) \cr} $$
Solving (i) and (ii), we get:
n = 4, x = 5
70. The average weight of 8 men is increased by 1.5 kg when one of the men, who weight 65 kg is replaced by a new man. The weight of the new man is
a) 70 kg
b) 74 kg
c) 76 kg
d) 77 kg
Explanation: Total weight increased
= (8 × 1.5) kg
= 12 kg
Weight of the new man
= (65 + 12) kg
= 77 kg
71. The average temperature of the town in the first four days of month was 58 degrees. The average for the second, third, fourth and fifth days was 60 degrees. If the temperature of the first and fifth days were in the ratio 7 : 8, then what is the temperature on the fifth day?
a) 64 degrees
b) 62 degrees
c) 56 degrees
d) None of these
Explanation: Sum of temperature on 1st, 2nd, 3rd and 4th days
= (58 × 4)
= 232 degrees
Sum of temperature on 2nd, 3rd, 4th and 5th days
= (60 × 4)
= 240 degrees
Subtracting (i) from (ii), we get :
Temperature on 5th day - Temperature on 1st day = 8 degrees
Let the temperature on 1st and 5th days be 7x and 8x degrees respectively.
Then,
⇒ 8x - 7x = 8
⇒ x = 8
∴ Temperature on the 5th day
= 8x
= 8 × 8
= 64 degree
72. 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 number is :
a) 11
b) 11.3
c) 11.4
d) 11.5
Explanation: Middle number
= [(10.5 × 6 + 11.4 × 6) - 10.9 × 11]
= (131.4 - 119.9)
= 11.5
73. 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. The number of pupils in the class is
a) 10
b) 20
c) 40
d) 73
Explanation: Let there be x pupils in the class
Total increase in marks
= x × $$\frac{1}{2}$$
= $$\frac{x}{2}$$
∴ $$\frac{x}{2}$$ = (83 - 63)
⇒ $$\frac{x}{2}$$ = 20
⇒ x = 40
74. 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 instead of the actual marks 72 and 61 respectively, what would be the correct average?
a) 66
b) 68.5
c) 69
d) 69.5
Explanation: Correct sum
= (68 × 20 + 72 + 61 - 48 - 65)
= 1380
∴ Correct average
= $$\frac{1380}{20}$$
= 69
75. Average of ten positive numbers is $$\overline x $$. If each number is increased by 10%, then $$\overline x $$
a) Remains unchanged
b) May decrease
c) May increase
d) Is increased by 10%
Explanation:
$$\eqalign{ & \Rightarrow \frac{{{x_1} + {x_2} + ..... + {x_{10}}}}{{10}} = \overline x \cr & \Rightarrow {x_1} + {x_2} + ..... + {x_{10}} = 10\overline x \cr & \Rightarrow \frac{{110}}{{100}}{x_1} + \frac{{110}}{{100}}{x_2} + ..... + \frac{{110}}{{100}}{x_{10}} = \frac{{110}}{{100}} \times 10\overline x \cr & \Rightarrow \frac{{\frac{{110}}{{100}}{x_1} + \frac{{110}}{{100}}{x_2} + ..... + \frac{{110}}{{100}}{x_{10}}}}{{10}} = \frac{{11}}{{10}}\overline x \cr} $$
⇒ Average is increased by 10%
76. A shop of electronic goods is closed on Monday. The average daily sales for remaining six days of a week is Rs. 15,640 and the average sale of Tuesday to Saturday is Rs. 14,124. The sales on Sunday is
a) Rs. 20,188
b) Data inadequate
c) Rs. 23,220
d) Rs. 21,704
Explanation: Average sales per day for six days of the week = Rs. 15640
Total sales of six days of the week
= 15640 × 6
= Rs. 93840
Average sales to Tuesday to Saturday = Rs. 14124
Total sales from Tuesday to Saturday
= 14124 × 5
= Rs. 70620
∴ Sales on Sunday
= (Rs. 93840 - 70620)
= Rs. 23,220
77. The average expenditure of a man for the first five months is Rs. 1200 and for the next seven months is Rs. 1300. If he saves Rs. 2900 in that year, his monthly average income is
a) Rs. 1500
b) Rs. 1600
c) Rs. 1700
d) Rs. 1400
Explanation: Average expenditure of a man for the first five month = Rs. 1200
Average expenditure of a man for the next seven month = Rs. 1300
Total annual expenditure of man
= Rs. (5 × 1200 + 7 × 1300)
= Rs. (6000 + 9100)
= Rs. 15100
Man saves = Rs. 2900
His total annual income
= Rs. (15100 + 2900)
= Rs. 18000
∴ Average monthly income
= $$\frac{18000}{12}$$
= Rs. 1500
78. The average weight of 21 boys was recorded as 64 kg. If the weight of the teacher was added, the average increased by 1 kg. What was the teacher’s weight ?
a) 86 kg
b) 64 kg
c) 72 kg
d) 84 kg
Explanation:Average weights of 21 boys = 64 kg
Total weights of 21 boys
= 64 × 21
= 1344 kg
The weight of the teacher was added then average increase by 1 kg
⇒ Total weight of teacher and 21 boys
= 65 × 22
= 1430 kg
∴ Weight of teacher
= 1430 - 1344
= 86 kg
79. In a school with 600 students, the average age of the boys is 12 years and that of the girls is 11 years. If the average of the school is 11 years 9 months, then the average the number of girls in the school is :
a) 150
b) 250
c) 350
d) 450
Explanation: Let the number of girls be x
Then, number of boys = (600 - x)
Then,
$$\left( {11\frac{3}{4} \times 600} \right)$$ = 11x + 12 (600 - x)
⇔ x = 7200 - 7050
⇔ x = 150
80. The arithmetic mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25% a mean score of 31. The mean score of remaining 55% is
a) 45
b) 50
c) 51.4 approx
d) 54.6 approx
Explanation: Let the required mean score be x
Then,
$$\eqalign{ & 20 \times 80 + 25 \times 31 + 55 \times x = 52 \times 100 \cr & \Leftrightarrow 1600 + 775 + 55x = 5200 \cr & \Leftrightarrow 55x = 2825 \cr & \Leftrightarrow x = \frac{{2825}}{{55}} \approx 51.4 \cr} $$
81. The average age of 3 children in a family is 20% of the average age of the father and the eldest child. The total age of the mother and the youngest child is 39 years. If the father's age is 26 years, what is the age of the second child ?
a) 15 years
b) 18 years
c) 20 years
d) Cannot be determined
Explanation: Since the total or average age of all the family members is not given, te given data is inadequate. So, the age of the second child cannot be determined.
82. Five years ago the average age of A, B, C, D was 45 years. By including X the present average of all the five is 49 years. Them the present age of X is
a) 40 years
b) 45 years
c) 48 years
d) 64 years
Explanation: Sum of the present ages of A, B, C and D
= (45 × 4 + 5 × 4) years
= 200 years
Sum of the present ages of A, B, C, D and X
= (49 × 5) years
= 245 years
∴ X's present age
= (245 - 200) years
= 45 years
83. A cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and thereby decreased his average by 0.4. The number of wickets taken by him the last match was
a) 64
b) 72
c) 80
d) 85
Explanation: Let the number of wickets taken till the last match be x
$$\eqalign{ & \Rightarrow \frac{{12.4x + 26}}{{x + 5}} = 12 \cr & \Rightarrow 12.4x + 26 = 12x + 60 \cr & \Rightarrow 0.4x = 34 \cr & \Rightarrow x = \frac{{34}}{{0.4}} \cr & \Rightarrow x = \frac{{340}}{4} \cr & \Rightarrow x = 85 \cr} $$
84. The average of 11 players of a cricket team is decreased by 2 months when two of them aged 17 years and 20 years are replaced by two new players. The average age of the new players is
a) 17 years 1 month
b) 17 years 7 month
c) 17 years 11 month
d) 17 years 3 month
Explanation: Total age decreased
= (11 × 2) months
= 22 months
= 1 year 10 months
∴ Average age of two new players
$$\left( {\frac{{{\text{35 years 2 months}}}}{2}} \right)$$
= 17 years 7 months
85. 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) Rs. 3500
b) Rs. 4000
c) Rs. 4050
d) Rs. 5000
Explanation: Let P, Q and R represent their respective monthly income
Then, we have
P + Q = (5050 × 2) = 10100.....(i)
Q + R = (6250 × 2) = 12500.....(ii)
P + R = (5200 × 2) = 10400.....(iii)
Adding (i), (ii) and (iii), we get
2 (P + Q + R) = 33000
⇒ P + Q + R = 16500
Subtracting (ii) from (iv), we get P = 4000
∴ P's monthly income = Rs. 4000
86. Ten years ago, the ages of the members of a joint family of eight people added up to 231 years. Three years later, one member died at the age of 60 years and a child was born during the same year. After another three years, one more member died, again at 60, and a child was born during the same year. The current average of this eight-member joint family is nearest to
a) 21 years
b) 22 years
c) 23 years
d) 24 years
Explanation: Sum of the ages of 8 members, 10 years ago = 231 years
Sum of the ages of all members, 7 years ago
= (231 + 8 × 3 - 60) years
= 195 years
Sum of the ages of all members, 4 years ago
= (195 + 8 × 3 - 60) years
= 159 years
Sum of the present ages of all 8 members
= (159 + 8 × 4) years
= 191 years
∴ Current average age = $$\frac{191}{8}$$ years
= 23.8 years $$ \approx $$ 24 years
87. Four years ago, the average age of a family of four persons was 18 years. During this period, a baby was born. Today if the average age of the family is still 18 years, the age of the baby is
a) 1.2 years
b) 2 years
c) 2.5 years
d) 3 years
Explanation: Sum of the ages of 4 members, 4 years ago
= (18 × 4) years
= 72 years
Sum of the ages of 4 members now
= (72 + 4 × 4) years
= 88 years
Sum of the ages of 5 members now
= (18 × 5) years
= 90 years
∴ Age of the baby
= (90 - 88) years
= 2 years
88. When the average age of a couple and their son was 42 years, the son married and got a child after one year. When the child was 5 years old, the average age of the family became 36 years. What was the age of daughter-in- law at the time of their marriage ?
a) 23 years
b) 24 years
c) 25 years
d) 26 years
Explanation: Sum of the ages of father, mother and son at the time of son's marriage
= (42 × 3) years
= 126 years
Sum of the present ages of father, mother and son
= (126 + 3 × 6)years
= 144 years
Sum of the present ages of father, mother, son and grandson
= (144 + 5) years
= 149 years
Sum of the present ages of father, mother, son, daughter-in-law and grandson
= (36 × 5) years
= 180 years
Daughter-in-law's present age
= (180 - 149) years
= 31 years
∴ Age of daughter-in-law at the time of marriage
= (31 - 6) years
= 25 years
89. 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
Explanation: Sum of the present ages of husband, wife and child
= (27 × 3 + 3 × 3) years
= 90 years
Sum of the present ages of wife and child
= (20 × 2 + 5 × 2) years
= 50 years
∴ Husband's present age
= (90 - 50) years
= 40 years
90. Four years ago, the average age of A and B was 18 years. At present the average age of A, B, and C is 24 years. What would be the age of C after 8 years?
a) 25 years
b) 28 years
c) 32 years
d) 36 years
Explanation: Sum of the present ages of A and B
= (18 × 2 + 4 × 2) years
= 44 years
Sum of the present ages of A, B and C
= (24 × 3) years
= 72 years
C's present age
= (72 - 44) years
= 28 years
∴ C's age after 8 years
= (28 + 8) years
= 36 years
91.In a primary school the average weight of male students is 65.9 kg and the average weight of female students is 57 kg. If the average weight of all the students ( both male and female ) is 60.3 kg and the number of male students in the school is 66, what is the number of female students in the school?
a) 162
b) 168
c) 180
d) 112
Explanation:Let the number of female students be x
Let weight of female students = 57x
Number of male students = 66
Total weights of male students = 65.9 × 66
Average weight of all the students = 60.3 kg
Total weights of all the students = 60.3 (66 + x)
According to given information,
Then,
⇒ 60.3 (66 + x) = 66 × 65.9 + 57x
⇒ 60.3 × 66 + 60.3x = 66 × 65.9 + 57x
⇒ 60.3x - 57x = 66 (65.9 - 60.3)
⇒ 3.3x = 66 × 5.6
∴ x = $$\frac{66 × 5.6}{3.3}$$
⇒ x = 2 × 56
⇒ x = 112
92. The average monthly income of P and Q is Rs. 6000; that of Q and R is Rs. 5250; and that P and R is Rs. 5500. What is P’s monthly income?
a) Rs. 3500
b) Rs. 4500
c) Rs. 6250
d) Rs. 4800
Explanation: Average monthly income of P and Q = Rs. 6000
Average monthly income of Q and R = Rs. 5250
Average monthly income of P and R = Rs. 5500
Total income of P + Q
= 2 × 6000
= Rs. 12000.....(i)
Total income of Q + R
= 2 × 5250
= Rs. 10500.....(ii)
Total income of R + P
= 2 × 5500
= Rs. 11000.....(iii)
On adding equation (i) (ii) and (iii), we have
2 (P + Q + R) = 12000 + 10500 + 11000
⇒ P + Q + R = $$\frac{33500}{2}$$
⇒ P + Q + R = Rs. 16750.....(iv)
By equation (iv) - (ii)
P's monthly income
= Rs. (16750 - 10500)
= Rs. 6250
93.The average weight of a group of 75 girls was calculated as 47 kgs. It was later discovered that the weight of one of the girls was read as 45 kgs. Whereas her actual weight was 25 kgs. What is the actual average weight of the group of 75 girls? ( Rounded off to two digits after decimal)
a) 34
b) 36
c) 30
d) None of these
Explanation: Average weight of 75 girls = 47 kgs
Total weight of 75 girls
= 47 × 75
= 3525 kgs
Actual weight of 75 girls = x
Correct weight of 75 girls
= 3525 - 45 + 25
= 3525 - 20
= 3505 kgs
∴ Required average weight
x =$$\frac{3505}{75}$$
x = 46.73 kgs
94. Find the average of the following sets of scores:
385, 441, 876, 221, 536, 46, 291, 428
a) 221
b) 403
c) 428
d) 536
Explanation:Average
$$ = \left( {\frac{{385 + 441 + 876 + 221 + 536 + 46 + 291 + 428}}{8}} \right){\text{kg}}$$
$$\eqalign{ & = \left( {\frac{{3224}}{8}} \right){\text{kg}} \cr & = 403{\text{ kg}} \cr} $$
95. If 25a + 25b = 115, what is the average of a and b?
a) 2.5
b) 3.4
c) 4.5
d) None of these
Explanation: 25a + 25b = 115
⇒ 25 (a + b) = 115
⇒ a + b = $$\frac{115}{25}$$
⇒ a + b = $$\frac{23}{5}$$
∴ Average of a and b
= $$\frac{a + b}{2}$$
= $$\frac{23}{5}$$ × $$\frac{1}{2}$$
= $$\frac{23}{10}$$
= 2.3
96. The average of six numbers is x and the average of three of these is y. If the average of the remaining three is z, then
a) x = y + z
b) 2x = y + z
c) x = 2y + 2z
d) None of these
Explanation: Clearly, we have :
$$\eqalign{ & \Rightarrow x = \left( {\frac{{3y + 3z}}{6}} \right) \cr & \Rightarrow 2x = y + z \cr} $$
97. The average of the first 100 positive integers is-
a) 49.5
b) 50.5
c) 51
d) 100
Explanation: Required average
$$\eqalign{ & = \left( {\frac{{1 + 2 + 3 + .... + 100}}{{100}}} \right) \cr & = \frac{1}{{100}} \times \frac{{100 \times 101}}{2} \cr & = 50.5 \cr} $$
98. After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member?
a) 2 years
b) 4 years
c) 8 years
d) 15 years
Explanation: Age decreased = (5 × 3) years = 15 years
So, the required difference = 15 years
99. The average weight of 8 men is increased by 1.5 kg when one of the men, who weight 65 kg is replaced by a new man. The weight of the new man is
a) 70 kg
b) 74 kg
c) 76 kg
d) 77 kg
Explanation: Total weight increased = (8 × 1.5) kg = 12 kg
Weight of the new man = (65 + 12) kg = 77 kg
100. Total expenses of a boarding house are partly fixed and partly varying linearly with the number of boarders. The average expense per boarder is Rs. 700 when there are 25 boarders and Rs. 600 when there are 50 boarders. What is the average expense per boarder when there are 100 boarders?
a) Rs. 540
b) Rs. 550
c) Rs. 570
d) Rs. 580
Explanation: Let the fixed cost be Rs. x and the variable cost be Rs. y per boarder.
Then,
x + 25y = 700 × 25
⇒ x + 25y = 17500.....(i)
x + 50y = 600 × 50
⇒ x + 50y = 30000.....(ii)
Subtracting (i) from (ii), we get:
25y = 12500
⇒ y = 500
Putting y = 500 in (i), we get:
x = 5000
∴ Total expenses of 100 boarders
= Rs. (5000 + 500 × 100)
= Rs. 55000
Hence, average expense
= Rs. $$\frac{55000}{100}$$
= Rs. 550
101.The mean high temperature of the first four days of a week is 25°C whereas the mean of the last four days is 25.5°C. If the mean of the whole week is 25.2°C, then the temperature of the 4th day is
a) 25.2°C
b) 25.5°C
c) 25.6°C
d) 25°C
Explanation: Average temperature of first four days = 25°C
Total temperature of first four days = 25° × 4 = 100°C
Average temperature last four days = 25.5°C
Total temperature of four days = 25.5° × 4 = 102°C
Total temperature of whole week = 25.2° × 7 = 176.4°C
∴ Temperature of the 4th day
= 100° + 102° - 176.4°
= 25.6°C
102. The average weight of A, B and C is 40 kgs. Weight of C is 24 kgs more than A’s weight and 3 kgs less than B’s weight. What will be the average weight of A, B, C and D, if D weights 15 kgs less than C ?
a) 42 kgs
b) 40 kgs
c) 36 kgs
d) 38 kgs
Explanation: Average weight of A, B and C = 40 kgs
Total weights of A , B and C = 40 × 3 = 120 kgs
Weight of C = (A + 24) and C = (B - 3)
∴ A + 24 = B - 3
⇒ B = A + 27
Now A + B + C = 120
⇒ A + A + 27 + A + 24 = 120
⇒ 3A + 51 = 120
⇒ A = $$\frac{69}{3}$$ = 23 kg
B = A + 27 = 23 + 27 = 50 kg
C = 120 - 23 - 50 = 47 kg
D = 47 - 15 = 32 kg
∴ Required average weight of A, B, C and D
= $$\frac{23 + 50 + 47 + 32}{4}$$
= $$\frac{152}{4}$$
= 38 kg
103. The average of 11 results is 60. If the average of first six results is 58 and that of last six is 63, find the 6th result
a) 66
b) 55
c) 64
d) 68
Explanation: The average of 11 results = 60
The total of 11 results = 60 × 11 = 660
Average of first six results = 58
Average of last six results = 63
Total of first six results = 58 × 6 = 348
Total of last six results = 63 × 6 = 378
∴ Sixth result = Total of first and last sixth results = Total of 11 results
= (348 + 378) - 660
= 726 - 660
= 66
104. The average age of students of a class is 15.8 years. The average age of boys in the class is 16.4 years and that of the girls is 15.4 years. The ratio of the number of boys to the number of girls in the class is
a) 1 : 2
b) 2 : 3
c) 3 : 4
d) 3 : 5
Explanation:Let the ratio be K : 1
Then,
⇔ K × 16.4 + 1 × 15.4 = (K + 1) × 15.8
⇔ (16.4 - 15.8) K = (15.8 - 15.4)
⇔ K = $$\frac{0.4}{0.6}$$
⇔ K = $$\frac{2}{3}$$
∴ Required ratio
= $$\frac{2}{3}$$ : 1
= 2 : 3
105. The average score of a class of boys and girls in an examination is A. The ratio of boys and girls in the class is 3 : 1. If the average score of the boys is A + 1, the average score of the girls is
a) A - 1
b) A - 3
c) A + 1
d) A + 3
Explanation: Let the number of boys and girls in the class be 3x and x respectively.
Let the average score of the girls be y.
Then,
⇒ 3x (A + 1) + xy = (3x + x) A
⇒ 3 (A + 1) y = 4A
⇒ y = A - 3
106. The average of two numbers is XY. If one number is X, the other is
a) $$\frac{Y}{2}$$
b) y
c) 2XY - X
d) X (Y - 1)
Explanation: Sum of numbers = 2XY
∴ Other number = 2XY - X
107. 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) 45 kg
b) 50 kg
c) 53 kg
d) 55 kg
Explanation: Weight of the teacher
= (35.4 × 25 - 35 × 24) kg
= 45 kg
108. The average age of an adult class is 40 years. 12 new students with an average age of 32 years join the class, thereby decreasing the average by 4 years. The original strength of the class was
a) 10
b) 11
c) 12
d) 15
Explanation: Sum of ages of the whole class = 40x years
Sum of ages of 12 now students
= (12 × 32) years
= 384 years
∴ $$\frac{40x + 384}{x + 12}$$ = 36
⇒ 40x + 384 = 36x + 432
⇒ 4x = 48
⇒ x = 12
Hence, the original strength of the class = 12
109. The average of the first five multiples of 3 is-
a) 3
b) 9
c) 12
d) 15
Explanation: Average
$$\eqalign{ & = \frac{{3\left( {1 + 2 + 3 + 4 + 5} \right)}}{5} \cr & = \frac{{45}}{5} \cr & = 9 \cr} $$
110. The average of X1, X2 and X3 is 14. Twice the sum of X2 and X3 is 30. What is the value of X1?
a) 12
b) 16
c) 20
d) 27
Explanation:
$$\eqalign{ & {{\text{X}}_1} + {{\text{X}}_2} + {{\text{X}}_3} \cr & = \left( {14 \times 3} \right) \cr & = 42 \cr} $$
According to the question,
$$\eqalign{ & 2\left( {{{\text{X}}_2} + {{\text{X}}_3}} \right) = 30 \cr & \Rightarrow {{\text{X}}_2} + {{\text{X}}_3} = 15 \cr & \therefore {{\text{X}}_1} = \left( {42 - 15} \right) = 27 \cr} $$