Question1 : A score of 50 in the 25th inning brings down a batsman's average score by 2. Find his average after the 25th inning.
Option 1 : 100
Option 2 : 98
Option 3 : 95
Option 4 : 90
Answer : B
Explanation : Score = Previous Average -(Decrease in Avg*No. of matches)
50 = x - (2*25)
x = 100
Average after 25th inning = 100-2 = 98
50 = x - (2*25)
x = 100
Average after 25th inning = 100-2 = 98
Question2 : In an English exam, the average marks of a class of 19 students is 75. The average of the first 10 students is 86 and the average of the last 10 students is 65. Find the mark scored by the tenth student.
Option 1 : 85
Option 2 : 75
Option 3 : 76
Option 4 : 75.5
Answer : A
Explanation : Sum of the marks of all students
= 19*75
= 1425
Sum of the marks of the first 10 students
= 860
Sum of the marks of the last 10 students
= 650
Marks scored by the 10th student
= 860+650-1425
= 85
= 19*75
= 1425
Sum of the marks of the first 10 students
= 860
Sum of the marks of the last 10 students
= 650
Marks scored by the 10th student
= 860+650-1425
= 85
Question3 : In a set of three numbers, the first is twice the second and the second is thrice the third. The average of the reciprocals of the three integers is 1/2. Find the three numbers.
Option 1 : 6,3,1
Option 2 : 12,6,2
Option 3 : 9,4.5,1.5
Option 4 : 3,1.5,0.5
Answer : A
Explanation : Let the integers be x, y and z.
y=3z
x=2y=6z
[(1/x)+(1/y)+(1/z)]÷3 = 1/2
(1/6z)+(1/3z)+(1/z) = 3/2
9/6z = 3/2
z = 1
x=6, y=3, z=1
y=3z
x=2y=6z
[(1/x)+(1/y)+(1/z)]÷3 = 1/2
(1/6z)+(1/3z)+(1/z) = 3/2
9/6z = 3/2
z = 1
x=6, y=3, z=1
Question4 : A set S contains eight positive integers. If each of these numbers is increased by 5%, the average of the set increases by
Option 1 : 40%
Option 2 : 8%
Option 3 : 5%
Option 4 : 20%
Answer : C
Explanation : If the average of all numbers in a set is increased equally,
the average of the set also increases equally.
So, the average of the set increases by 5%
the average of the set also increases equally.
So, the average of the set increases by 5%
Question5 : The area of a rectangle is 180 cm². If the average of three of its sides is 14 cm, find the fourth side which is the length.
Option 1 : 15 cm
Option 2 : 6 cm
Option 3 : 30 cm
Option 4 : 18 cm
Answer : C
Explanation : Traditional Method:
Average of 3 sides is 14cm.
(l+2b)/3 = 14 cm
l+2b = 42
l = 42-2b
lb = 180 cm²
(42-2b)*b = 180
42b-2b² = 180
b²-21b+90 = 0
(b-15)(b-6) = 0
b can take two values,
15cm and 6 cm
If b = 15 cm, l =12 cm
If b = 6 cm, l = 30cm
Breadth cant be larger than length.
So, b=6 cm, l=30cm
Trial and Error method is the quickest way.
If l=15cm,
b=180/15=12 cm
(l+2b)/3=39/3=13
But, the average is 14.
If l=6cm, b=180/6=30cm
b can't be greater than l
If l=30cm, b=180/30=6 cm
(l+2b)/3=42/3 = 14cm
Hence, l=30 cm
Average of 3 sides is 14cm.
(l+2b)/3 = 14 cm
l+2b = 42
l = 42-2b
lb = 180 cm²
(42-2b)*b = 180
42b-2b² = 180
b²-21b+90 = 0
(b-15)(b-6) = 0
b can take two values,
15cm and 6 cm
If b = 15 cm, l =12 cm
If b = 6 cm, l = 30cm
Breadth cant be larger than length.
So, b=6 cm, l=30cm
Trial and Error method is the quickest way.
If l=15cm,
b=180/15=12 cm
(l+2b)/3=39/3=13
But, the average is 14.
If l=6cm, b=180/6=30cm
b can't be greater than l
If l=30cm, b=180/30=6 cm
(l+2b)/3=42/3 = 14cm
Hence, l=30 cm
Question6 : After his 15th inning, the average score of a batsman increases by 1. His previous average was 80. Find his score in the 15th inning.
Option 1 : 81
Option 2 : 95
Option 3 : 90
Option 4 : 85
Answer : B
Explanation : Total runs scored by him in 14 innings
= 80*14
= 1120
Total runs scored by him in 15 innings
= 81*15
= 1215
Score in the 15th inning
= 1215-1120
= 95
To generalize this,
Score = Previous average + (Increase in average*No. of matches)
= 80*14
= 1120
Total runs scored by him in 15 innings
= 81*15
= 1215
Score in the 15th inning
= 1215-1120
= 95
To generalize this,
Score = Previous average + (Increase in average*No. of matches)
Question 7: The average temperature of a given week was 25°C. If the average temperature on Monday, Tuesday and Wednesday was 23°C and the average temperature of Thursday, Friday and Saturday was 26°C, find the temperature on Sunday.
Option 1 : 29°C
Option 2 : 27°C
Option 3 : 26°C
Option 4 : 28°C
Answer : D
Explanation : Sum of the temperatures of the week
= 25*7 = 175
Sum of the temperatures of Monday, Tuesday and Wednesday
= 23*3 = 69
Sum of the temperatures of Thursday, Friday and Saturday
= 26*3 = 78
Temperature on Sunday
= 175-69-78
= 28°C
= 25*7 = 175
Sum of the temperatures of Monday, Tuesday and Wednesday
= 23*3 = 69
Sum of the temperatures of Thursday, Friday and Saturday
= 26*3 = 78
Temperature on Sunday
= 175-69-78
= 28°C
Question8 : S={0,2,5,6,8,9} Which number can be removed from the S without affecting the average of S?
Option 1 : 0
Option 2 : 6
Option 3 : 5
Option 4 : 8
Answer : C
Explanation : If a number which equals the average of a set is removed, the average of the resulting set remains the same.
Average of S
= 30/6 = 5
If 5 is removed from the set, the average of S remains the same
Average of S
= 30/6 = 5
If 5 is removed from the set, the average of S remains the same
Question 9: If the average of x and y is 30 and the average of y and z is 20, find x-z.
Option 1 : 10
Option 2 : 15
Option 3 : 30
Option 4 : 20
Answer : D
Explanation : (x+y)/2 = 30
x+y=60
(y+z)/2 = 20
y+z=40
x-z = (x+y)-(y+z)
= 20
x+y=60
(y+z)/2 = 20
y+z=40
x-z = (x+y)-(y+z)
= 20
Question10 : If the average of 4,8,6,p and q is 6, find the average of p+1 and q-3.
Option 1 : 5
Option 2 : 6
Option 3 : 7
Option 4 : 8
Answer : A
Explanation : (4+6+8+p+q)/5 = 6
p+q = 12
(p+1+q-3)/2
= (p+q-2)/2
= (12-2)/2
= 5
p+q = 12
(p+1+q-3)/2
= (p+q-2)/2
= (12-2)/2
= 5
Question11 : The average of a two digit number and the number obtained by interchanging the digits is 55. What is the sum of the two digits of the number?
Option 1 : 8
Option 2 : 7
Option 3 : 5
Option 4 : 10
Answer : D
Explanation : Let the number be 10x+y.
Then the number obtained by interchanging the digits is 10y+x.
(10x+y+10y+x)/2 = 55
11x+11y = 110
x+y=10
Then the number obtained by interchanging the digits is 10y+x.
(10x+y+10y+x)/2 = 55
11x+11y = 110
x+y=10
Question12 : Find the average of 3x+4, 7-2x and 2x-11.
Option 1 : (5x-11)/3
Option 2 : x
Option 3 : 7x/3
Option 4 : 3x/4
Answer : B
Explanation : (3x+4+7-2x+2x-11)/3
= 3x/3
= x
= 3x/3
= x
Question13 : Vinu drove 350 kms from Chennai to Bengaluru at an average speed of 70 kmph and drove back a distance of 400 kms at an average speed of 100 kmph. Find his average speed for the whole journey.
Option 1 : 83.33 kmph
Option 2 : 85 kmph
Option 3 : 81.66 kmph
Option 4 : 84 kmph
Answer : A
Explanation : Time taken for the forward journey
= 350/70
= 5 hours
Time taken for the return journey
= 400/100
= 4 hours
Total time taken
= 9 hours
Total distance travelled
= 750 km
Average Speed
= 750/9
= 83.33 kmph
= 350/70
= 5 hours
Time taken for the return journey
= 400/100
= 4 hours
Total time taken
= 9 hours
Total distance travelled
= 750 km
Average Speed
= 750/9
= 83.33 kmph
Question14 : During the journey from place A to place B, Peter drove at an average of 70 kmph for 3 hours and 50 kmph for 2 hours. Find his average speed for the total journey.
Option 1 : 60 kmph
Option 2 : 61.66 kmph
Option 3 : 62 kmph
Option 4 : 62.5 kmph
Answer : C
Explanation : Distance travelled
= (70*3)+(50*2)
= 310 km
Time taken
= 5 hours
Average Speed
= 310/5
= 62 kmph
= (70*3)+(50*2)
= 310 km
Time taken
= 5 hours
Average Speed
= 310/5
= 62 kmph
Question15 : Find the average of the first six prime numbers.
Option 1 : 7
Option 2 : 5.167
Option 3 : 6.833
Option 4 : 6
Answer : C
Explanation : The first 6 prime numbers are
2,3,5,7,11,13
Their average
= (2+3+5+7+11+13)/6
= 41/6
= 6.833
2,3,5,7,11,13
Their average
= (2+3+5+7+11+13)/6
= 41/6
= 6.833
Question16 : The average mark of a class in a mathematics exam is 65. The sum of the marks of all students is 2210. How many students are there in the class?
Option 1 : 34
Option 2 : 44
Option 3 : 32
Option 4 : 42
Answer : A
Explanation : Number of Students
= Sum/Average
= 2210/65
= 34
= Sum/Average
= 2210/65
= 34
Question17 : A class is divided into two groups consisting of 20 and 30 students. If the average marks of the two groups in a test are 83 and 88 respectively, find the average mark of the whole class.
Option 1 : 85
Option 2 : 85.5
Option 3 : 86
Option 4 : 86.5
Answer : C
Explanation : Sum of the marks of the first group
= 20*83 = 1660
Sum of the marks of the second group
= 30*88 = 2640
Average of the class = (1660+2640)/50 = 86
= 20*83 = 1660
Sum of the marks of the second group
= 30*88 = 2640
Average of the class = (1660+2640)/50 = 86
Question18 : The average mark of a class in a science test is 75. Geetha was the topper with 96 marks. If her mark was wrongly entered as 69, the average of the class decreases by 1. Find the number of students in the class.
Option 1 : 45
Option 2 : 37
Option 3 : 27
Option 4 : 25
Answer : C
Explanation : Let the number of students be x.
Sum of all marks = 75x
(75x-96+69)/x = 74
x=27
Sum of all marks = 75x
(75x-96+69)/x = 74
x=27
Question19 : In a cricket match, a team loses all its wickets for 242 runs. Excluding the highest and lowest score, the average runs of the other 9 players is 16. The highest score exceeds the lowest score by 86 runs. Find the highest score.
Option 1 : 98
Option 2 : 90
Option 3 : 92
Option 4 : 96
Answer : C
Explanation : Let the highest score be x.
Sum of the scores of the other 9 players
= 16*9 = 144
x+x-86+144 = 242
Solving, x=92
Sum of the scores of the other 9 players
= 16*9 = 144
x+x-86+144 = 242
Solving, x=92
Question 20: The average weight of A and B is 47.5kg. The average weight of B and C is 52.5 kg. The average weight of all three of them is 50 kg. Find the average weight of A and C.
Option 1 : 45 kg
Option 2 : 50 kg
Option 3 : 55 kg
Option 4 : 52.5 kg
Answer : B
Explanation : (A+B)/2 = 47.5
(B+C)/2 = 52.5
(A+B+C)/3 = 50
Solving these, A=45, C=55
(A+C)/2 = 50
(B+C)/2 = 52.5
(A+B+C)/3 = 50
Solving these, A=45, C=55
(A+C)/2 = 50
Question 21: The average age of our family of four is 35. Excluding me, the average age is 40. Find my age.
Option 1 : 20
Option 2 : 25
Option 3 : 30
Option 4 : 15
Answer : A
Explanation : Sum of all 4 ages
= 35*4
= 140
(140-x)/3 = 40
x=20
= 35*4
= 140
(140-x)/3 = 40
x=20
Question22 : The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?
Option 1 : 10
Option 2 : 20
Option 3 : 30
Option 4 : 40
Answer : D
Explanation : The sum of the ages of the 10 members after 4 years increases by 40.
Since the average remains the same, the sum should remain the same.
Hence, the new member is 40 years younger than the replaced member.
Since the average remains the same, the sum should remain the same.
Hence, the new member is 40 years younger than the replaced member.
Question23 : The average height of a class of 30 students is 166 cm. The average height of the girls in the class is 158 cm and the average height of the boys is 170 cm. Find the number of boys in the class.
Option 1 : 10
Option 2 : 20
Option 3 : 18
Option 4 : 15
Answer : B
Explanation : Let the number of boys be x.
170x + 158*(30-x) = 166*30
Solving, x = 20
170x + 158*(30-x) = 166*30
Solving, x = 20
Question24 : In a T-20 match, India’s target is 180. After 12 overs, the run rate is 7.5. What is the required run for the remaining overs to achieve the target?
Option 1 : 12
Option 2 : 11.5
Option 3 : 11.25
Option 4 : 11.75
Answer : C
Explanation : The score after 12 overs
= 12*7.5 = 90.
To achieve another 90 runs in 8 overs,
the required rate = 90/8 = 11.25
= 12*7.5 = 90.
To achieve another 90 runs in 8 overs,
the required rate = 90/8 = 11.25
Question 25: Vinay scores 96 in the first internal and 82 in the second internal. How much should he score in the third internal in order to get an average of 90?
Option 1 : 94
Option 2 : 92
Option 3 : 90
Option 4 : 88
Answer : B
Explanation : (96+82+x)/3 = 90
x = 92
x = 92
Question26 : Find the average of the first ten multiples of 3.
Option 1 : 30
Option 2 : 15
Option 3 : 18
Option 4 : 16.5
Answer : D
Explanation : Sum of the first 10 multiples of 3
= 3+6+9+…+30
= 3*(1+2+..+10)
= 3*(10*11/2)
= 165
Average = 165/10 = 16.5
= 3+6+9+…+30
= 3*(1+2+..+10)
= 3*(10*11/2)
= 165
Average = 165/10 = 16.5
Question27 : Find the average of all composite numbers between 1 and 15.
Option 1 : 10
Option 2 : 10.5
Option 3 : 9.5
Option 4 : 9
Answer : D
Explanation : The composite numbers between 1 and 15 are 4, 6, 8, 9, 10, 12 and 14.
Their average is 63/7 = 9
Their average is 63/7 = 9
Question28 : The average of 5 consecutive odd numbers is 25. Find the smallest of these numbers.
Option 1 : 21
Option 2 : 19
Option 3 : 25
Option 4 : 29
Answer : A
Explanation : Let the 5 numbers be
x,x+2,x+4,x+6 and x+8
Avg of the five numbers
= (5x+20)/5 = 25
x=21
x,x+2,x+4,x+6 and x+8
Avg of the five numbers
= (5x+20)/5 = 25
x=21
Question29 : Find the average of the first 50 whole numbers.
Option 1 : 24
Option 2 : 24.5
Option 3 : 25
Option 4 : 25.5
Answer : B
Explanation : Sum of the first 50 whole numbers
= 0+1+2+….+48+49
= 49*50/2
Avg = Sum/50 = 49/2 = 24.5
= 0+1+2+….+48+49
= 49*50/2
Avg = Sum/50 = 49/2 = 24.5
Question30 : In a class test, the average mark of Leena and Meena is 80. The average of Meena and Reena is 85 and the average of Leena and Reena is 90. Find Meena’s mark.
Option 1 : 95
Option 2 : 85
Option 3 : 82.5
Option 4 : 75
Answer : D
Explanation : (L+M)/2 = 80
(M+R)/2 = 85
(L+R)/2 = 90
Solving the three equations,
M = 75
TAGS--
(M+R)/2 = 85
(L+R)/2 = 90
Solving the three equations,
M = 75
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