# average problems with solutions

**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

**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

**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

**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%

**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

**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)

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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

**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.

**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

**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

**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

**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

**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

**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

**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

**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

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