Average Quiz

Question 1

The average age of a group of 10 persons is 25 years. If the age of the youngest person is excluded, the average age becomes 27 years. What is the age of the youngest person?

Accepted Answer

Let the age of the youngest person be x. The sum of the ages of the 10 persons is 25*10=250. The sum of the ages of the remaining 9 persons is 27*9=243. Therefore, x = 250-243 = 7.

Question 2

A batsman scores 87 runs in the 17th innings and thus increases his average by 3. What is his average after the 17th innings?

Accepted Answer

Let the average runs before the 17th innings be 'x'.
Total runs in 16 innings = 16x.
In the 17th innings, he scores 87 runs.
Total runs in 17 innings = 16x + 87.
The new average after 17 innings is x + 3.
So, (16x + 87) / 17 = x + 3.
16x + 87 = 17(x + 3)
16x + 87 = 17x + 51
87 - 51 = 17x - 16x
x = 36.
Average after the 17th innings = x + 3 = 36 + 3 = 39.

Question 3

The average of 7 consecutive numbers is 20. The largest of these numbers is:

Accepted Answer

For a series of consecutive numbers (or numbers in an arithmetic progression), the average is the middle number.
Since there are 7 consecutive numbers, the 4th number will be the average.
So, the 4th number is 20.
The sequence of numbers is: x, x+1, x+2, 20, x+4, x+5, x+6.
If the 4th number is 20, then the numbers are:
17, 18, 19, 20, 21, 22, 23.
The largest number is 23.

Question 4

The average age of 30 students in a class is 15 years. If the age of the teacher is included, the average age increases by 1 year. What is the teacher's age?

Accepted Answer

Total age of 30 students = 30 * 15 = 450 years.
When the teacher's age is included, the number of people becomes 30 + 1 = 31.
The average age increases by 1 year, so the new average = 15 + 1 = 16 years.
Total age of (30 students + teacher) = 31 * 16 = 496 years.
Teacher's age = (Total age of students and teacher) - (Total age of students)
Teacher's age = 496 - 450 = 46 years.

Alternatively, using the formula: New member's age = New Average + (Number of old members * Increase in average)
Teacher's age = 16 + (30 * 1) = 16 + 30 = 46 years.

Question 5

The average weight of 25 students is 50 kg. If 5 students with an average weight of 40 kg leave the class, what will be the new average weight of the remaining students?

Accepted Answer

Total weight of 25 students = 25 * 50 kg = 1250 kg.
Total weight of 5 students who left = 5 * 40 kg = 200 kg.
Remaining total weight = 1250 kg - 200 kg = 1050 kg.
Number of remaining students = 25 - 5 = 20.
New average weight of remaining students = Remaining total weight / Number of remaining students
New average weight = 1050 kg / 20 = 52.5 kg.

Question 6

The average of 10 numbers is calculated as 15. It was discovered later that one number was wrongly read as 26 instead of 36. What is the correct average?

Accepted Answer

Incorrect sum of 10 numbers = 10 * 15 = 150.
The number wrongly read was 26, but it should have been 36.
Difference in value = Correct value - Incorrect value = 36 - 26 = 10.
The actual sum should be 10 more than the incorrect sum.
Correct sum = 150 + 10 = 160.
Correct average = Correct sum / Number of numbers = 160 / 10 = 16.

Question 7

The average marks of 40 students in an examination was 25. The average marks of the first 20 students was 20. The average marks of the remaining 20 students is:

Accepted Answer

Total marks of 40 students = Number of students * Average marks = 40 * 25 = 1000.
Total marks of the first 20 students = 20 * 20 = 400.
Marks of the remaining 20 students = Total marks of 40 students - Total marks of first 20 students
Marks of remaining 20 students = 1000 - 400 = 600.
Average marks of the remaining 20 students = Marks of remaining 20 students / Number of remaining students
Average marks = 600 / 20 = 30.

Question 8

What is the average of the first 5 multiples of 7?

Accepted Answer

The first 5 multiples of 7 are: 7, 14, 21, 28, 35.
Since these numbers are in an arithmetic progression, their average is the middle term.
In this case, the middle term is 21.
Alternatively, calculate the sum and divide by the count:
Sum = 7 + 14 + 21 + 28 + 35 = 105.
Average = Sum / Number of multiples = 105 / 5 = 21.

Question 9

The average age of A, B and C is 20 years. If the average age of A and B is 19 years, what is the age of C?

Accepted Answer

Total age of A, B, and C = Number of persons * Average age = 3 * 20 = 60 years.
Total age of A and B = Number of persons * Average age = 2 * 19 = 38 years.
Age of C = (Total age of A, B, C) - (Total age of A and B)
Age of C = 60 - 38 = 22 years.

Question 10

The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What is the weight of the new person?

Accepted Answer

Let the original average weight of 8 persons be 'x' kg.
Total weight of 8 persons = 8x kg.
When a person weighing 65 kg is replaced by a new person (let their weight be 'W' kg), the new total weight is 8x - 65 + W.
The new average weight is x + 2.5 kg.
So, (8x - 65 + W) / 8 = x + 2.5
8x - 65 + W = 8(x + 2.5)
8x - 65 + W = 8x + 20
-65 + W = 20
W = 20 + 65
W = 85 kg.

Alternatively, the increase in total weight is 8 persons * 2.5 kg/person = 20 kg.
This increase in total weight is due to the new person being heavier than the person who left.
Weight of new person = Weight of person replaced + Increase in total weight
Weight of new person = 65 kg + 20 kg = 85 kg.

Question 11

A student's average score in 5 tests is 75. If the student scored 70 in the first three tests and 80 in the fourth test, what was the score in the fifth test?

Accepted Answer

Total score in 5 tests = Number of tests * Average score = 5 * 75 = 375.
Score in the first three tests = 3 * 70 = 210.
Score in the fourth test = 80.
Total score in the first four tests = 210 + 80 = 290.
Score in the fifth test = Total score in 5 tests - Total score in first four tests
Score in the fifth test = 375 - 290 = 85.

Question 12

The average of 5 numbers is 27. If one number is excluded, the average becomes 25. What is the excluded number?

Accepted Answer

Sum of 5 numbers = 5 * 27 = 135.
When one number is excluded, there are 4 numbers left.
Sum of 4 numbers = 4 * 25 = 100.
Excluded number = (Sum of 5 numbers) - (Sum of 4 numbers) = 135 - 100 = 35.

Question 13

The average age of 24 students and their teacher is 15 years. If the teacher's age is excluded, the average age decreases by 1 year. What is the teacher's age?

Accepted Answer

Total number of people initially = 24 students + 1 teacher = 25.
Sum of ages of 25 people = 25 * 15 = 375 years.
When the teacher's age is excluded, number of people = 24 students.
New average age = 15 - 1 = 14 years.
Sum of ages of 24 students = 24 * 14 = 336 years.
Teacher's age = (Sum of ages of 25 people) - (Sum of ages of 24 students) = 375 - 336 = 39 years.

Question 14

A batsman has an average of 87 runs in 21 innings. How many runs must he score in his next innings to increase his average by 2 runs?

Accepted Answer

Total runs in 21 innings = 21 * 87 = 1827.
To increase the average by 2 runs, the new average should be 87 + 2 = 89.
After the next (22nd) innings, the total number of innings will be 22.
Required total runs in 22 innings = 22 * 89 = 1958.
Runs to be scored in the next innings = (Required total runs in 22 innings) - (Total runs in 21 innings) = 1958 - 1827 = 131.

Question 15

The average of 7 consecutive numbers is 20. The largest of these numbers is:

Accepted Answer

For a series of consecutive numbers (or numbers in an arithmetic progression), the average is the middle number.
Since there are 7 consecutive numbers, the 4th number will be the average.
So, the 4th number = 20.
The 7 consecutive numbers are: 17, 18, 19, 20, 21, 22, 23.
The largest number is 23.

Question 16

The average weight of 15 students in a class is 40 kg. The average weight of 10 students in another class is 35 kg. What is the average weight of all 25 students?

Accepted Answer

Total weight of 15 students = 15 * 40 kg = 600 kg.
Total weight of 10 students = 10 * 35 kg = 350 kg.
Total number of students = 15 + 10 = 25.
Total weight of all 25 students = 600 + 350 = 950 kg.
Average weight of all 25 students = Total weight / Total number of students = 950 / 25 = 38 kg.

Question 17

The average of 100 observations was calculated as 40. It was later found that one observation was misread as 53 instead of 83. The correct average is:

Accepted Answer

Incorrect sum of 100 observations = 100 * 40 = 4000.
The error was reading 53 instead of 83. The difference in value = Correct value - Incorrect value = 83 - 53 = 30.
Since the correct value is higher, the sum should be increased by 30.
Correct sum = Incorrect sum + Difference = 4000 + 30 = 4030.
Number of observations remains 100.
Correct average = Correct sum / Number of observations = 4030 / 100 = 40.3.

Question 18

What is the average of the first 5 multiples of 7?

Accepted Answer

The first 5 multiples of 7 are: 7, 14, 21, 28, 35.
These numbers form an arithmetic progression. For an odd number of terms in an AP, the average is the middle term.
The middle term is the 3rd term, which is 21.
Alternatively, Sum of the numbers = 7 + 14 + 21 + 28 + 35 = 105.
Number of terms = 5.
Average = Sum / Number of terms = 105 / 5 = 21.

Question 19

The average of 11 results is 50. If the average of the first six results is 49 and that of the last six results is 52, the sixth result is:

Accepted Answer

Total sum of 11 results = 11 * 50 = 550.
Sum of the first six results = 6 * 49 = 294.
Sum of the last six results = 6 * 52 = 312.
When we sum the first six and the last six results, the 6th result is included twice.
So, the 6th result = (Sum of first six results) + (Sum of last six results) - (Total sum of 11 results).
6th result = 294 + 312 - 550 = 606 - 550 = 56.

Question 20

A man travels from A to B at 30 km/hr and returns from B to A at 20 km/hr. What is his average speed for the whole journey?

Accepted Answer

Let the distance between A and B be 'd' km.
Time taken to travel from A to B = d/30 hours.
Time taken to travel from B to A = d/20 hours.
Total distance covered = d + d = 2d km.
Total time taken = (d/30) + (d/20) = (2d + 3d) / 60 = 5d / 60 = d / 12 hours.
Average speed = Total Distance / Total Time = 2d / (d/12) = 2d * (12/d) = 24 km/hr.
Alternatively, for two equal distances covered at speeds x and y, the average speed is given by the formula: 2xy / (x+y).
Here, x = 30 km/hr and y = 20 km/hr.
Average speed = (2 * 30 * 20) / (30 + 20) = 1200 / 50 = 24 km/hr.

Question 21

The average of 8 numbers is 20. The average of the first two numbers is 15.5 and the average of the next three numbers is 23. The 6th number is 4 less than the 7th and the 8th number is 7 more than the 7th number. Find the 8th number.

Accepted Answer

Sum of 8 numbers = 8 * 20 = 160.
Sum of the first two numbers = 2 * 15.5 = 31.
Sum of the next three numbers (3rd, 4th, 5th) = 3 * 23 = 69.
Sum of the first five numbers = Sum of (1st and 2nd) + Sum of (3rd, 4th, 5th) = 31 + 69 = 100.
Sum of the remaining three numbers (6th, 7th, 8th) = Sum of 8 numbers - Sum of first five numbers = 160 - 100 = 60.
Let the 7th number be x.
According to the problem, the 6th number is 4 less than the 7th, so 6th number = x - 4.
And the 8th number is 7 more than the 7th, so 8th number = x + 7.
Sum of 6th, 7th, 8th numbers = (x - 4) + x + (x + 7) = 3x + 3.
We know this sum is 60, so:
3x + 3 = 60
3x = 57
x = 19.
The 7th number is 19.
The 8th number = x + 7 = 19 + 7 = 26.

Question 22

The average age of 30 students and their teacher is 15 years. If the teacher's age is excluded, the average age of the students becomes 14 years. What is the teacher's age?

Accepted Answer

Total number of people (students + teacher) = 30 + 1 = 31.
Sum of ages of 31 people = 31 * 15 = 465 years.
Sum of ages of 30 students = 30 * 14 = 420 years.
Teacher's age = (Sum of ages of 31 people) - (Sum of ages of 30 students) = 465 - 420 = 45 years.

Question 23

The average score of a cricketer in 10 innings is 60 runs. How many runs must he score in his next innings so as to increase his average score by 2 runs?

Accepted Answer

Total runs in 10 innings = 10 * 60 = 600 runs.
New average score desired = 60 + 2 = 62 runs.
Number of innings will be 10 + 1 = 11.
Total runs needed in 11 innings = 11 * 62 = 682 runs.
Runs to be scored in the next innings = (Total runs in 11 innings) - (Total runs in 10 innings) = 682 - 600 = 82 runs.

Question 24

The average of 11 results is 50. If the average of the first six results is 49 and that of the last six results is 52, what is the sixth result?

Accepted Answer

Sum of 11 results = 11 * 50 = 550.
Sum of first six results = 6 * 49 = 294.
Sum of last six results = 6 * 52 = 312.
Sum of first six + Sum of last six = 294 + 312 = 606.
Since the 6th result is included in both sums, it is counted twice.
Sixth result = (Sum of first six + Sum of last six) - (Sum of 11 results) = 606 - 550 = 56.

Question 25

The average marks obtained by 120 students in an examination were 35. If the average marks of passed students were 39 and that of failed students were 15, what is the number of students who passed the examination?

Accepted Answer

Let 'P' be the number of passed students and 'F' be the number of failed students.
Total students = P + F = 120.
Total marks = 120 * 35 = 4200.
Total marks = (P * 39) + (F * 15).
So, 39P + 15F = 4200.
We know F = 120 - P.
Substitute F: 39P + 15(120 - P) = 4200.
39P + 1800 - 15P = 4200.
24P = 4200 - 1800.
24P = 2400.
P = 2400 / 24 = 100.
Number of students who passed = 100.

Question 26

A student travels from his home to school at 30 km/hr and returns from school to home at 20 km/hr. What is his average speed for the entire journey?

Accepted Answer

For average speed when distance is constant:
Average Speed = (2 * S1 * S2) / (S1 + S2)
Where S1 = speed from home to school = 30 km/hr
S2 = speed from school to home = 20 km/hr
Average Speed = (2 * 30 * 20) / (30 + 20) = (1200) / (50) = 24 km/hr.

Question 27

The average of 5 numbers is 40. If two numbers, 30 and 35, are removed, what is the average of the remaining numbers?

Accepted Answer

Sum of 5 numbers = 5 * 40 = 200.
Sum of the two removed numbers = 30 + 35 = 65.
Sum of the remaining 3 numbers = 200 - 65 = 135.
Average of the remaining 3 numbers = 135 / 3 = 45.

Question 28

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. What is the monthly income of P?

Accepted Answer

P + Q = 2 * 5050 = 10100  (Equation 1)
Q + R = 2 * 6250 = 12500  (Equation 2)
P + R = 2 * 5200 = 10400  (Equation 3)
Add all three equations: (P + Q) + (Q + R) + (P + R) = 10100 + 12500 + 10400
2(P + Q + R) = 33000
P + Q + R = 33000 / 2 = 16500 (Equation 4)
To find P, subtract Equation 2 from Equation 4:
P = (P + Q + R) - (Q + R) = 16500 - 12500 = 4000.
So, P's monthly income is Rs. 4000.

Question 29

The average of seven consecutive numbers is 20. Find the largest of these numbers.

Accepted Answer

For consecutive numbers, the average is the middle number.
Since there are 7 consecutive numbers, the 4th number (middle number) is 20.
The numbers are X, X+1, X+2, X+3, X+4, X+5, X+6.
The middle number is X+3, so X+3 = 20, which means X = 17.
The largest number is X+6 = 17+6 = 23.
Alternatively, if the average (middle number) is 20, then the 3 numbers after it are 21, 22, 23. So the largest is 23.

Question 30

The average marks of 40 students in a class is 65. The average marks of 20 boys is 70. What is the average marks of the girls in the class?

Accepted Answer

Total marks of 40 students = 40 * 65 = 2600.
Total marks of 20 boys = 20 * 70 = 1400.
Number of girls in the class = Total students - Number of boys = 40 - 20 = 20.
Total marks of girls = Total marks of students - Total marks of boys = 2600 - 1400 = 1200.
Average marks of girls = Total marks of girls / Number of girls = 1200 / 20 = 60.

Question 31

A person travels from point A to point B at a speed of 40 km/hr and returns from B to A at a speed of 60 km/hr. What is the average speed for the entire journey?

Accepted Answer

Let the distance between A and B be 'd' km.
Time taken from A to B (t1) = d/40 hours.
Time taken from B to A (t2) = d/60 hours.
Total distance = d + d = 2d km.
Total time = t1 + t2 = d/40 + d/60 = (3d + 2d)/120 = 5d/120 = d/24 hours.
Average speed = Total distance / Total time = 2d / (d/24) = 2d * (24/d) = 48 km/hr.
Alternatively, for two equal distances, the average speed formula is 2xy / (x+y) = (2 * 40 * 60) / (40 + 60) = (4800) / 100 = 48 km/hr.

Question 32

The average of 100 observations was calculated as 40. It was later found that one observation was misread as 83 instead of 53. What is the correct average?

Accepted Answer

Incorrect sum of 100 observations = 100 * 40 = 4000.
The incorrect observation was 83, and the correct one should have been 53.
The difference due to error = Incorrect value - Correct value = 83 - 53 = 30.
Since 83 was added instead of 53, the sum was overestimated by 30.
Correct sum = Incorrect sum - Error difference = 4000 - 30 = 3970.
Correct average = Correct sum / Number of observations = 3970 / 100 = 39.7.

Question 33

The average age of 8 persons in a committee is increased by 2 years when two men, aged 35 years and 45 years, are substituted by two women. What is the average age of these two women?

Accepted Answer

Total age of the two men who left = 35 + 45 = 80 years.
Increase in total age of the committee = Number of persons * Increase in average age = 8 * 2 = 16 years.
The sum of ages of the two women must be the sum of ages of the two men plus the total increase in age.
Sum of ages of the two women = 80 + 16 = 96 years.
Average age of the two women = Sum of ages of women / Number of women = 96 / 2 = 48 years.

Question 34

A cricketer has an average of 60 runs in 10 innings. How many runs must he score in his next inning so as to increase his average by 2 runs?

Accepted Answer

Total runs in 10 innings = 10 * 60 = 600 runs.
After the next inning (11th inning), the new average should be 60 + 2 = 62 runs.
Total runs needed in 11 innings to achieve the new average = 11 * 62 = 682 runs.
Runs to be scored in the 11th inning = (Total runs in 11 innings) - (Total runs in 10 innings) = 682 - 600 = 82 runs.

Question 35

The average temperature of Monday, Tuesday, and Wednesday was 30°C. The average temperature of Tuesday, Wednesday, and Thursday was 33°C. If the temperature on Thursday was 35°C, what was the temperature on Monday?

Accepted Answer

Let M, Tu, W, Th be the temperatures on Monday, Tuesday, Wednesday, and Thursday respectively.
From the first statement: (M + Tu + W) / 3 = 30 => M + Tu + W = 3 * 30 = 90°C (Equation 1).
From the second statement: (Tu + W + Th) / 3 = 33 => Tu + W + Th = 3 * 33 = 99°C (Equation 2).
We are given Th = 35°C.
Substitute Th in Equation 2: Tu + W + 35 = 99 => Tu + W = 99 - 35 = 64°C.
Now substitute (Tu + W) in Equation 1: M + 64 = 90.
M = 90 - 64 = 26°C.

Question 36

The average of five numbers is 18. If one number is excluded, their average is 16. The excluded number is:

Accepted Answer

Sum of five numbers = 5 * 18 = 90.
When one number is excluded, there are 4 numbers left, and their average is 16.
Sum of four numbers = 4 * 16 = 64.
The excluded number = (Sum of five numbers) - (Sum of four numbers) = 90 - 64 = 26.

Question 37

The average age of 30 students and a teacher is 16 years. If the teacher's age is excluded, the average age of the students becomes 15 years. What is the teacher's age?

Accepted Answer

Total age of 30 students and 1 teacher = 31 * 16 = 496 years.
Total age of 30 students = 30 * 15 = 450 years.
Teacher's age = (Total age of students + teacher) - (Total age of students)
Teacher's age = 496 - 450 = 46 years.

Question 38

The average of 5 consecutive odd numbers is 61. What is the difference between the highest and lowest numbers?

Accepted Answer

Let the 5 consecutive odd numbers be x-4, x-2, x, x+2, x+4.
Their average is x. So, x = 61.
The numbers are 57, 59, 61, 63, 65.
The highest number is 65 and the lowest number is 57.
Difference = 65 - 57 = 8.
Alternatively, for 5 consecutive odd numbers, the difference between highest and lowest is (n-1)*2 = (5-1)*2 = 4*2 = 8.

Question 39

The average weight of 15 students is 40 kg. If the weight of the teacher is included, the average weight increases by 2 kg. What is the weight of the teacher?

Accepted Answer

Total weight of 15 students = 15 * 40 = 600 kg.
When the teacher's weight is included, there are 16 people, and the average becomes 40 + 2 = 42 kg.
Total weight of 15 students + teacher = 16 * 42 = 672 kg.
Weight of the teacher = (Total weight of students + teacher) - (Total weight of students)
Weight of the teacher = 672 - 600 = 72 kg.

Question 40

The average of 11 results is 50. If the average of the first six results is 49 and that of the last six results is 52, what is the sixth result?

Accepted Answer

Sum of 11 results = 11 * 50 = 550.
Sum of first six results = 6 * 49 = 294.
Sum of last six results = 6 * 52 = 312.
Sum of (first six + last six) results = 294 + 312 = 606.
Since the sixth result is counted twice, the sixth result = (Sum of first six + Sum of last six) - (Sum of 11 results)
Sixth result = 606 - 550 = 56.

Question 41

A batsman has a certain average of 9 innings. In the 10th inning, he scores 100 runs, thereby increasing his average by 8 runs. What is his average after the 10th inning?

Accepted Answer

Let the average after 9 innings be 'x' runs.
Total runs in 9 innings = 9x.
In the 10th inning, he scores 100 runs.
Total runs in 10 innings = 9x + 100.
New average after 10 innings = (9x + 100) / 10.
According to the problem, the new average is x + 8.
So, (9x + 100) / 10 = x + 8.
9x + 100 = 10x + 80.
100 - 80 = 10x - 9x.
x = 20.
His average after the 9th inning was 20. His average after the 10th inning is x + 8 = 20 + 8 = 28 runs.

Question 42

The average monthly income of P and Q is ₹5050. The average monthly income of Q and R is ₹6250. The average monthly income of P and R is ₹5200. What is the monthly income of P?

Accepted Answer

P + Q = 2 * 5050 = 10100 (Eq 1)
Q + R = 2 * 6250 = 12500 (Eq 2)
P + R = 2 * 5200 = 10400 (Eq 3)
Adding all three equations:
2(P + Q + R) = 10100 + 12500 + 10400 = 33000
P + Q + R = 33000 / 2 = 16500 (Eq 4)
To find P, subtract (Eq 2) from (Eq 4):
P = (P + Q + R) - (Q + R) = 16500 - 12500 = ₹4000.

Question 43

The average of 7 numbers is 30. If the average of the first three numbers is 28 and that of the next two numbers is 32, what is the average of the remaining two numbers?

Accepted Answer

Sum of 7 numbers = 7 * 30 = 210.
Sum of first three numbers = 3 * 28 = 84.
Sum of next two numbers = 2 * 32 = 64.
Sum of first five numbers = 84 + 64 = 148.
Sum of remaining two numbers = Total sum - Sum of first five numbers = 210 - 148 = 62.
Average of remaining two numbers = 62 / 2 = 31.

Question 44

A student's average marks in 5 subjects is 75. If his marks in one subject were wrongly entered as 50 instead of 80, what is the correct average?

Accepted Answer

Total marks based on incorrect average = 5 * 75 = 375.
The correct marks for one subject should be 80 instead of 50. This is a difference of +30 (80-50).
Correct total marks = 375 + 30 = 405.
Correct average = 405 / 5 = 81.

Question 45

The average age of 24 boys and their teacher is 15 years. When the teacher's age is excluded, the average age of the boys decreases by 1 year. What is the teacher's age?

Accepted Answer

Total age of 24 boys and 1 teacher = 25 * 15 = 375 years.
When the teacher's age is excluded, the average age of 24 boys becomes 15 - 1 = 14 years.
Total age of 24 boys = 24 * 14 = 336 years.
Teacher's age = (Total age of boys + teacher) - (Total age of boys)
Teacher's age = 375 - 336 = 39 years.

Question 46

A car travels from point A to point B at a speed of 40 km/hr and returns from point B to point A at a speed of 60 km/hr. What is the average speed for the entire journey?

Accepted Answer

Let the distance between A and B be 'd' km.
Time taken to travel from A to B = d/40 hours.
Time taken to travel from B to A = d/60 hours.
Total distance = d + d = 2d km.
Total time = (d/40) + (d/60) = (3d + 2d)/120 = 5d/120 = d/24 hours.
Average speed = Total Distance / Total Time = (2d) / (d/24) = 2d * (24/d) = 48 km/hr.
Alternatively, for two equal distances traveled at different speeds x and y, the average speed is (2xy) / (x+y).
Average speed = (2 * 40 * 60) / (40 + 60) = (4800) / 100 = 48 km/hr.

Question 47

The average of 5 numbers is 30. If one number is excluded, the average of the remaining numbers becomes 28. What is the excluded number?

Accepted Answer

Sum of 5 numbers = 5 * 30 = 150.
Sum of 4 numbers (after exclusion) = 4 * 28 = 112.
Excluded number = (Sum of 5 numbers) - (Sum of 4 numbers) = 150 - 112 = 38.

Question 48

The average of five consecutive odd numbers is 27. What is the product of the smallest and the largest of these numbers?

Accepted Answer

For five consecutive odd numbers, the average is the middle number. So, the third number is 27.
The five consecutive odd numbers are: 27-4, 27-2, 27, 27+2, 27+4, which are 23, 25, 27, 29, 31.
Smallest number = 23.
Largest number = 31.
Product of smallest and largest = 23 * 31 = 713.

Question 49

The average age of 15 students in a class is 18 years. If the age of the teacher is included, the average age increases by 1 year. What is the age of the teacher?

Accepted Answer

Total age of 15 students = 15 * 18 = 270 years.
When the teacher's age is included, the total number of people becomes 15 + 1 = 16.
The new average age is 18 + 1 = 19 years.
Total age of 16 people (students + teacher) = 16 * 19 = 304 years.
Age of the teacher = (Total age of 16 people) - (Total age of 15 students) = 304 - 270 = 34 years.

Question 50

A batsman has an average of 87 runs in 21 innings. How many runs must he score in the next inning so as to increase his average by 3 runs?

Accepted Answer

Total runs in 21 innings = 21 * 87 = 1827.
New average desired = 87 + 3 = 90 runs.
Total innings after the next inning = 21 + 1 = 22 innings.
Total runs required in 22 innings to achieve the new average = 22 * 90 = 1980.
Runs to be scored in the next inning = (Total runs in 22 innings) - (Total runs in 21 innings) = 1980 - 1827 = 153 runs.

Question 51

The average weight of 20 boys in a class is 40 kg and the average weight of 30 girls in the same class is 30 kg. What is the average weight of the entire class?

Accepted Answer

Total weight of 20 boys = 20 * 40 kg = 800 kg.
Total weight of 30 girls = 30 * 30 kg = 900 kg.
Total weight of the entire class = 800 + 900 = 1700 kg.
Total number of students in the class = 20 + 30 = 50.
Average weight of the entire class = (Total weight) / (Total number of students) = 1700 / 50 = 34 kg.

Question 52

The average of 10 numbers was calculated as 40. Later it was found that one number, 36, was wrongly read as 63. What is the correct average?

Accepted Answer

Incorrect sum of 10 numbers = 10 * 40 = 400.
The number 36 was wrongly read as 63. The error in reading is 63 - 36 = 27 (overestimated).
To find the correct sum, subtract the overestimation from the incorrect sum: Correct sum = 400 - 27 = 373.
Correct average = (Correct sum) / (Number of numbers) = 373 / 10 = 37.3.

Question 53

The average price of 10 pens and 5 pencils is Rs. 12. If the average price of 5 pencils is Rs. 8, what is the average price of the 10 pens?

Accepted Answer

Total number of items (pens + pencils) = 10 + 5 = 15.
Total price of 15 items = 15 * Rs. 12 = Rs. 180.
Total price of 5 pencils = 5 * Rs. 8 = Rs. 40.
Total price of 10 pens = (Total price of 15 items) - (Total price of 5 pencils) = 180 - 40 = Rs. 140.
Average price of 10 pens = (Total price of 10 pens) / 10 = 140 / 10 = Rs. 14.

Question 54

The average temperature of Monday, Tuesday, and Wednesday was 35°C. The average temperature of Tuesday, Wednesday, and Thursday was 38°C. If the temperature on Thursday was 39°C, what was the temperature on Monday?

Accepted Answer

Sum of temperatures (Mon + Tue + Wed) = 3 * 35°C = 105°C  (Equation 1)
Sum of temperatures (Tue + Wed + Thu) = 3 * 38°C = 114°C  (Equation 2)
We are given that Temperature on Thursday (Thu) = 39°C.
Substitute Thu = 39°C into Equation 2:
Tue + Wed + 39 = 114
Tue + Wed = 114 - 39 = 75°C
Now substitute (Tue + Wed) = 75°C into Equation 1:
Mon + 75 = 105
Mon = 105 - 75 = 30°C.

Question 55

The average of A, B, and C is 45. The average of A and B is 40. The average of B and C is 43. What is the value of B?

Accepted Answer

Given:
1. (A + B + C) / 3 = 45 => A + B + C = 135 (Equation 1)
2. (A + B) / 2 = 40 => A + B = 80 (Equation 2)
3. (B + C) / 2 = 43 => B + C = 86 (Equation 3)

From Equation 1 and Equation 2:
C = (A + B + C) - (A + B) = 135 - 80 = 55.

From Equation 1 and Equation 3:
A = (A + B + C) - (B + C) = 135 - 86 = 49.

Now, substitute the value of A or C into the relevant equations:
Using A = 49 in Equation 2: 49 + B = 80 => B = 80 - 49 = 31.
(Alternatively, using C = 55 in Equation 3: B + 55 = 86 => B = 86 - 55 = 31.)
Thus, the value of B is 31.