AVERAGES
Note: Question numbers are numbers from the actual exam in the respective years mentioned below
- The monthly average salary paid to all the employees of a company was Rs. 5000. The monthly average salary paid to male and female employees was Rs. 5200 and Rs. 4200 respectively. Then the percentage of males employed in the company is (2015)
(A) 75%
(B) 80%
(C) 85%
(D) 90%
Answer-B
Let the total number of employees be x, and let the number of male employees be y. Then, the number of female employees is (x - y).
Total salary paid to all employees = 5000x
Total salary paid to male employees = 5200y
Total salary paid to female employees = 4200(x - y)
We know that the monthly average salary paid to all the employees of the company is Rs. 5000. Therefore,
5000x = (5200y + 4200(x - y))
Simplifying this equation, we get:
800x = 1000y
y = (4/5)x
Therefore, the percentage of male employees is (y/x) * 100, which is:
(4/5) * 100 = 80%
Hence, the answer is option (b) 80%.
- The average monthly income of a person in a certain family of 5 is Rs. 10,000. What will be the average monthly income of a person in the same family if the income of one person increased by Rs. 1,20,000 per year?(2016)
(A) Rs. 12,000
(B) Rs. 16,000
(C) Rs. 20,000
(D) Rs. 34,000
Answer-A
The average monthly income of a person in a family of 5 is Rs. 10,000. This means that the total monthly income of the family is Rs. 10,000 multiplied by 5, which is Rs. 50,000.
If the income of one person in the family increases by Rs. 1,20,000 per year, it means that their monthly income increases by Rs. 1,20,000 divided by 12, which is Rs. 10,000.
So, the new monthly income of that person is Rs. 10,000 + Rs. 10,000 = Rs. 20,000.
Now, the total monthly income of the family becomes Rs. 50,000 - Rs. 10,000 + Rs. 20,000 = Rs. 60,000.
Since there are still 5 members in the family, the average monthly income of a person in the family becomes Rs. 60,000 divided by 5, which is Rs. 12,000.
Therefore, the answer is (a) Rs. 12,000.
- Suppose the average weight of 9 persons is 50 kg. The average weight of the first 5 persons is 45 kg, whereas the average weight of the last 5 persons is 55 kg. Then the weight of the 5th person will be(2017)
(A) 45 kg
(B) 47.5 kg
(C) 50 kg
(D) 52.5 kg
Answer-C
The sum of the weights of the 9 persons is equal to the average weight multiplied by the number of persons, which is 50 kg x 9 = 450 kg.
The sum of the weights of the first 5 persons is equal to the average weight of the first 5 persons multiplied by the number of persons, which is 45 kg x 5 = 225 kg.
The sum of the weights of the last 5 persons is equal to the average weight of the last 5 persons multiplied by the number of persons, which is 55 kg x 5 = 275 kg.
Now, let's find the weight of the 5th person.
The sum of the weights of the 9 persons is equal to the sum of the weights of the first 5 persons plus the sum of the weights of the last 5 persons, minus the weight of the 5th person.
450 kg = 225 kg + 275 kg - weight of the 5th person
Simplifying the equation:
450 kg = 500 kg - weight of the 5th person
weight of the 5th person = 500 kg - 450 kg = 50 kg
Therefore, the weight of the 5th person is 50 kg.
The correct answer is (c) 50 kg.
- The average rainfall in a city for the first four days was recorded to be 0.40 inch. The rainfall on the last two days was in the ratio of 4:3. The average of six days was 0.50 inch. What was the rainfall on the fifth day?(2017)
(A) 0.60 inch
(B) 0.70 inch
(C) 0.80 inch
(D) 0.90 inch
Answer-C
Let the amount of the rainfall in first four days be a,b,c and d inches.
Average Rainfall in four days = 0.40 inches.
∴ (a + b + c + d) ÷ 4 = 0.40
⇒ (a + b + c + d) = 0.40 × 4
∴ a + b + c + d = 1.6 inches. --------(i)
Now, Let the rainfall in fifth and sixth (or last two) days be 4x and 3x respectively.
Also, According to the Question,
Average Rainfall of Six days = 0.50 inches.
∴ (a + b + c + d + 4x + 3x)/6 = 0.50
⇒ 1.6 + 7x = 0.50 × 6
⇒ 1.6 + 7x = 3
⇒ 3 - 1.6 = 7x
∴ 7x = 1.4
∴ x = 0.2 inches.
Rainfall in fifth days = 4x
= 4 × 0.2
= 0.8 inches.
Hence, the Rainfall in fifth day is 0.8 inches.
- In 2002, Meenu's age was one-third of the age of Meera, whereas in 2010, Meenu's age was half the age of Meera. What is Meenu's year of birth ?(2019)
(A) 1992
(B) 1994
(C) 1996
(D) 1998
Answer-B
Let's assume Meenu's age in 2002 was M years and Meera's age in 2002 was R years.
According to the given information, Meenu's age in 2002 was one-third of Meera's age, so we can write:
M = (1/3)R
Similarly, in 2010, Meenu's age was half the age of Meera, so we can write:
M + 8 = (1/2)(R + 8)
Simplifying the equations, we get:
M = (1/3)R
M + 8 = (1/2)(R + 8)
Let's solve these equations to find the values of M and R:
(1/3)R + 8 = (1/2)R + 4
(1/3)R - (1/2)R = 4 - 8
(2/6)R - (3/6)R = -4
(-1/6)R = -4
R = (-4) / (-1/6)
R = 24
Now that we know Meera's age in 2002 is 24, we can find Meenu's age in 2002:
M = (1/3)R
M = (1/3)(24)
M = 8
Meenu's age in 2002 was 8 years.
To find Meenu's year of birth, we subtract 8 from 2002:
Year of birth = 2002 - 8
Year of birth = 1994
Therefore, Meenu's year of birth is (b) 1994.
- The average marks of 100 students are given to be 40. It was found later that marks of one student were 53 which were misread as 83.The corrected mean marks are(2019)
(A) 39
(B) 39.7
(C) 40
(D) 40.3
Answer-B
To find the corrected mean marks, we need to adjust the incorrect value and recalculate the mean.
The average marks of 100 students is given as 40. This means the total sum of all the marks is 40 multiplied by 100, which is 4000.
However, one student's marks were misread as 83 instead of 53. To correct this, we need to subtract the incorrect value (83) and add the correct value (53) to the total sum.
The corrected total sum of all the marks is 4000 - 83 + 53 = 3970.
Now, we divide the corrected total sum by the number of students (100) to find the corrected mean marks:
Corrected mean marks = Corrected total sum / Number of students
Corrected mean marks = 3970 / 100
Corrected mean marks = 39.7
Therefore, the corrected mean marks are (b) 39.7.
- A family has two children along with their parents. The average of the weights of the children and their mother is 50 kg. The average of the weights of the children and their father is 52 kg. If the weight of the father is 60 kg, then what is the weight of the mother ?(2019)
(A) 48 kg
(B) 50 kg
(C) 52 kg
(D) 54 kg
Answer-D
Let's assume the weight of the mother is M kg.
According to the given information, the average of the weights of the children and their mother is 50 kg. This means the sum of the weights of the children and their mother is 50 multiplied by 3, which is 150 kg.
Similarly, the average of the weights of the children and their father is 52 kg. Since we know the weight of the father is 60 kg, the sum of the weights of the children and their father is 52 multiplied by 3 minus the weight of the father, which is 156 kg - 60 kg = 96 kg.
Now, let's set up an equation to find the weight of the mother:
Weight of the mother + Sum of the weights of the children = 150 kg
M + Sum of the weights of the children = 150
Since we know the sum of the weights of the children and their father is 96 kg, we can substitute this value into the equation:
M + 96 = 150
Now, we can solve the equation to find the weight of the mother:
M = 150 - 96
M = 54
Therefore, the weight of the mother is (d) 54 kg.
- Consider the following data:
|
|
Average marks in English |
Average marks in Hindi |
|
Girls |
9 |
8 |
|
Boys |
8 |
7 |
|
Overall average marks |
8.8 |
x |
What is the value of x in the above table?(2020)
(A) 7.8
(B) 7.6
(C) 7.4
(D) 7.2
Answer-A
To find the value of x, we need to calculate the overall average marks.
Let's assume the number of girls is G and the number of boys is B.
The total marks in English for girls is 9G, and the total marks in Hindi for girls is 8G.
The total marks in English for boys is 8B, and the total marks in Hindi for boys is 7B.
The overall average marks in English can be calculated as:
(9G + 8B) / (G + B) = 8.8
Similarly, the overall average marks in Hindi can be calculated as:
(8G + 7B) / (G + B) = x
To find the value of x, we need to solve these two equations.
From the first equation, we can write:
9G + 8B = 8.8(G + B)
9G + 8B = 8.8G + 8.8B
0.2G = 0.8B
G = 4B
Substituting G = 4B into the second equation, we get:
(8(4B) + 7B) / (4B + B) = x
(32B + 7B) / 5B = x
39B / 5B = x
7.8 = x
Therefore, the value of x is 7.8.
- In a class, there are three groups A, B and C. If one student from group A and two students from group B are shifted to group C, then what happens to the average weight of the students of the class?(2020)
(A) It increases.
(B) It decreases.
(C) It remains the same.
(D) No conclusion can be drawn due to insufficient data.
Answer-C
The average weight of the students of the class is calculated by dividing the total weight of the students by the number of students. When one student from group A and two students from group B are shifted to group C, the total weight of the students and the number of students remain the same. Therefore, the average weight of the students of the class remains the same.
- The average score of a batsman after his 50th innings was 46.4. After 60th innings, his average score increases by 2.6. What was his average score in the last ten innings?(2020)
(A) 122
(B) 91
(C) 62
(D) 49
Answer-C
Let's solve this problem step by step.
We are given that after the 50th innings, the average score of the batsman is 46.4. Therefore, the total score after 50 innings would be 50 * 46.4.
After the 60th innings, the average score increases by 2.6. So, the new average score is 46.4 + 2.6 = 49.
To find the total score after the 60th innings, we need to multiply the new average score by the number of innings played. So, the total score after 60 innings would be 60 * 49.
Now, let's find the total score in the last ten innings. We can subtract the total score after 50 innings from the total score after 60 innings to get the score in the last ten innings: (60 * 49) - (50 * 46.4).
Calculating this, we find:
(60 * 49) - (50 * 46.4) = 2940 - 2320 = 620.
Therefore, the average score in the last ten innings is 620 / 10 = 62.
So, the correct answer is (c) 62.
- There are two Classes A and B having 25 and 30 students respectively. In Class A the highest score is 21 and lowest score is 17. In Class-B the highest score is 30 and lowest score is 22. Four students are shifted from Class-A to Class-B.
Consider the following statements:
- The average score of Class-B will definitely decrease.
- The average score of Class-A will definitely increase.
Which of the above statements is/are correct?(2021)
(a) 1 only
(b) 2 only
(c) Both 1 and 2
(d) Neither 1 nor 2
Answer-A
In Class-A the highest score is 21 and lowest score is 17.
Hence class A average lies between 17 and 21
In Class-B the highest score is 30 and lowest score is 22.
Hence class B average lies between 22 and 30
Four students are shifted from Class A to Class B.
4 Students will have average between 17 and 21
Hence Average of Class B which is more than 21 will definitely decrease
Average of class A can remain same , decrease or increase depending upon the average of Class A and average of 4 students shifting to other class
- The average weight of A, B, C is 40 kg, the average weight of B, D, E is 42 kg and the weight of F is equal to that of B. What is the average weight of A, B, C, D, E and F?(2022)
(A) 40.5 kg
(B) 40.8 kg
(C) 41 kg
(D) Cannot be determined as data is inadequate
Answer-C
The average weight of A, B, C is 40 kg.
The average weight of B, D, E is 42 kg.
The weight of F is equal to that of B.
From the given information, we can find the weight of B. Since the average weight of A, B, C is 40 kg, and assuming each person's weight is counted equally in the average calculation, we have:
(A + B + C) / 3 = 40
A + B + C = 120
Similarly, for B, D, E, we have:
(B + D + E) / 3 = 42
B + D + E = 126
Since the weight of F is equal to that of B, we can replace F with B in the above equation:
F + D + E = 126
Now, let's add the equations A + B + C = 120 and F + D + E = 126:
(A + F) + (B + D + E) + C = 246
We know that (A + F) represents the total weight of A, B, C, D, E, and F. Therefore, the average weight of A, B, C, D, E, and F is (A + F + B + D + E + C) / 6.
Substituting the value of (A + F) + (B + D + E) + C from above, we get:
(A + F + B + D + E + C) / 6 = 246 / 6
(A + F + B + D + E + C) / 6 = 41
Therefore, the average weight of A, B, C, D, E, and F is 41 kg.
Hence, the correct answer is (c) 41 kg.
