# Solved Example

**The average of 10 consecutive numbers starting from 21 is:**

** **

**If a male with age 45 joins a group of 5 male with an average age of 39 years. What will be the new average age of the group?**

Total age will be 45 + 5 Ã— 39 = 240. And there will be 6 persons now.

So the average will be 240/6 = 40.

(or)

Since 45 is 6 more than 39, by joining the new person, the total will increase by 6 and so the average will increase by 1.

So, the average is 39 + 1 = 40

**Two students with marks 50 and 54 leave class VIII A and move to class VIII B. As a result the average marks of the class VIII A fall from 48 to 46. How many students were there initially in the class VIII A?**

** **

**The average of x successive natural numbers is N. If the next natural number is included in the group, the average increases by:**

- Depends on x
- Depends on the starting number of the series
- Both (1) and (2)
- Â½
- None

The average of consecutive numbers is the middle number. If one more number is added to the list, the middle number moves 0.5 towards right. So the answer is (d).

**The average marks of 30 students in a section of class X are 20 while that of 20 students of second section is 30. Find the average marks for the entire class X?**

Simple average = Sum of marks of all students / Total number of students

= (20 Ã— 30 + 30 Ã— 20) / (30 + 20) = 24

By the weighted mean method,

Average = (3/5) x 20 + (2/5) x 30 = 24.

Some Things to Remember:

**There are two classes A and B., each has 20 students. The average weight of class A is 38 and that of class B is 40. X and Y are two students of classes A and B respectively. If they interchange their classes, then the average weight of both the classes will be equal. If weight of x is 30 kg, what is the weight of Y?**

**Method 1:**

Total weight of class A = 38 Ã— 20, and class B = 40 Ã— 20, if X & Y are interchanged, then

the total weight of both the classes are equal.

So 38 Ã— 20 - x + y = 40 Ã— 20 - y + x

2(y - x) = 2 x 20

y = x + 20 = 50

(OR)

**Method 2:**

Since both the classes have same number of students, after interchange, the average weight of each class will be 39. Since the average of class â€˜Aâ€™ is increasing by 1, the total should increase by 20. So, x must be replaced by â€˜yâ€™, who must be 20 Kg heavier than â€˜xâ€™. So, y must be 50 Kg.

**The average weight of 10 apples is 0.4 kg. If the heaviest and lightest apples are taken out, the average is 0.41 kg. If the lightest apple weights 0.2 kg, what is the weight of heaviest apple?**

Weight of apples except heaviest & lightest = 0.41 Ã— 8 = 3.28 kg

Weight of Rest apples = Total weight â€“ without the weight of Heaviest and Lightest

Heaviest + lightest = 4 - 3.28 = 0.72 kg.

It is given lightest = 0.2 kg.

Heaviest is 0.72 - 0.2 = 0.52 kg.

**While finding the average of â€˜9â€™ consecutive numbers starting from X; a student interchanged the digits of second number by mistake and got the average which is 8 more than the actual. What is X?**

Since the average is 8 more than the actual, the second no will increase by 72 (9 Ã— 8) by interchanging the digits.

If ab is the second no,

then 10a + b + 72 = 10b + a

i.e. 9 (b - a) = 72

b - a = 8.

The possible number ab is 19. Since the second no is 19.

The first no is 18

So X = 18

**There are 30 consecutive numbers. What is the difference between the averages of first and last 10 numbers?**

**What is the highest 3 digit number, which is exactly divisible by 3, 5, 6 and 7?**

The least no. which is exactly divisible by 3, 5, 6, & 7 is LCM (3, 5, 6, 7) = 210. So, all the multiples of 210 will be exactly divisible by 3, 5, 6 and 7. So, such greatest 3 digit number is 840. (210 Ã— 4).

**In a farewell party, some students are giving pose for photograph, If the students stand at 4 students per row, 2 students will be left if they stand 5 per row, 3 will be left and if they stand 6 per row 4 will be left. If the total number of students are greater than 100 and less than 150, how many students are there?**

**There are some students in the class. Mr. X brought 130 chocolates and distributed to the students equally, then he was left with some chocolates. Mr. Y brought 170 chocolates and distributed equally to the students. He was also left with the same no of chocolates as Mr. X was left. Mr. Z brought 250 chocolates, did the same thing and left with the same no of chocolates. What is the max possible no of students that were in the class?**

The question can be stated as, what is the highest number, which divides 130, 170 and 250 gives the same remainder, i. e. HCF ((170 - 130), (250 - 170), (250 - 130)). i.e. HCF (40, 80, 120) = 40.

**The present age of a father is 3 years more than three times the age of his son. Three years hence, fatherâ€™s age will be 10 years more than twice the age of the son. Find the present age of the father.**

Then father's present age is 3x + 3 years.

Three years hence â†’ (3x + 3) + 3 = 2(x + 3) + 10; x = 10; Hence father's present age = 3x + 3 = 33 years.

**The total age of A & B is 12 years more than the total age of B & C. C is how many year younger than A.**

From the given data

A + B = 12 + (B + C)

A + B - (B + C) = 12 â†’ A - C = 12 years

C is 12 years younger than A

**Gaurav's age is 1/6th of her father's age. Gaurav's father's age will be twice the age of Vimal's age after 10 years. If Vimalâ€™s eight birthday was celebrated two years before, then what is Gaurav's present age.**

Gauravâ€™s age is 1/6 of her fathers age i.e. x = y/6.

Fatherâ€™s age will be twice of Vimal's age after 10 years.

i.e. y + 10 = 2(V + 10) (where 'V' is the Vimal's age)

Vimal's eight birthday was celebrated two years before, Then the Vimal's present age is 10 years.

y + 10 = 2(10 + 10) thus, y = 30 years.

Gaurav's present age x = y/6 thus, x = 30/6 = 5 years. Thus Gaurav's present age is 5 years.