The average age of a teacher and three students is 20 years. If all the three students are of same age and the difference between the age of the…

Correct Option (3)

Let 'x' denote the age of each student. As there are three students of the same age, their combined age is 3x.

The problem states that the difference between the teacher's age and each student's age is 20 years. Therefore, the teacher's age is (x + 20) years.

The group consists of one teacher and three students, totaling four individuals. The average age of this group is given as 20 years. The formula for average age is:

Average Age = (Sum of Ages) / (Number of Individuals)

Substituting the known values into the formula:

$$20 = \frac{(x + 20) + 3x}{4}$$20=4(x+20)+3x​

Simplifying the numerator:

$$20 = \frac{4x + 20}{4}$$20=44x+20​

Multiplying both sides by 4:

80 = 4x + 20

Subtracting 20 from both sides:

4x = 60

Dividing by 4:

x = 15

This value 'x' represents the age of each student, which is 15 years.

To find the teacher's age, substitute x = 15 into the expression for the teacher's age (x + 20):

Teacher's age = 15 + 20 = 35 years.

Thus, the age of the teacher is 35 years.

Incorrect Options:

Options 1, 2, and 4 are inconsistent with the problem's conditions, specifically the given average age of 20 years, when tested against the derived student age and the specified age difference.

• If the teacher's age were 25 years (Option 1), the student's age would be 5 years (25 - 20). The average age of the group would then be $\frac{25 + (3 \times 5)}{4} = \frac{25 + 15}{4} = \frac{40}{4} = 10$425+(3×5)​=425+15​=440​=10 years, which contradicts the stated average of 20 years.
• If the teacher's age were 30 years (Option 2), the student's age would be 10 years (30 - 20). The average age of the group would be $\frac{30 + (3 \times 10)}{4} = \frac{30 + 30}{4} = \frac{60}{4} = 15$430+(3×10)​=430+30​=460​=15 years, which contradicts the stated average of 20 years.
• If the teacher's age were 45 years (Option 4), the student's age would be 25 years (45 - 20). The average age of the group would be $\frac{45 + (3 \times 25)}{4} = \frac{45 + 75}{4} = \frac{120}{4} = 30$445+(3×25)​=445+75​=4120​=30 years, which contradicts the stated average of 20 years.

Only the teacher's age of 35 years satisfies all the conditions provided in the question.