# The mean of three numbers is 40. All the three numbers are different natural numbers. If lowest is 19, what could be highest possible number of remaining two numbers?

(a) 81

(b) 40

(c) 100

(d) 71

**Solution:**

Given, the mean of three numbers is 40.

All the three numbers are different __natural numbers__.

Lowest number is 19.

We have to determine the highest possible number of the remaining two numbers.

We know, __mean__ = sum of observations/Number of observation

Let the remaining two numbers be x and y.

Sum of observations = 19 + x + y

Number of observation = 3

40 = 19+x+y/3

19+x+y = 40(3)

19+x+y = 120

x + y = 120 - 19

x + y = 101

Let the highest number be y.

According to the question,

So, y = 101 - x

Since the lowest number is 19, x cannot be less than 19.

i.e., x > 19

Let x = 20

Now, y = 101 - 20

y = 81

Therefore, the highest number is 81.

**✦ Try This: **The mean of three numbers is 30. All the three numbers are different natural numbers. If lowest is 9, what could be highest possible number of remaining two numbers

**☛ Also Check: **NCERT Solutions for Class 7 Maths Chapter 3

**NCERT Exemplar Class 7 Maths Chapter 3 Problem 13**

## The mean of three numbers is 40. All the three numbers are different natural numbers. If lowest is 19, what could be highest possible number of remaining two numbers? (a) 81 (b) 40 (c) 100 (d) 71

**Summary:**

The mean of three numbers is 40. All the three numbers are different natural numbers. If lowest is 19, the highest possible number of remaining two numbers could be 81.

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