There are three positive numbers. One third of the average of all the three numbers is 8 less than the value of the highest number. The average of the lowest and the second lowest number is 8 . What is the highest number?

Correct option is A

Given:

1. There are three positive numbers.
2. One third of the average of all three numbers is 8 less than the value of the highest number.
3. The average of the lowest and the second-lowest number is 8.
4. Options:
A) 11, B) 14, C) 10, D) 9.

Formula Used:

1. Average of n numbers:
​$$\text{Average} = \frac{\text{Sum of the numbers}}{n}$$​
2. From the problem:
​$$\frac{1}{3} \times \text{Average of all three numbers} = \text{Highest number} - 8$$​.

Solution:

1. Let the three numbers be a (lowest), b (second-lowest), and c (highest)
From the problem, we know:
​$$\frac{1}{3} \times \frac{a + b + c}{3} = c - 8$$​

Simplify:
​$$\frac{a + b + c}{9} = c - 8$$​
Multiply through by 9:
a + b + c = 9(c - 8).
Expand:
a + b + c = 9c - 72.
Rearrange:
a + b = 8c - 72. 

2. From the second condition, the average of the lowest and second-lowest numbers is 8:
​$$\frac{a + b}{2} = 8$$​​
Simplify:
a + b = 16 

3. Equating a + b from equations (1) and (2):
16 = 8c - 72.
Solve for c :
8c = 88 $$\implies$$​ c = 11

Final Answer:

The highest number is 11 .
**Option A: 11**