There are three positive numbers, one-third of average of all the three numbers is 8 less than the value of the highest number. Average of the lowest and the second lowest number is 8. Which is the highest number? |
A) 11
B) 14
C) 10
D) 9
E) 13
Correct Answer: A
Solution :
Let the three positive numbers are x, y and z in increasing order, and average of three numbers be A. |
According to the question, |
\[\Rightarrow \] \[\frac{x+y+z}{3}=A\] (i) |
Given, \[z-\frac{A}{3}=8\] |
\[z-\frac{x+y+z}{9}=8\] ... (ii) |
Also given that, \[\frac{x+y}{2}=8\] |
\[x+y=16\] ... (iii) |
Put the value of \[(x+y)\] in Eq. (ii), we get |
\[z-\frac{16+z}{9}=8\] |
\[\Rightarrow \] \[\frac{9z-16-z}{9}=8\] |
\[\Rightarrow \] \[8z=72+16=88\] |
\[\Rightarrow \] \[z=\frac{88}{8}=11\] |
So, highest number is 11. |
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