Search any question & find its solution
Question:
Answered & Verified by Expert
If the median and the range of four numbers $\{x, y, 2 x+y, x-y\}$, where $0 < y < x < 2 y$, are 10 and 28 respectively, then the mean of the numbers is :
Options:
Solution:
2218 Upvotes
Verified Answer
The correct answer is:
14
14
Since $0 < y < x < 2 y$
$$
\begin{aligned}
& \therefore \quad y>\frac{x}{2} \Rightarrow x-y < \frac{x}{2} \\
& \therefore x-y < y < x < 2 x+y
\end{aligned}
$$
Hence median $=\frac{y+x}{2}=10$
$\Rightarrow x+y=20$
And range $=(2 x+y)-(x-y)=x+2 y$
But range $=28$
$$
\therefore x+2 y=28
$$
From equations (i) and (ii),
$$
\begin{aligned}
& x=12, y=8 \\
& \therefore \text { Mean } \\
& =\frac{(x-y)+y+x+(2 x+y)}{4}=\frac{4 x+y}{4}
\end{aligned}
$$
$$
=x+\frac{y}{4}=12+\frac{8}{4}=14
$$
$$
\begin{aligned}
& \therefore \quad y>\frac{x}{2} \Rightarrow x-y < \frac{x}{2} \\
& \therefore x-y < y < x < 2 x+y
\end{aligned}
$$
Hence median $=\frac{y+x}{2}=10$
$\Rightarrow x+y=20$
And range $=(2 x+y)-(x-y)=x+2 y$
But range $=28$
$$
\therefore x+2 y=28
$$
From equations (i) and (ii),
$$
\begin{aligned}
& x=12, y=8 \\
& \therefore \text { Mean } \\
& =\frac{(x-y)+y+x+(2 x+y)}{4}=\frac{4 x+y}{4}
\end{aligned}
$$
$$
=x+\frac{y}{4}=12+\frac{8}{4}=14
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.