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The following sequence of numbers is arranged in increasing order: $1,x,x,x,y,y,9,16,18$. Given that the mean and median are equal, and are also equal to twice the mode, the value of $y$ is

1. $5$
2. $6$
3. $7$
4. $8$

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Migrated from GO Civil 6 months ago by Arjun

$1,x,x,x,y,y,9,16,18$

The mean

$$\bar{x}=\dfrac{\sum x}{N}$$

Here,

• $\sum$ represents the summation
• $x$ represents scores
• $N$ represents number of scores

Mean $=\dfrac{1+x+x+x+y+y+9+16+18}{9}=\dfrac{3x+2y+44}{9}$

The Median

$(1)$ If the total number of numbers$(n)$ is an odd number, then the formula is given below$:$

$\text Median=\left(\dfrac{n+1}{2}\right)^{th}\text{term}$

$(2)$ If the total number of numbers(n) is an even number , then the formula is given below$:$

$\text Median=\dfrac{\left(\dfrac{n}{2}\right)^{th}\text{term}+\left(\dfrac{n}{2}+1\right)^{th}\text{term}}{2}$

Here in our question $n=9,$ which is odd. So, we apply the first formula and get the median

Write sequence in ascending or descending order

$1,x,x,x,y,y,9,16,18$

Median $=\left(\dfrac{9+1}{2}\right)^{th}\text{term}=\left(\dfrac{10}{2}\right)^{th}\text{term}=5^{th}$ $\text{term}$

Median $=y$

The Mode

The mode is the most frequency occurring score or value.

$1,x,x,x,y,y,9,16,18$

Here mode $=x$

According to the question

$\dfrac{3x+2y+44}{9}=y$

$\implies3x+2y+44=9y$

$\implies 3x-7y=-44\qquad \to (1)$

and $y=2x\qquad \to (2)$

From $(1)$ and $(2)$ we get

$3x- 14x=-44$

$\implies -11x=-44$

$\implies x=4$

$y=2x=2\times 4 =8$

The correct answer is $(D)$

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