1. Use the mean to get one equation

There are $6$ observations: $a,\ b,\ 68,\ 44,\ 48,\ 60$

Given mean $=55$, so $\frac{a+b+68+44+48+60}{6}=55$

First add the known terms: $68+44+48+60=220$

Hence, $\frac{a+b+220}{6}=55$ $a+b+220=330$ $a+b=110 \qquad ...(1)$

2. Use the variance formula

Given variance $=194$.

For observations $x_1,x_2,\dots,x_6$ with mean $\bar x=55$, $\text{Variance} = \frac{1}{6}\sum (x_i-55)^2$

So, $\frac{1}{6}\left[(a-55)^2+(b-55)^2+(68-55)^2+(44-55)^2+(48-55)^2+(60-55)^2\right]=194$

Now compute the known squared deviations: $68-55=13 \Rightarrow 13^2=169$ $44-55=-11 \Rightarrow (-11)^2=121$ $48-55=-7 \Rightarrow (-7)^2=49$ $60-55=5 \Rightarrow 5^2=25$

Their sum is $169+121+49+25=364$

Thus, $\frac{(a-55)^2+(b-55)^2+364}{6}=194$ $(a-55)^2+(b-55)^2+364=1164$ $(a-55)^2+(b-55)^2=800 \qquad ...(2)$

3. Simplify using $a+b=110$

From (1), $b=110-a$

Then, $b-55=(110-a)-55=55-a=-(a-55)$

So, $(b-55)^2=(a-55)^2$

Using in (2): $2(a-55)^2=800$ $(a-55)^2=400$ $a-55=\pm 20$

Hence, $a=75 \text{ or } 35$

Since $a>b$ and $a+b=110$:

• If $a=75$, then $b=35$
• If $a=35$, then $b=75$ (not allowed since $a>b$)

Therefore, $a=75,\quad b=35$

4. Find $a+3b$

$a+3b=75+3(35)=75+105=180$

5. Check the options

Option A: $180$ ✅

So the correct answer is A.