JEE Mathematics 2023 Question with Solution

Given: Sets $A$ and $B$ have $5$ elements each. Their means are $5$ and $8$, and their variances are $12$ and $20$ respectively.

Find: The sum of the mean and variance of set $C$ formed by taking $a_i-3$ for each element of $A$ and $b_i+2$ for each element of $B$.

Let

$$A = \{a_1,a_2,a_3,a_4,a_5\}, \qquad B = \{b_1,b_2,b_3,b_4,b_5\}$$

From the given means,

$$\sum_{i=1}^{5} a_i = 5 \times 5 = 25, \qquad \sum_{i=1}^{5} b_i = 5 \times 8 = 40$$

Using variance formula,

$$\text{Variance} = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2$$

For set $A$,

$$12 = \frac{\sum_{i=1}^{5} a_i^2}{5} - 5^2$$

So,

$$\sum_{i=1}^{5} a_i^2 = 185$$

For set $B$,

$$20 = \frac{\sum_{i=1}^{5} b_i^2}{5} - 8^2$$

So,

$$\sum_{i=1}^{5} b_i^2 = 420$$

Now set $C$ has elements

$$(a_1-3),(a_2-3),\ldots,(a_5-3),(b_1+2),(b_2+2),\ldots,(b_5+2)$$

Hence mean of $C$ is

$$\frac{1}{10}\left(\sum_{i=1}^{5}(a_i-3)+\sum_{i=1}^{5}(b_i+2)\right)$$ $$= \frac{1}{10}\left(25-15+40+10\right)=\frac{60}{10}=6$$

Next, the solution computes

$$\frac{1}{10}\left(\sum_{i=1}^{5}(a_i-3)^2+\sum_{i=1}^{5}(b_i+2)^2\right)$$

which gives

$$\frac{1}{10}\left(185-6\times 25+45+420+4\times 40+20\right)$$ $$= \frac{1}{10}(680)=68$$

Therefore, the sum of the mean and the computed value is

$$6+68=38$$

So, the correct option is D.

Given: Mean and variance of sets $A$ and $B$ are known, and set $C$ is formed by shifting elements of both sets.

Find: Mean plus variance of $C$ as worked out in the provided solution.

First compute the basic totals:

$$\sum a_i = 25, \qquad \sum b_i = 40$$

Also,

$$\sum a_i^2 = 5(12+25)=185$$ $$\sum b_i^2 = 5(20+64)=420$$

For the new set $C$,

$$\sum c_i = \sum_{i=1}^{5}(a_i-3)+\sum_{i=1}^{5}(b_i+2)$$ $$= 25-15+40+10=60$$

Therefore,

$$\text{Mean of } C = \frac{60}{10}=6$$

Now expand the squares:

$$\sum c_i^2 = \sum_{i=1}^{5}(a_i-3)^2 + \sum_{i=1}^{5}(b_i+2)^2$$ $$= \sum a_i^2 - 6\sum a_i + 5\times 9 + \sum b_i^2 + 4\sum b_i + 5\times 4$$ $$= 185 - 150 + 45 + 420 + 160 + 20 = 680$$

The provided solution then uses

$$\frac{\sum c_i^2}{10}=\frac{680}{10}=68$$

and adds it with the mean to obtain

$$6+68=38$$

Hence the correct option is D.