CAT 2023 Slot 1 Solved Paper

Given that the means of the ratings given by $\mathit{R}1,\mathit{R}2,\mathit{R}3,\mathit{R}4$ and $\mathit{R}5$ were $3.4,2.2,3.8,2.8$ and 3.4 respectively.
⇒ The sum of ratings given by $\mathrm{R}1,\mathrm{R}2,\mathrm{R}3\mathrm{R}4,\mathrm{R}5$ are $5^{\star }$ means $=17,11,19,14$, and 17 respectively.
Similarly the sum of ratings received by $\mathit{U},\mathit{V},\mathit{W},\mathit{X}$ and $\mathit{Y}$ are $5^{\star }$ means $=11,19,17,18$, and 13 respectively.
Also capturing the absolute data given in the partial information (a) and (b) and representing as a table, we get:Now,
Consider U
Given median $=2$, mode $=2$ and range $=3$
$\Rightarrow$ His ratings should be of the form $1,\mathit{a},2,\mathit{b},4\Rightarrow 1+2+4+\mathit{a}+\mathit{b}=11\Rightarrow \mathit{a}+\mathit{b}=4$. For mode $=2\Rightarrow \mathit{a}=\mathit{b}=2$
$\Rightarrow$ U's ratings are $1,2,2,2,4$.
Consider V
Given median $=4$, mode $=4$ and range $=3$
$\Rightarrow$ His ratings should be of the form $2,\mathit{a},4,\mathit{b},5\Rightarrow 2+4+5+\mathit{a}+\mathit{b}=19\Rightarrow \mathit{a}+\mathit{b}=8\Rightarrow$ For mode $=4\Rightarrow \mathit{a}=\mathit{b}=$ 4
$\Rightarrow$ V's ratings are 2, 4, 4, 4, 5 .
Consider W
Given median $=4$, mode $=5$ and range $=4$
⇒ His ratings should be of the form $1,\mathit{a},4,5,5\Rightarrow 1+\mathit{a}+4+5+5=17\Rightarrow \mathit{a}=2$
⇒ W's ratings are 1, 2, 4, 5 , 5 .
Consider X
Given median $=4$, mode $=5$ and range $=4$
$\Rightarrow$ His ratings should be of the form $1,\mathit{a},4,5,5\Rightarrow \mathit{a}+1+4+5+5=18\Rightarrow \mathit{a}=3$
⇒ X's ratings are 1, 3, 4, 5, 5
Consider $\mathit{Y}$
Given median $=3$, mode , Range $=3$
⇒ His ratings are $1,1,3,4,4$.
Capturing this data in the table, we get:Given median $=4$, mode $=5$ and range $=4$
$\Rightarrow$ His ratings should be of the form $1,\mathit{a},4,5,5\Rightarrow \mathit{a}+1+4+5+5=18\Rightarrow \mathit{a}=3$
⇒ X's ratings are 1, 3, 4, 5, 5
Consider $\mathit{Y}$
Given median $=3$, mode , Range $=3$
⇒ His ratings are $1,1,3,4,4$.
Capturing this data in the table, we get:
Now, consider column R3 $\Rightarrow$ The two missing entries should add up to $19-1-5-5=8$, (only possibility is $4+4$ )⇒ We can fill the row "U" and 4 in the row "V"Now, consider column R2 ⇒ Missing entry should be$11-2-1-5-1=2$
Consider column R1, the missing elements should add up to $17-5-4-1=7\left(3+4\mathit{o}\mathit{r}4+3\right)$ ----(1)Consider R5, the missing elements should add up to $10\Rightarrow 2+4+4\mathit{o}\mathit{r}4+3+3$ (not possible) as (1) requires a 3.
Now, we can fill column R1 as 3 + 4 and the remaining in column R4 and we can get the complete table
$\Rightarrow$ Ratings give by R3 are $1,4,4,5,5\Rightarrow$ Median $=4$.