Question

There are $25$ trays on a table in the cafeteria. Each tray contains a cup only, a plate only, or both a cup and a plate. If $15$ of the trays contain cups and $21$ of the trays contain plates, how many contain both a cup and a plate?

• A. $10$
• B. $11$
• C. $12$
• D. $13$

Answer 1

Answer: (B)

Let ${}^{\prime }{A}^{\prime }$ denote the set with a cup

${}^{\prime }{B}^{\prime }$ denote the set with a plate

${}^{\prime }A$ $\cap$ $B^{\prime }$ denote the set with both a cup and a plate

${}^{\prime }A$ $\cup$ $B^{\prime }$ denote the set with the trays either a cup or a plate

$n($ $X)$ denote the number of elements in the set ${}^{\prime }{X}^{\prime }$

Given, total number of trays $n(A$ $\cup$ $B)$ $=$ $25$

Number of trays that contain cups $n($$A)$ $=$ $15$

Number of trays that contain cups $n($$B)$ $=$ $21$

To find the trays with both a cup and a plate $n(A$ $\cap$ $B)$,

We know that

$n(A$ $\cup$ $B)$ $=$ $n($$A)$$+$ $n($$B)$ $-$ $n(A$ $\cap$ $B)$

Rearranging the terms, we get

$n(A$ $\cap$ $B)$ $=$ $n($$A)$$+$ $n($$B)$ $-$ $n(A$ $\cup$ $B)$

From the above,

$n(A$ $\cap$ $B)$ $=$ $15$ $+$ $21$ $-$ $25$

$=$ $11$

$n(A$ $\cap$ $B)$ $=$ $11$

Therefore, number of trays with both a cup and a plate is ${}^{\prime }{11}^{\prime }$.