<p>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?</p>
<p>If you add the trays with cups (15) and trays with plates (21), you get 36. Obviously over the limit of 25 trays.</p>
<p>Subtract the “limit” from the sum. 36 - 25 = 11.</p>
<p>There are 11 trays with both a cup and a plate.</p>
<hr>
<p>Explanation:</p>
<p>When you add the cup tray and the plate tray together, you go over the limit because you double count the trays with cups and plates. By finding the excess, you find the number of trays you double count.</p>
<p>x= both cups and plates
15-x= cups only
21-x=plates only
(x)+(15-x)+(21-x)=total
36-x=25
x=11.</p>
<p>I did this kind of problem a gazilllion times. lmao</p>
<p>Here’s an easy formula to remember the next time you have a problem like this.</p>
<p>It’s the Group Formula:</p>
<p>Total = Group 1 (only) + Group 2 (only) + Neither - Both</p>
<p>In this problem, there is no neither as the only possibilities are Cup only (Group 1), plate only (group 2) and Cup and plate (both).</p>
<p>plugging in the info, you get:</p>
<p>25 = 15 + 21 + 0 - both</p>
<p>25 = 36 - both</p>
<p>both = 11</p>