Question 1172823
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At a luncheon party for 37 eminent people, 25 requested for fried rice, 15 requested for salad, 
while 20 requested for shredded beef. 
7 people requested for fried rice and salad, 13 people requested for fried rice and shredded beef 
and 8 people requested for salad and shredded beef. Also 10 people requested for fried rice only , 
while 5 requested for salad only . Using a venn diagram, find

1) the number of people that requested for all the three dishes
2) the number of people that requested for two different dishes
3) the number of people that requested for only one dish. (Assume that each person requested for only one dish.)
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            When you use Venn diagram,  the solution is simple,  but you will learn nothing (or almost nothing).


            Instead,  I  will provide  Algebra solution,  which is much more interesting and educative.


            You can make an associated Venn  diagram on your own to visualize my Algebra solution for yourself.



<pre>
(1)  You have three sets

         R of 25 elements (people requested Rice)           

         D of 15 elements (people requested salaD)          

         B of 20 elements (people requested Beef)


    You have their in-pair intersections

        RD of  7 elements

        RB of 13 elements

        DB of  8 elements


    You also are given

        The subset Ro ("Rice  only") contains 10 elements,
    and
        the subset Do ("salaD only") contains  5 elements.


    Let x be the number of elements in the triple intersection RDB.


    Then you have this equation

        Ro = R - RS - RB + x,   or, substituting the given numbers

        10 = 25 - 7 - 13 + x,   which implies

        10 - 25 + 7 + 13 = x,   or  x = 5.


    So, the triple intersection RDB has 5 elements.  It means that exactly 5 people requested all 3 dishes.    Question 1) is answered.



(2)  Now it is easy to answer the second question.


         The number of elements in RDo (two dishes "Rice and salaD only") is

             RDo = RD - RDB = 7 - 5 = 2.


         The number of elements in RBo (two dishes "Rice and Beef only") is

             RBo = RB - RDB = 13 - 5 = 8.


         The number of elements in DBo (two dishes "Salad and Beef only") is

             SBo = SB - RDB = 8 - 5 = 3.


    Hence, the number of those, who ordered only 2 dishes, is RDo + RB0 + SBo = 2 + 8 + 3 = 13.    Question 2) is answered.



3)  We are just given Ro = 10 ("Rice only"), Do = 5 ("salaD only"). 

    We need to determine Bo (the number of elements in the subset "Beef only").


    We have  Bo = B \ (RB) \ (DB) + RDB (here the symbol " \ " means "subtract a narrow subset from the larger subset").


    THEREFORE,  the number of elements Bo is equal to  20 - 13 - 8 + 5 = 4.


    Thus the partial answers for this part are  Ro = 10 ("Rice only"); Do = 5 ("salaD only") and Bo = 4 ("Beef only").

    These partial answers produce the final answer for this part Ro + Do + Bo = 10 + 5 + 4 = 19
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