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400Solution :

Let <br> U be the set of all students who took part in the survey <br> T be the set of students who like tea , and <br> C be the set of the students who like coffee . <br> According to the question , <br> `n(u ) = 800 ,n( T ) = 250 , n( c) = 300 , n ( T cap C ) = 150 ` <br> To find the number of students taking neither tea nor coffee.ie, <br> we have to find `n( T cap C ) ` <br> `n( T cap C) = n ( T cup C) ` <br> `n( U)- n( T cup C ) ` <br> `=n( U) -n ( T cup C) ` <br> `= n( U) - [n ( T)+ n(C)- n( T cap C )]` <br> =800-[250+300-150]=400 <br> Hence , 400 student like neither tea nor coffee. <br> Alternative method : ltbr gt <img src="https://d10lpgp6xz60nq.cloudfront.net/physics_images/CEN_ALG_C01_E01_011 _S01.png" width="80%"> <br> We can use Venn diagram as shown in the figure. <br> In the above figure regions are <br> `r rarr` students liking tea and coffee both`=n(T cap C)` <br> `p rarr `students liking tea only `n(T)-n( T cap C)` <br> `q rarr ` students liking coffee only .`= bn( C) - n ( T cap C)` <br> `s rarr `students liking neither tea neither tea noe coffee `= n( T cap C)` <br> Here r= 150 ( given ) <br> Now , p+ r student liking tea = 250 (given) <br> `rArr p= 250 - 150 = 100` <br> q+r = student liking coffee = 300 ( given ) <br> `rArr q = 300 - 150 = 150 ` <br> p+ q+r = student liking at least one of tea and coffee <br> = 100 + 150 + 150 = 400 <br> s = student liking neither tea nor coffee = 800 - 400= 400 **What is sets in general collection of things**

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