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There are 97 students in South High School. South High School offers only Chinese and Spanish. There are 5 more students in Chinese than in Spanish, and every student takes at least one language. If 48 students take only Spanish, then how many take both languages?

 May 1, 2022
 #1
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Well, 48+5=53.
53+48=101
101-97=4
So 4 students take both.

Test of correctness: Spanish=44                                                44+49+4=97
                                Chinesse=49
                                 Both:4
 

 May 1, 2022
 #2
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Not sure this is possible

 

To see why

Consider that no one takes both

Then 48 take Spanish and 49 take Chinese.....but we're told that 5 more take Chinese than Spanish

 

Consider that only  1 takes both

Then  48 + 1  =  49 take Spanish

So  53  take only Chinese and  1 takes both  =  54

But this gives us     53 + 1 + 48    =   102 total students (too  many)

 

Conslder that  2 take both

Then 50 take Spanish   and  55 would take Chinese   (53 taking only Chinese and 2  taking both)

But this gives us 53 + 2 + 48  = 103

 

The same thing happens if 3 take both

51 take Spanish   and 56 take Chinese  (53 taking only Chinese and 3 taking both)   =  53 + 3 + 48 = 104

 

So.....it appears impossible that 48 take only Spanish

 

 

cool cool cool

 May 1, 2022
 #3
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Well.... I was very wrong. But it isn't impossible, although the question is very poorly written.

 

First I will explain some notations:

T(insert the launguage) is the number of students that take only that language +that take both

B(insert language) is the ammount of students that only take the language

 

And x is the unknown ammount of students that are taking both.

 

 

So we have :B(span.) + B(chin.) + x = 97

48+ ((48+x+5)-x)+x=97

So 91+x= 97

x = 6

 

 

 

T(span.) = 48+6 = 54

T(chin.) 53+6=59

So chin. is 5 more.

Guest May 3, 2022

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