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?

Guest May 1, 2022

#1**0 **

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

Guest May 1, 2022

#2**+1 **

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

CPhill May 1, 2022

#3**0 **

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