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?
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
+130081 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
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.
