Mixing
PIE — # of students taking Spanish
$begingroup$
A school with $100$ students offers French and Spanish as its language courses. Twice as many students are in the French class as the Spanish class. Three times as many students are in both classes as are in neither class. The number of students in both classes is even, and fewer than $10$ students are in neither class. How many students are taking Spanish?
Let the number of students taking French be $f$, the number of those taking Spanish be $s$, the number of students taking both be $x$, and the number of students taking neither be $n$. Then we have $100 = f+s-x+n$ (by PIE). We are also given that $f=2s$ and $x=3n$, where $n<10$.
Substituting some values, $100 = 3s-2n$ or alternatively, $100+2n = 3s$. Since $n<10$, we can see which values work: $n=1, 4$ or $7$.
When $n=1$, $102=3s$ and thus $s=34$, and $f=68$.
When $n=4$, $108=3x$ and thus $s=36$ and $f=72$.
When $n=7$, $114=3s$ and thus $s=38$ and $f=76$.
All of these situations satisfy the condition that $fequiv 0mod 2$ & $sequiv 0mod 2$, and satisfy that $x=3n$ (when calculated using PIE) as well. However, it seems that there should only be one set of $(f,s)$ values that should work. What am I missing or what have I done wrong?
probability inclusion-exclusion
asked Jan 18 at 15:32
jjhhjjhh
2,13911122
$endgroup$
add a comment |
$begingroup$
A school with $100$ students offers French and Spanish as its language courses. Twice as many students are in the French class as the Spanish class. Three times as many students are in both classes as are in neither class. The number of students in both classes is even, and fewer than $10$ students are in neither class. How many students are taking Spanish?
Let the number of students taking French be $f$, the number of those taking Spanish be $s$, the number of students taking both be $x$, and the number of students taking neither be $n$. Then we have $100 = f+s-x+n$ (by PIE). We are also given that $f=2s$ and $x=3n$, where $n<10$.
Substituting some values, $100 = 3s-2n$ or alternatively, $100+2n = 3s$. Since $n<10$, we can see which values work: $n=1, 4$ or $7$.
When $n=1$, $102=3s$ and thus $s=34$, and $f=68$.
When $n=4$, $108=3x$ and thus $s=36$ and $f=72$.
When $n=7$, $114=3s$ and thus $s=38$ and $f=76$.
All of these situations satisfy the condition that $fequiv 0mod 2$ & $sequiv 0mod 2$, and satisfy that $x=3n$ (when calculated using PIE) as well. However, it seems that there should only be one set of $(f,s)$ values that should work. What am I missing or what have I done wrong?
probability inclusion-exclusion
asked Jan 18 at 15:32
jjhhjjhh
2,13911122
$endgroup$
1
$begingroup$
$n$ must be even as $3n$ is even.
$endgroup$
– lulu
Jan 18 at 15:44
$begingroup$
Oh I had thought that "the number of students in both classes is even" meant that $f$ and $s$ had to be even, but from what you've told me, it seems that they meant $x$ was even. Thank you!
$endgroup$
– jjhh
Jan 18 at 18:20
add a comment |
$begingroup$
A school with $100$ students offers French and Spanish as its language courses. Twice as many students are in the French class as the Spanish class. Three times as many students are in both classes as are in neither class. The number of students in both classes is even, and fewer than $10$ students are in neither class. How many students are taking Spanish?
Let the number of students taking French be $f$, the number of those taking Spanish be $s$, the number of students taking both be $x$, and the number of students taking neither be $n$. Then we have $100 = f+s-x+n$ (by PIE). We are also given that $f=2s$ and $x=3n$, where $n<10$.
Substituting some values, $100 = 3s-2n$ or alternatively, $100+2n = 3s$. Since $n<10$, we can see which values work: $n=1, 4$ or $7$.
When $n=1$, $102=3s$ and thus $s=34$, and $f=68$.
When $n=4$, $108=3x$ and thus $s=36$ and $f=72$.
When $n=7$, $114=3s$ and thus $s=38$ and $f=76$.
All of these situations satisfy the condition that $fequiv 0mod 2$ & $sequiv 0mod 2$, and satisfy that $x=3n$ (when calculated using PIE) as well. However, it seems that there should only be one set of $(f,s)$ values that should work. What am I missing or what have I done wrong?
probability inclusion-exclusion
asked Jan 18 at 15:32
jjhhjjhh
2,13911122
$endgroup$
A school with $100$ students offers French and Spanish as its language courses. Twice as many students are in the French class as the Spanish class. Three times as many students are in both classes as are in neither class. The number of students in both classes is even, and fewer than $10$ students are in neither class. How many students are taking Spanish?
Let the number of students taking French be $f$, the number of those taking Spanish be $s$, the number of students taking both be $x$, and the number of students taking neither be $n$. Then we have $100 = f+s-x+n$ (by PIE). We are also given that $f=2s$ and $x=3n$, where $n<10$.
Substituting some values, $100 = 3s-2n$ or alternatively, $100+2n = 3s$. Since $n<10$, we can see which values work: $n=1, 4$ or $7$.
When $n=1$, $102=3s$ and thus $s=34$, and $f=68$.
When $n=4$, $108=3x$ and thus $s=36$ and $f=72$.
When $n=7$, $114=3s$ and thus $s=38$ and $f=76$.
All of these situations satisfy the condition that $fequiv 0mod 2$ & $sequiv 0mod 2$, and satisfy that $x=3n$ (when calculated using PIE) as well. However, it seems that there should only be one set of $(f,s)$ values that should work. What am I missing or what have I done wrong?
probability inclusion-exclusion
probability inclusion-exclusion
asked Jan 18 at 15:32
jjhhjjhh
2,13911122
asked Jan 18 at 15:32
jjhhjjhh
2,13911122
asked Jan 18 at 15:32
jjhhjjhh
2,13911122
asked Jan 18 at 15:32
jjhhjjhh
2,13911122
asked Jan 18 at 15:32
jjhhjjhh
2,13911122
2,13911122
1
$begingroup$
$n$ must be even as $3n$ is even.
$endgroup$
– lulu
Jan 18 at 15:44
$begingroup$
Oh I had thought that "the number of students in both classes is even" meant that $f$ and $s$ had to be even, but from what you've told me, it seems that they meant $x$ was even. Thank you!
$endgroup$
– jjhh
Jan 18 at 18:20
add a comment |
1
$begingroup$
$n$ must be even as $3n$ is even.
$endgroup$
– lulu
Jan 18 at 15:44
$begingroup$
Oh I had thought that "the number of students in both classes is even" meant that $f$ and $s$ had to be even, but from what you've told me, it seems that they meant $x$ was even. Thank you!
$endgroup$
– jjhh
Jan 18 at 18:20
1
1
$begingroup$
$n$ must be even as $3n$ is even.
$endgroup$
– lulu
Jan 18 at 15:44
$begingroup$
$n$ must be even as $3n$ is even.
$endgroup$
– lulu
Jan 18 at 15:44
$begingroup$
Oh I had thought that "the number of students in both classes is even" meant that $f$ and $s$ had to be even, but from what you've told me, it seems that they meant $x$ was even. Thank you!
$endgroup$
– jjhh
Jan 18 at 18:20
$begingroup$
Oh I had thought that "the number of students in both classes is even" meant that $f$ and $s$ had to be even, but from what you've told me, it seems that they meant $x$ was even. Thank you!
$endgroup$
– jjhh
Jan 18 at 18:20
add a comment |
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1
$begingroup$
$n$ must be even as $3n$ is even.
$endgroup$
– lulu
Jan 18 at 15:44
$begingroup$
Oh I had thought that "the number of students in both classes is even" meant that $f$ and $s$ had to be even, but from what you've told me, it seems that they meant $x$ was even. Thank you!
$endgroup$
– jjhh
Jan 18 at 18:20