1991 AIME Problems/Problem 13: Difference between revisions
Gabiloncho (talk | contribs) |
Gabiloncho (talk | contribs) |
||
| Line 10: | Line 10: | ||
P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}. | P=\frac{r(r-1)}{(r+b)(r+b-1)}+\frac{b(b-1)}{(r+b)(r+b-1)}=\frac{r(r-1)+(t-r)(t-r-1)}{t(t-1)}=\frac{1}{2}. | ||
</math> | </math> | ||
Solving the quadratic equation for <math>r</math> in terms of <math>t</math>, one obtains | |||
Solving the resulting quadratic equation <math>r^{2}-rt+t(t-1)/4=0</math> for <math>r</math> in terms of <math>t</math>, one obtains that | |||
<math> | |||
r=\frac{t\pm\sqrt{t}}{2} | |||
</math> | |||
== See also == | == See also == | ||
{{AIME box|year=1991|num-b=12|num-a=14}}</math> | {{AIME box|year=1991|num-b=12|num-a=14}}</math> | ||
Revision as of 18:22, 18 April 2007
Problem
A drawer contains a mixture of red socks and blue socks, at most 1991 in all. It so happens that, when two socks are selected randomly without replacement, there is a probability of exactly 
that both are red or both are blue. What is the largest possible number of red socks in the drawer that is consistent with this data?
Solution
Let 
and 
denote the number of red and blue socks, respectively. Also, let 
.
The probability 
that when two socks are drawn without replacement, both are red or both are blue is given by
Solving the resulting quadratic equation 
for in terms of 
, one obtains that
See also
| 1991 AIME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
</math>
