Math Problems, Probability Problems

Problem of the Day: 1/4/13

Melissa has 50 pennies. 16 of the pennies are green on both the heads and tails sides, and 6 are shiny copper on both the heads and tails sides. If the number of coins with only green heads and the number with only green tails are equal, what is the probability that Melissa will pick a coin with a shiny copper heads and a green tails?

Solution to yesterday’s problem:
Let’s represent Sally’s age as s and Tommy’s age as t. Their ages 6 years ago are s — 6 and t — 6, so we can write the first piece of information as this equation:

s-6=\frac{1}{2}(t-6)

Using the same method, we can write the second piece of information like this:

s-2=\frac{3}{4}(t-2)

Easy, right? Now let’s solve this system of equations by isolating the s on both equations.

s=\frac{1}{2}(t-6)+6
s=\frac{3}{4}(t-2)+2

We can now set these equations equal to each other and solve:

\frac{1}{2}(t-6)+6=\frac{3}{4}(t-2)+2

\frac{1}{2}t-3+6=\frac{3}{4}t-\frac{3}{2}+2

3=\frac{1}{4}t+\frac{1}{2}

\frac{5}{2}=\frac{1}{4}t

t=10

So Tommy is 10 years old. Now let’s substitute 10 for t in one of the original equations:

s-2=\frac{3}{4}(10-2)
s-2=\frac{3}{4}(8)
s-2=6
s=8

Finally, Sally is 8 years old, and Tommy is 10 years old.

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s