Mike N. answered • 10/24/14
Tutor
5
(3)
Professional Mathematician with homeschool experience
Hi Jasmine,
My answer is so long, I'll have to split it into two!
There may be a clever way to do this, but I don't see it, so I'm going to just solve it brute force. Let's draw ourselves a Venn Diagram to show all possible cases. We'll use the letters B for brown hair, T for short tail, and H for long hair. In the Venn diagram, we'll label the sections by including a letter if it's true, and leave it out if not. So, for example, BH would be brown dogs with long hair but not short tails.
_______________________
| |
| |
| |
| B | | |
| | BT | |
| | | T |
| | | |
| | | | | |
| | | BHT | HT | |
| | | | | |
| | | | | |
| | BH | |
| | | |
| | | |
| H |
| |
| |
So what are we given?
There are 28 dogs
B + H + T + BT + BH + HT + BHT = 28
11 of the dogs are brown
B + BH + BT + BHT = 11
8 of the dogs have short tails
T + BT + HT + BHT = 8
17 of the dogs have long hair
H + BH + HT + BHT = 17
3 of the dogs are brown with short tails
BT + BTH = 3
3 of the dogs have short tails and no long hair
BT + T = 3
How many dogs are brown with long hair but don't have short tails?
BH = ?
So, let's assemble our facts:
1) B + H + T + BT + BH + HT + BHT = 28
2) B + BH + BT + BHT = 11
3) T + BT + HT + BHT = 8
4) H + BH + HT + BHT = 17
5) BT + BTH = 3
6) BT + T = 3
BH = ?
2) B + BH + BT + BHT = 11
3) T + BT + HT + BHT = 8
4) H + BH + HT + BHT = 17
5) BT + BTH = 3
6) BT + T = 3
BH = ?
So now we're down to a set of simultaneous equations.
I'll finish up in the comment.
I'll finish up in the comment.
Mike N.
B + H + T + BT + BH + HT + BHT - (B + BH + BT + BHT) = 28 - 17
H + T + HT = 17
I did that so that equation 2 would be the only place a B occurs. Now I can solve everything else, and if I ever wanted to know B, which I won't, I would substitute the values into equation 2 to find B. And in fact, let's proceed along those lines. I will now forget equation 2 and move on to solve the rest of the equations.
1) H + T + HT = 17
2) B + BH + BT + BHT = 11
3) T + BT + HT + BHT = 8
4) H + BH + HT + BHT = 17
5) BT + BTH = 3
6) BT + T = 3
BH = ?
Let's subtract 1 from 4:
H + BH + HT + BHT - (H + T + HT) = 17 - 17
BH + BHT - T = 0
Now equation 1 is the only place an H occurs...
1) H + T + HT = 17
3) T + BT + HT + BHT = 8
4) BH + BHT - T = 0
5) BT + BTH = 3
6) BT + T = 3
BH = ?
Ok, now let's get rid of T. We'll add equation 4 to equations 3 and 6:
3:
T + BT + HT + BHT + (BH + BHT - T) = 8 + 0
BT + HT + 2BHT + BH = 8
6:
BT + T + (BH + BHT - T) = 3 + 0
BT + BH + BHT = 3
3) BT + HT + 2BHT + BH = 8
4) BH + BHT - T = 0
5) BT + BHT = 3
6) BT + BH + BHT = 3
BH = ?
Ok, now we can spot the final move. Subtract equation 5 from equation 6.
BT + BH + BHT - (BT + BHT) = 3 - 3
BH = 0
So, the final answer is zero. There are no brown dogs with long hair that don't have short tails.
I hope that helped. If there's a trick to spot, I'm sorry I missed it.
Regards,
Mike N.
10/24/14