Hello Math Wizards! I Need Help!?

Question

Okay, so it has been 20 years since I have been in high school algebra. My son has a problem I can't solve.

There were a number of bats in a cave. 2 bats could see out of the right eye, 3 could see out of the left eye, 4 could not see out of the left eye, and 5 could not see out of the right eye. What is the least number of bats in the cave and what might their eyesight have been? Were they totally blind, blind only in the right,, etc.?

So far I have
x = total bats
x-2 = +right eye bat
x-3 = + left eye
x-4 = -left eye
x-5 = -right eye

Does that make any sense? I was going to line up tha equation and solve but now I am stuck. Am I even going in the right direction with this?

PLEASE HELP ME.

Talk to me like I am 5 years old. I still don't understand.

The sheet says that there can be 3 possible answers.

This is Algebra 1. I think this question is abstract and really hard. My son is only 14 years old. He's in 9th grade.

Answer 1

7 bats

two can see right, five can not 7 bats
three can see left, four can not 7 bats

three bats are blind
one bat sees in both eyes
one bat see only right
two bats see only left

or

Four bats are blind
two can see both
One can see only left

or

2 bats are blind
2 only see right
3 only see left

You have four variables.
right eye
left eye
both
blind.

You do not have enough equations to solve. You need four independant equations. This is why you get three answers.

How I solved this was by arbitrary setting varibles and then solving for one unknown. I knew I had only seven bats, so I only took solutions that arrived at seven.

I started out with zero, one, two bats that could see both. Putting that in to your equations I solved for the left and right, then I solved for blind.

right+both=2
left+both=3
left+blind=5
right+blind=4
right+both+left+blind=7

What is needed is how many could see out both eyes? or how many were completely blind.

Answer 2

Wow this is a tough question. What grade of math homework is this cause this is tough and I am in Geometry Honors.

I think the least amount of bats in the cave is 9. I found that 2 could see out of their right eye and 4 couldn't see out of the left so 2 of those 4 could only see out of their right and the other 2 were blind.
Then if 3 could see out of their left eye and 5 couldn't see out of their right that means that 3 could just see out of the right eye and the other 2 are totally blind.
So 4 were totally blind and there were 9 bats in the cave.

This is what I found.

I hope it is right and it helped get the homework done.

Answer 3

The people above suggested Venn diagrams, and that is a good idea. One way to do it rhetorically is to consider the conditions involved. If we think of right eye sight, 2 could see and 5 can't see, so there are at least seven bats total. If we think of left eye sight, 3 could see and 4 could not, so there are at least 7 bats total.

There is a solution with only seven bats; two can see fully, one can see only out of its left eye, and four totally blind.

So there are at least seven bats.

Answer 4

I'm a teacher and I'm going to save you a lot of time on this. You can solve this by simply making an old fashioned Venn Diagram

4 can't see out of left and 5 can't see out of the right=9

2 can see out of right and 3 could see out of left= min. of 3 (can't they see out of both eyes0

Minimum of 12

Answer 5

Think about the right eye first. You have two that can see out of their right eye and five that cannot. Those two categories cannot overlap, so that's seven minimum right there. The question now is, can we fit the others into those seven.

We're told that three could see out of their left eye and four could not - seven again.

One way to work it:

Five cannot see out of their right eye, four cannot see out of their left. That gives us four blind and one blind on the right side. We have three that can see out of their left eye - one of those is the one blind on the right. The other two can be the ones who can see out of their right eye, so they have perfect vision.

So:

Four blind - no left, no right.
One blind only on the right.
Two that can see out of both eyes.

Total is seven. To check it: five are blind on the right, two can see out of their right eye. Four are blind on the left, three can see on the left. It checks..

Answer 6

Set up a venn diagram. YOu know the intersecting circle thing. One circle would be "Not able to see out of" right eye then left

Answer 7

Seven Bats:
Right Eye Min: 5 (can't) +2 (can) = 7
Left Eye MIn: 4 (can't) + 3 (can) = 7
Min = 7

RL
YY
YY
NY
NN
NN
NN
NN

Answer 8

7 bats
4 cannot see out of either eye
1 sees from the left but not the right
2 see out of both eyes

Answer 9

NO trick question, Bats are blind.