ninety two legs and thirthy three heads how many kids and dogs are there?

Question

I need help on this question.
There are 92 legs and 33 heads how many kids and dogs are they?

Answer 1

Hmmm... let's say kids=x and dogs=y.

33 heads...33 kids and dogs, combined.

x+y=33

Legs=2x+4y

2x+4y=92

Change the first equation to represent one variable.

x+y=33
x=33-y

Insert that (what we have determined x equals) in the second equation, as x

2(33-y)+4y=92
66-2y+4y=92
2y=26
y=13

There are 13 dogs and 20 kids.

Answer 2

There are 20 kids and 13 dogs. This is how you would do it:

Dogs have four legs and kids have two.

x = # of kids
y = # of dogs

92 = 2x + 4y (this is the "legs"equation)
33 = x + y (this is the "heads" equation)

The best way to do this is with substitution.
First you should get x by itself in the second equation.
33 - y = x

Then you substitute 33 -y for x in the other equation.
92 = 2(33-y) +4y

Then you solve for you in that equation.
92 = 2(33-y) +4y
92 = 66 - 2y +4y
92 = 66 + 2y
26 = 2y
y = 13

Then you plug the y value back into one of the equations to find x.
33 = x + y
33 = x + (13)
x = 20

So, there are 20 kids and 13 dogs.

You could also use combination when you have the two equations. It would look like this:

Multiply the second equation by -2 so the x's will cancel when you combine.
92 = 2x + 4y
(33 = x + y) -2

Combine the equations and solve.
92 = 2x + 4y
-66 = -2x + -2y
----------------------
26 = 2y
y = 13

33 = x + y
33 = x + (13)
x = 20

Either way works.

Answer 3

13 dogs and 20 kids
13 dogs = 13 heads, 52 legs
20 kids = 20 heads, 40 legs

For a total of 33 heads and 92 legs

Answer 4

This is a two operation problem, since there are two variables.

Kids should be represented by "a" and dogs by "b".

Since there are 33 of them combinded, it would imply that

a + b = 33 (i)

If you then know that for each person there are two legs (hopefully) and for each dog there are 4, it would imply that

2a + 4b = 92

Divide both sides by 2.
a + 2b = 46 (ii)

If you are good enough at math, you would then know that b is equal to 13, but for the purpose of this demonstration, I will work it out:

Subtract both sides of equation (ii) by 2b.
a = -2b + 46

Subtract both sides of equation (i) by b.
a= -b + 33

Subtract equation (ii) by equation (i).
0 = -b + 13

b = 13

a + 13 = 33

a = 20


a = 13, b = 20

Hope I could help!

Answer 5

x = number of kids
y = number of dogs

So, for head's number, you have
x + y = 33 (since every kid and every dog have onle one head, the sum of kids and dogs have to match the number of heads)

For legs, every kid has 2 legs, and every dog have 4. So
2x + 4y = 92 (the number of kid's legs and the number of dog's legs have to be 92)

so, you have to solve two simultaneous equations:
x + y = 33
2x + 4y = 92

Solving, you have that there are 20 kids and 13 dogs.

Answer 6

20 kids & 13 dogs including Patricia G.

Answer 7

Okay I have been working on it for a little while now and I think I have it down to 20 kids and 13 dogs. It took me a while but I feel good after finally getting it. Thanks for the mental workout. I guess it's been a while and I kind of needed it. Good one.

Answer 8

13 dogs, 20 kids

Answer 9

Well, obviously there are 33 total.

20 kids
13 dogs

Answer 10

20 Kids and 13 Dogs

Solution: x=number of kids, y=number of dogs
x+y=13
2x+4y=92

x=(33-y)
2(33-y)+4y=92
66-2y+4y=92

y=13
x=20