Math Help?

cat • 16 December 2012 at 6:17 PM

I need help with math... Here's the problem.

I need to find out who will win the third round of tug-of-war.

Round 1: On one side, there are four acrobats, each of equal strength. On the other side, there is five grandmas, all of equal strength. The result is even.

Round 2: On one side, there is a dog. The dog is against two grandmas and one acrobat. Again, it's a draw.

Round 3: The dog and three grandmas are on one side, and four acrobats are on the other. Who will win?


I need a detailed explanation about how you know and what you did to find the answer. Help? Dx

trish • 16 December 2012 at 6:25 PM

http://answers.yahoo.com/question/index?qid=20111016202556AAL86RK

;D

dreamer • 16 December 2012 at 6:29 PM

@Cat Da heck? O_o What grade are you in? That's not challenging, but it sure as heck makes NO sense.

alannah • 16 December 2012 at 6:33 PM

@cat
The dog and three grandmas would win. I can't really explain it.
R1: 4a=5g
R2: d=2g+a
R3: d+2g (?) 4a


2g+a+3g>4a

You should substitue the dog or round 2 into the last equation.
That would give you 5 grandmas and one acrobat vs. 4 acrobats. So, the first side would win.

cat • 16 December 2012 at 7:42 PM

@trish Thanks, that totally helped me 😃

@dreamer I'm in sixth 😋 I know the answer, but I just can't explain it Dx

@alannah Thanks ❤️

2elle2 • 16 December 2012 at 7:52 PM

@cat

Let:
x= one acrobat
y= one grandma
z= one dog

We know that:

4x = 5y
z = 2y + x

We need to figure out:

(3y + z) is stronger or weaker than 4x.

We can use substitution to figure this out.
I solved this by finding x, y, and z in terms of y. You don't have to, I just did it this way ;D

From the equation z = 2y + x, I subtracted 2y from both sides.
z = 2y + x
-2y -2y
x = z - 2y

Now we substitute z - 2y for x in the first equation.
4x = 5y
4(z - 2y) = 5y
4z - 8y = 5y
+8y +8y
4z = 13y
z = 13/4 y

Now we know z = 13/4 y. To find x in terms of y, we use the first equation and just divide both sides by 4.

4x = 5y
x = 5/4 y

x = 5/4 y, y = y, and z = 13/4 y. We can substitute the quantities into the last equation.

(3y + z) is stronger or weaker than 4x.
3y + 13/4 y is stronger or weaker than 4( 5/4 y)
4[3y +13/4 y] is stronger or weaker than [5y]4
12y + 13y is stronger or weaker than 20y
25y > 20y

Therefore, the dog and 3 grandmas would win. 😸 Sorry if its too confusing >.<
EDIT- Whoops, you already figured it out xD

twilight_raptor • 16 December 2012 at 8:23 PM

@cat

Round 1: On one side, there are four acrobats, each of equal strength. On the other side, there is five grandmas, all of equal strength. The result is even.

4A = 5G
4/5 A = G

*12/5 A = 3G
*8/5 A = 2G

Round 2: On one side, there is a dog. The dog is against two grandmas and one acrobat. Again, it's a draw.

1D = 2G + 1A
1D = 8/5 A + 1A
1D = 13/5A

Round 3: The dog and three grandmas are on one side, and four acrobats are on the other. Who will win?

1D + 3G ~ 4A

13/5 A + 12/5 A ~ 4A
25/5 A ~ 4A
5A ~ 4A
5A > 4A

the dog and the grannies would win 😸 I know you already got the answer but I'm bored so though I'd try xP

cat • 16 December 2012 at 8:44 PM

@twilight_raptor @2elle2 How would you be able to check your answer for this?

twilight_raptor • 16 December 2012 at 9:11 PM

@cat hmm I'm not sure. the method I used is, since we wanted to find the strength of the dog and grannies in proportion to the acrobats, I found that using the first and second rounds by solving the equations (4A = 5G and 1D = 2G + 1A) for G and D.
Round one:
4/5 A = G

Round 2:
1D = 13/5A

Knowing the strengths of 1 granny and 1 dog, I substituted these to the final equation:

1D + 3G ~ 4A (the sign '~' could be a = or > or < and this would signify if round 3 is a draw or a win on either side)

Substitution gave 1D + 3G equal to 5A which is larger than 4A on the other side, signifying a win for the dog and grannies.