IN A TUG OF WAR, 5 DONKEYS ARE EXACTLY EQUAL TO 2 ELEPHANTS. IN ANOTHER TUG OF WAR, 3 ELEPHANTS ARE EQUAL TO 1 CAR. WHICH TEAM SHOULD WIN IF A CAR AND 3 DONKEYS ARE MATCHED AGAINST 4 ELEPHANTS?

Answer 1

Try this.
5D = 2E
3E = 1 car
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Since E are common, let's make them equal (by multiply 1 by 3 and 2 by 2)
3*5D = 3*2E and
2*3E = 2*1 car
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So we have
15D=6E and
6E = 2 car
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Now the 6E are common; therefore,
15D = 2 Car or
7.5D = 1 Car
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The question asks about
3D ----- 1 Car
Which do you think?

Answer 2

The donkeys and the car win because 2.5 donkeys equal 1 elephant

Answer 3

X has a 3 digit s the ten digit is half the hundred digits the number is a odd the sum of the digit is 9

Answer 4

1car + 3 donkeys = 4.2 elephants, so they win.

Answer 5

5d = 2e
3e = 1c
1c+3d ? 4e
*************************************************************************************
d = (2/5)e
c = (3/1)e
1(3e)+3((2/5e)) = 4.2e > 4e

So, 1 car + 3 donkeys equals to 4.2 elephants, so they win.

Answer 6

so it is a tie