tricky maths

[aaman1]

Can someone help with these questions:

Q1. We have ducks and pigs in a field ? There are 30 creatures in all. All the legs add up to 92. a)how many ducks? b) How many pigs?

Q2. 30 x 65= (v x 30) + (v x 35) find v?

Many thanks.

[hermanmunster]

q

Can someone help with these questions:

Q1. We have ducks and pigs in a field ? There are 30 creatures in all. All the legs add up to 92. a)how many ducks? b) How many pigs?

Q2. 30 x 65= (v x 30) + (v x 35) find v?

Many thanks.

Q2 is actually 30x65=30V + 35V
--> 30x65=65v
divide both sides by 65 gives v=30

[xyzzy]

Can someone help with these questions:

Q1. We have ducks and pigs in a field ? There are 30 creatures in all. All the legs add up to 92. a)how many ducks? b) How many pigs?

16 pigs and 14 ducks (16 x 4 + 14 x 2 = 92). I did it graphically (draw 2x + 4y = 92 and x + y = 30, where they cross is the answer) but you can do it algebraically as well:

4p + 2d = 92
p + d = 30

2p + 2d = 60 (multiply second equation by 2)
2p = 32 (subtract from first equation)
so p = 16, and so d = 14.

As it happens, you can do it very quickly by trial and error, too. Suppose there's 15 of each: 90 legs. So we need two more legs. Swap a duck for a pig and there you are.

[Guest55]

We have ducks and pigs in a field ? There are 30 creatures in all. All the legs add up to 92. a)how many ducks? b) How many pigs

30 creatures - if they were all ducks then we would have 60 legs ... ohh we've got 92 which is 32 more - these must be for 16 pigs (2 extra each)

Algebra is not needed for this type of question - and, in fact, makes it harder

[minubht80]

Let x be the number of 4 legged creatures, then 2 legged will be 30-x(as total is 30)

4X+2(30-X)=92

2x=92-60=32
X=16 -> 4 legged animal, 30-16=14 - > 2 legged animal

[Guest55]

As I said - algebra is NOT needed -

These questions are to test logical thought