Plees solved iq test?

Question

CHICKEN 0.05 CENT, GOAT 1 DOLLAR , COW 5 DOLLAR. I GIVE U 100 DOLLARS TO BUY THIS THEE THINGS. U NEED TO BUY 100 THINGS USED THIS 1OO DOLLARS. QUANTITY SHOULD BE 100 IF U PLUS THE QUANTITY OF THIS THREE THINGS. AT THE SAME TIME THE MONEY U USED CANNOT MORE OR LESSTHEN 100 DOLLARS

Answer 1

There are 2 ways to work this out

1) You have 2 equations for 3 unknowns so you can solve for one in terms of the other

let x be the number of chickens, y be the number of goats and z the number of cows, then we know

0.05X + Y + 5Z = 100 and

X + Y + Z = 100

Therefore

0.05X + Y + 5Z = X + Y + Z

0.05X -1X +5Z - Z =0

4Z = 0.95 X

Z = 0.95/4 X

or since we have to be in whole numbers

Z = 95/400 X = 19/80X

Since 19 is a prime, this is the minimum number of cows there can be, and there must be 80 chickens to go along with this.

That leaves 1 Goat. Check back and we find we have 80 chickens (4 $) 1 goat (1 $) and 19 cows (95 $) and both animals and money add up to 100

2) Chickens cost 0,05, and since the cost is a whole number of $, the number of chickens must be a multiple of 20 and therefore there can only be 20, 40 60 or 80 chickens

20 chickens : there are 99 $ left with which to buy things and 80 aminals left to buy.

40 chickens : there are 98 $ left with which to buy things and 60 aminals left to buy.

60 chickens : there are 97 $ left with which to buy things and 40 aminals left to buy.

80 chickens : there are 96 $ left with which to buy things and 20 aminals left to buy.

Lets look at the 20 chicken case

Y + 5Z = 98 : Y = 98 - 5Z
Y + Z = 80 : Y = 80 - Z
98 - 5Z = 80 - Z
therefore 4Z = 18, Z = 18/4 = 4.5 not feasible

You can do the same with the 40 and 60 chicken cases, you will find they are also un feasible.

Lets take the 80 chicken case.

Y + 5Z = 96 : Y = 96 - 5Z
Y + Z = 20 : Y = 20 -Z
96 - 5Z = 20 - Z
76 = 4Z
Z= 76/4= 19

Therefore Y = 1

The same 80 chickens, 19 cows and a goat

Okay?

Answer 2

By inspection the solution will be linearly dependent and contain infinitely many solutions in a mathematical sense, but not in a real sense. So first we should define our solution set:
0 <= x <100
0 <= y <= 100
0 <=z < 20

and x,y,z belong to the Whole Numbers (you pry won't buy 2/3 a chicken or 1/2 a cow)

Now consider the matrix representation of the problem:
.05 1 5 100
1 1 1 100

Then a row reduction yields

1 0 -80/19 0
0 1 99/19 100

Now from the constraints before we can make a conclusion, luckily 19 is prime, so in the 2nd equation (the bottom row), x is "free" and y + 99/19 = 100. By inspection, z = 19, y = 1 and necessarily x = 80 by the constraints. Then in the first equation, y is "free" and by inspection z = 19, x=80, and necessarily y=1 by our constraints. Also considering the case, of x and / or y and / or z=0, yields y =100.

So the solution sets are (80 , 1 , 19) and (0, 100, 0). There may be more a few more solutions if you row reduce differently, I'm not sure and I'm not going to try, but you can try.

Answer 3

There is more information hidden here,

1) you must buy at least one of each animal, so you cannot just go home with 100 goats!

2) you cannot buy fractions of animals, they must be whole numbers, so you must buy multiples of 20 chickens to use the whole dollars remaining after buying whole goats.... (20 chickens = $1 = 1goat)

3) Your spend on chickens and goats must total to multiples of $5 to use the money remaining after buying cows.... (5 goats = $5 = 1cow)

4) $5 minimum must be spent as...
20chickens + 4goats = 24 animals
40chickens + 3goats = 43 animals
60chickens + 2goats = 62 animals
80chickens + 1goat = 81 animals

5) $100 buys a maximum of 20 cows, but a minimum of $5 needed for chickens and goats, so 19 cows maximum, leaving $5 for 81 other animals

Ah ha!!! A solution is noticed.....!!!

81 animals for $5 is possible, so I can buy 100 animals for $100

19 cows @ $95
1 goat @ $ 1
80 chickens @ $ 4

There may be other solutions, and there may be rigorous solutions, but I leave that to others LOL

Answer 4

Let x be the number of chickens, y be the number of goats and z be the number of cows.

0.05x + y + 5z = 100 ---(1)
x + y + z = 100
y = 100 - x - z --- (2)

Substitute (2) into (1),
0.05x + 100 - x - z + 5z = 100
4z - 0.95x = 0
4z = 0.95x
x = (4/0.95)z

0.05x + y + 5z = 100
0.05(4/0.95)z + y + 5z = 100
(99/19)z + y = 100 --- (3)

x + y + z = 100
(4/0.95)z + y + z = 100
(80/19)z + y = 100 --- (4)

From (3) and (4), we find that y = 100 and z = 0
Since z = 0, x = (4/0.95)z => x = 0

The only combination that's possible is 100 goats and no chickens or cows.

Answer 5

100 goats for $100

Answer 6

With $100 you can buy:
2000 chickens, or
100 goats, or
20 cows.

The only answer to this problem is to buy 100 goats for $100 and buy no chickens and no cows.

Answer 7

0.05x + 1y + 5z = 100
x + y+ z =100

you need another equation to solve

add: should have been paid more attention.. great job guys/ gals

Answer 8

i am thoroughly impressed with 'the otter's solution. i just dismissed it as having 3 variables with only two equations and thought it couldnt be done.

i applaud you!!!

Answer 9

first answer is true..