# Please help!?

A farm has pigs, cows and chickens. The number of cows is one more than the number of pigs. The number of chickens is three times the number of cows. If the farm has between 201 and 210 animals, what is the maximum number of chickens on the farm?

### 5 Answers

- PhilomelLv 71 month ago
cows+cows-1+cows*3=201>210

excel

cows=41, pigs =40, ckn=123= 204

or

cows=42, pigs=41, ckns=126 = 209

max ckn =126

- KrishnamurthyLv 71 month ago
A farm has pigs, cows and chickens.

The number of cows is one more than the number of pigs.

The number of chickens is three times the number of cows.

If the farm has between 201 and 210 animals, what is the maximum number of chickens on the farm?

- AshLv 71 month ago
Cows = Pigs + 1

Pigs = Cows - 1 ......(1)

Chicken = 3 Cows

Cows = ⅓Chicken .....(2)

plug (2) in (1)

Pigs = ⅓Chicken - 1 ......(3)

201≤ Total animals ≤ 210

201≤ Cows + Pigs + Chicken ≤ 210

201≤ ⅓Chicken + ⅓Chicken - 1 + Chicken ≤ 210

201≤ (5/3)Chicken - 1 ≤ 210

Add 1

201+1≤ (5/3)Chicken - 1+1 ≤ 210+1

202 ≤ (5/3)Chicken ≤ 211

Multiply by ⅗

202*⅗ ≤ (5/3)Chicken*⅗ ≤ 211*⅗

121.2 ≤ Chicken*⅗ ≤ 126.6

The max number of chicken are 126

Check:

Chicken = 126

Cows = ⅓Chicken = ⅓(126) = 42

Pigs = Cows - 1 = 42 - 1 = 41

Total 126+42+41 = 209 !!!

- llafferLv 71 month ago
Let p = number of pigs

Let w = number of cows

Let n = number of chickens

The number of cows is 1 more than the number of pigs:

w = p + 1

The number of chickens is 3 times the number of cows:

n = 3w

There are between 201 and 210 total animals:

p + w + n = x where 201 ≤ x ≤ 210

We know that the there variables must all be integers. So let's do some substitutions to see what we end up with and compare the result to the possible values.

Let's leave the equation in terms of x as-is for now. I'll start with substituting the second equation into it and simplifying:

p + w + n = x

p + w + 3w = x

p + 4w = x

Now substitute the first equation into this and simplify:

p + 4(p + 1) = x

p + 4p + 4 = x

5p + 4 = x

Solve this for p:

5p = x - 4

p = (x - 4) / 5

So in order for p to be an integer, x must be 4 more than a multiple of 5 (or one less than a multiple of 5).

So that gives us possilbe values of x of:

204 and 209

Let's start with substituting those values into x and solving for p, n, and w and see if any of them don't make sense.

p = (x - 4) / 5 and p = (x - 4) / 5

p = (204 - 4) / 5 and p = (209 - 4) / 5

p = 200 / 5 and p = 205 / 5

p = 40 and p = 41

w = p + 1

w = 40 + 1 and w = 41 + 1

w = 41 and w = 42

n = 3w

n = 3w and n = 3w

n = 3(41) and n = 3(42)

n = 123 and n = 126

If we then add up the three values in each possible solution we get:

40 + 41 + 123 and 41 + 42 + 126

204 and 209

Both of these sets are possible solutions.

We are asked to find the maximum possible number of chickens so the answer is the larger of the two n's:

The maximum number of chickens that satisfies this criteria is 126.

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- fcas80Lv 71 month ago
Let K=cows, C=chickens, P=pigs.

I used trial and error and got C=126 when K=42, P=41.