please answer all three questions im very confused. Show all your work. Answers

ID: 141342 • Letter: P

Question

please answer all three questions im very confused. Show all your work. Answers are to be carried to two full decimal places. When working with micronutrients like calcium or phosphorus, carry your answer to 4 decimal places

1. Brome hay contains 10.4% moisture and 13.9% CP on a DM basis.

(a) What is the % DM of the brome hay? __________________________

(b) How many pounds of DM are there in an 89 lb bale of hay? ___________________

(d) On an as fed basis? __________________________

2. Alfalfa hay contains 19% CP on a DM basis. If a 1,000 lb cow consumes 2% of her body weight in DM alfalfa per day, how many lbs of fresh alfalfa would she need to consume if fresh alfalfa had 75% moisture? ________________________________

How many lb CP on a DM basis would she eat per day? ___________________

How many lb CP as fed per day? ____________________

Extra Credit (2 points) Calculate the % CP and the % TDN of the ration below on an as fed and on a DM basis 2 You will need to add additional columns to the existing table to complete it. Remember that to find DM pounds of a nutrient, you need to know the DM lb of the feed and the % nutrient on a DM basis. Likewise, to calculate the as fed pounds of nutrient, you need to know the as fed lb of the feed and the % nutrient on an as fed basis. REMEMBER the two glasses of milk used in the class demonstration, percent nutrient changes with the addition of water due to the "dilution effect," but amount does not! Ib CP %DM DM DM as fed Ingredient % CP Ib CP % TDN Cotton seed hulls DM basis % CP DM basis %TDN As fed basis % CP As fed basis % TDN 0 Item Number Description 1,739.24 200.00 Alfalfa Meal Yellow Greas 0 onium chloride Sheep Salt Bevate. 5.65g Vitamin A Pmx 30 KIU/gm Vitamin D Pmx 4 MIUlb 8 Page 3 of 4 Vitamin E-50 (227 KIU/lb) Manganese Oxide Ingredients 2,000.00

Explanation / Answer

Q1. Brome hay contains 10.4% moisture and 13.9% CP on a DM basis.

(a) What is the % DM of the brome hay?

Given 10.4 % moisture,

We know that, % of moisture + % DM = 100

So, % DM = 100-10.4 = 89.6 %

(b) How many pounds of DM are there in an 89 lb bale of hay?

As % DM is 89.6 i.e.
In 100 lb of brome hay, DM = 89.6 lb

In 89 lb of brome hay, DM = (89.6 x 89)/100 = 79.74 lb or 79.74 pounds

On DM basis, % CP = 13.9.

So, actual amount of CP (in lbs)= (13.9 x amount of DM)/100 = (13.9 x 79.74)/100 = 11.08 lb

(d) On an as fed basis?

% CP(as fed) = (% CP(DM) x % DM)/100 = 12.45 %

Now, So, actual amount of CP (in lbs)= (12.45 x amount of feed)/100 = (12.45 x 89)/100 = 11.08 lbs

Hence Actual amount of ingredient remains same whether it is on as fed or DM basis. So, it is 11.08 lb

Q2 Alfalfa hay contains 19% CP on a DM basis. If a 1,000 lb cow consumes 2% of her body weight in DM alfalfa per day, how many lbs of fresh alfalfa would she need to consume if fresh alfalfa had 75% moisture?

Wt of cow = 1000 lb

DM (in lb) required = 2% of 1000 = 20 lb

% DM of alfa alfa = 100 -75 = 25

If 100% DM, feed requred = 20 lb

If 25 % DM, feed required = (20 x 25)/100 = 80lb

How many lb CP on a DM basis would she eat per day?

Given % CP = 19

Amount of CP (DM basis) = (19 x amount of DM)/100 = (19 x 20)/100 = 3.8 lb

How many lb CP as fed per day?

As we know from Q1, Actual amount of ingredient remains same whether it is on as fed or DM basis. So, Amount of CP (as fed basis) = 3.8 lb

Q3 As per table provided, the values were calculated and by formula given in table. New column DM (lb) has been added to the table. All values in bold have been calculated and added in table.

Values of TDN and CP are same in both as fed and DM cases.

Ingredient

As fed (lb)

% DM

DM (lb) = (%DM x As fed in lb)/100

% CP (DM basis)

CP (lb, DM basis) = (%CP x DM in lb)/100

CP (lb, as fed)

% TDN (DM basis)

TDN (lb, DM basis) = (%TDN x DM in lb)/100

TDN (lb, as fed)

Alfa alfa hay

90.5

(90.5 x 50)/100 = 45.25

22.6

(22.6 x 45.25)/100 = 10.23

10.23

57.2

(57.2 x 45.25)/100 = 25.88

25.88

Barley

88.6

(88.6 x 20)/100 = 17.72

13.0

(13 x 17.72)/100 = 2.30

2.30

83.2

(83.2 x 17.72)/100 = 14.74

14.74

Cotton seed hulls

90.4

(90.4 x 30)/100 = 27.12

4.2

(4.2 x 27.12)/100 = 1.14

1.14

44.9

(44.9 x 27.12)/100 = 12.18

12.18

Total

100

---

90.09

----

13.67

-----

52.80

% CP (DM)= { Total CP (in lb, DM basis) / Total DM (in lbs)} x 100 = {13.67/90.09} x 100 = 15.17 %

% CP (as fed)= { Total CP (in lb, as fed) / Total feed (in lbs)} x 100 = {13.67/100} x 100 = 13.67 %

% TDN (DM)= { Total TDN (in lb, DM basis) / Total DM (in lbs)} x 100 = {52.80/90.09} x 100 = 58.61 %

% TDN (as fed)= { Total TDN (in lb, as fed) / Total feed (in lbs)} x 100 = {52.80/100} x 100 = 52.80 %