A researcher performs an experiment to test a hypothesis that involves the nutrients niacin and retinol. She feeds one group of laboratory rats a daily diet of precisely 125.6 units of niacin and 5,400 units of retinol. She uses two types of commercial pellet food. Food A contains 0.86 units of niacin and 30 units of retinol per gram. Food B contains 0.66 units of niacin and 40 units of retinol per gram. How many grams of each food does she feed this group of rats each day

Respuesta :

Answer:

- 100 grams of food A (x)

- 60 grams of food B (y)

Explanation:

According to this question, the group of laboratory rats are being fed a daily diet of:

- 125.6 units of niacin and 5,400 units of retinol

Let x represent the number of grams needed for food A.

Let y represent the number of grams needed for food B.

This means that:

Food A: Niacin = 0.86x, Retinol = 30x

Food B: Niacin = 0.66y, Retinol = 40y

Since there are 125.6 units of niacin fed to the rats in total, then;

0.86x (A) + 0.66y (B) = 125.6

Since there are 5400 units of retinol fed to the rats in total, then;

30x (A) + 40y (B) = 5400

Solving these two equations simultaneously using elimination method:

0.86x (A) + 0.66y (B) = 125.6 ........(eqn 1)

30x (A) + 40y (B) = 5400 ............ (eqn 2)

Multiply eqn 1 by 30 and eqn 2 by 0.86

30 × 0.86x + 0.66y = 125.6

0.86 × 30x + 40y = 5400

25.8x + 19.8y = 3768 ............ (eqn 3)

25.8x + 34.4y = 4644 ........... (eqn 4)

Subtract eqn 3 from eqn 4

14.6y = 876

Divide both sides by 14.6

y = 876/14.6

y = 60

Substitute value for y (60) into eqn 2

30x + 40y = 5400

30x + 40(60) = 5400

30x + 2400 = 5400

30x = 5400 - 2400

30x = 3000

x = 3000/30

x = 100

Therefore, she feeds the group of rats 100grams of food A (x) and 60 grams of food B (y).