Question

6. A hospital needs a supply of an expensive medicine. Company A has the most

Answer 1

We are required to find the total milligrams if medicine the hospital can get from the three companies.

The total quantity of medicine the hospital can get from the three companies is 2.75 milligrams

There are there companies:

Company A

Company B

Company C

Company A = 1.3 milligrams

Company A is 0.9 milligrams more than Company C's

A = 0.9 + C

1.3 = 0.9 + C

1.3 - 0.9 = C

0.4 = C

C = 0.4 milligrams

Company A is also twice the difference between the weight of

Company B's supply and Company C's

A = 2(B - C)

1.3 = 2(B - 0.4)

1.3 = 2B - 0.8

1.3 + 0.8 = 2B

2.1 = 2B

Divide both sides by 2

B = 2.1/2

B = 1.05 milligrams

Therefore,

A + B + C

= 1.3 + 1.05 + 0.4

= 2.75 milligrams

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Answer 2

Answer:

The hospital gets $2.75\:\rm{mg}$ of medicines from these three companies.

Step-by-step explanation:

Company A has $1.3$ milligrams, which is $0.9$ milligram more than Company C's.

If $A,B,C$, are the medicines produced by Companies A,B,C, respectively,

then, $A=1.3\:\rm{mg}$, and its expression is given as,

$\begin{aligned}A&=0.9+C\\1.3&=0.9+C\\C&=1.3-0.9\\C&=0.4\:\rm{mg}\end{aligned}$

Company A has $1.3$ milligrams, which is twice the difference between the weight of Company B's supply and Company C's.

So, its expression is given as,

$A=2(B-C)$

Substituting values of $A,C$, the expression becomes,

$\begin{aligned}1.3&=2(B-0.4)\\0.65&=B-0.4\\B&=0.65+0.4\\B&=1.05\:\rm{mg}\end{aligned}$

So, the hospital gets,

$\begin{aligned}A+B+C&=1.3+0.4+1.05\\&=2.75\:\rm{mg}\end{aligned}$

of medicines.