Four devices (A, B, C, D) contain a total 225GB

Question

Four devices (A, B, C, D) contain a total 225GB of data.
- The device A has half as much data as device B.
- The device C has five times as much data as device B and device D combined.
- The sum of twice the amount of data on device C and four times the amount of data on the
device D is equal to the difference between 500GB and the amount of data on device B.
Determine the amount of data (in GB) on each device. (4 m

Answer 1

A+B+C+D = 225

 

B = 2A

 

C = 5(B+D) = 5B + 5D

 

2C + 4D = 500 - B --->  B + 2C + 4D  = 500

 

Substitutes B=2A into the other 3 equations, thus eliminating B:

 

3A + C + D = 225

 

C = 10A + 5D

 

2A + 2C + 4D = 500

 

 

Substitutes C = 10A + 5D into the first and third equation to eliminate C

 

3A + (10A + 5D) + D = 225

 

2A + 2(10A + 5D) + 4D = 500

 

 

which simplifies to:

 

13A + 6D = 225  <---- please label this equation ALPHA: we will need it later

 

22A + 14D = 500

 

 

Elimination method finishes it off...

Since the LCD of 6 and 14 is 42,

Multiplies top equation by -7 and bottom by 3:

 

-91A + -42D = -1575

 66A + 42D =  1500

 

Adding them together:

-25A = -75

So,  A = 3

Then B = 2(3) = 6

 

By equation ALPHA in bold above:

13(3) + 6D = 225

39 + 6D = 225

6D = 225 - 39

6D = 186

D = 31

 

Plugging into the original first equation

3 + 6 + C + 31 = 225

C + 40 = 225

C = 185

 

 

The check is left for you.