Expert: Ahmed Salami - 2/23/2011

Question
QUESTION: Here is an example calculation: (12/6)/4 X 10 = 2/4 X 10 = 20/4 = 5

Can the same calculation be solved if two of the numbers, 6 and 10, are missing?

(12/y)/4 X z = 5

Try to use a simple solution if possible.

I thank you for your reply.

ANSWER: Hi Kenneth,
(12/y)/4 x z = 5
(12/4y) x z = 5
(3/y) x z = 5
3z/y = 5
z/y = 5/3
which shows that we can choose our z and y values as we please as long as the ratio of the numbers is 5:3 e.g z = 15, y = 9.

Regards

---------- FOLLOW-UP ----------

QUESTION: Hello:

I want to thank you for your reply. I found it very helpful. I do have a follow-up question.

Do the following have a similar ratio relationship as in my first example? If so, what are they?

(12/y)/z X 10 = 5

(12/6)/y X z = 5

(y/z)/4 X 10 = 5

(y/6)/4 X z = 5

(y/6)/z X 10 = 5


I thank you for your follow-up reply.

Answer
Hi Kenneth,
a) (12/y)/z x 10 = 5
(12/yz) x 10 = 5
120/yz = 5
yz = 120/5
yz = 24

b) (12/6)/y x z = 5
(12/6y) x z = 5
(2/y) x z = 5
2z/y = 5
z/y = 5/2

c) (y/z)/4 x 10 = 5
(y/4z) x 10 = 5
(10y/4z)  = 5
5y/2z = 5
y/z = 2

d) (y/6)/4 x z = 5
(y/24) x z = 5
yz/24 = 5
yz = 24 x 5
yz = 120

e) (y/6)/z x 10 = 5
(y/6z) x 10 = 5
10y/6z = 5
5y/3z = 5
y/z = 3


Regards