Question 1201935
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Answer: <font color=red size=4>Choice (A)  29/12</font>




Explanation:


I'll label the equations eq1, eq2, and eq3


Divide eq1 over eq2.
The LHS (left hand sides) and RHS (right hand sides) will divide separately.


eq1/eq2 
(2ab)/(3bc) = 1/2
(2a)/(3c) = 1/2
a/c = (1/2)*(3/2)
a/c = 3/4
4a = 3c


Now multiply both sides by 'c' so the 4a becomes 4ca.


c*4a = c*3c
4ca = 3c^2 
3 = 3c^2 ... replace 4ca with 3; valid because of eq3
3c^2 = 3 
c^2 = 3/3
c^2 = 1
c = sqrt(1) ... since c is positive
c = 1


Go back to eq3 to determine 'a'.
4ca = 3
4*1a = 3
4a = 3
a = 3/4


Now use either eq1 or eq2 to find b
2ab = 1
2*(3/4)*b = 1
(6/4)*b = 1
(3/2)*b = 1
b = 1*(2/3)
b = 2/3
or
3bc = 2
3b*1 = 2
3b = 2
b = 2/3


We found these values
a = 3/4
b = 2/3
c = 1


Add them up.
a + b + c
(3/4) + (2/3) + 1
(3/4)*(3/3) + (2/3)*(4/4) + 1*(12/12)
(9/12) + (8/12) + (12/12)
(9+8+12)/12
<font color=red>29/12</font>


Extra info:
The improper fraction 29/12 converts to the mixed number 2 & 5/12 because of the scratch work below.
29/12 = (24+5)/12
29/12 = (24/12)+(5/12)
29/12 = 2+(5/12)
29/12 = 2 & 5/12
Or you could use long division to find that 29/12 = 2 remainder 5. Both of these values are found in "2 & 5/12".
The template is quotient & remainder/12.


29/12 = 2.41667 approximately
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