yashii9 wrote:
Vips0000 wrote:
yashii9 wrote:
A.
q(R+S) =(R+S)
Let the value of r and s be anything q will always be equal to 1.
Not correct approach.
what if R=S=0 then?
Q(R+S) = 1*(R+S) => Q=1 ?
as well as Q(R+S)=10*(R+S) =>Q=10 ?
and it can be shown to be true for any other number... So first you can not claim Q=1 this way.
Second, the conclusion u reached
Q(R+S) =(R+S) => Q=1
this is arrived when you've divided both sides by (R+S). Which can not be done if R+S = 0, as it will be illegal operation to divide by 0. Unless you are certain about r+s <>0, you can not reach to conclusion.
Dear vips - if r or s is 0... the equation will be 0 on both side..and u can not get the value of q.
if either r or s is 0 u still get q=1
isnt it?
Also if we are to consider this as an unanswerable case - then even point 2 wont help. r not equal to -s...could well mean r =s =0 again no answer and then reach the conclusion to be E.
What do you say?
Try to follow these steps:
Q(R+S) = 1*(R+S) => Q(R+S) - (R+S) = 0
=>
(Q-1)(R+S)=0
Apply number property here.. This statement (Q-1)(R+S)=0 means
it is possible only if,
case a. Q-1 =0
or
case b. R+S = 0
or
case c. both are 0
We dont have any clue as to which one of these 3 is true.
Now if use statement 2 in question R <> -S which means R+S <> 0, hence if we combine it to the 3 possiblities found in statement 1 : We get choce b is not possible , choice c is not possible. Hence we are left with choice a and Q-1 has to be 0.
So combining the statement gives u ans.. not individual statements.
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