The question can be reduced to Is PZ = 0?
St1: p=0
Sufficient (PZ will always be 0)
St2: z=0
Sufficient (PZ will always be 0)
Ans D
PZ
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srcc25anu
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No It CANT.greenwich wrote:Can the question be further reduced to p+pz=p ---> p(1+z)=p ---> 1+z=1 ---> z=0?
P + PZ = P => P - P + PZ = 0 OR PZ = 0
The way you are approaching this, P(1+Z) = P => P(1+Z-1)=0 THAT FURTHER IMPLIES EITHER P = 0 OR (1 + Z - 1) OR Z = 0. Same as above solution.
If you strike out P from your calculations, you are eliminating one variable altogether which is Not right.
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I just wanted to clear up one thing about the above comment.srcc25anu wrote:No It CANT.greenwich wrote:Can the question be further reduced to p+pz=p ---> p(1+z)=p ---> 1+z=1 ---> z=0?
P + PZ = P => P - P + PZ = 0 OR PZ = 0
The way you are approaching this, P(1+Z) = P => P(1+Z-1)=0 THAT FURTHER IMPLIES EITHER P = 0 OR (1 + Z - 1) OR Z = 0. Same as above solution.
If you strike out P from your calculations, you are eliminating one variable altogether which is Not right.
Eliminating p (by dividing both sides by p) is not incorrect because it results in eliminating one variable altogether.
For example, if the target question asked, "Is p+z=p?" it would be perfectly acceptable to subtract p from both sides to get the equivalent target question, "Is z=0?" [even though we are eliminating one variable altogether]
The reason has to do with dividing both sides by a variable, because may be inadvertently dividing by zero, which can yield bizarre results.
Consider the equation (2)(0) = (3)(0) [true equation]
Divide both sides by 0: 2 = 3 [not true]
Of course dividing both sides by 0 would not yield 2 = 3, but this is what's happening if we take p(1+z)= p and divide both sides by p to get 1 = 1+z. Notice that, on the left side, we're saying that p/p = 1. This is true for most values of p, however p/p does not equal 1 when p=0.
Takeaway: you can add or subtract variables from both sides of an equation and there will be no possibilities for unintended results. However, if you multiply or divide both sides of an equation by a variable you should ensure that there's no possibility that the variable equals zero, otherwise this can lead to unintended results.
Cheers,
Brent
So what is the appropriate rephrasing? and answer?
if subtract get
p + pz = p
p + pz - p = 0
does pz = 0
statement 1 p=0 will definite yes
statement 2 z=0 will definite yes
if factor out
p + pz = p
P(1+z) = p
1+z=1
does z=0
statement 1 p=0 does not apply to rephrasing
statement 2 z=0 will definite yes
without rephrasing both statements work
0+0=0
P+0=P
if subtract get
p + pz = p
p + pz - p = 0
does pz = 0
statement 1 p=0 will definite yes
statement 2 z=0 will definite yes
if factor out
p + pz = p
P(1+z) = p
1+z=1
does z=0
statement 1 p=0 does not apply to rephrasing
statement 2 z=0 will definite yes
without rephrasing both statements work
0+0=0
P+0=P
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I should first mention that we're not required to rephrase any target question. If we leave it as is, both statements are sufficient (as you suggest).J N wrote:So what is the appropriate rephrasing? and answer?
if subtract get
p + pz = p
p + pz - p = 0
does pz = 0
statement 1 p=0 will definite yes
statement 2 z=0 will definite yes
if factor out
p + pz = p
P(1+z) = p
1+z=1
does z=0
statement 1 p=0 does not apply to rephrasing
statement 2 z=0 will definite yes
without rephrasing both statements work
0+0=0
P+0=P
However, we cannot rephrase this target question by taking the equation p + pz = p and factoring out the p to get 1 + z = 1.
This is not acceptable because p + pz = p is not necessarily equal to 1 + z = 1.
For example, if p=0 and z=1, we can see that the equation p + pz = p holds true, however the equation 1 + z = 1 does not hold true. Given this we can't replace the equation p + pz = p with the equation 1 + z = 1.
You'll find a more detailed explanation in my earlier earlier post.
So, IF we're going to rephrase the target question, the only valid way to do so is to take "Is p + pz = p?" and subtract p from both sides to get the equivalent target question, "Is pz = 0?"
Cheers,
Brent













