For real numbers a, b, and c, is ab = bc + 3?

C is the correct option: Below is the solution

Statement 1: ab=3b i.e b(a-3)=0 . Either b=0 or a=3 : We leave this, as no solution we get
Statement 2: b+3=c No info about anything else, so leave this too

Combining both 1 & 2

If we take b=0 , then c=3 according to statement 2 .
Try inputting this in ab= bc+3 . 0 is not equal to 3 Therefore ruled out

Now take a=3 and b+3 =c
Input these in ab=bc+3 ; b(a-c)=3 ; (c-3)(3-b-3) =3 ; -bc+3b =3 ; b(3-c)=3 : This matches the required where 3=a.

Hence combining the statements gave the answer .

Therefore Option C :D

ab = bc + 3?
ab - bc = 3 ?
b(a-c) = 3?

Statement 1:
ab - 3b = 0
=> b = 0 OR a = 3
if b = 0 => b(a-c) not equals to 3
if a = 3 => we dont know the value of c to say if b(a-c) equals to 3

Statement 2:
b + 3 = c
b(a-c) = 3? => (c-3)(a-3)= 3 ? => don't know about a to conclude => insuff

1 + 2
if b = 0, we saw from statement 1, b(a-c) is not equal to 3
if a = 3 and b + 3 = c, b(a-c) = 3 ? => (c-3)(3-c) = 3? => (3-c)(3-c) = -3? => (3-c)^2 = -3? definitely not equal to -3, as squared values are >= 0
sufficient to answer the question as NO => (C)

Forget conventional ways of solving math questions. For DS problems, the VA (Variable Approach) method is the quickest and easiest way to find the answer without actually solving the problem. Remember that equal numbers of variables and independent equations ensure a solution.

Since we have 3 variables and 0 equations, E is most likely to be the answer. So, we should consider 1) & 2) first.

Conditions 1) & 2):
ab = 3b
⇔ ab - 3b = 0
⇔ (a-3)b = 0
⇔ a = 3 or b = 0

Case 1: \(a = 3\)
\(ab = 3b\) and \(bc + 3 = b(b+3)+3\)
\(ab = bc + 3\)
\(⇔ 3b = b(b+3)+3\)
\(⇔ 3b = b^2 + 3b + 3\)
\(⇔ 0 = b^2 + 3\), which is false.
The answer is "No".

Case 2: \(b = 0\)
\(ab = bc + 3\)
\(⇔ 0a = 0c + 3\)
\(⇔ 0 = 0 + 3\), which is false.
The answer is "No".

The answer is always "No".

They are sufficient by CMT (Common Mistake Type) 1, since the answer "no" also means the condition is sufficient.

The answer is C.

In cases where 3 or more additional equations are required, such as for original conditions with “3 variables”, or “4 variables and 1 equation”, or “5 variables and 2 equations”, conditions 1) and 2) usually supply only one additional equation. Therefore, there is an 80% chance that E is the answer, a 15% chance that C is the answer, and a 5% chance that the answer is A, B or D. Since E (i.e. conditions 1) & 2) are NOT sufficient, when taken together) is most likely to be the answer, it is generally most efficient to begin by checking the sufficiency of conditions 1) and 2), when taken together. Obviously, there may be occasions on which the answer is A, B, C or D.

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