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For real numbers a, b, and c, is ab = bc + 3?
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23 Aug 2016, 09:10
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57% (01:40) correct 43% (02:14) wrong based on 144 sessions
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Re: For real numbers a, b, and c, is ab = bc + 3?
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23 Aug 2016, 10:47
(1) ab = 3b => It gives no information about 'c' > Insufficient
(2) b + 3 = c => It gives no information about 'a' > Insufficient
Both (1) and (2) together .. Lets substitute these two conditions (1) and (2) in the main eqn (to be proved)
ab = b.(b+3) + 3 = b^2 + 3b + 3
3b = b^2 + 3b + 3 => b^2 = 3 > For this eqn to be equal, 'b' should be imaginary.
But, we are given that a, b and c are real numbers => These numbers cannot be imaginary ..
Hence, b^ 2 = 3 CANNOT be true for any values of a,b and c.. Option (C)



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Re: For real numbers a, b, and c, is ab = bc + 3?
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23 Aug 2016, 10:55
Bunuel wrote: For real numbers a, b, and c, is ab = bc + 3?
(1) ab = 3b (2) b + 3 = c Consider 1) \(ab = 3b\) there are 2 possibilities, \(b = 0\) or \(b\neq{0}\) which means \(a = 3\) or \(a\neq{3}\) and also \(c\) is not known. Not Sufficient , Eliminate A and DConsider 2) \(b + 3 = c\) Substituting on the given equation: \(ab = b(b+3) + 3 = b^2 + 3b + 3\) . Still \(a\) and \(b\) are not known. Not Sufficient , Eliminate BConsider 1) and 2) together (include that 1) has 2 possibilities based on \(b = 0\) and \(b\neq{0}\)) when \(b = 0\), then \(c=3\)and \(ab = bc +3\) becomes \(0 = 0 + 3\) and Hence FAILSwhen \(b\neq{0}\), then \(a = 3\) and \(ab = bc + 3\) becomes \(3b = b(b+3) + 3=> b^2 = 3\) giving values for b which are not real and the Hence Fails, so answer is C, since when 1) and 2) considered together, the equality FAILS. PS: this took more than 2 mins, anyone with faster approach please. +1 for kudos.



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Re: For real numbers a, b, and c, is ab = bc + 3?
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03 Sep 2016, 07:35
C is the correct option: Below is the solution
Statement 1: ab=3b i.e b(a3)=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(ac)=3 ; (c3)(3b3) =3 ; bc+3b =3 ; b(3c)=3 : This matches the required where 3=a.
Hence combining the statements gave the answer .
Therefore Option C :D



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For real numbers a, b, and c, is ab = bc + 3?
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16 Jan 2018, 10:56
ab = bc + 3? ab  bc = 3 ? b(ac) = 3?
Statement 1: ab  3b = 0 => b = 0 OR a = 3 if b = 0 => b(ac) not equals to 3 if a = 3 => we dont know the value of c to say if b(ac) equals to 3
Statement 2: b + 3 = c b(ac) = 3? => (c3)(a3)= 3 ? => don't know about a to conclude => insuff
1 + 2 if b = 0, we saw from statement 1, b(ac) is not equal to 3 if a = 3 and b + 3 = c, b(ac) = 3 ? => (c3)(3c) = 3? => (3c)(3c) = 3? => (3c)^2 = 3? definitely not equal to 3, as squared values are >= 0 sufficient to answer the question as NO => (C)



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Re: For real numbers a, b, and c, is ab = bc + 3?
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20 Jan 2018, 22:13
Bunuel wrote: For real numbers a, b, and c, is ab = bc + 3?
(1) ab = 3b (2) b + 3 = 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 ⇔ (a3)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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