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Am I missing something here. My ans was C. Stat 1 gives two results; x= 11,-17 Stat 2 gives two results; x= 11,-15
Combining these two, we get that x=11 satisfies the statements and answers the question. I have read this reasoning somewhere. Pls suggest whether the logic is correct or not, or what should be the strategy behind these type of questions when one variable satisfies both statements and in turn becomes sufficient to answer the question. Thanks
A. gives us two values of x i.e. x = 11 or x = -11. However when you substitute these values back into equation given in A, x = 11 is the only valid value. Hence we can find mod ( x + 7 ) = 18
Hence A sufficient.
B. also gives us two values of x, i.e. 11 and -11. Resubstitute both values and validate the equation given in B. It holds true only for x = 11. Hence we are able to find mod (x + 7).
A. gives us two values of x i.e. x = 11 or x = -11. However when you substitute these values back into equation given in A, x = 11 is the only valid value. Hence we can find mod ( x + 7 ) = 18
Hence A sufficient.
B. also gives us two values of x, i.e. 11 and -11. Resubstitute both values and validate the equation given in B. It holds true only for x = 11. Hence we are able to find mod (x + 7).
Hence B is sufficient as well.
Hence D is correct answer. Thank You.
Thanks, Akhil M.Parekh
B gives 2 values as 11 and -15
\((x+2)^2= 13 ^2\)
\((x+2)=+-13\)
x=11 , -15
Hence 2 different values
Can you explain how did you get values as 11 and -11 in statement B?
A. gives us two values of x i.e. x = 11 or x = -11. However when you substitute these values back into equation given in A, x = 11 is the only valid value. Hence we can find mod ( x + 7 ) = 18
Hence A sufficient.
B. also gives us two values of x, i.e. 11 and -11. Resubstitute both values and validate the equation given in B. It holds true only for x = 11. Hence we are able to find mod (x + 7).
Hence B is sufficient as well.
Hence D is correct answer. Thank You.
Thanks, Akhil M.Parekh
I think its C.
1) This raises two values, X=11 and -17. Therefore, insufficient. 2) This again raises to two values, X=11 and -15. Therefore, insufficient.
Considering 1 and 2, the common value X=11 is considered. Therefore, sufficient.
Statement (1) gives two values, x = 11, and x = -17. Both of these values adequately satisfy the equation |x+3|=14. Plug them into |x+7| to get |11+7|=18 and |-17+7|=10. Two different values, thus we can't determine a single solution for x.
Statement (2) also gives two values, x = 11 and x = -15. Both of these values satisfy the equation (x+2)^2=169. Plug them into |x+7| to get |11+7|=18 and |-15+7|=8. Two different values again, thus we can't determine a single solution for x.
Combine (1) and (2) to get x = 11, which results in a single value for |x+7|.
(1) \(|x+3|=14\) --> \(x=11\) or \(x=-17\), so \(|x+7|=18\) or \(|x+7|=10\). Not sufficient.
(2) \((x+2)^2=169\) --> \(x=11\) or \(x=-15\), so \(|x+7|=18\) or \(|x+7|=8\). Not sufficient.
(1)+(2) \(|x+7|=18\). Sufficient.
Answer: C.
The working is fairly simple here. My only doubt is we getting lX+7l=18 in each of the statement 1 and 2. So, the answer is C But we also getting lX+7l = 10 (from statement 1 ) and lX+7l=8 (statement 2). So why we ignoring this values of lX+7l. And both of these are different values of lX+7l Thanks
(1) \(|x+3|=14\) --> \(x=11\) or \(x=-17\), so \(|x+7|=18\) or \(|x+7|=10\). Not sufficient.
(2) \((x+2)^2=169\) --> \(x=11\) or \(x=-15\), so \(|x+7|=18\) or \(|x+7|=8\). Not sufficient.
(1)+(2) \(|x+7|=18\). Sufficient.
Answer: C.
The working is fairly simple here. My only doubt is we getting lX+7l=18 in each of the statement 1 and 2. So, the answer is C But we also getting lX+7l = 10 (from statement 1 ) and lX+7l=8 (statement 2). So why we ignoring this values of lX+7l. And both of these are different values of lX+7l Thanks
In DS when you have say a = 1 or a = 2 in (1) and a = 1 or a = 3 in (2), then when considering (1)+(2) you are taking the common value, so a = 1.
The same in the given question, both (1) and (2) give two possible values of |x+7|. When considering (1)+(2) |x+7| could only be one of those value, the one which is common for (1) and (2).
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67% (01:29) correct 
