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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.
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The first step of the VA (Variable Approach) method is to modify the original condition and the question. We then recheck the question. We should simplify conditions if necessary.
p = a/(a+b)(a+c)+b/(b+c)(b+a)+c/(c+a)(c+b)
= a(b+c)/(a+b)(a+c)(b+c)+b(c+a)/(b+c)(b+a)(c+a)+c(a+b)/(c+a)(c+b)(a+b) (getting a common denominator)
= ab+ca+bc+ab+ca+bc/(a+b)(b+c)(c+a) (multiplying through the brackets and combining in one fraction)
= 2(ab+ca+bc)/(a+b)(b+c)(c+a) (adding like terms and taking out a common factor of 2)
q = b+c/a+c+a/b+a+b/c
= b+c+a-a/a+c+a+b-b/b+a+b+c-c/c (adding and subtracting variables in order to get a numerator of a + b + c)
= a+b+c/a-a/a+a+b+c/b-b/b+a+b+c/c-c/c (separating fractions)
= a+b+c/a-1+a+b+c/b-1+a+b+c/c-1 (simplifying fractions)
= (a+b+c)(1/a+1/b+1/c)-3 (adding the constants and taking out a common factor of a + b + c)
Condition 2)
1/a+1/b+1/c=0
=> bc/abc+ca/abc+ab/abc=0 (getting a common denominator)
=> ab+bc+ca/abc=0 (combining into one fraction)
=> ab + bc + ca = 0 (multiplying both sides by abc)
Then, we have p = 2(ab+ca+bc)/(a+b)(b+c)(c+a) = 0/(a+b)(b+c)(c+a) = 0 and q = (a+b+c)(1/a+1/b+1/c)-3 = (a+b+c)*0-3 = -3.
Thus, p – q = 0 – (-3) = 3.
Since condition 2) yields a unique solution, it is sufficient.
Condition 1)
Since condition 1) does not yield a unique solution, it is not sufficient.
Therefore, B is the answer.
Answer: B
