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29 Jan 2016, 03:28
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C
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If a, b, and c are not equal to zero, what is the difference between the maximum and minimum value of S?
\(S=1 + \frac{|a|}{a} + \frac{2|b|}{b} + \frac{3|ab|}{ab} −\frac{4|c|}{c}\)
A. 12
B. 14
C. 18
D. 20
E. 22
\(S=1 + \frac{|a|}{a} + \frac{2|b|}{b} + \frac{3|ab|}{ab} −\frac{4|c|}{c}\)
A. 12
B. 14
C. 18
D. 20
E. 22
29 Jan 2016, 05:31
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ajmalsyeed987
\(S=1 + \frac{|a|}{a} + \frac{2|b|}{b} + \frac{3|ab|}{ab} −\frac{4|c|}{c}\)
To maximize S , a and b need to be positive and c needs to be negative
s(max) = 1 + 1 + 2 + 3 + 4 = 11
To minimize S , a needs to be positive , b needs to be negative and c needs to be positive
s(min) = 1 + 1 -2 -3 - 4 = - 7
Difference between max and min value = 11 - (-7) = 18
Answer C
ENGRTOMBA2018
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Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
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WE:Engineering (Aerospace and Defense)
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29 Jan 2016, 06:16
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ajmalsyeed987
You are missing the fact that when a,b both are <0, then the product ab > 0 and hence |ab| > 0 making your "minimum" sum = 1-1-2+3-4 = -3 , not a minimum.
If a, b, and c are not equal to zero, what is the difference between the maximum and minimum value of S?
\(S=1+\frac{|a|}{a}+2\frac{|b|}{b}+3\frac{|ab|}{ab}−4\frac{|c|}{c}\)
1-1-2+3-4 = -4 and NOT -7
In order to minimize S, you need to have opposite signs for ab in order to make |ab| negative as its coefficient (=3) is the highest. Lets say a>0 and b<0 ---> |ab| = -ab, giving you minimum value = 1+1-2-3-4 = -7 and with a<0, b>0 you get, 1-1+2-3-4 = -5.
Thus the minimum value of S = -7
The maximum value of S will be when a>0, b>0 and c<0,
\(S=1+\frac{|a|}{a}+2\frac{|b|}{b}+3\frac{|ab|}{ab}−4\frac{|c|}{c}=1+1+2+3+4 = 11\)
Thus the difference = 11-(-7)=18
Hence C is the correct answer.
Hope this helps.
