58.In the xy-plane, at what two points does the graph of y = (x+a)(x+b) intersect the x-axis? 1) a+b = -1 2) The graph intersects the y-axis at (0,-6)
Solution:
1. Insufficient. All we have it x^2 -x + ab = 0. Cannot solve for x 2. insufficient. y = x^2 + (a+b)x + ab implies ab = -6. Cannot solve for x Using 1 and 2, x^2 - x - 6 = 0 x = 3 & -2 So, C is the correct answer
59.If k#0, 1, or -1, is 1/k>0? 1. 1/(k-1)>0 2. 1/(k+1)>0
Solution:
1 - If 1/(k-1) > 0, then k > 1 and hence positive. So, 1/k > 0.Sufficient 2 - If 1/(k+1) > 0, then k > -1. K is not given to be an integer and can take values -0.5, 0.5, 2 etc. So, 1/k can be positive or negative. Insufficient. Hence A
60.If q is a positive integer less than 17 and r is the remainder when 17 is divided by q, what is the value of r? 1. q>10 2. q=2^k, where k is a positive integer
Solution:
1 - Insufficient. q Can be 11...16 and each yields a different value for r 2 - sufficient. Values for q are 2, 4, 8 and 16 17 / 2 remainder is 1 17 / 4 remainder is 1 17 / 8 remainder is 1 17 / 16 remainder is 1 Hence B
Solution:
1. Insufficient. All we have it x^2 -x + ab = 0. Cannot solve for x 2. insufficient. y = x^2 + (a+b)x + ab implies ab = -6. Cannot solve for x Using 1 and 2, x^2 - x - 6 = 0 x = 3 & -2 So, C is the correct answer
59.If k#0, 1, or -1, is 1/k>0? 1. 1/(k-1)>0 2. 1/(k+1)>0
Solution:
1 - If 1/(k-1) > 0, then k > 1 and hence positive. So, 1/k > 0.Sufficient 2 - If 1/(k+1) > 0, then k > -1. K is not given to be an integer and can take values -0.5, 0.5, 2 etc. So, 1/k can be positive or negative. Insufficient. Hence A
60.If q is a positive integer less than 17 and r is the remainder when 17 is divided by q, what is the value of r? 1. q>10 2. q=2^k, where k is a positive integer
Solution:
1 - Insufficient. q Can be 11...16 and each yields a different value for r 2 - sufficient. Values for q are 2, 4, 8 and 16 17 / 2 remainder is 1 17 / 4 remainder is 1 17 / 8 remainder is 1 17 / 16 remainder is 1 Hence B
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