Aptitude Tests 4 Me

Basic Numeracy/Quantitative Aptitude

Detailed Solution

170. Test choice C: x = 21.
• Is the average of 2, 7, and 21 equal to 12?

171. • If 146 is the largest of the five integers, the integers are 146, 144, 142, 140, and 138. Quickly add them on your calculator. The sum is 710.
• Since 710 is too small, eliminate C, D.
• If you noticed that the amount by which 710 is too small is 30, you should realize that each of the five numbers needs to be increased by 6; therefore, the largest is 152 (B).
• If you didn’t notice, just try 152, and see that it works.
This solution is easy, and it avoids having to set up and solve the required equation: n + (n + 2) + (n + 4) + (n + 6) + (n + 8) = 740.

172. Solution.. Test choice C: \$100,000.
• If the largest prize is \$100,000, the second largest is \$50,000
(they are in the ratio of 6:3 = 2:1). The third prize is much less than \$50,000, so all three add up to less than \$200,000.
• Eliminate A, B, and C; and, since \$100,000 is way too small, try D.
• Test choice E. The prizes are \$150,000, \$75,000, and \$25,000 (one-third of \$75,000). Their total is \$250,000. The answer is D.
Again, TACTIC 5 lets you avoid the algebra if you can’t do it or just don’t want to. Here is the correct solution. By TACTIC D1 the three prizes are x, 3x, and 6x. Therefore, x + 3x + 6x = \$250,000; 10x = \$250,000.
So, x = \$25,000 and 6x = \$150,000.

173.

174. Since you want the largest value of n for which 112/2n is an integer, start by testing 4 choice D, the largest of the choices

175. Here, you have to test each of the choices until you find one that satisfies the condition that it is not equal to 3/5 . If, as you glance at the choices to see if any would be easier to test than the others, you happen to notice that 60% = 0.6, then you can immediately eliminate choices B and C, since it is impossible that both are correct.

176.

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