Aptitude Tests 4 Me

Basic Numeracy/Quantitative Aptitude

Detailed Solution

326. d: All possible ouctcomes, E = S, give a sum less than 13, hence. P(E) = n(E) / n(S) = 36 / 36 = 1

327. b: The sample space S of the experiment described in question 5 is as follows
S = { (1,H),(2,H),(3,H),(4,H),(5,H),(6,H)
(1,T),(2,T),(3,T),(4,T),(5,T),(6,T)}
Let E be the event "the die shows an odd number and the coin shows a head". Event E may be described as follows
E={(1,H),(3,H),(5,H)}
The probability P(E) is given by
P(E) = n(E) / n(S) = 3 / 12 = 1 / 4

328. c: We first construct a table of frequencies that gives the marbles color distributions as follows color frequency
red 3
green 7
white 10
We now use the empirical formula of the probability
Frequency for white color P(E)= ________________________________________________ Total frequencies in the above table
= 10 / 20 = 1 / 2

329.

330.

331. Many students would use their calculators on each step of this problem.
Electric bill: \$150 x 0.85 = \$127.50
Gift for mother: \$150 x 0.05 = \$7.50
Total spent: \$127.50 + \$7.50 = \$135
Amount left: \$150 – \$135 = \$15
Good test-takers would have proceeded as follows, finishing the problem in less time than it takes to calculate the first percent: John used 90% of his money, so he had 10% left; and 10% of \$150 is \$15.

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