Aptitude Tests 4 Me

Basic Numeracy/Quantitative Aptitude

Detailed Solution

320.

321. c: Let the basic fare for the child be \$X.
Therefore, the basic fare for an adult = \$2X.
Let the reservation charge per ticket be \$Y
Hence, an adult ticket will cost 2X + Y = \$216
And ticket for an adult and a childe will cost 2X + Y + X + Y = 3X + 2Y = 327
Solving for X, we get X = 105.
The basic fare of an adult ticket = 2X = 2*105 = \$210

322. c: The sample space S is given by.
S = {(H,T),(H,H),(T,H),(T,T)}
Let E be the event "two heads are obtained".
E = {(H,H)}
We use the formula of the classical probability.
P(E) = n(E) / n(S) = 1 / 4

323. d: A probability is always greater than or equal to 0 and less than or equal to 1, hence only a) and c) above cannot represent probabilities: -0.00010 is less than 0 and 1.001 is greater than 1.

324. c: The sample space S of two dice is shown below.
S = { (1,1),(1,2),(1,3),(1,4),(1,5),(1,6)
(2,1),(2,2),(2,3),(2,4),(2,5),(2,6)
(3,1),(3,2),(3,3),(3,4),(3,5),(3,6)
(4,1),(4,2),(4,3),(4,4),(4,5),(4,6)
(5,1),(5,2),(5,3),(5,4),(5,5),(5,6)
(6,1),(6,2),(6,3),(6,4),(6,5),(6,6) }
Let E be the event "sum equal to 1".
There are no outcomes which correspond to a sum equal to 1, hence P(E) = n(E) / n(S) = 0 / 36 = 0

325. d: Three possible ouctcomes give a sum equal to 4: E = {(1,3),(2,2),(3,1)}, hence. P(E) = n(E) / n(S) = 3 / 36 = 1 / 12

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