MHT-CET Full Test-9 Mathematics Que-35 Solution

Question-35 A number x is chosen at random from the set {1,2,3,4 \ldots \ldots, 100}. Define the event : A such that the chosen number x satisfies \frac{(x-10)(x-50)}{(x-30)} \geq 0. Then P(A) is

A) 0.51

B) 0.70

C) 0.71

D) 0.20

Answer: C) 0.71

Explanation

Given \frac{(x-10)(x-50)}{(x-30)} \geq 0

Let x \geq 10, x \geq 50 equation will be true \forall x \geq 50

as \left(\frac{x-50}{x-30}\right) \geq 0, \forall x \in[10,30)

\frac{(x-10)(x-50)}{x-30} \geq 0 \forall x \in[10,30)

Total value of \boldsymbol{x} between \mathbf{1 0} to \mathbf{3 0} is \mathbf{2 0}.

Total values of x between \mathbf{5 0} to \mathbf{1 0 0} including \mathbf{5 0} and \mathbf{1 0 0} is \mathbf{5 1}

Total values of x=51+20=71

P(\mathrm{~A})=\frac{71}{100}=0.71
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