NIMCET 2021 Question Paper with solutions

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Question : 15 of 120

Marks: +1, -0

A number

x

is selected from natural numbers

1,2, \......, 100

. Then the probability for

x+{100}/{x}>29

is

${37}/{50}$
${39}/{50}$
${41}/{50}$
${43}/{50}$

Solution:

x+{100}/{x}>29

Here

{x^{2}+100}/{x}>29

x^{2}+100>29 x

x^{2}-29 x+100>0

(x-4)(x-25)>0

x

can takes values less than

4 \&

greater than 25

x < 4 \text "or "x>25

1,2,3 \text "or "26,27 ....... 100

Here total number,

x

can take is 78
Then Required probability

\⇒ {78}/{100} \⇒{39}/{50}

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