Question

The probabilities of a randomly selected integer N after some algebraic operation on it are given in Column-II. The algebraic operation on N is given in Column-I. Match the entries.

Column I		Column II
A	Squaring the number $N$ ending in 4	P	$\frac{9}{100}$
B	Raising the number $N$ to the fourth power, ending in 1.	Q	$\frac{12}{100}$
C	Multiplying the number $N$ by an arbitrary integer and units place of the product is 5	R	$\frac{20}{100}$
D	Multiplying the number $N$ by an arbitrary integer and units place of the product is 6	S	$\frac{30}{100}$
		T	$\frac{40}{100}$

• A. (A) P; (B) T ; (C) P ; (D) Q
• B. (A) S; (B) T ; (C) P ; (D) Q
• C. (A) R; (B) T ; (C) P ; (D) Q
• D. (A) Q; (B) T ; (C) P ; (D) Q

Answer 1

Answer: (C)

(A) Square of a number can end in $0,1,4,5,6,9$. Now $N ^2$ can end in 4 , if $N$ ends in 2 or 8 .
Hence $P(A)=\frac{2}{10}=\frac{1}{5}=\frac{5}{25}=\frac{20}{100}$ Ans.
(B) $4^{\text {th }}$ power of a number can end in $0,1,4,5,6$.
Now, $N ^4$ can end in 1 if $N$ ends in $1,3,7,9 \Rightarrow P ( B )=\frac{4}{10}=\frac{2}{5}=\frac{10}{25}=\frac{40}{100}$ Ans.
(C) Given $CN$ ends in 5 . This is possible if c ends in 1 and $N$ in 5 and hence total number of ordered pairs $(C, N)$ is $(1,5),(5,1),(3,5),(5,3),(5,7),(7,5),(9,5),(5,9),(5,5)$ (Total 9 ordered pairs)
$\left(\frac{1}{100}\right)^{\cdot 9}=\frac{9}{100} \text { Ans. }$
(D) Given $CN$ ends in 6 and hence total number of ordered pairs $( C , N )$ is
$(1,6),(6,1),(2,3),(3,2),(2,8),(8,2),(4,4),(6,6),(4,9),(9,4),(8,7),(7,8) \rightarrow 12 \text { cases } $
$\Rightarrow \frac{12}{100}=\frac{3}{25}=\frac{12}{100}$