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2613.41, 43, 47, 53, 61, 71, 73, 81
61
71
73
81
Explanation:

Each of the numbers except 81 is a prime number.

2621.Find the odd man out. 23, 27, 36, 52, 77, 111, 162
162
111
52
27
Explanation:

23 + 22 = 27

27 + 32 = 36

36 + 42 = 52

52 + 52 = 77

77 + 62 = 113

113 + 72 = 162

Hence, 113 should have come in place of 111

2633.Which of the following number is divisible by 24 ?
35718
63810
537804
3125736
Explanation:

24 = 3 x8, where 3 and 8 co-prime.

Clearly, 35718 is not divisible by 8, as 718 is not divisible by 8.

Similarly, 63810 is not divisible by 8 and 537804 is not divisible by 8.

Consider option (D),

Sum of digits = (3 + 1 + 2 + 5 + 7 + 3 + 6) = 27, which is divisible by 3.

Also, 736 is divisible by 8.

$\therefore$ 3125736 is divisible by 3 x 8, i.e., 24.

2645.How many natural numbers are there between 23 and 100 which are exactly divisible by 6 ?
8
11
12
13
Explanation:

Required numbers are 24, 30, 36, 42, ..., 96

This is an A.P. in which $ a $ = 24, $ d $ = 6 and $ l $ = 96

Let the number of terms in it be $ n $.

Then tn = 96  $\Rightarrow$  $ a $ + $\left(n - 1\right)$d = 96

$\Rightarrow$ 24 + $\left( n - 1\right)$ x 6 = 96

$\Rightarrow$ $\left( n - 1\right)$ x 6 = 72

$\Rightarrow$ $\left( n - 1\right)$ = 12

$\Rightarrow n $ = 13

Required number of numbers = 13.

2838.From a pack of 52 cards, two cards are drawn together at random. What is the probability of both the cards being kings?
$ \dfrac{1}{15} $
$ \dfrac{25}{57} $
$ \dfrac{35}{256} $
$ \dfrac{1}{221} $
Explanation:

Let S be the sample space.

Then, $ n \left(S\right)$ = 52C2 =$ \dfrac{(52 \times 51)}{(2 \times 1)} $= 1326.

Let E = event of getting 2 kings out of 4.

$\therefore n \left(E\right)$ = 4C2 =$ \dfrac{(4 \times 3)}{(2 \times 1)} $= 6.

$\therefore P\left(E\right)$ =$ \dfrac{n(E)}{n(S)} $=$ \dfrac{6}{1326} $=$ \dfrac{1}{221} $.

2848.Two cards are drawn together from a pack of 52 cards. The probability that one is a club and one is a diamond?
13/51
1/52
13/102
1/26
Explanation:

$n\left(S\right)$ = Total number of ways of drawing 2 cards from 52 cards = 52C2

Let E = event of getting 1 club and 1 diamond.

We know that there are 13 clubs and 13 diamonds in the total 52 cards.

$Hence, n\left(E\right)$ = Number of ways of drawing one club from 13 and one diamond from 13

= 13C1 × 13C1

$\text{P(E) = }\dfrac{\text{n(E)}}{\text{n(S)}} $= $\dfrac{13_{C_1} \times 13_{C_1}}{52_{C_2}}$

= $\dfrac{13 \times 13}{\left( \dfrac{52 \times 51}{2}\right)}$=$ \dfrac{13 \times 13}{ 26 \times 51}= \dfrac{13}{ 2\times 51}$=$ \dfrac{13}{102}$

10721.22 – (1/4) {–5 – (– 48) ÷ (–16)}
19
20
24
21
Explanation:

22 – (1/4) {–5 – (– 48) ÷ (–16)}

= 22 – (1/4) {–5 – 3}

= 22 – (1/4) × –8

= 22 – 1 × –2

= 22 + 2

= 24

10726.Find the value of 1/(3+1/(3+1/(3–1/3)))
3/10
10/3
27/89
89/27
Explanation:

1/(3+1/(3+1/(3–1/3)))

= 1/(3 + 1/(3 + 1/(8/3)))

= 1/(3 + 1/(3 + 3/8))

= 1/(3 + 8/27)

= 1/(89/27)

= 27/89

43870.(25)7.5 x (5)2.5 ÷ (125)1.5 = 5?
8.5
13
16
17.5
Explanation:

Let (25)7.5 x (5)2.5 ÷ (125)1.5 = 5x

Then,$\dfrac{(5^{2})^{7.5}\times(5)^{2.5}}{(5^{3})^{1.5}}=5^{x}$

$\Rightarrow\dfrac{5^{(2\times7.5)}\times5^{2.5}}{5^{(3\times1.5)}}= 5^{x}$

$\Rightarrow\dfrac{5^{15}\times5^{2.5}}{5^{4.5}}= 5^{x}$

$\Rightarrow 5^{x}$=5(15+2.5-4.5)

$\Rightarrow 5^{x}$=5(13)

$\therefore$ x = 13.
44471.$\dfrac{243^{\dfrac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}=?$
1
3
9
27
Explanation:

Given Expression $=\dfrac{243^{\dfrac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}$

$=\dfrac{\left(3^{5}\right)^{\dfrac{n}{5} }\times 3^{2n+1}}{\left(3^{2}\right)^{n} \times 3^{n-1}}$

$=\dfrac{3^{\left(5 \times \dfrac{n}{5} \right)} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$

$=\dfrac{3^{n} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$

$=\dfrac{ 3^{n+2n+1}}{ 3^{2n+n-1}}$

$=\dfrac{ 3^{3n+1}}{ 3^{3n-1}}$

$=3^{\left(3n+1-3n+1\right)}$

$=3^{2}=9$

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