Quantitative Aptitude for Competitive Exams
Each of the numbers except 81 is a prime number.
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
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.
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.
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} $.
$n\left(S\right)$ = 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}$
22 – (1/4) {–5 – (– 48) ÷ (–16)}
= 22 – (1/4) {–5 – 3}
= 22 – (1/4) × –8
= 22 – 1 × –2
= 22 + 2
= 24
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
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.
Given Expression $=\dfrac{243^{\dfrac{n}{5} \times} 3^{2n+1}}{9^{n} \times 3^{n-1}}$
$=\dfrac{\left(3^{5}\right)^{\dfrac{n}{5} }\times 3^{2n+1}}{\left(3^{2}\right)^{n} \times 3^{n-1}}$
$=\dfrac{3^{\left(5 \times \dfrac{n}{5} \right)} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$
$=\dfrac{3^{n} \times 3^{2n+1}}{3^{2n}\times 3^{n-1}}$
$=\dfrac{ 3^{n+2n+1}}{ 3^{2n+n-1}}$
$=\dfrac{ 3^{3n+1}}{ 3^{3n-1}}$
$=3^{\left(3n+1-3n+1\right)}$
$=3^{2}=9$