8! = $8\times 7\times 6\times 5\times 4\times 3\times 2\times 1$

so 8! is divisible by 8 consecutive number

Sum of digits at even places = 4 + 2 + 4 = 10

Sum of digits at odd places = 21

Differences = 21 – 10 = 11 so number is divisible by 11

If 932P25Q is divisible by 88 then it is divisible by 8 and 11 both. For divisibility by 8 we know Q 88 must be divisible by 8 so Q = 6. For divisibility by 11

[(9 + P + 5) – (3+ 3 + 2)] should be either 0 or multiple of 11

$\therefore$ (9 + P + 5) – (3 + 2 + Q) = 14 + P – 11

$\Rightarrow$ P = 8

If a > 0 and b < 0 then a – b > 0.

x = y = 8 $\Rightarrow$ y = 8 – x

$\frac{x+2}{y+2}=\frac{x}{y} +\frac{4}{35}\Rightarrow \frac{x+2}{8- x+2}=\frac{x}{8-x}+\frac{4}{35}$

$\Rightarrow \frac{2-x}{10-x}-\frac{x}{8-x}=\frac{4}{35}$

$\Rightarrow \frac{\left ( 8-x \right )(2+x)-x(10-x)}{(10-x)\left ( 8-x \right )}=\frac{4}{35}$

$\Rightarrow \frac{16+6x-x^{2}-10x+x^{2}}{80-18x+x^{2}}=\frac{4}{35}$

$\Rightarrow \frac{16-4x}{80-18x+x^{2}}=\frac{4}{35}$

$\Rightarrow 16\times 35-4\times 35x=4\left ( 80-18x+x^{2} \right )$

$\Rightarrow 560-140x=320-72x+4x^{2}$

$\Rightarrow 4x^{2}+68x-240=0$

$\Rightarrow x^{2}+20x-3x-60=0$

$\Rightarrow x\left ( x+20 \right )-3\left ( x+20 \right )=0$

$\Rightarrow \left ( x-20 \right )-3\left ( x+20 \right )=0$

$\Rightarrow$ (x – 3) (x + 20)

$\Rightarrow$ x = 3, y = 5

So fraction is $\frac{3}{5}$

If x = 7 then number 3773 is divisible by 7

$\left ( x^{n} +a^{n}\right )$, will be divisible by (x + a) if n is odd, so common factor is (13 + 11) i.e. 24

$\left ( 1-\frac{1}{3} \right )\left ( 1-\frac{1}{4} \right )\left ( 1-\frac{2}{5} \right ).....\left ( 1-\frac{1}{n} \right )$

= $\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{n-1}{n}=\frac{2}{n}$

x + y = 12 and xy = 35

$\Rightarrow \frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{12}{35}$

Unit place digit of $\left ( 264 \right )^{102}\:is\:6\:and\:for\:\left ( 264 \right )^{103}$ is 4.

So unit place of $\left [ \left ( 264 \right )^{102}+\left ( 264 \right )^{103} \right ]$ is 6 + 4 = 10 digit at so it is 0

$\frac{9|3-5|-5|4|\div 10}{-3\left ( 5 \right )-2\times 4\div 2}=\frac{9\times \left ( 2 \right )-5\times \left ( 4 \right )\div 10}{-3\times \left ( 5 \right )-8\div 2}$

= $\frac{18-2}{-15-4}=-\frac{16}{19}$

Let number be 10x + y as equation is not solvable for y if n = 6 so (c) is right option

When 2851, 2862 and 2873 are divided by 23 then remainders 22, 10 and 21

So remainder R = $\frac{22\times \left ( 10 \right )^{2}\times \left ( 21 \right )^{3}}{23}=\frac{2640}{23}$

$\Rightarrow$ Remainder = 18

When number formed by last 4 digits 1920 is divided by 16 when remainder is 0

899 has four factors as 1, 29, 31 and 899

digit at unit place is [1 + 1 – 6 + 5 – 6 + 9] = 4

$\frac{1}{\frac{1}{-2\frac{7}{9}}}+\frac{17}{22}=\frac{1}{3+\frac{1}{-2\frac{9}{7}}}+\frac{17}{22}$

$\frac{1}{3+\frac{1}{\frac{5}{7}}}+\frac{17}{22}=\frac{1}{3+\frac{7}{5}}+\frac{17}{22}=\frac{1}{\frac{22}{5}}+\frac{17}{22}$

= $\frac{5}{22}+\frac{17}{22}=1$

$2^{3}+4^{3}.... +20^{3}=2^{3}\left [ 1^{3}+2^{3}+... +10^{3} \right ]$

= $8\left [ \frac{10\times 11}{2} \right ]^{2}=24200$

2 = $x+\frac{1}{1+\frac{1}{3+\frac{1}{4}}}=x+\frac{1}{1+\frac{4}{13}}=x+\frac{13}{17}$

$\Rightarrow x=2-\frac{13}{17}=\frac{21}{17}$

Let numbers be 2x and (2x + 2) then

$\left ( 2x+2 \right )^{2}-\left ( 2x \right )^{2}=84$

$\Rightarrow 8x+4=84\Rightarrow x=10\therefore 2x+2x+2=42$

$1\times 2\times 3\times 4\times 5\times .....\times 8\times 9$

$1\times 3\times 4\times 6\times .....\times \left ( 5\times 2 \right )=1\times 3\times 4\times ....\times 10$

$\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=\frac{3+2+1}{6}=1$

$\frac{7}{3+\sqrt{n}}\times \frac{\left ( 3-\sqrt{n} \right )}{3-\sqrt{n}}=\frac{7\left ( 3-\sqrt{n} \right )}{9-n}$

$\Rightarrow \frac{7}{9-n}=1\Rightarrow n=2$

$1^{1},2^{2},3^{3},.....,6^{6},7^{7}i.e.,823543$

No. of zeroes will be equal to power of 5

Powers of 5 in $1^{1}\times 2^{2}\times 3^{3}\times 4^{4}\times5^{5} ...\times 10^{10}\times ....\times 15^{15}\times ...\times 100^{100}\:will\:be\:5^{5}\times 10^{10}\times 15^{15}\times ......\times 100^{100}$

$\therefore Total\:powers\:of\:5=5+10+15+20+\left ( 25\times 2 \right )+30+...+\left ( 50\times 2 \right )+....+\left ( 100\times 2 \right )$

= (5 + 10 + 15 + 20 + 25 + …… + 50 + ……. + + 100) + (25 + 50 + 75 + 100)

When $\left ( 17 \right )^{200}=\left ( 18-1 \right )^{200}$ is divided by 18 remainder is 0

3, 10, 36, 172, 885, 5346, 37471

$3\times 2+2^{2}=10$

$10\times 3+3^{2}=39$

$39\times 4^{2}=172$

$172\times 5+5^{2}=885$ and so on

$1^{2}-1=0$

$2^{2}-1=3$

$3^{2}-1=8$

$4^{2}-1=15$

……………………….

……………………….

$7^{2}-1=48$

$\frac{1}{2},\frac{3}{4},\frac{5}{8},\frac{7}{16},?.$

Taking numerator 1, 3, 5, 7, 9

Taking denominator 2, 4, 8, 16, 32

$\therefore Fraction\: is\: \frac{9}{32}$

$999\frac{995}{999}\times 999=\left ( 999+\frac{995}{999} \right )\times 999$

$999^{2}+995=\left ( 1000-1 \right )^{2}+995$

= 1000000 - 2000 + 1 + 995 = 998996

$\sqrt{\frac{16}{36}+\frac{1}{4}}=\sqrt{\frac{4}{9}+\frac{1}{4}}=\sqrt{\frac{16+9}{36}}=\sqrt{\frac{25}{36}}=\frac{5}{6}$

$\left [ \sqrt[3]{\sqrt[6]{2^{9}}}\:\:\right ]^{4}\times \left [ \sqrt[6]{\sqrt[3]{2^{9}}}\:\:\right ]^{4}=2^{9\times \frac{1}{6}\times \frac{1}{3}\times 4}\times 2^{9\times \frac{1}{3}\times \frac{1}{6}\times 4}$

= $2^{2}\times 2^{2}=2^{4}$

$\frac{52}{x}=\sqrt{\frac{169}{289}}=\frac{13}{17}\Rightarrow x=\frac{52\times 17}{13}=68$

$\frac{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}+\sqrt{8-\sqrt{28}}}}\times \frac{\sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}}}{\sqrt{8+\sqrt{28}+\sqrt{8-\sqrt{28}}}}$

= $\frac{\left [ \sqrt{8+\sqrt{28}}-\sqrt{8-\sqrt{28}} \:\:\right ]^{2}}{(\sqrt{8+\sqrt{28}})^{2}-(\sqrt{8-\sqrt{28}})^{2}}$

= $\frac{\left ( 8+\sqrt{28} \:\right )+\left ( 8-\sqrt{28} \:\right )-2\sqrt{8+\sqrt{28}}\times \sqrt{8-\sqrt{28}}}{\left ( 8+\sqrt{28} \:\right )-\left ( 8-\sqrt{28} \right )}$

= $\frac{16-2\sqrt{64-28}}{2\sqrt{28}}=\frac{16-2\times 6}{2\sqrt{28}}=\frac{4}{2\sqrt{28}}=\frac{1}{\sqrt{7}}$

$\left ( \frac{1}{5} \right )^{3y}=0.008=\frac{8}{1000}=\frac{1}{125}=\left ( \frac{1}{5} \right )^{3}$

$\Rightarrow 3y=2\Rightarrow y=1\therefore \left ( 0.25 \right )^{y}=0.25$

$\frac{\left ( 225 \right )^{0.2}\times \left ( 225 \right )^{0.3}}{\left ( 225 \right )^{0}\times \left ( 225 \right )^{0.2}}=\left ( 225 \right )^{0.2+0.3-0.8-0.2}$

= $\left ( 225 \right )^{0.5-1.0}$

= $\left ( 225 \right )-0.5=15^{2\times \left ( -\frac{1}{2} \right )}=\frac{1}{15}$

$\left ( \frac{1}{64} \right )^{0}+\left ( 64 \right )^{-\frac{1}{2}}+\left ( 32 \right )^{\frac{4}{5}}-32^{-\frac{4}{5}}$

= $1+8^{2\times -\frac{1}{2}}+2^{5\times \frac{4}{5}}-2^{5\times -\frac{4}{5}}$

= $1+8^{-1}+2^{4}-2^{-4}$

= $1+\frac{1}{8}+16-\frac{1}{16}=17+\frac{1}{16}=17\frac{1}{16}$

$\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{154+\sqrt{225}}}}}$

= $\sqrt{10+\sqrt{25+\sqrt{108+\sqrt{\left ( 154+15 \right )}}}}$

= $\sqrt{10+\sqrt{25+\sqrt{108+13}}}$

= $\sqrt{10+\sqrt{25+11}}$

= $\sqrt{10+6}=4$

$\frac{(0.55)^{2}+(0.007)^{2}+(0.027)^{2} }{(0.055)^{2}+(0.007)^{2}+(0.0027)^{2} }$

$\frac{(0.55)^{2}+(0.007)^{2}+(0.027)^{2} }{ \frac{1}{10^{2} }[(0.55)^{2}+(0.027)^{2}+(0.07)^{2} ] } =100$

$\frac{ \sqrt{x} }{0.0064} = \sqrt[3]{0.008}$

$\Rightarrow \frac{ \sqrt{x} }{0.08}= \sqrt[3]{ \frac{8}{1000} } \Rightarrow \sqrt{x}=0.08 \times \frac{2}{10}$

$\Rightarrow \sqrt{x} =0.016 \Rightarrow x= (0.016)^{2} =0.000256$

$A=5+2 \sqrt{6}$

$\sqrt{A}= \sqrt{5+2 \sqrt{6} }= \sqrt{( \sqrt{2}+ \sqrt{3})^{2} }=( \sqrt{2}+ \sqrt{3})$

$\frac{1}{ \sqrt{A} }= \frac{1}{ \sqrt{2}+ \sqrt{3} }= \frac{ \sqrt{3}- \sqrt{2} }{( \sqrt{3})^{2}-( \sqrt{2} )^{2} }=( \sqrt{3}- \sqrt{2} )$

$\therefore \sqrt{A}+ \frac{1}{ \sqrt{A} }=( \sqrt{3}+ \sqrt{2})+( \sqrt{3}- \sqrt{2} )=2 \sqrt{3}$

$\begin{bmatrix} \big(2 \frac{10}{27} )^{ -\frac{2}{3} } \times \big(11 \frac{1}{9} )^{-0.5}\end{bmatrix}+[(16.25)^{0.5} \div (4)^{-1}]$

= $\begin{bmatrix} \big( \frac{64}{27} )^{- \frac{2}{3} } \times \big( \frac{9}{100} )^{-0.5} \end{bmatrix}+ \begin{bmatrix}2.5 \div \big(- \frac{1}{4} ) \end{bmatrix}$

= $\begin{bmatrix} \big( \frac{27}{64} )^{\frac{2}{3} } \times \big( \frac{100}{9} )^{ \frac{1}{2} } \end{bmatrix}+ \begin{bmatrix}2.5 \times(-4) \end{bmatrix}$

= $\begin{bmatrix} \big( \frac{3}{4} )^{2} \times \frac{10}{3} \end{bmatrix} +(-10)= \begin{bmatrix} \frac{9}{16} \times \frac{10}{3} \end{bmatrix}-10$

= $\frac{15}{8} -10=- \frac{65}{8}=-8 \frac{1}{8}$

$2 \sqrt{5}= \sqrt{4 \times5 }= \sqrt{20},6 \sqrt{3}= \sqrt{36 \times3 } = \sqrt{108}$

$3 \sqrt{7}= \sqrt{9 \times7 }=63,8 \sqrt{2}= \sqrt{64 \times 2}= \sqrt{128}$

$\therefore8 \sqrt{2}$ is the largest number.

$\sqrt{4+ \sqrt{44+ \sqrt{10000} } }= \sqrt{4+ \sqrt{44+100} }$

$\sqrt{4+ \sqrt{144} }= \sqrt{4+12}= \sqrt{16}=4$

$\frac{1+ \sqrt{2} }{ \sqrt{5}+ \sqrt{3} }+ \frac{ \sqrt{1- \sqrt{2} } }{ \sqrt{5}- \sqrt{3} }$

= $\frac{(1+ \sqrt{2} )( \sqrt{5}- \sqrt{3} )+(1- \sqrt{2} )( \sqrt{5} )+ \sqrt{3} }{5-3}$

= $\frac{2 \sqrt{5}-2 \sqrt{6} }{6}= \sqrt{5}- \sqrt{6}$

$\frac{1}{ \sqrt{2}+ \sqrt{3}- \sqrt{5} }= \frac{ \sqrt{2}+ \sqrt{3}+ \sqrt{5} }{( \sqrt{2}+ \sqrt{3} )^{2}-( \sqrt{5} )^{2} }$

= $\frac{ \sqrt{2}+ \sqrt{3}+ \sqrt{5} }{2 \sqrt{6} }$

$\frac{1}{ \sqrt{2}- \sqrt{3}- \sqrt{5} }= \frac{ \sqrt{2}- \sqrt{3}+ \sqrt{5} }{( \sqrt{2}- \sqrt{3} )^{2}-( \sqrt{5} )^{2} }$

= $\frac{ \sqrt{2}- \sqrt{3}+ \sqrt{5} }{-2 \sqrt{6} }$

$\therefore \frac{1}{ \sqrt{2}+ \sqrt{3}- \sqrt{5} }+ \frac{1}{ \sqrt{2}- \sqrt{3}- \sqrt{5} }$

= $\frac{ \sqrt{2}+ \sqrt{3}+ \sqrt{5} }{2 \sqrt{6} }- \frac{ \sqrt{2} - \sqrt{3}+\sqrt{5} }{2 \sqrt{6} }$

= $\frac{2 \sqrt{3} }{2 \sqrt{6} }= \frac{1}{ \sqrt{2} }= \frac{ \sqrt{2} }{2}$

6.4 + .64 + .064 + .0064 = 7.11

$\sqrt{- \sqrt{3+ \sqrt{3+8 \sqrt{7+4 \sqrt{3} } } } }$

= $\sqrt{- \sqrt{3+ \sqrt{3+8 \sqrt{(2+ \sqrt{3} )^{2}}}}}$

= $\sqrt{- \sqrt{3+ \sqrt{(19+8 \sqrt{3} )} } }$

= $\sqrt{- \sqrt{3+(4+ \sqrt{3}) } }= \sqrt{4}=2$

$1+ \frac{1}{1+ \big\{ \frac{1}{ \big(1+ \frac{1}{x} \big) } \big\} }= \frac{11}{7} \Rightarrow \frac{1}{1+ \big\{ \frac{x}{x+1} \big\} }$

$\Rightarrow 1+ \frac{x}{x+1}= \frac{7}{4} \Rightarrow \frac{x}{x+1}= \frac{7}{4}-1= \frac{3}{4}$

$\Rightarrow 4x=3x+3 \Rightarrow x=3$

After squaring of all these we get

$\big( \sqrt{7}- \sqrt{5} \big)^{2}=12- \sqrt{35}$ = 12 – 6 (Approx) = 6

$\big( \sqrt{5}- \sqrt{3} \big)^{2}=8- \sqrt{15}$ = 8 – 4 (Approx) = 4

$\big( \sqrt{9}- \sqrt{7} \big)^{2}=16- \sqrt{63}$ = 16 – 8 (Approx) = 8

$\big( \sqrt{11}- \sqrt{9} \big)^{2}=20- \sqrt{99}$ = 20 – 10 (Approx) = 10

So $\big( \sqrt{11}- \sqrt{9} \big)$ is the largest number

$\frac{7}{6}=1.166, \frac{7}{9}=0.777, \frac{4}{5}0.8, \frac{5}{7}=0.714$

$\sqrt{15}=3.88$

$\sqrt{ \frac{5}{3} }= \sqrt{ \frac{5}{3} \times \frac{3}{3} }= \sqrt{ \frac{15}{3} }= \frac{3.88}{3}=1.29$

$\frac{-2}{3}=-0.666,\frac{3}{4}=0.75,\frac{-3}{4}=0.75$

$\therefore Ascending\:order\:is\: \frac{-3}{4}< \frac{-2}{3}< \frac{3}{4}$

$6. \overline{46}=6+ \overline{0.46}=6 \frac{46}{99}= \frac{640}{99}$

$\frac{1}{2}=0.500,\frac{5}{6}=0.833..., \frac{7}{8}=0.875$

Correct descending order is $\frac{7}{8}, \frac{5}{6},\frac{1}{2}$

$9-1 \frac{2}{9}of3\frac{3}{11} \div5 \frac{1}{7}of\frac{7}{9}$

Using Bodmas

$9- \big( \frac{11}{9} \times \frac{36}{11} \big) \div \big( \frac{36}{7} \times \frac{7}{9} \big)$

= $9-4 \div 4$

= 9 – 1

= 8

$99 \times \frac{2}{3}-2 \times \frac{3}{4}$

= $33 \times 2-13 \times 3$

= 66 – 39 = 27

$LCM \times HCF =a \times b$

$\Rightarrow 4125 \times 25=375 \times b$

$\Rightarrow b=275$

LCM (5, 15, 20) = 60, HCF (5, 15, 20) = 5

$\therefore$ LCM:HCF = 60:5 = 12:1.

HCF (250, 150) = 50 cm

$\therefore No\:of\:tiles\:= \frac{250 \times 150}{50 \times 50}=15$

LCM (2, 4, 7, 8, 10, 12) = 840 sec

30 min = $30 \times 60=1800sec$

They will ring 2 times at the same time in 30 min. as $\begin{bmatrix} \frac{1800}{840} \end{bmatrix} =2$

$\frac{LCM(5,8,110}{HCF(2,9,14)}= \frac{440}{1}=440$

LCM (1000 cm, 85 cm) = 17000 cm = 17 cm.

LCM = 28 HCF …… (1)

And LCM + HCF = 1740 m

Or LCM + 28 HCF = 1740 [From (1)]

Or HCF = $\frac{1740}{29}=60$

$\therefore$ HCF = 60,

LCM = $28 \times 60=1680$

$HCF \times LCM=ab$

$\Rightarrow b= \frac{1680 \times 60}{240}=420$

We know that LCM of two numbers is always greater than their HCF so HCF is 11

Factors of 36 are 1, 2, 3, 4, 6, 9, 12, 18 and 36 and sum of 12 and 18 is 30 and their LCM is 36.

LCM (7, 9, 12) = 252

$\therefore$ Number = LCM (7, 9, 12) + 1 = 253