(xyz , xyz) is number
⇒ 100000x + 10000y + 1000z + 100x + 10y + z
⇒ 1001 (100x + 10y + z)
Since 1001 has factors i.e., 7, 11 and 13
So, the answer is 1001

$$\eqalign{ & {\text{By option }}\frac{8}{9},{\text{ }} \cr & {\text{Proper fraction = }}\frac{8}{9} \cr & {\text{Its reciprocal = }}\frac{9}{8} \cr & {\text{According to question}} \cr & = \frac{9}{8} - \frac{8}{9} \cr & = \frac{{81 - 64}}{{72}} \cr & = \frac{{17}}{{72}}{\text{ satisfied}} \cr} $$

$${\text{0}}{\text{.1 = }}\frac{1}{{10}}\left( {{\text{mud}}} \right),\,\,\,\,\frac{5}{8}\left( {{\text{water}}} \right)$$
Let total length of bamboo = 40x ( LCM of 8, 10 )
$$\eqalign{ & {\text{Mud part :}} \cr & {\text{ = }}\frac{1}{{10}} \times 40x = 4x \cr & {\text{Water part :}} \cr & {\text{ = }}\frac{5}{8} \times 40x = 25x \cr & {\text{Remaining part :}} \cr & = (40x - 25x - 4x) \cr & = 11x \cr & 11x{\text{ }} \to {\text{ }}2.75m \cr & {\text{ }}x{\text{ }} \to {\text{ }}0.25m \cr & 40x{\text{ }} \to {\text{ }}10m \cr} $$

$$\eqalign{ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{(1001)^{2008}} + (1002) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \cr & {\text{unit digit}} \to {\text{ }}{{\text{1}}^{2008}} + 1002 \cr} $$
Unit digit will be 1 in case of 1 respective of power
⇒ 1 + 1002 = 1003
⇒ 3 unit digit (last digit)

$$\frac{3}{5},\frac{7}{9},\frac{{11}}{{13}}$$
$$\eqalign{ & \frac{3}{5} = 0.60, \cr & \frac{7}{9} = 0.7, \cr & \frac{{11}}{{13}} = 0.8 \cr & \frac{{11}}{{13}}{\text{ > }}\frac{7}{9} > \frac{3}{5} \cr} $$

Clearly, M = 0,
since 304 × 4 = 1216

Let the 3 consecutive number is
x, x + 1, x + 2
According to question,
$$ \Rightarrow {x^2} + {\left( {x + 1} \right)^2} + {\left( {x + 2} \right)^2}$$     = 365
$$ \Rightarrow {x^2} + {x^2} + 1 + 2x + {x^2} + 4 + 4x$$       = 365
$$\eqalign{ & \Rightarrow 3{x^2} + 6x = 360 \cr & \Rightarrow {x^2} + 2x - 120 = 0 \cr & \Rightarrow \left( {x - 10} \right)\left( {x + 12} \right) = 0 \cr & \Rightarrow x = 10 \cr & {\text{Sum of numbers :}} \cr & = 10 + 11 + 12 \cr & = 33 \cr} $$

= 142² - 1
= (142 - 1) (142 + 1) [a² - b² = (a - b) (a + b)]
= 141 × 143
= 141 × 11 × 13

$$\eqalign{ & {\text{0}}{\text{.121212}}........ \cr & \Rightarrow 0.\overline {12} = \frac{{12}}{{99}} = \frac{4}{{33}} \cr} $$

$$\eqalign{ & \frac{6}{{\frac{7}{8}}} = \frac{{6 \times 8}}{7} = \frac{{48}}{7} \cr & \frac{{\frac{6}{7}}}{8} = \frac{6}{{7 \times 8}} = \frac{6}{{56}} = \frac{3}{{28}} \cr & {\text{Difference - }} \cr & = \frac{{48}}{7} - \frac{3}{{28}} \cr & = \frac{{192 - 3}}{{28}} \cr & = \frac{{189}}{{28}} \cr & = \frac{{27}}{4} \cr & = 6\frac{3}{4} \cr} $$

Let the number is a, b, c
a + b + c = 2
a = $$\frac{1}{2}$$ b.....(i)
c = $$\frac{1}{4}$$ b.....(ii)
$$\frac{a}{b}$$ = $$\frac{1}{2}$$
$$\frac{b}{c}$$ = $$\frac{4}{1}$$

a	:	b	:	c
1	:	2	↘
↘		4	:	1
4	:	8	:	2

a : b : c = 2 : 4 : 1
(2 + 4 + 1) = 7
7 units → 2
1 unit → $$\frac{2}{7}$$
4 units → $$\frac{2}{7}$$ × 4 = $$\frac{8}{7}$$
So, 2^nd number is $$\frac{8}{7}$$

Let the numbers are 7x and 9x
According to the question,
$$\eqalign{ & \Rightarrow 7x \times 9x = 1575 \cr & \Rightarrow 63{x^2} = 1575 \cr & \Rightarrow {x^2} = 25 \cr & \Rightarrow x = 5 \cr} $$
Then greater number
= 9x
= 9 × 5
= 45

According to the question,
$$ \Rightarrow \left( {2 - \frac{1}{3}} \right)\left( {2 - \frac{3}{5}} \right)\left( {2 - \frac{5}{7}} \right).....$$       $$\left( {2 - \frac{{997}}{{999}}} \right)$$
$$\eqalign{ & \Rightarrow \frac{5}{3} \times \frac{7}{5} \times \frac{9}{7}.....\frac{{1001}}{{999}} \cr & \Rightarrow \frac{1}{3} \times 1001 \cr & \Rightarrow \frac{{1001}}{3} \cr} $$

Smallest number in case of decimal = 0.001

Let the number is a, b, c
a = 3b       $$\frac{a}{b}$$ = $$\frac{3}{1}$$
c = $$\frac{2}{3}$$a       $$\frac{c}{a}$$ = $$\frac{2}{3}$$

c	a	b
2	3
	3	1

c : a : b = 2 : 3 : 1
So, (2 + 3 + 1) = 252 units
6 units → 252
1 unit → 42
3 units → 42 × 3
So, second number is = 42

$$\left( {1 - \frac{1}{3}} \right)\left( {1 - \frac{1}{4}} \right)\left( {1 - \frac{1}{5}} \right)$$     . . . . .$$\left( {1 - \frac{1}{{99}}} \right)$$ $$\left( {1 - \frac{1}{{100}}} \right)$$
$$\eqalign{ & = \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times ..... \times \frac{{98}}{{99}} \times \frac{{99}}{{100}} \cr & = \frac{2}{{100}} \cr & = \frac{1}{{50}} \cr} $$

If first number (729) is completely divide by second number (27) then we divide remainder from second number then remaining remainder will be answer.
$$\frac{729}{27}$$ = 27 times
Now, $$\frac{56}{27}$$ = 2 × 27 + 2 → Remainder

Let the numbers = a, b
According to the question,
a + b = 3 (a - b)
$$\frac{a + b}{a - b}$$   = $$\frac{3}{1}$$
By componendo and dividendo
$$\frac{2a}{2b}$$   = $$\frac{4}{2}$$
$$\frac{a}{b}$$   = $$\frac{2}{1}$$
a : b = 2 : 1

Let number is N
N is divided by the sum of 555 and 445, the quotient is 2 times difference
Quotient = 2 (555 - 445) = 220
Remainder = 30 (given)
N = 1000 × 220 + 30
    = 220030

Let 414 × x × 7 = 127512
Then,
$$\eqalign{ & x = \frac{{127512}}{{414 \times 7}} \cr & \,\,\,\,\,\, = \frac{{18216}}{{414}} \cr & \,\,\,\,\,\, = \frac{{2024}}{{46}} \cr & \,\,\,\,\,\, = \frac{{1012}}{{23}} \cr & \,\,\,\,\,\, = 44 \cr} $$

Dividend = (divisor × quotient) + remainder
According to the question
$$\eqalign{ & {\text{Divisor :}} \cr & \Rightarrow {\text{16}} \times {\text{25 = 5}} \times {\text{remainder}} \cr & \Rightarrow {\text{R = }}\frac{1}{5} \times 16 \times 25 \cr & {\text{Dividend :}} \cr & {\text{ = }}\left[ {\left( {{\text{16}} \times {\text{25}}} \right) \times 16} \right] + \frac{1}{5} \times 16 \times 25 \cr & = \left[ {16 \times 25 \times 16} \right] + 80 \cr & = 6480 \cr} $$

Remaining distance to be run
$$\eqalign{ & = 5 - \frac{5}{4} \cr & = \frac{{20 - 5}}{4} \cr & = \frac{{15}}{4}\,{\text{laps}} \cr} $$

Let the total length of rod is 1 unit
According to question,
$$ \Rightarrow 1 - \frac{1}{{10}} - \frac{1}{{20}} - \frac{1}{{30}} - \frac{1}{{40}} - \frac{1}{{50}}$$       $$ - \frac{1}{{60}}$$

$$ \Rightarrow \frac{{600 - 60 - 30 - 20 - 15 - 12 - 10}}{{600}}$$
$$\eqalign{ & \Rightarrow \frac{{453 \leftarrow {\text{ Violet part}}}}{{600 \leftarrow {\text{ Total part}}}} \cr & 453{\text{ part }} \to {\text{ 12}}{\text{.08 mtr}} \cr & {\text{1 part }} \to \,\,\,\,\,\,\,\,\frac{{12.08}}{{453}} \cr & 600{\text{ part }} \to \frac{{12.08}}{{453}} \times 600 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 16{\text{mtr}} \cr} $$
So total length of rod is 16 metres.

z = 3 × 31 + 2 = 95
y = 7 × z + 4 = 7 × 95 + 4 = 669
x = 8 × y + 3 = 8 × 669 + 3 = 5355
∴ Required number = 5355

Divisor :
= 7 × quotient
= 5 × remainder and dividend
= 6 × remainder

Let remainder be x
Then, divisor = 5x and dividend = 6x
On dividing 6x by 5x, we get 1 as quotient and x as remainder.
∴ Quotient = 1

No. →	I	II	III
	3x	6x	2x	= 11x

$$\eqalign{ & \frac{{11x}}{3} = 44 \Rightarrow x = 12 \cr & {\text{Largest number = 6x = 72}} \cr} $$

$$\eqalign{ & \frac{{{\text{Remainder of number }}}}{{17}} = \frac{{45}}{{17}} \cr & \Rightarrow {\text{Remainder = 11}} \cr} $$

$$\eqalign{ & {\text{Let the total number of picnic party = x}} \cr & {\text{Each member contributed = 2x}} \cr & {\text{Accroding to question, }} \cr & (2x){\text{ }}(x) = 3042 \cr & 2{x^2} = 3042 \cr & {x^2} = 1521 \cr & x = 39 \cr} $$

$$\eqalign{ & \left( {1 + \frac{1}{2}} \right)\left( {1 + \frac{1}{3}} \right)\left( {1 + \frac{1}{4}} \right).....\left( {1 + \frac{1}{{120}}} \right) \cr & = \frac{3}{2} \times \frac{4}{3} \times \frac{5}{4}.........\frac{{121}}{{120}} \cr & = \frac{{121}}{2} \cr & = 60.5{\text{ Ans}}. \cr} $$

$$\eqalign{ & 0.\overline 2 + 0.\overline 3 + 0.\overline {32} \cr & = \frac{2}{9} + \frac{3}{9} + \frac{{32}}{{99}} \cr & = \frac{{22 + 33 + 32}}{{99}} \cr & = \frac{{87}}{{99}} \cr & = 0.\overline {87} \cr} $$

$$\eqalign{ & = 0.\overline {142857} \div 0.\overline {285714} \,{\text{ }} \cr & = \frac{{142857}}{{999999}} \div \frac{{285714}}{{999999}} \cr & = \frac{{142857}}{{285714}} \cr & = \frac{1}{2} \cr} $$

$$\eqalign{ & {\text{Accroding to question}} \cr & ({\text{P are the pages)}} \cr & \Rightarrow {\text{P}} - \frac{2}{5}{\text{P}} - \frac{8}{{15}}{\text{P}} = 15, \cr & \because \left[ {\frac{8}{{15}} = \frac{2}{5} \times \frac{4}{3}} \right] \cr & \Rightarrow \frac{{15{\text{P}} - {\text{6P}} - 8{\text{P}}}}{{15}} = 15 \cr & \Rightarrow {\text{P}} = 225{\text{ pages}} \cr} $$

Sum = 5 + 10 + 15 + 20 + .....+ 95 , It is an arithmetic progression
Number of terms (n)
= $$\frac{{{\text{Last term }} - {\text{ First term}}}}{{{\text{Differnce}}}}$$     + 1
⇒ n = $$\frac{95 - 5}{5}$$ + 1 = 19
So, sum = $$\frac{n}{2}$$ [2a + (n - 1)d]
Here, a = 5, n = 19, d = 5
Sum = $$\frac{19}{2}$$ [10 + (19 - 1)5]
        = 950

$$\eqalign{ & {{\text{a}}^3} + {b^3} = (a + b)({{\text{a}}^2} + {b^2} - ab) \cr & {\text{Take option (A)}} \cr & {\text{ = }}{{\text{2}}^{96}} + 1{\text{ }} \cr & = {({2^{32}})^3} + {(1)^3} \cr & = ({{\text{2}}^{32}} + 1)[{({2^{32}})^2}{\text{ + }}{{\text{1}}^2}{\text{ - }}{{\text{2}}^{32}}] \cr & = ({{\text{2}}^{32}} + 1)[{2^{64}} + 1 - {2^{32}}] \cr & {\text{Clearly, }} \cr & {{\text{2}}^{32}} + 1{\text{ is a factor of }}{{\text{2}}^{96}} + 1 \cr} $$

25⁶²⁵¹ = 5⁶²⁵¹ = 5 (because unit digit of 5 is always 5)
36⁵²⁸ = 6⁵²⁸ = 6 (because unit digit of 6 is always 6)
73⁵⁴ = 3⁵⁴ = 9(unit digit of 3⁵⁴ = 3² ($$\frac{{54}}{4}$$ = 2 remainder) = 9)
now add all the unit digit 5 + 6 + 9 = 20 = 0

We keep on dividing 33333... by 7 till we get 0 as remainder.

∴ Required number = 47619

$$\eqalign{ & {\text{Let thr number be x}} \cr & {\text{According to question}} \cr & \frac{3}{4}x - \frac{3}{{14}}x = 150 \cr & \frac{{21x - 6x}}{{28}} = 150 \cr & 15x = 150 \times 28 \cr & x = 280 \cr} $$

\[\begin{gathered} {\text{ unit place}} \hfill \\ \left[ \begin{gathered} {7^1} \to 7 \to 7 \hfill \\ {7^2} \to 49 \to 9 \hfill \\ {7^3} \to 343 \to 3 \hfill \\ {7^4} \to 2401 \to 1 \hfill \\ \end{gathered} \right]{\text{cycle }} \hfill \\ \end{gathered} \]
$$\eqalign{ & {\left( {4387} \right)^{245}}\,\,\,\,\, \times \,\,\,\,\,{\left( {621} \right)^{72}} \cr & \,\,\,\,\mathop {{{\left( 7 \right)}^{245}}\,\,\,\,\,\,\,\,}\limits^ \downarrow \times \,\,\,\,\,\,\,\,\,{\left( 1 \right)^{72}} \cr & \,\,\,\,\mathop {{{\left( 7 \right)}^{4 \times 61 + 1}}\,\,\,}\limits^ \downarrow \times \,\,\,\,\,\,\,\,\,\,\,1 \cr & \,\mathop {{{\left( 7 \right)}^{4 \times 61}} \times 7\,\,}\limits^ \downarrow \times \,\,\,\,\,\,\,\,\,\,\,1 \cr & \,\,\,\,\,\,\mathop {{{\left( 1 \right)}^{61 }\times 7}}\limits^ \downarrow \,\,\,\,\, \times \,\,\,\,\,\,\,\,\,\,\,1 \cr} $$
unit digit = 7

$$\eqalign{ & \frac{6}{7},\frac{5}{6},\frac{7}{8},\frac{4}{5} \cr & \frac{6}{7} = 0.85, \cr & \frac{5}{6} = 0.83, \cr & \frac{7}{8} = 0.87, \cr & \frac{4}{5} = 0.80 \cr & \frac{7}{8}\,\,{\text{is}}\,{\text{largest}} \cr} $$

Explanation :
When the fractions are given in this order when the difference between the numerator and denominator is same so largest fraction will be largest numerator fraction and the smallest will be smallest numerator fraction.
Note: - Fraction should be in the order where denominator will be larger than numerator.
In this question : largest is $$\frac{7}{8}$$and smallest is $$\frac{4}{5}$$

$$\eqalign{ & {\text{ }}\frac{2}{5},\frac{5}{6},\frac{{11}}{{15}}{\text{ & }}\frac{7}{8} \cr & \frac{7}{8}\,{\text{is largest among }}\frac{2}{5},\frac{5}{6},\frac{7}{8} \cr & {\text{now compare}} \cr } $$

(105)		(88)
$$\frac{7}{8}$$	⤧	$$\frac{{11}}{{15}}$$

So $$\frac{{7}}{{8}}$$ is largest

4x² - 12x + k
for perfect square must be
b² = 4ac
12 × 12 = 4 × 4 × k
9 = k
k = 9

$$\eqalign{ & {\text{Let the unit digit }} \cr & {\text{Ten digit }} = x - 2 \cr & \therefore {\text{ Number :}} \cr & = {\text{ }}10(x - 2){\text{ }} + {\text{ }}x \cr & = {\text{ }}10x{\text{ }} - {\text{ }}20{\text{ }} + {\text{ }}x \cr & = {\text{ }}11x{\text{ }} - {\text{ }}20 \cr} $$
New number obtained after reversing the digits
$$\eqalign{ & = 10x + x - 2{\text{ }} \cr & = {\text{ }}11x - 2 \cr & {\text{According to the question,}} \cr & {\text{3}}\left( {11x - 20} \right) + \frac{6}{7}\left( {11x - 2} \right) = 108 \cr & 7\left( {11x - 20} \right) + 2\left( {11x - 2} \right) = 36 \times 7 \cr & 77x - 140 + 22x - 4 = 252 \cr & 99x = 252 + 144 \cr & x = \frac{{396}}{{99}} = 4 \cr & {\text{Number :}} \cr & = {\text{ }}11x - 20{\text{ }} \cr & = {\text{ }}11 \times 4 - 20 \cr & = {\text{ }}24 \cr & {\text{Sum of digit }} = 2 + 4 = 6 \cr} $$

Since 13 is factor of 52. So divide its remainder directly by 13.
$$\eqalign{ & \Rightarrow \frac{{{\text{remainder}}}}{{13}} = \frac{{45}}{{13}} \cr & \Rightarrow {\text{remainder }} = 6 \cr} $$

Let the two - digit number be 10a + b where a > b
According to the question,
a + b = 12 ..... (i)
a - b = 6 ..... (ii)
On adding equation (i) and (ii)
2a = 18
⇒ a = 9
From equation (i),
9 + b = 12
⇒ b = 12 - 9 = 3
∴ Number is 10a + b
= 9 × 10 + 3
= 93
∴ When a < b
Then required number is = 39

Let the numerator and denominator br x and 11 - x
According to question,
$$\eqalign{ & {\text{Fraction }} = \frac{x}{{11 - x}} \cr & \Rightarrow \frac{{x + 2}}{{11 - x + 2}} = \frac{x}{{11 - x}} + \frac{1}{{24}} \cr & \Rightarrow \frac{{x + 2}}{{13 - x}} - \frac{x}{{11 - x}} = \frac{1}{{24}} \cr} $$
$$ \Rightarrow \frac{{11x + 22 - {x^2} - 2x - 13x + {x^2}}}{{\left( {13 - x} \right)\left( {11 - x} \right)}}$$       $$ = \frac{1}{{24}}$$
$$\eqalign{ & \Rightarrow \frac{{22 - 4x}}{{143 - 24x + {x^2}}} = \frac{1}{{24}} \cr & \Rightarrow 528 - 96x = 143 - 24x + {x^2} \cr & \Rightarrow {x^2} + 72x - 385 = 0 \cr & \Rightarrow \left( {x + 77} \right)\left( {x - 5} \right) = 0 \cr & \Rightarrow x = 5 \cr & {\text{Numerator }}(x) = 5 \cr & {\text{Denominator }} = 11 - 5 = 6 \cr & {\text{Difference }} = 6 - 5 = 1 \cr} $$

	(1570)2	+	(1571)2	+	(1572)2	+	(1573)2
unit digit →	↓		↓		↓		↓
	02	+	12	+	22	+	32
	↓		↓		↓		↓
	0	+	1	+	4	+	9 = 14

unit digit = 4

$$\eqalign{ & \frac{{{\text{remainder of number}}}}{{16}} = \frac{{54}}{{16}} \cr & \Rightarrow {\text{Remainder = 6}} \cr} $$

$$\eqalign{ & {{\text{7}}^{105}} \Rightarrow \cr & {{\text{7}}^1} \to 7 \to 7 \cr & {{\text{7}}^2} \to 49 \to 9 \cr & {{\text{7}}^3} \to 343 \to 3 \cr & {{\text{7}}^4} \to 2401 \to 1 \cr & {\text{Divided power of 7 by 4}} \cr & \frac{{105}}{4} \to {\text{ remainder = 1}} \cr & \Rightarrow {{\text{7}}^1}\,{\text{is left unit digit = 1}} \times 7 = 7 \cr} $$

Let us assume that he answered x question correctly. Marks scored by him in x question = 4x
Wrong answer would be 200 - x

According to question,
4x - (200 - x) = 200
or 4x - 200 + x = 200
or 5x = 200 + 200 = 400
or x = 80

Number of pants can be made = $$\frac{{252}}{{\frac{5}{2}}}$$ = 100

Number of shirts can be made = $$\frac{{141}}{{\frac{7}{4}}}$$ = 80
Number of pants and shirts = 100, 80