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

Let the numerator be n and the denominator be n + 3
$$\eqalign{ & {\text{According to question}} \cr & \frac{{n + 7}}{{\left( {n + 3} \right) - 2}} = \frac{2}{1} \cr & \frac{{n + 7}}{{n + 1}} = \frac{2}{1} \cr & 2n + 2 = n + 7 \cr & n = 5 \cr & {\text{Fraction = }}\frac{5}{{5 + 3}} = \frac{5}{8} \cr & {\text{Sum = 5 + 8 = 13}} \cr} $$

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

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

No. →  I II III        3x       6x       2x    = 11x
$$\eqalign{ & \frac{{11x}}{3} = 44 \Rightarrow x = 12 \cr & {\text{Largest number = 6x = 72}} \cr} $$

$$\eqalign{ & {\left( {124} \right)^{372}}\,\,\,\,\,\,\,\,\,\,\,\,\,{\left( {124} \right)^{373}} \cr & \,\,\,\mathop {{4^{372}}}\limits^ \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\mathop {{4^{373}}}\limits^ \downarrow \cr} $$
When 4 has odd power then unit digit is 4, when 4 has even power then unit digit is 6

41 →   4 →   4   42 → 16 → 6   43 → 64 → 4   44 →  256  → 6   4372        4373    ↓       ↓   6   +   4 = 10 = 0
Last (unit) digit = 0

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

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

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

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

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

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

Let the unit digit of the number be x
And ten's place number be y
Number = 10y + x
Interchange digit's = 10x + y
Adding result
= 10y + x + 10x + y
= 11x + 11y
= 11(x + y)
So number is divisible by 11

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

a + b + 2ab is even
Explanation : Sum of two odd number is even number and multiplication of two odd number with 2 always given an even number.
Alternate :
Put , a = 3,   b = 5, option (D) will satisfy

Five digit least value = 10000
10000 is divided by 333
therefore, remainder = 10
Number = 10000 + 333 - 10 = 10323

Take L.C.M. of and 6 = 12 and numbers should be divisible by 12.
First number (after 100) = 108
Last of integers
$$\eqalign{ & = \frac{{{\text{ Last term }} - {\text{ First term}}}}{{{\text{Difference}}}} + 1 \cr & = \frac{{588 - 108}}{{12}} + 1 \cr & = \frac{{480}}{{12}} + 1 \cr & = 40 + 1 \cr & = 41 \cr} $$

$$\eqalign{ & \Rightarrow \left( {99\frac{{95}}{{99}}} \right) \times 99 \cr & = \left( {99 + \frac{{95}}{{99}}} \right) \times 99 \cr & = \left\{ {\left( {100 - 1} \right) + \frac{{95}}{{99}}} \right\} \times 99 \cr & = 9900 - 99 + 95 \cr & = 9896 \cr} $$

Let x and y are two numbers (x > y)
According to the question,
$$\eqalign{ & xy = {\text{ }}11520 \cr & \frac{x}{y} = \frac{9}{5} \Rightarrow \boxed{x = \frac{{9y}}{5}} \cr & {\text{put the value of x in eq}}{\text{. (i)}} \cr & \frac{{9{\text{y}}}}{5} \times {\text{y = 11520 }} \cr & \Rightarrow {\text{ }}{{\text{y}}^2} = 6400 \cr & \Rightarrow \boxed{{\text{y = 80}}} \cr & 80{\text{x = 11520 }} \cr & \Rightarrow \boxed{{\text{x = 144}}} \cr & {\text{x - y = 144 - 80 = 64 Ans}}{\text{.}} \cr} $$

$$\eqalign{ & {\left( {22} \right)^{23}} \cr & {\text{ Result unit digit}} \cr & {2^1}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\,\,\,\,\,\,\,\,\,\,\,2\, \leftarrow | \cr & {2^2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,\,\,4\,\,\,\,\,\,\,\,\,\,\,|\,\,{\text{Cycle}} \cr & {2^3}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,\,\,8\,\,\,\,\,\,\,\,\,\,\,|\,\,\,{\text{Completes}} \cr & {2^4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,16\,\,\,\,\,\,\,\,\,\,\,\,6 \leftarrow \,| \cr & {2^5}\,\,\,\,\,\,\,\,\,\,\,\,\,32\,\,\,\,\,\,\,\,\,\,\,\,2\,\,\, \cr & {\text{So divided power of 22 by 4 }} \cr & \frac{{23}}{4} = {\text{ remainder 3}} \cr & {2^3}\,\,\,\, = \,\,\,\,\,8\, \cr & {\text{unit digit = 8}} \cr} $$