Among all options only option C has unit digit 5, and in given equation unit digit will also be 5.
So only 155 can divide the given equation completely.

The least number which when divided by 25, 40 and 60 leaves the remainder 7

= L.C.M. (25,40,60) + 7

= 600 + 7 = 607

1+4=5
5+(4+3)=12
12+(4+3+5)=24
24+(4+3+5+7)=43
43+(4+3+5+7+9)=71

let's say one number is n and another number is p
so acc. to question sum will be 18n+21p
and this number will be divisible by 3 so answer will be (A)

$$a^2-b^2 = 19$$
19 is a prime number 
(a + b)(a - b) = 19 x 1
So a + b = 19
     a - b = 1
Adding both the equations we get 2a = 20
                                                   a = 10

21+x=3x-7
or 2x=28
x = 14

For an equation $$ax^{2} + bx + c = 0,

the product of the roots is given by = $$\frac{c}{a}$$

Comparing, $$x^{2}-\sqrt{3}=0$$ from above equation,

product of roots = $$\frac{-\sqrt{3}}{1} = -\sqrt{3}$$

$$x = \sqrt{3} + \sqrt{2}$$

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

=> $$\frac{1}{x}$$ = $$\frac{\sqrt{3} - \sqrt{2}}{3 - 2}$$ = $$\sqrt{3} - \sqrt{2}$$

Now, $$x - \frac{1}{x}$$ = $$\sqrt{3} + \sqrt{2} - \sqrt{3} + \sqrt{2}$$

= 2$$\sqrt{2}$$

Using, $$(x - \frac{1}{x})^{3}$$ = $$x^{3} - \frac{1}{x^{3}} - 3(x - \frac{1}{x})$$

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

= $$16\sqrt{2} + 6\sqrt{2} = 22\sqrt{2}$$

The numbers 6,9,12,15,18 leaves same remainder 2 in each case.

So, what we need to do is find the L.C.M. of these numbers and add 2 to it

=> L.C.M. of 6,9,12,15,18 = 180

=> Required no. = 180+2 = 182

Expression : 2 × (2.11 + 2.23 + 2.16)

= 2 * (6.50)

= 13

Expression : $$(3^2 - 2^2 )^2 + (5^2 - 4^2 )^2 + (6^2 - 5^2 )^2$$

= $$[(3-2) (3+2)]^2 + [(5-4) (5+4)]^2 + [(6-5) (6+5)]^2$$

= $$25 + 81 + 121 = 227$$

It is given that $$aΘ b =a^{2}\frac{b}{{3}}$$

Applying the same rule for $$2Θ {3Θ(-1)}$$

= 2 Θ $${\frac{3^2 \times (-1)}{3}}$$

= 2 Θ -3

= $$\frac{2^2 \times (-3)}{3} = -4$$

$$\frac{1 + (a-1)(a+1)}{a^2}$$ = 1

and as $$\frac{1+876542\times876544}{876543\times876543}$$ is of the same form so it is equal to 1 

$$2+x\sqrt{3}$$ = $$\frac{1}{2+\sqrt{3}}$$

Rationalizing the R.H.S.

=> $$2+x\sqrt{3}$$ = $$\frac{1}{2+\sqrt{3}}$$ * $$\frac{2-\sqrt{3}}{2-\sqrt{3}}$$

=> $$2+x\sqrt{3}$$ = $$\frac{2-\sqrt{3}}{4-3}$$

=> $$2+x\sqrt{3}$$ = $$2-\sqrt{3}$$

Comparing both sides, we get $$x$$ = -1

The least number which is divisible by all natural numbers from 1 to 10

= L.C.M.(1,2,3,4,5,6,7,8,9,10)

= 2520

$$x^{8}-1$$ can be written as $$(x^{4}-1)(x^{4}+1)$$, which, in turn, can be written as $$(x^{2}-1)(x^{2}+1)(x^{4}+1)$$

$$x^{4}+2x^{3}-2x-1$$ can be written as $$(x^{2}-1)(x^{2}+2x +1)$$.

Hence, we can see that $$x^{2} - 1$$ is the HCF.

NOTE :- Sum of first 'n' odd natural numbers = n$$^{2}$$

Sum of first 'n' even natural numbers = n(n+1)

Sum of first 20 odd natural numbers = $$20^{2}$$ = 400

Arithmetic mean = 400/20 = 20

Let the sum = $$x$$

rate = 10% and time = 2 years

Amount after compound interest = $$x (1 + \frac{10}{100})^{2}$$ = 12100

=> $$\frac{121x}{100}$$ = 12100

=> $$x$$ = 10,000

Let A = $$4x$$ and B = $$5x$$

Difference in their squares = $$(5x)^{2} - (4x)^{2}$$ = 81

=> 25$$x^{2}$$ - 16$$x^{2}$$ = 81

=> $$x^{2}$$ = $$\frac{81}{9}$$

=> $$x$$ = 3

value of A = 4*3 = 12

$$\frac{x^{24}+1}{x^{12}}$$ = 7

We need to find, $$\frac{x^{72}+1}{x^{36}}$$ = $$x^{36} + \frac{1}{x^{36}}$$

=> $$x^{12} + \frac{1}{x^{12}}$$ = 7

Cubing both sides, and using the formula $$(a+b)^{3}$$ = $$a^{3}+b^{3}$$+ 3ab(a+b) , we get :

=> $$x^{36} + \frac{1}{x^{36}} + 3*1*(x^{12}+\frac{1}{x^{12}})$$ = 343

=> $$x^{36} + \frac{1}{x^{36}}$$ + 21 = 343

=> $$x^{36} + \frac{1}{x^{36}}$$ = 343-21 = 322

n = 5k+2 (where k is quotient )
so $$n^2 = 25k^2 + 4 + 20k$$
Now when $$n^2$$ will divided by 5 , remainder will be 4.

The remainder obtained by dividing 2055 by 27 = 3

So, the least number that should be 'subtracted' from 2055 to make it perfectly divisible by 27 = 3

and the least number that should be added = 27-3 = 24

243 = 3 * 3 * 3 * 3 * 3

= $$3^3 * 3^2$$

Thus, required number to be multiplied = 3

$$121a^{2}+64b^{2}$$

= $$(11a)^{2} + (8b)^{2}$$

$$\because (x+y)^{2} = x^{2} + y^{2} + 2xy$$

$$\therefore$$ Required expression = $$2 * 11a * 8b$$

= $$176ab$$

$$x=\frac{\cos\theta}{1-\sin\theta}$$ = $$\frac{\cos\theta (1 + \sin\theta)}{1-\sin\theta (1 + \sin\theta)}$$

=> $$x=\frac{\cos\theta (1 + \sin\theta)}{1-\sin^{2}\theta}$$

=> $$x=\frac{\cos\theta (1 + \sin\theta)}{\cos^{2}\theta}$$

=> $$x=\frac{1 + \sin\theta}{\cos\theta}$$

$$\therefore$$ $$\frac{\cos\theta}{1 + \sin\theta} = \frac{1}{x}$$

$$\frac{1}{\sqrt{7}-\sqrt{6}}-\frac{1}{\sqrt{6}-\sqrt{5}}+\frac{1}{\sqrt{5}-2}-\frac{1}{\sqrt{8}-\sqrt{7}}+\frac{1}{3-\sqrt{8}}$$

Rationalizing each term, we get, the denominator of each term will be 1, we get :

= $$\sqrt{7}$$ + $$\sqrt{6}$$ - ($$\sqrt{6}$$ + $$\sqrt{5}$$) + $$\sqrt{5}$$ + 2 - ($$\sqrt{8}$$ + $$\sqrt{7}$$) + 3 + $$\sqrt{8}$$

= 2+3 = 5

As we know a number with unit digit 2 have repeating cycle of 2,4,8,6 after every fourth power
as power is 173 or (172+1) where till 172 , 43rd cycle will get complete and next unit digit will be 2.

$$\sqrt{2}$$ = 1.414

$$\sqrt{3}$$ = 1.732

$$\sqrt{5}$$ = 2.236

$$\sqrt{6}$$ = 2.449

=> $$\sqrt{2}$$ + $$\sqrt{6}$$ = 3.863

and $$\sqrt{5}$$ + $$\sqrt{3}$$ = 3.968

=> $$\sqrt{2}$$ + $$\sqrt{6}$$ < $$\sqrt{5}$$ + $$\sqrt{3}$$

Thus, (ii) is correct.

675 = $$5^2 \times 3^3$$

and hence in order to make it a perfect cube we need to multiply it with 5 .

675 x 5 = $$5^3 \times 3^3$$

Both of numbers have unit digit as 4 and it has a repeating cycle of 2 with unit digits as 4 and 6 
so in first number power is 372 which is exactly divisible by 2 hence unit digit of first number will be 6.
and in second number power is 373 which exceeds one with the reapeating cycle of 2 hence its unit digit will be 4.

now unit digit of the sum will be 6+4 = 10

$$17^{200}=(18-1)^{200}$$
Hence, when it is divided by 18, the reminder equals $$(-1)^{200}=1$$

We know that, 41 * 41 = 1681

and 42 * 42 = 1764

So, the least number to be added in 1728 to make it a perfect square = 1764-1728

= 36

$$\frac{75070}{65}=1154.92$$

so the nearest multiple of 65 is $$65\times1155=75075$$

so the answer is option B.

Let's say number is N
So according to student result is $$112= \frac{N+12}{6} $$
or N = 660
Correct answer will be = $$\frac{660}{6} +12 = 110+12 = 122$$

if the number N is divided by 899 and leaves a remainder 65 then N = 899K + 65

and hence when N will be divided by 31

remainder of $$\frac{899K + 65}{31}$$ = Remainder of $$\frac{65}{31}$$ as 899 is completely divisible by 31

and hence remainder is 3

Let sum of money = $$100x$$

rate = 5%, time = 2 years

=> S.I. = $$\frac{100x * 5 * 2}{100} = 10x$$

=> C.I. = $$100x(1 + \frac{5}{100})^{2} - 100x$$

= $$100x(\frac{441}{400} - 1)$$

= $$100x(\frac{41}{400})$$ = $$\frac{41x}{4}$$

Required difference = $$\frac{41x}{4} - 10x$$ = 25

= $$\frac{x}{4}$$ = 25 => $$x$$ = 100

=> Sum = 100*100 = 10,000

let the missing term be y 

here the given pattern is  ­-1, 6,25, 62,123, 214 , y 

- 1    6     25       62    123      214           y 

     7    19     37       61      91        y-214

        12    18       24      30      36 

 and hence y - 214 - 91 = 36 

y = 305 + 36 = 341

Number will be equal to 420t +3 = 13M
put values of M and t accordingly and find least value of it.

21 = 3 x 7

36 = $$3^2 x 2^2$$

66 = 2 x 3 x 11

and hence minimum number divisible by these numbers and perfect square is $$3^2 \times 2^2 \times 7^2 \times 11^2$$

= 213444

here the given sequence is 3, 7, 13, 21, 33, 43, 57

7 - 3 = 4

13 - 7 = 6

21 - 13 = 8

33 - 21 = 12

43 - 33 = 10

57 - 43 = 14

here we can see that difference between consecutive terms is increasing by 2 every time but due to presence of 33 the pattern is not getting formed and hence 31 is the right number in place of 33 and hence odd one out is 33

Given:
 Sequence of numbers 0,3,8,15,24,35
To know:
 9th term of the sequence
As we can see difference of sequence is in arthmatic progression of 3,5,7,9,11,13 and so on.
6th term of sequence is 35
and adding common difference for it (that is 13) , hence 7th term will be 48
accordingly 9th term will be 80

$$p+\frac{1}{4}\sqrt{p}+k^{2}$$

= $$(\sqrt{p})^{2} + 2 * \sqrt{p} * \frac{1}{8} + (\frac{1}{8})^{2} - (\frac{1}{8})^{2} + k^{2}$$

=> $$k^{2} = (\frac{1}{8})^{2}$$

=> $$k = \pm\frac{1}{8}$$

In the expression, $$\frac{x^{4}-\frac{1}{x^{2}}}{3x^{2}+5x-3}$$

Dividing numerator and denominator by $$x$$, we get

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

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

Now, putting $$x-\frac{1}{x}$$ = 1

we get, = $$\frac{1+3}{3+5} = \frac{4}{8} = \frac{1}{2}$$

$$\frac{b-c}{a}+\frac{a+c}{b}+\frac{a-b}{c}=1$$

=> $$\frac{b-c}{a}+\frac{a-b}{c}+\frac{a+c}{b} - 1 = 0$$

=> $$\frac{b-c}{a}+\frac{a-b}{c}+\frac{a+c-b}{c} = 0$$

=> $$\frac{c-b}{a}+\frac{b-a}{c} = \frac{a+c-b}{b}$$

=> $$\frac{c^{2}-bc+ab-a^{2}}{ac} = \frac{a+c-b}{b}$$

=> $$\frac{(c^{2}-a^{2}) - (bc-ab)}{ac} = \frac{a+c-b}{b}$$

=> $$\frac{(c-a)(c+a) -b(c-a)}{ac} = \frac{a+c-b}{b}$$

=> $$\frac{(c-a)(c+a-b)}{ac} = \frac{a+c-b}{b}$$

=> $$\frac{c-a}{ac} = \frac{1}{b}$$

=> $$\frac{c}{ac} - \frac{a}{ac} = \frac{1}{b}$$

=> $$\frac{1}{a} - \frac{1}{c} = \frac{1}{b}$$

15% of $$x = \frac{15}{100} * x = \frac{3x}{20}$$

20% of $$y = \frac{20}{100} * y = \frac{y}{5}$$

Now, both are equal

=> $$\frac{3x}{20}$$ = $$\frac{y}{5}$$

=> $$\frac{x}{y}$$ = $$\frac{20}{5*3}$$

= 4 : 3

here we need to find least number which when divided by 35, 45, 55 leaves the remainder 18, 28, 38

35 - 18 = 17

45 - 28 = 17

55 - 38 = 17

the number will be of the form = LCM(35,45,55)K - 17

N = 3465k - 17

for least value put k = 1

N = 3448

Let the same remainder in every case be y

hence we need to find HCF of 19-y ,35-y and 59- y

using difference method ,

35 - y - 19 + y = 16

59 - y - 35 + y = 24

HCF of 16 and 24 is 8

hence 8 is the highest number which on dividing 19, 35 and 59 will leave same remainder

Let the given number be x
Let a be the quotient when x is divided by 114
So $$\frac{x}{114}$$ = a$$\frac{21}{114}$$
so x = 114a + 21
when x is divided by 19 it can be written as
$$\frac{x}{19} = \frac{114a + 21}{19}$$
114 is divisible by 19 and 21 leaves a remainder of 2.

Given: Numbers- First = 5834
                         Second = x (Suppose)
                         And number (5834 - x) is divisible by each of 20,28,32,35
                         Let's say it is y
                         Hence 5834 - x = y
                         or x = 5834 - y
Now for x to be greatest y should be least
hence y should be least common multiple of 20,28,32,35
y = 1120
now x = 5834 - 1120
       x = 4714

SInce solving this problem algebraically is a very tedious process, let us put some values for a,b, and c. Then, we will try to match the options.

Let a =1, b = 2 and c=3.

$$\frac{m-1}{13}+\frac{m-4}{10}+\frac{m-9}{5}=3$$

Taking LCM, we get,

$$\frac{10m-10+13m-52+26m-234}{130}=3$$

$$49m = 390 + 296$$

$$49m = 686$$

$$m = 14$$

Substituting a,b and c in options, only option C gives 14 as the answer. Hence, option C is the right answer.