Expression : $$X=0.3\times0.3$$

=> $$X=0.09$$

=> Ans - (C)

As c=0, and substituting the point (0,0) in the equation, we get ax+by+c = 0 at the point (0,0).
Hence, the line passes through origin.

The $$n^{th}$$ term of an A.P. = $$a + (n - 1) d$$, where 'a' is the first term , 'n' is the number of terms and 'd' is the common difference.

5th term, $$A_5 = a + (5 - 1) d = 9$$

=> $$a + 4d = 9$$ -----------------(i)

Similarly, 12th term, $$A_{12} = a + 11d = -26$$ ------------------(ii)

Subtracting equation (i) from (ii), we get :

=> $$(11d - 4d) = -26 - 9$$

=> $$d = \frac{-35}{7} = -5$$

Substituting it in equation (i), => $$a - 20 = 9$$

=> $$a = 9 + 20 = 29$$

Let $$n^{th}$$ term = -6

=> $$a + (n - 1) d = -6$$

=> $$29 + (n - 1) (-5) = -6$$

=> $$(n - 1) (-5) = -6 - 29 = -35$$

=> $$(n - 1) = \frac{-35}{-5} = 7$$

=> $$n = 7 + 1 = 8$$

 $$t_{1}=1$$, $$t_{2}=2$$

 $$t_{n+2}$$ = $$t_{n}+t_{n+1}$$

put n=3, then  $$t_{5}$$ = $$t_{3}+t_{4}$$

$$t_{3}$$ = $$t_{1}+t_{2}$$ = 1+2 = 3

$$t_{4}$$ = $$t_{2}+t_{3}$$ = 2+3 = 5

$$t_{5}$$ = $$t_{3}+t_{4}$$ = 3+5 = 8

so the answer is option D.

I : $$(\sqrt{15}+\sqrt{7})<(2\sqrt{22})$$

Squaring both sides, we get :

L.H.S. = $$(\sqrt{15}+\sqrt{7})^2=15+7+2\sqrt{105}=(22+2\sqrt{105})\approx(22+2\times10)=42$$

R.H.S. = $$(2\sqrt{22})^2=88$$

Thus, L.H.S. < R.H.S., which is correct.

II : $$(\sqrt{17}+\sqrt{5})<(\sqrt{20}+\sqrt{2})$$

Squaring both sides, we get :

L.H.S. = $$(\sqrt{17}+\sqrt{5})^2=17+5+2\sqrt{85}=(22+2\sqrt{85})$$

R.H.S. = $$(\sqrt{20}+\sqrt{2})^2=20+2+2\sqrt{40}=(22+2\sqrt{40})$$

$$\because$$ $$\sqrt{85}>\sqrt{40}$$, then L.H.S. > R.H.S.

Thus, only I is correct.

=> Ans - (A)

Given : $$x=\frac{\sqrt{5}+1}{\sqrt{5}-1}$$ and $$y=\frac{\sqrt{5}-1}{\sqrt{5}+1}$$

To find : $$x-y=?$$

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

= $$\frac{(\sqrt5+1)^2-(\sqrt5-1)^2}{(\sqrt5-1)(\sqrt5+1)}$$

= $$\frac{(6+2\sqrt5)-(6-2\sqrt5)}{(5-1)}$$

= $$\frac{4\sqrt5}{4}=\sqrt5$$

=> Ans - (B)

Given : $$z=6-2\sqrt3$$ -----------------(i)

To find : $$(\sqrt{z}-\frac{1}{\sqrt{z}})^2$$

= $$(\frac{z-1}{\sqrt{z}})^2=\frac{(z-1)^2}{z}$$ ------------(ii)

From equation (i), => $$z-1=5-2\sqrt3$$

Squaring both sides, we get :

=> $$(z-1)^2=(5-2\sqrt3)^2$$

=> $$(z-1)^2=25+12-20\sqrt3=(37-20\sqrt3)$$ ----------(iii)

Substituting values from equations (i) and (iii) in equation (ii), 

=> $$\frac{(z-1)^2}{z}=\frac{37-20\sqrt3}{6-2\sqrt3}$$

Rationalizing the denominator, we get :

= $$\frac{37-20\sqrt3}{6-2\sqrt3}\times\frac{(6+2\sqrt3)}{(6+2\sqrt3)}$$

= $$\frac{37(6+2\sqrt3)-20\sqrt3(6+2\sqrt3)}{36-12}$$

= $$\frac{222+74\sqrt3-120\sqrt3-120}{24}$$

= $$\frac{102-46\sqrt3}{24}$$

=> Ans - (C)

Terms : $$\sqrt[4]{5},\sqrt[3]{4}$$ and $$\sqrt[4]{6}$$

Multiplying the exponents by L.C.M. (4,3,4) = 12

=> $$(5)^{\frac{12}{4}}$$ , $$(4)^{\frac{12}{3}}$$ and $$(6)^{\frac{12}{4}}$$

= $$5^3,4^4,6^3$$

= $$125,256,216$$

Thus, in descending order = $$256>216>125$$

$$\equiv$$ $$\sqrt[3]{4}>\sqrt[4]{6}>\sqrt[4]{5}$$

=> Ans - (D)

The $$n^{th}$$ term of an A.P. = $$a + (n - 1) d$$, where 'a' is the first term , 'n' is the number of terms and 'd' is the common difference.

3rd term, $$A_3 = a + (3 - 1) d = 17$$

=> $$a + 2d = 17$$ -----------------(i)

Similarly, 17th term, $$A_{17} = a + 16d = -25$$ ------------------(ii)

Subtracting equation (i) from (ii), we get :

=> $$(16d - 2d) = -25 - 17$$

=> $$d = \frac{-42}{14} = -3$$

Substituting it in equation (i), => $$a - 6 = 17$$

=> $$a = 17 + 6 = 23$$

Let $$n^{th}$$ term = -1

=> $$a + (n - 1) d = -1$$

=> $$23 + (n - 1) (-3) = -1$$

=> $$(n - 1) (-3) = -1 - 23 = -24$$

=> $$(n - 1) = \frac{-24}{-3} = 8$$

=> $$n = 8 + 1 = 9$$

In an arithmetic progression with first term, $$a = 37$$ , last term, $$l = -18$$

Let number of terms = $$n$$

$$\therefore$$ Sum of A.P. = $$\frac{n}{2} (a + l) = 114$$

=> $$\frac{n}{2} (37 - 18) = 114$$

=> $$19n = 114 \times 2 = 228$$

=> $$n = \frac{228}{19} = 12$$

=> Ans - (B)

Expression : (81 + 82 + 83 + ……… +130)

This is an arithmetic progression with first term, $$a = 81$$ , last term, $$l = 130$$ and common difference, $$d = 1$$

Let number of terms = $$n$$

Last term in an A.P. = $$a + (n - 1)d = 130$$

=> $$81 + (n - 1)(1) = 130$$

=> $$n - 1 = 130 - 81 = 49$$

=> $$n = 49 + 1 = 50$$

$$\therefore$$ Sum of A.P. = $$\frac{n}{2} (a + l)$$

= $$\frac{50}{2} (81 + 130)$$

= $$25 \times 211 = 5275$$

Let the number be $$x$$

=> 5.2% of $$x$$ = 39

=> $$\frac{5.2}{100} \times x = 39$$

=> $$x = \frac{3900}{5.2} = \frac{300}{0.4}$$

=> $$x=750$$

=> Ans - (B)

Let the number be $$x$$

According to ques,

=> $$\frac{9}{4} \times \frac{7}{2} \times x = 126$$

=> $$\frac{63}{8} x = 126$$

=> $$x = \frac{126}{63} \times 8$$

=> $$x = 2 \times 8 = 16$$

$$\therefore (\frac{7}{2})^{th}$$ of the number = $$\frac{7}{2} \times 16$$

= $$7 \times 8 = 56$$

=> Ans - (A)

Expression : $$\frac{2714}{5074}$$

Dividing both numerator and denominator by 2, we get = $$\frac{1357}{2537}$$

Similarly, dividing by 59, we get :

= $$\frac{23}{43}$$

=> Ans - (C)

Expression : $$\frac{2530}{1430}$$

Dividing both numerator and denominator by 10, we get = $$\frac{253}{143}$$

Similarly, dividing by 11, we get :

= $$\frac{23}{13}$$

=> Ans - (B)

Let the number be $$x$$

According to ques, 3.6% of $$x$$ = 27

=> $$\frac{3.6}{100} \times x = 27$$

=> $$x = \frac{2700}{3.6} = \frac{300}{0.4}$$

=> $$x = 750$$

=> Ans - (A)

Expression : $$4x+\frac{\sqrt{x}}{6}+\frac{m^2}{4}$$

Let $$\sqrt{x}=y$$

= $$4y^2+\frac{y}{6}+\frac{m^2}{4}$$

= $$(2y)^2+2(2y)(\frac{1}{24})+(\frac{m}{2})^2$$

Using, $$a^2+2ab+b^2=(a+b)^2$$

=> $$\frac{m}{2}=\frac{1}{24}$$

=> $$m=\frac{2}{24}=\frac{1}{12}$$

=> Ans - (B)

Expression : $$2\sqrt[3]{243}+3\sqrt[3]{9}-\sqrt[3]{1125}$$

= $$2\sqrt[3]{27\times9}+3\sqrt[3]{9}-\sqrt[3]{125\times9}$$

= $$6\sqrt[3]9+3\sqrt[3]9-5\sqrt[3]9$$

= $$4\sqrt[3]9$$

=> Ans - (B)

Expression : $$\sqrt[3]{512}+\sqrt{169}+\sqrt[3]{216}+\sqrt{225}$$

= $$\sqrt[3]{8\times8\times8}+13+\sqrt[3]{6\times6\times6}+15$$

= $$8+13+6+15=42$$

=> Ans - (C)

Expression : $$[25+4\sqrt{39}]$$

= $$[25+2\sqrt{4\times39}]=[25+2\sqrt{156}]$$

= $$13+12+2\sqrt{13\times12}$$

= $$(13)^2+(12)^2+2\sqrt{13\times12}$$

Now, we know that $$a^2+b^2+2ab=(a+b)^2$$

= $$(\sqrt{13}+\sqrt{12})^2$$

Thus, square root is = $$\sqrt{13}+\sqrt{12}$$

= $$\sqrt13+2\sqrt3$$

=> Ans - (A)

Expression : $$[(\sqrt{529})+(\sqrt{5.29})+(\sqrt{0.0529})]$$

= $$23+2.3+0.23=25.53$$

=> Ans - (B)

Expression : $$\sqrt{4x^2-4x+1}$$

= $$\sqrt{(2x-1)^2}=(2x-1)$$ -------------(i)

It is given that $$x=1.1$$, substituting it in equation (i), 

=> $$2(1.1)-1=2.2-1=1.2$$

=> Ans - (D)

Expression : $$\sqrt{1-(\frac{x}{529}})=(\frac{16}{23})$$

=> $$\sqrt{\frac{529-x}{529}}=\frac{16}{23}$$

=> $$\frac{529-x}{529}=(\frac{16}{23})^2$$

=> $$\frac{529-x}{529}=\frac{256}{529}$$

=> $$529-x=256$$

=> $$x=529-256=273$$

=> Ans - (C)

In an arithmetic progression with first term, $$a = 33$$ , last term, $$l = -57$$

Number of terms = $$n = 16$$

$$\therefore$$ Sum of A.P. = $$\frac{n}{2} (a + l)$$

= $$\frac{16}{2} (33 - 57)$$

= $$8 \times (-24) = -192$$

=> Ans - (B)

Given : $$p+\frac{1}{p}=\sqrt{10}$$

Squaring both sides, we get :

=> $$(p+\frac{1}{p})^2=(\sqrt{10})^2$$

=> $$p^2+\frac{1}{p^2}+2(p)(\frac{1}{p})=10$$

=> $$p^2+\frac{1}{p^2}=10-2=8$$

Again squaring both sides, we get :

=> $$(p^2+\frac{1}{p^2})^2=(8)^2$$

=> $$p^4+\frac{1}{p^4}+2(p^2)(\frac{1}{p^2})=64$$

=> $$p^4+\frac{1}{p^4}=64-2=62$$

=> Ans - (C)

Let the number be $$x$$

According to ques,

=> $$\frac{4}{5} \times \frac{6}{7} \times x = 216$$

=> $$x = 216 \times \frac{35}{24} = 9 \times 35$$

$$\therefore$$ 8/9th of the number = $$\frac{8}{9} \times (35 \times 9)$$

= $$8 \times 35 = 280$$

=> Ans - (B)

Let the number be $$x$$

According to ques, 0.4% of x = 26

=> $$\frac{0.4}{100} \times x = 26$$

=> $$\frac{1}{250} \times x = 26$$

=> $$x = 26 \times 250 = 6500$$

=> Ans - (A)

According to question, => $$(\frac{5}{9} \times x)-(\frac{2}{5} \times \frac{9}{4})=\frac{-4}{5}$$

=> $$\frac{5x}{9} - \frac{9}{10} = \frac{-8}{10}$$

=> $$\frac{5x}{9} = \frac{9}{10}-\frac{8}{10} = \frac{1}{10}$$

=> $$x = \frac{9}{50} = 0.18$$

=> Ans - (A)

The $$n^{th}$$ term of an A.P. = $$a + (n - 1) d$$, where 'a' is the first term , 'n' is the number of terms and 'd' is the common difference.

4th term, $$A_4 = a + (4 - 1) d = 15$$

=> $$a + 3d = 15$$ -----------------(i)

Similarly, 15th term, $$A_{15} = a + 14d = -29$$ ------------------(ii)

Subtracting equation (i) from (ii), we get :

=> $$(14d - 3d) = -29 - 15$$

=> $$d = \frac{-44}{11} = -4$$

Substituting it in equation (i), => $$a - 12 = 15$$

=> $$a = 15 + 12 = 27$$

$$\therefore$$ 10th term, $$A_{10} = a + (10 - 1)d$$

= $$27 + (9 \times -4) = 27 - 36 = -9$$

=> Ans - (D)

Let the number be $$x$$

=> $$\frac{7}{8} \times \frac{5}{4} \times x = 315$$

=> $$x = 9 \times 8 \times 4 = 32 \times 9$$

$$\therefore$$ 5/9th of the number = $$\frac{5}{9} \times 32 \times 9$$

= 160

Given : $$\sqrt{21}=4.58$$

To find : $$(8\sqrt{\frac{3}{7}}-3\sqrt{\frac{7}{3}})$$

= $$\frac{(8\times3)-(3\times7)}{\sqrt{21}}$$

= $$\frac{(24-21)}{\sqrt{21}}=\frac{3}{\sqrt{21}}$$

= $$\frac{3}{4.58}\approx0.655$$

=> Ans - (C)

Expression : $$x=\sqrt{15\sqrt{15\sqrt{15\sqrt{15\sqrt{15....\infty}}}}}$$

=> $$x=\sqrt{15x}$$

Squaring both sides, we get :

=> $$x^2=15x$$

=> $$x=15$$

To find : $$x^2+4$$

= $$(15)^2+4=225+4=229$$

=> Ans - (D)

Expression : $$4^x= \sqrt[7]{1024}$$

=> $$4^x=\sqrt[7]{2^{10}}$$

=> $$4^x=\sqrt[7]{4^5}$$

=> $$4^x=4^{\frac{5}{7}}$$

=> $$x=\frac{5}{7}$$

=> Ans - (A)

Given : $$x = 4 + \sqrt{15}$$ -------------(i)

=> $$\frac{1}{x} = \frac{1}{4 + \sqrt{15}}$$

=> $$\frac{1}{x} = \frac{1}{4 + \sqrt{15}}\times(\frac{4-\sqrt{15}}{4-\sqrt{15}})$$

=> $$\frac{1}{x}=\frac{4-\sqrt{15}}{(16-15)}=4-\sqrt{15}$$ -------------(ii)

To find : $$[x^2 + (\frac{1}{x^2})]$$ 

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

Substituting values from equations (i) and (ii), we get :

= $$[(4+\sqrt{15})+(4-\sqrt{15})]^2-2$$

= $$(8)^2-2=64-2=62$$

=> Ans - (A)

In an arithmetic progression with first term, $$a = 29$$ , last term, $$l = -49$$

Let number of terms = $$n$$

$$\therefore$$ Sum of A.P. = $$\frac{n}{2} (a + l) = -140$$

=> $$\frac{n}{2} (29 - 49) = -140$$

=> $$\frac{-20n}{2} = -140$$

=> $$n = \frac{(-140) \times 2}{-20} = 7 \times 2$$

=> $$n=14$$

=> Ans - (B)

First term of AP, $$a=-32$$ and last term, $$l=43$$

Let there be $$n$$ terms

Sum of AP = $$\frac{n}{2}(a+l) = 88$$

=> $$\frac{n}{2}(-32+43)=88$$

=> $$\frac{11n}{2}=88$$

=> $$n=88 \times \frac{2}{11}$$

=> $$n=8 \times 2=16$$

=> Ans - (A)

Product of : 6ab, 3b and 4a

= $$(6 \times 3 \times 4) \times (ab \times b \times a)$$

= $$72a^2b^2$$

=> Ans - (A)

Expression : 4*(-0.1(5.5 + 8.5)) of 0.5 = ?

= $$4 \times 0.5 [-0.1(14)]$$

= $$2 \times (-1.4)$$

= $$-2.8$$

=> Ans - (A)

In an arithmetic progression with first term, $$a = -23$$ , last term, $$l = 42$$

Number of terms = $$n = 14$$

$$\therefore$$ Sum of A.P. = $$\frac{n}{2} (a + l)$$

= $$\frac{14}{2} (-23 + 42)$$

= $$7 \times 19 = 133$$

=> Ans - (B)

Terms in arithmetic progression : $$(2x - 4) , (4x + p) , (6x - 12)$$

=> Difference between first two terms is equal to the difference between last two terms

=> $$(4x + p) - (2x - 4) = (6x - 12) - (4x + p)$$

=> $$2x + p + 4 = 2x - 12 - p$$

=> $$2p = -12 - 4 = -16$$

=> $$p = \frac{-16}{2} = -8$$

Terms in arithmetic progression : $$(-3x - 11) , (x + p) , (5x + 7)$$

=> Difference between first two terms is equal to the difference between last two terms

=> $$(x + p) - (-3x - 11) = (5x + 7) - (x + p)$$

=> $$4x + p + 11 = 4x + 7 - p$$

=> $$2p = 7 - 11 = -4$$

=> $$p = \frac{-4}{2} = -2$$

=> Ans - (D)

The $$n^{th}$$ term of an A.P. = $$a + (n - 1) d$$, where 'a' is the first term , 'n' is the number of terms and 'd' is the common difference.

3rd term, $$A_3 = a + (3 - 1) d = 13$$

=> $$a + 2d = 13$$ -----------------(i)

Similarly, 13th term, $$A_{13} = a + 12d = 47$$ ------------------(ii)

Subtracting equation (i) from (ii), we get :

=> $$(12d - 2d) = 47 - 13 = 34$$

=> $$d = \frac{34}{10} = 3.4$$

Substituting it in equation (i), => $$a + 2 \times 3.4 = 13$$

=> $$a = 13 - 6.8 = 6.2$$

Let $$n^{th}$$ term = 30

=> $$a + (n - 1) d = 30$$

=> $$6.2 + (n - 1) (3.4) = 30$$

=> $$(n - 1) (3.4) = 30 - 6.2 = 23.8$$

=> $$(n - 1) = \frac{23.8}{3.4} = 7$$

=> $$n = 7 + 1 = 8$$

Given : $$(a - b) = -6$$ and $$a^2 + b^2 = 44$$

Using $$(a - b)^2 = a^2 + b^2 - 2ab$$

=> $$(-6)^2 = 44 - (2 \times ab)$$

=> $$2 ab = 44 - 36 = 8$$

=> $$ab = \frac{8}{2} = 4$$

=> Ans - (C)

Given : $$x=\frac{1}{\sqrt{3}}$$ and $$y=\frac{1}{\sqrt{5}}$$

=> $$x^2=\frac{1}{3}$$ and $$y^2=\frac{1}{5}$$

Again, squaring both terms, we get : $$x^4=\frac{1}{9}$$ and $$y^4=\frac{1}{25}$$ -------------(i)

To find : $$(6x^2 - 5y^2)(6x^2 + 5y^2)$$

Using, $$(a-b)(a+b)=a^2-b^2$$ and substituting values from equation (i),

= $$36x^4-25y^4$$

= $$(36\times\frac{1}{9})-(25\times\frac{1}{25})$$

= $$4-1=3$$

=> Ans - (B)

Expression : $$\frac{\sqrt{5}+\sqrt{4}}{\sqrt{5}-\sqrt{4}}+\frac{\sqrt{5}-\sqrt{4}}{\sqrt{5}+\sqrt{4}}$$

= $$\frac{(\sqrt5+\sqrt4)^2+(\sqrt5-\sqrt4)^2}{(\sqrt5-\sqrt4)(\sqrt5+\sqrt4)}$$

= $$\frac{(5+4+2\sqrt{20})+(5+4-2\sqrt{20})}{5-4}$$

= $$\frac{(9+9)}{1}=18$$

=> Ans - (C)

I : $$\sqrt{676}+\sqrt{6.76}+\sqrt{0.0676}=27.76$$

L.H.S. = $$26+2.6+0.26=28.86\neq$$ R.H.S.

II : $$\sqrt{339+\sqrt{36}+\sqrt{49}+\sqrt{81}}=19$$

L.H.S. = $$\sqrt{339+6+7+9}=\sqrt{361}=19=$$ R.H.S.

Thus, only II is correct.

=> Ans - (B)

Expression : $$(cosec 60^\circ - \frac{1}{2})$$

= $$\frac{2}{\sqrt3}-\frac{1}{2}$$

= $$\frac{(4-\sqrt{3})}{2\sqrt{3}}$$

=> Ans - (A)

Expression : $$(\frac{1}{\sqrt3} - sin 45^\circ)$$

= $$\frac{1}{\sqrt3}-\frac{1}{\sqrt2}$$

= $$\frac{(\sqrt{2}-\sqrt{3})}{\sqrt{6}}$$

=> Ans - (B)

Expression : $$p=\sqrt{72-\sqrt{72-\sqrt{72-\sqrt{72-......\infty}}}}$$

=> $$p=\sqrt{72-p}$$

Squaring both sides, we get :

=> $$p^2=72-p$$

=> $$p^2+p-72=0$$

=> $$p^2+9p-8p-72=0$$

=> $$p(p+9)-8(p+9)=0$$

=> $$(p+9)(p-8)=0$$

=> $$p=-9,8$$

But $$p$$ cannot be negative, thus $$p=8$$

To find : $$2p^2+1$$

= $$2(8)^2+1=128+1=129$$

=> Ans - (C)

Given : $$\sqrt5=2.236$$

To find : $$\frac{\sqrt{5}}{(5-\sqrt5)}$$

= $$\frac{\sqrt{5}}{\sqrt5(\sqrt5-1)}=\frac{1}{\sqrt5-1}$$

Rationalizing the denominator, we get :

= $$\frac{1}{(\sqrt5-1)}\times\frac{(\sqrt5+1)}{(\sqrt5+1)}$$

= $$\frac{\sqrt5+1}{5-1}=\frac{\sqrt5+1}{4}$$

= $$\frac{2.236+1}{4}=\frac{3.236}{4}=0.809$$

=> Ans - (A)