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# Surds and Indices Questions for SSC CHSL PDF

For Previous year Surds and Indices Questions of SSC CHSL 2018-2019 Tier I exam download PDF. Go through the video of Repeatedly asked Surds and Indices questions explanations most important for the CHSL exam.

Question 1:Â Arranging the following in descending order, we get
$\sqrt[3]{4},\sqrt{2},\sqrt[6]{3},\sqrt[4]{5}$

a)Â $\sqrt[3]{4}>\sqrt[4]{5}>\sqrt{2}>\sqrt[6]{3}$

b)Â $\sqrt[4]{5}<\sqrt[3]{4}>\sqrt[6]{3}>\sqrt{2}$

c)Â $\sqrt{2}>\sqrt[6]{3}>\sqrt[3]{4}>\sqrt[4]{5}$

d)Â $\sqrt[6]{3}>\sqrt[4]{5}>\sqrt[3]{4}>\sqrt{2}$

Question 2:Â Number of digits in the square root of 62478078 is:

a)Â 4

b)Â 5

c)Â 6

d)Â 3

Question 3:Â The approx value of $5\frac{1}{3}+1\frac{2}{9}\times \frac{1}{4}(10+\frac{3}{1-\frac{1}{5}})$ is

a)Â $10$

b)Â $\frac{67}{25}$

c)Â $\frac{128}{11}$

d)Â $\frac{128}{99}$

Question 4:Â Which is the largest of the following fractions ?
$\frac{2}{5},\frac{3}{5},\frac{8}{11},\frac{11}{17}$

a)Â $\frac{8}{11}$

b)Â $\frac{3}{5}$

c)Â $\frac{11}{17}$

d)Â $\frac{2}{3}$

Question 5:Â The simplified value of $999\frac{1}{7}+999\frac{2}{7}+999\frac{3}{7}+999\frac{4}{7}+999\frac{5}{7}+999\frac{6}{7}$ is

a)Â $10009\frac{2}{7}$

b)Â $5994\frac{6}{7}$

c)Â $9999\frac{2}{7}$

d)Â 5997

SSC CHSL Study Material (FREE Tests)

Question 6:Â If $x=6+2\sqrt{6}$, then what is the value of $\sqrt{x-1}+\frac{1}{\sqrt{x-1}}$ ?

a)Â $2\sqrt{3}$

b)Â $3\sqrt{2}$

c)Â $2\sqrt{2}$

d)Â $3\sqrt{3}$

Question 7:Â If $x=\frac{2\sqrt{15}}{\sqrt{3}+\sqrt{5}}$, then what is the value of $\frac{x+\sqrt{5}}{x-\sqrt{5}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}$

a)Â $\sqrt{5}$

b)Â $\sqrt{3}$

c)Â $\sqrt{15}$

d)Â $2$

Question 8:Â Which value among $\sqrt[3]{5},\sqrt[4]{6},\sqrt[6]{12},\sqrt[12]{276}$ is the largest ?

a)Â $\sqrt[3]{5}$

b)Â $\sqrt[4]{6}$

c)Â $\sqrt[6]{12}$

d)Â $\sqrt[2]{12}$

Question 9:Â If $x-y-\sqrt{18}=-1$ and $x + y – 3\sqrt{2} = 1$, then what is the value of $12xy(x^{2} – y^{2})$ ?

a)Â $0$

b)Â $1$

c)Â $512\sqrt{2}$

d)Â $612\sqrt{2}$

Question 10:Â If $\frac{p}{q}=\frac{r}{s}=\frac{t}{u}=\sqrt{5}$, then what is the value of $[\frac{(3p^{2} + 4r^{2} + 5t^{2})}{(3q^{2} + 4s^{2} + 5u^{2})}]$ Â ?

a)Â 1/5

b)Â 5

c)Â 25

d)Â 60

Question 11:Â If $x+[\frac{1}{(x+7)}]=0$, then what is the value of $x-[\frac{1}{(x+7)}]$ ?

a)Â $3\sqrt{5}$

b)Â $3\sqrt{5}-7$

c)Â $3\sqrt{5}+7$

d)Â $8$

Question 12:Â If $\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=194$, then what is the value of x?

a)Â 7/2

b)Â 4

c)Â 7

d)Â 14

Question 13:Â If $x=8+2\sqrt{15}$, then what is the value of $\sqrt{x}+\frac{1}{\sqrt{x}}$ ?

a)Â $2\sqrt{5}$

b)Â $2\sqrt{3}$

c)Â $\frac{(3\sqrt{5}+\sqrt{3})}{2}$

d)Â $\frac{(3\sqrt{3}-\sqrt{5})}{2}$

Question 14:Â What is the value of $\frac{1+a}{a^{\frac{1}{2}}+a^{\frac{-1}{2}}}-\frac{a^{\frac{1}{2}}+a^{\frac{-1}{2}}}{1+a}+a^{\frac{-1}{2}}$ ?

a)Â $\sqrt{a}$

b)Â $\frac{1}{\sqrt{a}}$

c)Â $\sqrt{a}+1$

d)Â $\sqrt{a}-1$

Question 15:Â If $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$, then what is the value of x?

a)Â 5/2

b)Â 25/3

c)Â 4

d)Â 3

Expression : $\sqrt[3]{4},\sqrt{2},\sqrt[6]{3},\sqrt[4]{5}$

= $4^{\frac{1}{3}} , 2^{\frac{1}{2}} , 3^{\frac{1}{6}} , 5^{\frac{1}{4}}$

Now, L.C.M. of the powers i.e. 3,2,4,6 = 12

Multiplying the powers by 12 in each of the numbers, we get :

= $4^4 , 2^6 , 3^2 , 5^3$

= $256 , 64 , 9 , 125$

Now arranging them in descending order,

=> $256 > 125 > 64 > 9$

$\equiv$ $\sqrt[3]{4} > \sqrt[4]{5} > \sqrt{2} > \sqrt[6]{3}$

When the number of digits in a no. is 7 or 8 , then no. of digits in square root will be 4.

62478078 has 8 digits, => Its square root has 4 digits.

$5\frac{1}{3}+1\frac{2}{9}\times \frac{1}{4}(10+\frac{3}{1-\frac{1}{5}})$

$5\frac{1}{3}+1\frac{2}{9}\times \frac{1}{4}(10+\frac{15}{4})$

$\frac{16}{3}+\frac{11}{9}\times \frac{55}{16}$ = 9.9 ~ 10

Â

Fractions : $\frac{2}{5},\frac{3}{5},\frac{8}{11},\frac{11}{17}$

L.C.M. of 5,11,17 = 935

Now, multiplying each fraction by 935, we get :

=> 374 , 561 , 680 , 605

Since, among these numbers, 680 is the largest $\equiv \frac{8}{11}$

=> $\frac{8}{11}$ is the largest.

we need to find value of $999\frac{1}{7}+999\frac{2}{7}+999\frac{3}{7}+999\frac{4}{7}+999\frac{5}{7}+999\frac{6}{7}$

$6000 – \frac{1+2+3+4+5+6}{7}$

= 6000 – $\frac{21}{7}$

= 6000 – 3 = 5997

We need to calculate $\sqrt{x-1}+\frac{1}{\sqrt{x-1}}$
This equals $\frac{x-1 + 1}{\sqrt{x-1}} = \frac{x}{\sqrt{x-1}}$
$x-1 = 5+2\sqrt{6} = (\sqrt{3} + \sqrt{2})^2$
Therefore, $\sqrt{x-1} = \sqrt{3} + \sqrt{2}$

Hence, the required expression becomes $\frac{6+2\sqrt{6}}{\sqrt{2}+\sqrt{3}}$
This equals $2*\frac{3+\sqrt{6}}{\sqrt{2}+\sqrt{3}}=2\sqrt{3}$

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

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

By C-D rule,

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

and

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

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

$\frac{x+\sqrt{5}}{x-\sqrt{5}}+\frac{x+\sqrt{3}}{x-\sqrt{3}}$

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

=Â  $\frac{2\sqrt{3}-2\sqrt{5}}{\sqrt{3}-\sqrt{5}}$

= $2$

so the answer is option D.

Values :Â $\sqrt[3]{5},\sqrt[4]{6},\sqrt[6]{12},\sqrt[12]{276}$

Taking L.C.M. of exponents, => L.C.M.(3,4,6,12) = 12

Now, multiplying all the exponents by 12, we getÂ :

ValuesÂ : $(5)^4,(6)^3,(12)^2,(276)^1$

=Â $625,216,144,276$

Thus, $625\equiv \sqrt[3]{5}$ is the largest.

=> Ans – (A)

GivenÂ :Â $x-y-\sqrt{18}=-1$

=> $x-y=\sqrt{18}-1$ ————-(i)

Squaring both sides,

=> $(x-y)^2=(\sqrt{18}-1)^2$

=> $x^2+y^2-2xy=18+1-2\sqrt{18}$

=> $x^2+y^2-2xy=19-2\sqrt{18}$ ————–(ii)

Also,Â $x + y – 3\sqrt{2} = 1$

=> $x+y=\sqrt{18}+1$ ————-(iii)

Squaring both sides,

=> $(x+y)^2=(\sqrt{18}+1)^2$

=> $x^2+y^2+2xy=18+1+2\sqrt{18}$

=> $x^2+y^2+2xy=19+2\sqrt{18}$ ————–(iv)

Subtracting equation (ii) from (iv),

=> $4xy=4\sqrt{18}$

=> $12xy=12\sqrt{18}$ ————(v)

Multiplying equations (i) and (iii),

=> $(x-y)(x+y)=(\sqrt{18}-1)(\sqrt{18}+1)$

=> $x^2-y^2=18-1=17$ ———–(vi)

Now, multiplying equations (v) and (vi), we getÂ :

=>Â $12xy(x^{2} – y^{2})= (12\sqrt{18})\times17$

=Â $204\sqrt{18}=612\sqrt2$

=> Ans – (D)

GivenÂ :Â $\frac{p}{q}=\frac{r}{s}=\frac{t}{u}=\sqrt{5}$

=> $p=\sqrt5q$ ,Â $r=\sqrt5s$ ,Â $t=\sqrt5u$

To findÂ :Â $[\frac{(3p^{2} + 4r^{2} + 5t^{2})}{(3q^{2} + 4s^{2} + 5u^{2})}]$

= $\frac{3(\sqrt5q)^2+4(\sqrt5s)^2+5(\sqrt5u)^2}{3q^2+4s^2+5u^2}$

=Â $\frac{15q^2+20s^2+25u^2}{3q^2+4s^2+5u^2}$

=Â $\frac{5(3q^2+4s^2+5u^2)}{3q^2+4s^2+5u^2}=5$

=> Ans – (B)

GivenÂ :Â $x+[\frac{1}{(x+7)}]=0$ ———–(i)

=> $\frac{x^2+7x+1}{x+7}=0$

=> $x^2+7x+1=0$

=> $x=\frac{-7\pm\sqrt{49-4}}{2}$

=> $x=\frac{3\sqrt{5}-7}{2}$ ———–(ii)

From equation (i), => $\frac{1}{(x+7)}=-x$ —————(iii)

To findÂ :Â $x-[\frac{1}{(x+7)}]$

Substituting values from equations (ii) and (iii), we getÂ :

= $x-(-x)=2x$

= $2\times\frac{3\sqrt5-7}{2}$

= $3\sqrt5-7$

=> Ans – (B)

GivenÂ :Â $\frac{x+\sqrt{x^2-1}}{x-\sqrt{x^2-1}}+\frac{x-\sqrt{x^2-1}}{x+\sqrt{x^2-1}}=194$

=> $\frac{(x+\sqrt{x^2-1})^2+(x-\sqrt{x^2-1})^2}{(x-\sqrt{x^2-1})(x+\sqrt{x^2-1})}=194$

=> $\frac{(x^2+x^2-1+2x\sqrt{x^2-1})+(x^2+x^2-1-2x\sqrt{x^2-1})}{x^2-(x^2-1)}=194$

=> $\frac{4x^2-2}{1}=194$

=> $4x^2=194+2=196$

=> $x^2=\frac{196}{4}=49$

=> $x=\sqrt{49}=7$

=> Ans – (C)

GivenÂ :Â $x=8+2\sqrt{15}$

=> $x=5+3+2\sqrt{(5)(3)}$

=> $x=(\sqrt5)^2+(\sqrt3)^2+2(\sqrt5)(\sqrt3)$

=> $x=(\sqrt5+\sqrt3)^2$

=> $\sqrt{x}=\sqrt5+\sqrt3$ ————(i)

Now,Â $\frac{1}{\sqrt{x}}=\frac{1}{\sqrt5+\sqrt3}$

=> $\frac{1}{\sqrt{x}}=\frac{1}{\sqrt5+\sqrt3}\times\frac{(\sqrt5-\sqrt3)}{(\sqrt5-\sqrt3)}$

=> $\frac{1}{\sqrt{x}}=\frac{\sqrt5-\sqrt3}{5-3}=\frac{\sqrt5-\sqrt3}{2}$ ———–(ii)

Adding equations (i) and (ii), we getÂ :

=> $\sqrt{x}+\frac{1}{\sqrt{x}}=(\sqrt5+\sqrt3)+(\frac{\sqrt5-\sqrt3}{2})$

= $\frac{2\sqrt5+2\sqrt3+\sqrt5-\sqrt3}{2}=\frac{3\sqrt5+\sqrt3}{2}$

=> Ans – (C)

ExpressionÂ :Â $\frac{1+a}{a^{\frac{1}{2}}+a^{\frac{-1}{2}}}-\frac{a^{\frac{1}{2}}+a^{\frac{-1}{2}}}{1+a}+a^{\frac{-1}{2}}$

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

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

= $\sqrt{a}-\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{a}}=\sqrt{a}$

=> Ans – (A)

ExpressionÂ :Â $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}=3$

Rationalizing the denominator,

=>Â $\frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}-\sqrt{5-x}}\times \frac{\sqrt{5+x}+\sqrt{5-x}}{\sqrt{5+x}+\sqrt{5-x}}=3$

=> $\frac{[(\sqrt{5+x})+(\sqrt{5-x})]^2}{(\sqrt{5+x})^2-(\sqrt{5-x})^2}=3$

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

=> $\frac{10+2\sqrt{25-x^2}}{2x}=3$

=> $5+\sqrt{25-x^2}=3x$

=> $3x-5=\sqrt{25-x^2}$

Squaring both sides, we getÂ :

=> $(3x-5)^2=(\sqrt{25-x^2})^2$

=> $9x^2+25-30x=25-x^2$

=> $10x^2-30x=0$

=> $10x(x-3)=0$

=> $x=0,3$ Â  Â [But $x$ can’t be zero because the denominator can’t be zero]

=> Ans – (D)