XAT Surds and Indices Questions PDF [Important]
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Question 1: If x = 1 + $\surd{2}$+ $\surd{3}$, then the value of 2x^{4}- 8x^{3}- 5x^{2} + 26x – 28 is
a) 2$\surd{2}$
b) 3$\surd{3}$
c) 5$\surd{5}$
d) 6 $\surd{6}$
1) Answer (D)
Solution:
x = 1 + $\surd{2}$+ $\surd{3}$
=> $ (x-1)^2 = ( \surd{2}+ \surd{3})^{2} $
=> $ x^2 + 1 -2x = 5 + 2\surd{6} $
=> $ x^2-2x =4 + 2\surd{6} $ ———– (1)
Squaring on both sides
=> $ (x^2-2x)^2 = x^4 + 4x^2 – 4x^3 = 40 + 16\surd{6} $ —— (2)
Now,
$2x^{4}- 8x^{3}- 5x^{2} + 26x – 28 = 2(x^{4}-4x^{3})- 5x^{2} + 26x – 28 $ —- (3)
Substituting values in (1) & (2) in equation (3), we get value as $ 6 \surd {6} $
Question 2: if x+ $\frac{1}{x}$=$\surd{3}$ then the value of $x^{18}+x^{12}+x^{6}+1$
a) 0
b) 1
c) 2
d) 3
2) Answer (A)
Solution:
Given that x+ $\frac{1}{x}$=$\surd{3}$
Squaring on both sides, we get
$(x+ \frac{1}{x})^{3}=(\surd{3})^{3}$
=> $x^{3}+\frac{1}{x^3}+3\surd{3}=3\surd{3}$
=> $x^{3}+\frac{1}{x^3} = 0 $
=> $x^{3}= – \frac{1}{x^3} $
=> $x^{6}= -1 $
Squaring on both sides
=> $x^{12}= 1 $
$ (x^{6})^{3} = (-1)^{3} = -1 $
Therefore,
$x^{18}+x^{12}+x^{6}+1$ = $ -1 + 1 -1 + 1 = 0 $
Question 3: If $P = 2^{29}\times3^{21} \times 5^8,Q = 2^{27} \times 3^{21} \times 5^8, R = 2^{26} \times 3^{22} \times 5^8$ and $S = 2^{25} \times 3^{22} \times 5^9$, then which of the following is TRUE?
a) $P > S > R > Q$
b) $S > P > R > Q$
c) $P > R > S > Q$
d) $S > P > Q > R$
3) Answer (A)
Solution:
Let say, $M=2^{25}\times3^{21}\times5^8.$
So, by rearranging above equation ,we can say that :
$P=2^4\times M=16M.$
$Q=2^2\times M=4M.$
$R=2\times3\times M=6M.$
$S=3\times5\times M=15M.$
So, $P > S > R > Q$.
A is correct choice.
Question 4: Which of the following statement(s) is/are TRUE?
I. $\surd1+\surd2+\surd3+\surd4+\surd5+\surd6>10$
II. $\surd(10)+\surd(12)+\surd(14)>3\surd(12)$
a) only I
b) only II
c) Niether I nor II
d) Both I and II
4) Answer (A)
Solution:
$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{\ 5}+\sqrt{6}=10.83.$
it means that (I) is correct.
And, $\sqrt{10}+\sqrt{12}+\sqrt{14}=10.36.$
but, $3\sqrt{12}=10.39.$
So, (II) is not correct .
So, A is correct choice.
Question 5: If $N = (12345)^2 + 12345 +12346$, then what is the value of $\surd N$ ?
a) 12346
b) 12345
c) 12344
d) 12347
5) Answer (A)
Solution:
$N=(12345)^2+12345+12346=12345^2+12345+12345+1=12345^2+2\times12345\times1+1^2$
So, $N=(12345+1)^2$
So, $\sqrt{N}=12346\ .$
A is correct choice.
Question 6: $\alpha$ and $\beta$ are the roots of quadratic equation. If $\alpha + \beta = 8$ and $\alpha – \beta = 2\surd5$, then which of the following equation will have roots $\alpha^4$ and $\beta^4$?
a) $x^2 – 1522x + 14641 = 0$
b) $x^2 + 1921x + 14641 = 0$
c) $x^2- 1764x + 14641 = 0$
d) $x^2+ 2520x + 14641 = 0$
6) Answer (A)
Solution:
According to question :
$2\alpha\ =8+\sqrt{5}\ \ or\ \ \ \alpha=4+\sqrt{5}\ .$
And, $2\beta=8-\sqrt{5}\ \ or\ \ \ \beta=4-\sqrt{5}\ .$
So, $\alpha^2=\left(4+\sqrt{5}\right)^2=\left(21+8\sqrt{5}\right).$
And, $\beta^2=\left(4-\sqrt{5}\right)^2=\left(21-8\sqrt{5}\right).$
Again ,
$\alpha^4=\left(\alpha^2\right)^2=\left(21+8\sqrt{5}\right)^2=\left(761+336\sqrt{5}\right).$
$\beta^4=\left(\beta^2\right)^2=\left(21-8\sqrt{5}\right)^2=\left(761-336\sqrt{5}\right).$
So, new equation whose roots are above two :
$x^2-\left(\alpha^4+\beta^4\right)x+\left(\alpha^4\beta^4\right)=0\ .$
or, $x^2-\left(761+336\sqrt{5}+761-336\sqrt{5}\right)x+\left(761+336\sqrt{5}\right)\left(761-336\sqrt{5}\right)=0\ .$
or, $x^2-1522x+14641=0\ .$
A is correct choice.
Question 7: Which of the following statement(s) is/are TRUE?
I. $\surd5 + \surd5 > \surd7 + \surd3$
II. $\surd6 + \surd7 > \surd8 + \surd5$
III. $\surd3 + \surd9 > \surd6 + \surd6$
a) Only I
b) Only I and II
c) Only II and III
d) Only I and III
7) Answer (B)
Solution:
Statement I :
$\sqrt{5}+\sqrt{5}=4.47\ .$ and $\sqrt{7}+\sqrt{3}=4.37\ .$
So, Statement I is correct .
Statement II :
$\sqrt{6}+\sqrt{7}=5.09\ \ and\ \ \sqrt{8}+\sqrt{5}=5.06\ .$
II is also correct .
Statement III:
$\sqrt{3}+\sqrt{9}=4.73\ \ and\ \ \sqrt{6}+\sqrt{6}=4.89\ .$
So, III is not correct .
B is correct choice.
Question 8: Which of the following statement(s) is/are TRUE?
I. $\surd(64) + \surd(0.0064) + \surd(0.81) + \surd(0.0081) = 9.07$
II. $\surd(0.010201) + \surd(98.01) + \surd(0.25) = 11.51$
a) Only I
b) Only II
c) Both I and II
d) Neither I nor II
8) Answer (A)
Solution:
$\surd(64)+\surd(0.0064)+\surd(0.81)+\surd(0.0081)=8+0.08+0.9+0.09=9.07\ .$
So, I is correct .
$\surd(0.010201)+\surd(98.01)+\surd(0.25)=10.501\ .$
II is not correct .
A is correct choice.
Question 9: Which of the following statement(s) is/are true?
$I. (65)^{\frac{1}{6}} > (17)^{\frac{1}{4}} > (12)^{\frac{1}{3}}$
$II. (17)^{\frac{1}{4}} > (65)^{\frac{1}{6}} > (12)^{\frac{1}{3}}$
$III. (12)^{\frac{1}{3}} > (17)^{\frac{1}{4}} > (65)^{\frac{1}{6}}$
a) Only I
b) Only III
c) Only II
d) None of these
9) Answer (B)
Solution:
$(65)^{\frac{1}{6}} , (17)^{\frac{1}{4}} , (12)^{\frac{1}{3}}$
$(12)^{\frac{1}{3}}$=$(144)^{\frac{1}{6}}$
So $(144)^{\frac{1}{6}}$>$(65)^{\frac{1}{6}} $
$(65)^{\frac{1}{6}} $=$(4225)^{\frac{1}{12}} $
$(17)^{\frac{1}{4}}$=$(4913)^{\frac{1}{12}} $
Therefore $(12)^{\frac{1}{3}}$ > $(17)^{\frac{1}{4}}$ >$(65)^{\frac{1}{6}} $
Question 10: Which of the following is TRUE?
$I. \sqrt[3]{11} > \sqrt{7} > \sqrt[4]{45}$
$II. \sqrt{7} > \sqrt[3]{11} > \sqrt[4]{45}$
$III. \sqrt{7} > \sqrt[4]{45} > \sqrt[3]{11}$
$IV. \sqrt[4]{45} > \sqrt{7} > \sqrt[3]{11}$
a) only $I$
b) only $II$
c) only $III$
d) only $IV$
10) Answer (C)
Solution:
$\sqrt[3]{11} , \sqrt{7}, \sqrt[4]{45}$
Here $\sqrt{7}=\sqrt[4]{49}$
$sqrt[4]{45}<\sqrt[4]{49}$
Now $sqrt[4]{45}>$\sqrt[3]{11} since 3rd power of 45 will be greater than 4th power of 11 i.e 14631
an so $\sqrt{7} > \sqrt[4]{45} > \sqrt[3]{11}$
Question 11: Which of the following is TRUE?
$I. \frac{1}{\sqrt[3]{12}} > \frac{1}{\sqrt[4]{29}} > \frac{1}{\sqrt{5}}$
$II. \frac{1}{\sqrt[4]{29}} > \frac{1}{\sqrt[3]{12}} > \frac{1}{\sqrt{5}}$
$III. \frac{1}{\sqrt{5}} > \frac{1}{\sqrt[3]{12}} > \frac{1}{\sqrt[4]{29}}$
$IV. \frac{1}{\sqrt{5}} > \frac{1}{\sqrt[4]{29}} > \frac{1}{\sqrt[3]{12}}$
a) only $I$
b) only $II$
c) only $III$
d) only $IV$
11) Answer (C)
Solution:
$\frac{1}{\sqrt[3]{12}} , \frac{1}{\sqrt[4]{29}} , \frac{1}{\sqrt{5}}$
Here if we write $\frac{1}{\sqrt{5}}$ in terms of fourth root we have $\frac{1}{\sqrt[4]{25}}$
So $\frac{1}{\sqrt[4]{29}} > \frac{1}{\sqrt[4]{5}}$
$\sqrt[3]{12}$=$\sqrt[12]{144*144}$
$\frac{1}{\sqrt[4]{29}}$=$\frac{1}{\sqrt[12]{29*29*29}}$
Therefore ${\sqrt[4]{29}}$>$\sqrt[3]{12}$
As these are given in the denominators order will be reversed.
$\frac{1}{\sqrt{5}} > \frac{1}{\sqrt[3]{12}} > \frac{1}{\sqrt[4]{29}}$
Question 12: Determine the value of ‘a’ which satisfies the equation $9^{\sqrt{a}}+40^{\sqrt{a}}=41^{\sqrt{a}}$
a) 1
b) 2
c) 3
d) 4
12) Answer (D)
Solution:
Expression : $9^{\sqrt{a}}+40^{\sqrt{a}}=41^{\sqrt{a}}$
Since equation consists of natural numbers, then ‘a’ must be an integer. Thus, substituting ‘a’ by perfect square numbers :
Put $a=1$
=> $9^{\sqrt{1}}+40^{\sqrt{1}}=41^{\sqrt{1}}$
L.H.S. = $9+40=49\neq$ R.H.S.
Now, putting $a=4$
=> $9^{\sqrt{4}}+40^{\sqrt{4}}=41^{\sqrt{4}}$
L.H.S. = $9^2+40^2=81+1600=1681$
R.H.S. = $41^2=1681$
Thus, $a=4$
=> Ans – (D)
Question 13: Calculate the value of $\frac{\sqrt{3}}{(3+\sqrt{3})}$ if $\sqrt{3}=1.7320$
a) 0.366
b) 0.566
c) 0.356
d) 0.346
13) Answer (A)
Solution:
Expression : $\frac{\sqrt{3}}{(3+\sqrt{3})}$ if $\sqrt{3}=1.7320$
Rationalizing the denominator :
= $\frac{\sqrt{3}}{(3+\sqrt{3})}\times\frac{(3-\sqrt3)}{(3-\sqrt3)}$
= $\frac{\sqrt3(3-\sqrt3)}{(3)^2-(\sqrt3)^2}=\frac{3\sqrt3-3}{9-3}$
= $\frac{\sqrt3-1}{2}=\frac{(1.7320-1)}{2}$
= $\frac{0.7320}{2}=0.366$
=> Ans – (A)
Question 14: If $4x=\sqrt{5}+2$, then $x-\frac{1}{16x}$ ?
a) 1
b) -1
c) 4
d) 2√5
14) Answer (A)
Solution:
Given : $4x=\sqrt{5}+2$
=> $x=\frac{\sqrt{5}+2}{4}$
To find : $x-\frac{1}{16x}$
= $\frac{\sqrt{5}+2}{4} – \frac{4}{16(\sqrt{5}+2)}$
= $\frac{\sqrt{5}+2}{4}-\frac{1}{4(\sqrt{5}+2)}$
= $\frac{(\sqrt{5}+2)^2-1}{4(\sqrt{5}+2)}$
= $\frac{5+4+4\sqrt{5}-1}{4(\sqrt{5}+2)}$
= $\frac{8+4\sqrt{5}}{8+4\sqrt{5}}=1$
=> Ans – (A)
Question 15: The value of $(1-\sqrt{2})+(\sqrt{2}-\sqrt{3})+(\sqrt{3}-\sqrt{4})+…..+(\sqrt{15}-\sqrt{16})$ is
a) 0
b) 1
c) -3
d) 4
15) Answer (C)
Solution:
Expression : $(1-\sqrt{2})+(\sqrt{2}-\sqrt{3})+(\sqrt{3}-\sqrt{4})+…..+(\sqrt{15}-\sqrt{16})$
= $1 + (-\sqrt{2}+\sqrt{2})+(-\sqrt{3}+\sqrt{3})+…..+(-\sqrt{14}+\sqrt{14})+(-\sqrt{15}+\sqrt{15})-\sqrt{16}$
= $1-\sqrt{16}=1-4=-3$
=> Ans – (C)
Question 16: The value of $\sqrt{9-2\sqrt{16}+3\sqrt[3]{512}}$ is
a) 6
b) 5
c) 2√8
d) 3√6
16) Answer (B)
Solution:
Expression : $\sqrt{9-2\sqrt{16}+3\sqrt[3]{512}}$
= $\sqrt{9-(2 \times 4)+(3 \times 8)}$
= $\sqrt{9-8+24}=\sqrt{25}$
= $5$
=> Ans – (B)
Question 17: If $\sqrt{1+\frac{x}{144}}=\frac{13}{12}$ then x equals to
a) 1
b) 13
c) 27
d) 25
17) Answer (D)
Solution:
Expression : $\sqrt{1+\frac{x}{144}}=\frac{13}{12}$
Squaring both sides,
=> $(1+\frac{x}{144})=(\frac{13}{12})^2$
=> $\frac{144+x}{144}=\frac{169}{144}$
=> $144+x=169$
=> $x=169-144=25$
=> Ans – (D)
Question 18: If $x=\sqrt{a}+\frac{1}{\sqrt{a}},y=\sqrt{a}-\frac{1}{\sqrt{a}}, (a>o)$ the the value of $x^{4}+y^4-2x^2y^2$ is
a) 16
b) 20
c) 10
d) 5
18) Answer (A)
Solution:
Given : $x=\sqrt{a}+\frac{1}{\sqrt{a}}$
Squaring both sides
=> $(x)^2=(\sqrt{a}+\frac{1}{\sqrt{a}})^2$
=> $x^2=(\sqrt{a})^2+(\frac{1}{\sqrt{a}})^2+2.\sqrt{a}.\frac{1}{\sqrt{a}}$
=> $x^2=a+\frac{1}{a}+2$ ————-(i)
Similarly, $y^2=a+\frac{1}{a}-2$ ———(ii)
To find : $x^{4}+y^4-2x^2y^2$
= $(x^2-y^2)^2$
Substituting values from equations (i) and (ii), we get :
= $[(a+\frac{1}{a}+2)-(a+\frac{1}{a}-2)]^2$
= $(2+2)^2=(4)^2=16$
=> Ans – (A)
Question 19: If $x=\sqrt[3]{28},y=\sqrt[3]{27}$, then the value of $x+y-\frac{1}{x^2+xy+y^2}$ is
a) 8
b) 7
c) 6
d) 5
19) Answer (C)
Solution:
Given : $x=\sqrt[3]{28},y=\sqrt[3]{27}$
=> $x^3=28$ and $y^ 3=27$ ———-(i)
=> $y=3$ ———–(ii)
To find : $x+y-\frac{1}{x^2+xy+y^2}$
= $(x+y)-(\frac{(x-y)}{(x-y)(x^2+xy+y^2)})$ [Multiply and divide by $(x-y)$]
Using, $(x-y)(x^2+xy+y^2)=(x^3-y^3)$
= $(x+y)-(\frac{x-y}{x^3-y^3})$
= $(x+y)-(\frac{x-y}{28-27})$ [Using (i)]
= $(x+y)-(x-y)=2y$
= $2 \times 3=6$ [Using (ii)]
=> Ans – (C)
Question 20: $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=0$, then the value of $\frac{1}{a}+\frac{1}{b}$ is
a) $\frac{1}{\sqrt{ab}}$
b) ${\sqrt{ab}}$
c) $\frac{2}{\sqrt{ab}}$
d) $\frac{1}{2\sqrt{ab}}$
20) Answer (C)
Solution:
Given : $\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}}=0$
Squaring both sides, we get :
=> $(\frac{1}{\sqrt{a}}-\frac{1}{\sqrt{b}})^2=0$
=> $(\frac{1}{\sqrt{a}})^2+(\frac{1}{\sqrt{b}})^2-2(\frac{1}{\sqrt{a}})(\frac{1}{\sqrt{b}})=0$
=> $\frac{1}{a}+\frac{1}{b}-\frac{2}{\sqrt{a}\sqrt{b}}=0$
=> $\frac{1}{a}+\frac{1}{b}=\frac{2}{\sqrt{ab}}$
=> Ans – (C)
