SNAP Surds and Indices Questions PDF [Most Important]

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_Surds and Indices Questions
_Surds and Indices Questions

Surds and Indices Questions for SNAP

Surds and Indices is an important topic in the Analytical & Logical Reasoning section of the SNAP Exam. You can also download this Free Surds and Indices Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Surds and Indices questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practising.

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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: 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

3) 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 4: If $N = (12345)^2 + 12345 +12346$, then what is the value of $\surd N$ ?

a) 12346

b) 12345

c) 12344

d) 12347

4) 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 5: $\alpha$ 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) $x^2+ 2520x + 14641 = 0$

5) 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.

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Question 6: What is the remainder when $(127^{97} + 97^{97})$ is divided by 32?

a) 4

b) 2

c) 7

d) 0

6) Answer (D)

Solution:

$(127^{97} + 97^{97})$ is divided by 32 So,
$\frac{(127^{97} + 97^{97})}{32}$
=$\frac{((128 – 1)^{97} + (96 + 1)^{97})}{32}$
Remainder = -1 + 1 = 0

Question 7: If $x = (164)^{169} + (333)^{337} – (727)^{726}$, then what is the units degit of x?

a) 5

b) 7

c) 8

d) 9

7) Answer (C)

Solution:

$x = (164)^{169} + (333)^{337} – (727)^{726}$
For the units digit of x,
= (unit digit)$^{remainder}$
Remainder = 169/4 = 1, 337/4 = 1, 726/4 = 2
Units digit of x = $(4)^1 + (3)^1 – (7)^2$
= 4 + 3 – 9 = -2
= 10 – 2 = 8
Units digit of x = 8

Question 8: Let $x = (633)^{24} – (277)^{38} + (266)^{54}$. What is the units digit of x ?

a) 7

b) 6

c) 4

d) 8

8) Answer (D)

Solution:

$x = (633)^{24} – (277)^{38} + (266)^{54}$For the unit digit,
24 = 4 $\times 6 + 0(remainder)
38 = 4 $\times 9 + 2(remainder)
54 = 4 $\times 13 + 2(remainder)
Now,
(Base number unit digit)$^{remainder}$
= $(3)^0 – (7)^2 + (6)^2$
On consider unit digit,
= 1 – 9 + 6 = 7 – 9
or 17 – 9 = 8
8 is the units digit of x.

Question 9: Which ofthe following is correct ?

a) $ \frac{2}{3} < \frac{3}{5}< \frac{11}{5} $

b) $ \frac{3}{5} < \frac{2}{3}< \frac{11}{5} $

c) $ \frac{11}{5} < \frac{3}{5}< \frac{2}{3} $

d) $ \frac{3}{5} < \frac{11}{15}< \frac{2}{3} $

9) Answer (B)

Solution:

$ \frac{2}{3} , \frac{3}{5} , \frac{11}{5} $

Take denominator as 15 ,

$ \frac{10}{15} , \frac{9}{15} , \frac{33}{15} $

So, the correct order will be ,$ \frac{3}{5} < \frac{2}{3}< \frac{11}{5} $

So, the answer would be option b)$ \frac{3}{5} < \frac{2}{3}< \frac{11}{5} $.

Question 10: 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) $S > P > Q > R$

10) 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.

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Question 11: 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

11) 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 12: If $A = 2^{32}, B = 2^{31} + 2^{30} + 2^{29} + … + 2^0$ and $C = 3^{15} + 3^{14} + 3^{13} + … +3^0$, then which of the following option is TRUE?

a) $C > B > A$

b) $C > A > B$

c) $A > B > C$

d) $A > C > B$

12) Answer (C)

Solution:

$B=2^{31}+2^{30}+2^{29}+\dots+2^0$

or, $B=2^0.\ \frac{2^{32}-1}{2-1}=\left(2^{32}-1\right)\ .$

And , $C = 3^{15} + 3^{14} + 3^{13} + … +3^0$

or, $C=3^0.\ \frac{3^{16}-1}{3-1}=\frac{1}{2}\left(3^{16}-1\ \right).$

So, $A>B>C\ .$

C is correct choice.

Question 13: Which of the following statement(s) is/are TRUE?
I. $\frac{3}{71} < \frac{5}{91} < \frac{7}{99}$
II. $\frac{11}{135} > \frac{12}{157} > \frac{13}{181}$

a) Only I

b) Only II

c) Both I and II

d) Neither I nor II

13) Answer (C)

Solution:

Statement I :

$\frac{7}{99}=0.0707\ ,\ \frac{5}{91}=0.0549,\ \frac{3}{71}=0.0422\ .$

So, I is correct .

Statement II :

$\frac{11}{135}=0.0814\ ,\ \frac{12}{157}=0.0764,\ \frac{13}{181}=0.0718\ ..$

So, II is also correct .

C is correct choice.

Question 14: 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

14) 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 15: Arranging the following in descending order, we get
$\sqrt[3]{4},\sqrt{2},\sqrt[6]{3},\sqrt[4]{5}$

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

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

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

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

15) Answer (A)

Solution:

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,

=> $256 > 125 > 64 > 9$

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

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