**Surds & Indices Questions for RRB NTPC PDF**

Download RRB NTPC Surds & Indices Questions PDF. Top 10 RRB NTPC questions based on asked questions in previous exam papers very important for the Railway NTPC exam.

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**Question 1: **The two numbers $4^{30}$ and $25^{30}$ are written next to each other. What is the total number of digits written down?

a) 30

b) 59

c) 60

d) None of these

**Question 2: **If $x = 9 + 4\sqrt{5}$, what is $x + \frac{1}{x}$

a) 17.83

b) 18.45

c) 18.00

d) None of these

**Question 3: **What is the value of x for which $x^{2/3}$ + $3x^{1/3} – 4<0$?

a) -64 < x < 1

b) -1 < x < 64

c) -64 < x < 64

d) 1 < x < 64

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**Question 4: **$ \frac{1}{{3-2}\sqrt{2}}$ – $ \frac{1}{{3+2}\sqrt{2}}$ = $2^x$. Find x?

a) 3/2

b) 5/2

c) 3

d) 7/2

**Question 5: **Which of the following surds is the greatest?

a) $ 4-\sqrt{7}$

b) $ 5-\sqrt{10}$

c) $ 8-\sqrt{15}$

d) Cannot be determined

**Question 6: **If $ x=\sqrt{17}-\sqrt{13}$, what is $ \frac{30-\sqrt{884}}{\sqrt{17}+\sqrt{13}}$?

a) $\frac{x^2}{x-1}$

b) $\frac{x^3}{x-1}$

c) $x^3$

d) $\frac{x^3}{4}$

**Question 7: **If $\sqrt{28+5\sqrt{12}} = a+\sqrt{b}$, where a and b are positive rational numbers. Find a+b?

a) 2

b) 8

c) 13/2

d) Cannot be determined

**Question 8: **Which of the following surds is the greatest?

a) $\sqrt{1}+\sqrt{21}$

b) $\sqrt{2}+\sqrt{20}$

c) $\sqrt{4}+\sqrt{18}$

d) All of them are equal

**Question 9: **Simplify: $\sqrt{19+4\sqrt{21}}$

a) $2+\sqrt{26} $

b) $3-\sqrt{15} $

c) $\sqrt{5}+\sqrt{26} $

d) $\sqrt{12}+\sqrt{7}$

**Question 10: **Which of the following surds is the greatest?

a) $\sqrt{2}+\sqrt{14}$

b) $\sqrt{3}+\sqrt{13}$

c) $\sqrt{5}+\sqrt{11}$

d) $\sqrt{7}+\sqrt{8}$

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**Answers & Solutions:**

**1) Answer (D)**

Let $\log_{10}{2}$ be x.

So, the number of digits in $4^{30}$ is [60x] + 1.

So, $\log_{10}{5}$ is 1-x and the number of digits of $25^{30}$ is [60-60x]+1.

Total number of digits is 2+[60x]+[60-60x] which is 62 + [60x]+[-60x].

As 60x is not an integer, the value of [60x]+[-60x] = -1. So, value is 61

**2) Answer (C)**

$x = 9 + 4\sqrt{5}$. So, $\frac{1}{x}$ = $9 – 4\sqrt{5}$. So, $x+ \frac{1}{x}$ = 18

**3) Answer (A)**

-4 < $x^{1/3}$ < 1 or -64 < x < 1

**4) Answer (B)**

$\frac{1}{3-2\sqrt{2}}-\frac{1}{3+2\sqrt{2}} = \frac{3+2\sqrt{2}-3+2\sqrt{2}}{3^2-(2\sqrt{2})^2} = 4\sqrt{2}/1 = 2^{5/2}$. Hence x=5/2.

**5) Answer (C)**

The value of $\sqrt{7}$ is between 2 and 3. Hence, $4-\sqrt{7}$ is between 1 and 2. Similarly, the value of b is between 1 and 2 and c is between 4 and 5. Hence, c) is the greatest.

**6) Answer (D)**

$\frac{30-\sqrt{884}}{\sqrt{17}+\sqrt{13}} = \frac{(\sqrt{17}-\sqrt{13})^2}{\sqrt{17}+\sqrt{13}} = \frac{(\sqrt{17}-\sqrt{13})^3}{(\sqrt{17}+\sqrt{13})(\sqrt{17}-\sqrt{13})}=\frac{x^3}{17-13} = \frac{x^3}{4}$.

**7) Answer (B)**

$\sqrt{28+5\sqrt{12}} = a+\sqrt{b} \rightarrow 28+5\sqrt{12} = a^2+b+2a\sqrt{b}$.

Hence $a^2+b=28$ and $4 a^2 b =300$.

Hence $a^2=25$ and b=3. As a is positive, a=5.

Hence a+b=8.

**8) Answer (C)**

$(\sqrt{a}+\sqrt{b})^2 = a+b+2\sqrt{ab}$.

As a+b is equal for all three of them we need to compare which has the highest value for $\sqrt{ab}$. So the term with highest value of ab will be the greatest.

ab values for the three options are 21, 40 and 72.

Hence c) is the greatest.

**9) Answer (D)**

Let $\sqrt{19+4\sqrt{21}} = \sqrt{a}+\sqrt{b}$

$\rightarrow a+b+2\sqrt{ab} = 19+4\sqrt{21}$.

Hence, a+b=19 and ab=84. Hence a=12, b=7.

**10) Answer (C)**

On squaring the four options we get $16+2\sqrt{28}, 16+2\sqrt{39}, 16+2\sqrt{55}, 15+2\sqrt{56}$.

Out of a-c options, c is clearly the greatest.

Similarly b is also rejected.

Now between c and d, let d>c

$15+2\sqrt{56}>16+2\sqrt{55}$

$2[\sqrt{56}-\sqrt{55}]>1$

Multiply both sides of the equation by $[\sqrt{56}+\sqrt{55}]$

$2[\sqrt{56}-\sqrt{55}][\sqrt{56}+\sqrt{55}]>[\sqrt{56}+\sqrt{55}]$

2>$[\sqrt{56}+\sqrt{55}]$

which is false as the value of each term of RHS lies between 7 and 8.

This contradicts our assumption that d>c

Hence c>d.

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