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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