Olympiad Mathematics - Number Systems
Exam Duration: 45 Mins Total Questions : 30
Which of the following is not a rational number?
- (a)
\(\sqrt { 2 } \)
- (b)
\(\sqrt { 4 } \)
- (c)
\(\sqrt { 9 } \)
- (d)
\(\sqrt { 16 } \)
\(\sqrt { 2 } \) is not a rational number.
Since, 2 cannot be expressed as a square of any rational number.
\(\sqrt { 4 } =\sqrt { { 2 }^{ 2 } } =2;\sqrt { 9 } =\sqrt { { 3 }^{ 2 } } =3;\)
\(\sqrt { 16 } =\sqrt { { 4 }^{ 2 } } =4\)
What type of a number is \(\left( 6+\sqrt { 2 } \right) \left( 6-\sqrt { 2 } \right) \)?
- (a)
Rational number
- (b)
Irrational number
- (c)
Prime number
- (d)
Negative Integer
What type of a number is \(\left( \sqrt { 2 } +{ \sqrt { 3 } } \right) ^{ 2 }\)?
- (a)
A rational number
- (b)
An irrational number
- (c)
A fraction
- (d)
A decimal number
Given X : Every fraction is a rational number. and
Y: Every rational number isafraction.
Which of the following is correct?
- (a)
X is False and Y is True.
- (b)
X is True and Y is False
- (c)
Both X and Yare True.
- (d)
Both X and Yare False.
What is the rationalizing factor of \(\sqrt [ 5 ]{ { a }^{ 2 }{ b }^{ 3 }{ c }^{ 4 } } \)?
- (a)
\(\sqrt [ 5 ]{ { a }^{ 3 }{ b }^{ 2 }{ c } } \)
- (b)
\(\sqrt [ 4 ]{ { a }^{ 3 }{ b }^{ 2 }c } \)
- (c)
\(\sqrt [ 3 ]{ { a }^{ 2 }{ b }^{ 2 }c } \)
- (d)
\(\sqrt { { a }^{ 3 }{ b }^{ 2 }c } \)
Rationalizing factor of \(\sqrt [ n ]{ { a }^{ p } } \ is \ \sqrt [ n ]{ { a }^{ n-p } } \ or\ { \ a }^{ 1\frac { p }{ n } }\)
\({ a }^{ 1\frac { p }{ n } }where\ n>p\)
\(\therefore R.F.\ of\ \sqrt [ 5 ]{ { a }^{ 2 }{ b }^{ 3 }{ c }^{ 4 } } =\sqrt [ 5 ]{ { a }^{ 3 }{ b }^{ 2 }c } \)
What isthe quotient when \(\sqrt [ 6 ]{ 12 } \) is divided by \(\sqrt { 3 } \sqrt [ 3 ]{ 2 } \)?
- (a)
\(\frac { 1 }{ \sqrt [ 2 ]{ 3 } } \)
- (b)
\(\frac { 1 }{ \sqrt [ 3 ]{ 3 } } \)
- (c)
\(\frac { 1 }{ \sqrt [ 4 ]{ 3 } } \)
- (d)
\(\frac { 1 }{ \sqrt [ 5 ]{ 3 } } \)
Find the value of x- yx-y when x = 2 and y = -2.
- (a)
18
- (b)
-18
- (c)
14
- (d)
-14
If 10x = 64, what is the value of \({ 10 }^{ \frac { x }{ 2 } +1 }\)?
- (a)
18
- (b)
42
- (c)
80
- (d)
81
If \(\frac { x }{ { x }^{ 1.5 } } \)=8x-1 and x > 0, find x
- (a)
\(\frac { \sqrt { 2 } }{ 4 } \)
- (b)
\(2\sqrt { 2 } \)
- (c)
4
- (d)
64
x-1/2=8x-1
squaring on both sides, we get
x = 82 = 64
If 4x - 4x-1 = 24, what is the value of (2x)X?
- (a)
\(5\sqrt { 5 } \)
- (b)
\(\sqrt { 5 } \)
- (c)
\(25\sqrt { 5 } \)
- (d)
125
4x-4x-1=24 \(\Rightarrow { 4 }^{ x }-\frac { { 4 }^{ x } }{ 4 } =24\)
\(\Rightarrow { 4 }^{ x }=8\times 4\Rightarrow { 2 }^{ 2x }={ 2 }^{ 5 }\Rightarrow x=\frac { 5 }{ 2 } \)
\(\therefore { (2x) }^{ x }={ \left( 2\times \frac { 5 }{ 2 } \right) }^{ 5/2 }={ 5 }^{ 5/2 }={ 5 }^{ 2 }{ 5 }^{ 1/2 }\)
\(=25\sqrt { 5 }\)
If \(x=\frac { 2 }{ 3-\sqrt { 7 } } \), find the value of (x-3)2.
- (a)
1
- (b)
3
- (c)
6
- (d)
7
\({ (x-3) }^{ 2 }={ \left[ \frac { 2 }{ 3-\sqrt { 7 } } -3 \right] }^{ 2 }={ \left[ \frac { 2\sqrt { 7 } }{ 2 } \right] }^{ 2 }=7\)
Identify a rational number between -5 and 5.
- (a)
0
- (b)
-7.3
- (c)
-5.7
- (d)
1.101100110001..
0 is theonly rational number among the given options between -5 and 5.
The sum of the digits of a number is subtracted from the number, the resulting number is always divisible by which of the following numbers?
- (a)
2
- (b)
5
- (c)
8
- (d)
9
The resulting number is always divisible by 9.
e.g., Consider 47. Sum of its digits is 4 + 7 = 11. So,47 - 11 = 36 which is exactly divisible by 9
Find the difference between 1.\(\bar { 4 } \) and 0.\(\bar { 2 } \).
- (a)
0.\(\bar { 2 } \)
- (b)
1.\(\bar { 3 } \)
- (c)
11/9
- (d)
12/9
\(1.\bar { 4 } =\frac { 13 }{ 9 } ,0.\overline { 2 } =\frac { 2 }{ 9 } \)
So, \(1.\overline { 4 } -0.\overline { 2 } =\frac { 13 }{ 9 } -\frac { 2 }{ 9 } =\frac { 11 }{ 9 } \)
If \(\sqrt { a } \) is an irrational number, what is a?
- (a)
Rational
- (b)
Irrational
- (c)
0
- (d)
Real
If a = 3 and b = 5, find the value of aa + bb.
- (a)
512
- (b)
251
- (c)
513
- (d)
152
Since a = 3 and b = 5, aa + bb = 33 + 55 = 27 + 125 = 152
Find the value of \(\frac { 1 }{ \sqrt { 5 } -\sqrt { 4 } } \).
- (a)
\(\sqrt { 5 } -\sqrt { 4 } \)
- (b)
\(\sqrt { 5 } +\sqrt { 4 } \)
- (c)
\(\frac { \sqrt { 5 } +\sqrt { 4 } }{ 9 } \)
- (d)
\(\sqrt { 4 } -\sqrt { 5 } \)
\(\frac { 1 }{ \sqrt { 5 } -2 } \times \frac { \sqrt { 5 } +2 }{ \sqrt { 5 } +2 } \)
\(=\frac { \sqrt { 5 } +2 }{ { \left( \sqrt { 5 } \right) }^{ 2 }-{ \left( 2 \right) }^{ 2 } } =\sqrt { 5 } +2\ or=\sqrt { 5 } +\sqrt { 4 } \)
Find the rational number for 0.\(\overline { 5 } \).
- (a)
\(\frac { 9 }{ 5 } \)
- (b)
\(\frac { 5 }{ 10 } \)
- (c)
\(\frac { 5 }{ 9 } \)
- (d)
\(\frac { 10 }{ 5 } \)
Let x = 0.5 = 0.555... ... (i)
Multiply (i) by 10 on both sides
10x = 5.555 ... (ii)
Subtract (i) from (ii), to get x = \(\frac { 5 }{ 9 } \)
If 2.5252525...............=\(\frac { p }{ q } \) (in the lowest form) what is the value of \(\frac { q }{ p } \)?
- (a)
0.4
- (b)
0.42525
- (c)
0.0396
- (d)
0.396
If \({ x }^{ \sqrt [ x ]{ x } }={ \left( x\sqrt { x } \right) }^{ x },\) find the value of x.
- (a)
\(\frac { 3 }{ 2 } \)
- (b)
\(\frac { 2 }{ 9 } \)
- (c)
\(\frac { 9 }{ 4 } \)
- (d)
\(\frac { 4 }{ 9 } \)
Which symbol is ,used to denote a collection of all positive Integers?
- (a)
N
- (b)
W
- (c)
Z
- (d)
Q
The set of all positive integers is the set of natural numbers. It is denoted by the symbol N.
If x = \(\sqrt [ 3 ]{ 2+\sqrt { 3 } } \), find the value of x3+\(\frac { 1 }{ { x }^{ 3 } } \)
- (a)
2
- (b)
4
- (c)
8
- (d)
9
\(x=\sqrt [ 3 ]{ 2+\sqrt { 3 } } \)
\(\Rightarrow { x }^{ 3 }=2+\sqrt { 3 } \ and\ \frac { 1 }{ { x }^{ 3 } } =2-\sqrt { 3 } \)
\(\therefore { x }^{ 3 }+\frac { 1 }{ { x }^{ 3 } } =2+\sqrt { 3 } +2-\sqrt { 3 } =4\)
Rational number \(\frac { -18 }{ 5 } \) lies between consecutive integers________.
- (a)
-2 and -3
- (b)
-3 and -4
- (c)
-4 and -5
- (d)
-5 and -6
We have, \(\frac { -18 }{ 5 } \)
On expressing it in decimal form, we get - 3.6
∴ The number lies between - 3 and - 4
An irrational number between \(\frac { 1 }{ 7 } \) and \(\frac { 2 }{ 7 } \) is_______.
- (a)
\(\frac { 1 }{ 2 } \left( \frac { 1 }{ 7 } +\frac { 2 }{ 7 } \right) \)
- (b)
\( \left( \frac { 1 }{ 7 } \times \frac { 2 }{ 7 } \right) \)
- (c)
\(\sqrt { \frac { 1 }{ 7 } \times \frac { 2 }{ 7 } } \)
- (d)
None of these
An irrational number between \(\frac { 1 }{ 7 } \) and \(\frac { 2 }{ 7 } \) is \(\sqrt { \frac { 1 }{ 7 } \times \frac { 2 }{ 7 } } \)
The denominator of \(\frac { a+\sqrt { a^{ 2 }-{ b }^{ 2 } } }{ a-\sqrt { a^{ 2 }-{ b }^{ 2 } } } +\frac { a-\sqrt { a^{ 2 }-{ b }^{ 2 } } }{ a+\sqrt { a^{ 2 }-{ b }^{ 2 } } } \)is_____.
- (a)
a2
- (b)
b2
- (c)
a2-b2
- (d)
\(\frac { 4a^{ 2 }-2b^{ 2 } }{ { b }^{ 2 } } \)
we have,\(\frac { a+\sqrt { a^{ 2 }-{ b }^{ 2 } } }{ a-\sqrt { a^{ 2 }-{ b }^{ 2 } } } +\frac { a-\sqrt { a^{ 2 }-{ b }^{ 2 } } }{ a+\sqrt { a^{ 2 }-{ b }^{ 2 } } } \)
=\(\frac { (a+\sqrt { a^{ 2 }-{ b }^{ 2 } } )^{ 2 }+(a-\sqrt { a^{ 2 }-{ b }^{ 2 } } )^{ 2 } }{ (a^{ 2 }-\sqrt { a^{ 2 }-{ b }^{ 2 } } )^{ 2 } } \)
a2+a2-b2+2a\(\sqrt { a^{ 2 }-{ b }^{ 2 } } \)+a2+a2
=\(\frac { -b2-2a\sqrt { a^{ 2 }-{ b }^{ 2 } } }{ a^{ 2 }-a^{ 2 }+b^{ 2 } } \)=\(\frac { 4a^{ 2 }-2b^{ 2 } }{ { b }^{ 2 } } \)
∴ The denominator of the given expression is b2
The ascending order of the surds \(\sqrt [ 3 ]{ 2 } \) ,\(\sqrt [ 6 ]{ 3 } \) ,\(\sqrt [ 9 ]{ 4 } \) is ________.
- (a)
\(\sqrt [ 9 ]{ 4 } \),\(\sqrt [ 6 ]{ 3 } \),\(\sqrt [ 3 ]{ 2 } \)
- (b)
\(\sqrt [ 9 ]{ 4 } \),\(\sqrt [ 3 ]{ 2 } \),\(\sqrt [ 6 ]{ 3 } \)
- (c)
\(\sqrt [ 3 ]{ 2 } \),\(\sqrt [ 6 ]{ 3 } \),\(\sqrt [ 9 ]{ 4 } \)
- (d)
\(\sqrt [ 6 ]{ 3 } \),\(\sqrt [ 9 ]{ 4 } \),\(\sqrt [ 3 ]{ 2 } \)
LCMof 3, 6, 9 = 18
Now, reduce the given surds in surds of order 18.
∴ \(\sqrt [ 3 ]{ 2 } =(2)^{ \frac { 1 }{ 3 } \times \frac { 6 }{ 6 } }({ 2 }^{ 6 })^{ \frac { 1 }{ 18 } }=(64)^{ \frac { 1 }{ 18 } }=\sqrt [ 18 ]{ 64 } \)
\(\sqrt [ 6 ]{ 3 } =(3)^{ \frac { 1 }{ 6 } \times \frac { 3 }{ 3 } }({ 3 }^{ 3 })^{ \frac { 1 }{ 18 } }=(27)^{ \frac { 1 }{ 18 } }=\sqrt [ 18 ]{ 27 } \)
\(\sqrt [ 9 ]{ 4 } =(2)^{ \frac { 1 }{ 9 } \times \frac { 2 }{ 2 } }({ 4 }^{ 2 })^{ \frac { 1 }{ 18 } }=(16)^{ \frac { 1 }{ 18 } }=\sqrt [ 18 ]{ 16 } \)
So, the order is,\(\sqrt [ 18 ]{ 16 } \)<\(\sqrt [ 18 ]{ 27 } \)<\(\sqrt [ 18 ]{ 64 } \)
\(\Rightarrow \)\(\sqrt [ 9 ]{ 4 } \)<\(\sqrt [ 6 ]{ 3 } \)<\(\sqrt [ 3 ]{ 2 } \)
If x = 1-\(\sqrt { 2 } \) , then find the value of \(\left( x-\frac { 1 }{ x } \right) ^{ 2 }\).
- (a)
2
- (b)
3
- (c)
4
- (d)
5
We have, x = 1 - \(\sqrt { 2 } \)
\(\therefore \frac { 1 }{ x } =\frac { 1 }{ 1-\sqrt { 2 } } \times \frac { 1+\sqrt { 2 } }{ 1+\sqrt { 2 } } =\frac { 1+\sqrt { 2 } }{ 1-\sqrt { 2 } } =(1+\sqrt { 2 } )\)
Now,\(\left( x-\frac { 1 }{ x } \right) ^{ 2 }\)=(1-\(\sqrt { 2 } \)+1+\(\sqrt { 2 } \))2=(2)2=4
The value of \(\frac { 1 }{ 1-\sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } +\sqrt { 3 } } +\frac { 1 }{ \sqrt { 3 } +\sqrt { 4 } } +\frac { 1 }{ \sqrt { 4 } +\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } +\sqrt { 6 } } +\frac { 1 }{ \sqrt { 6 } +\sqrt { 7 } } +\frac { 1 }{ \sqrt { 7 } +\sqrt { 8 } } +\frac { 1 }{ \sqrt { 8 } +\sqrt { 9 } } \) is________.
- (a)
0
- (b)
1
- (c)
2
- (d)
4
We have,\(\frac { 1 }{ 1-\sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } +\sqrt { 3 } } +\frac { 1 }{ \sqrt { 3 } +\sqrt { 4 } } +\frac { 1 }{ \sqrt { 4 } +\sqrt { 5 } } +\frac { 1 }{ \sqrt { 5 } +\sqrt { 6 } } +\frac { 1 }{ \sqrt { 6 } +\sqrt { 7 } } +\frac { 1 }{ \sqrt { 7 } +\sqrt { 8 } } +\frac { 1 }{ \sqrt { 8 } +\sqrt { 9 } } \)
On rationalising each of the above number separately, we get
\(\frac { 1-\sqrt { 2 } }{ 1-2 } +\frac { \sqrt { 2 } -\sqrt { 3 } }{ 2-3 } +\frac { \sqrt { 3 } -\sqrt { 4 } }{ 3-4 } +\frac { \sqrt { 4 } -\sqrt { 5 } }{ 4-5 } +\frac { \sqrt { 5 } -\sqrt { 6 } }{ 5-6 } +\frac { \sqrt { 6 } -\sqrt { 7 } }{ 6-7 } +\frac { \sqrt { 7 } -\sqrt { 8 } }{ 7-8 } +\frac { \sqrt { 8 } -\sqrt { 9 } }{ 8-9 } \)
\(=-(1-\sqrt { 2 } )-(\sqrt { 2 } -\sqrt { 3 } )-(\sqrt { 3 } -\sqrt { 4 } )-(\sqrt { 4 } -\sqrt { 5 } )-(\sqrt { 5 } -\sqrt { 6 } )-(\sqrt { 6 } -\sqrt { 7 } )-(\sqrt { 7 } -\sqrt { 8 } )-(\sqrt { 8 } -\sqrt { 9 } )\)
=-1+\(\sqrt { 9 } \)=-1+3=2
Read the statements carefully.
Statement-1: Every point on the number line represent a unique real number.
Statement-2: Irrational numbers cannot be represented on a number line. Which of the following options hold?
- (a)
Both Statement-1 and Statement-2 are true
- (b)
Statement-1 is true but Statement-2 is false.
- (c)
Statement-1 is false but Statement-2 is true.
- (d)
Both Statement-1 and Statement-2 are false.
Statement-1 is true and statement-2 is false as irrational numbers can be represented on number line.
Match the following:
Column-I | Column-II |
If \(\frac { 3 }{ x+8 } =\frac { 4 }{ 6-x } ,\)then x is________. | 3 |
If \(\frac { 2^{ x-1 }.4^{ 2x+1 } }{ 8^{ x-1 } } =64,\)then x is_____. | 55/2 |
If 4x-4x-1=24,then (2x)x is______. | -2 |
If 4x+1=256 ,then x is_____. | 1 |
- (a)
(a)\(\rightarrow \)(i); (b)\(\rightarrow \)(ii); (c)\(\rightarrow \)(iii); (d)\(\rightarrow \)(iv)
- (b)
(a)\(\rightarrow \)(iii); (b)\(\rightarrow \)(iv); (c)\(\rightarrow \)(i); (d)\(\rightarrow \)(ii)
- (c)
(a)\(\rightarrow \)(i); (b)\(\rightarrow \)(iv); (c)\(\rightarrow \)(ii); (d)\(\rightarrow \)(iii)
- (d)
(a)\(\rightarrow \)(iii); (b)\(\rightarrow \)(iv); (c)\(\rightarrow \)(ii); (d)\(\rightarrow \)(i)
(a)We have,\(\frac { 3 }{ x+8 } =\frac { 4 }{ 6-x } \)
\(\Rightarrow \)18-3x=4x+32\(\Rightarrow \)7x=-14\(\Rightarrow \)x=-2
(b)We have,\(\frac { 2^{ x-1 }.4^{ 2x+1 } }{ 8^{ x-1 } } =64\)\(\Rightarrow \frac { 1 }{ { 4 }^{ x } } .4^{ 2x }.16=64\Rightarrow { 4 }^{ x }=\frac { 64 }{ 16 } =4\)
On comparing, we get x = 1
(c) We have, 4x- 4x-1= 24
\(\Rightarrow 4^{ x }-\frac { 4^{ x } }{ 4 } =24\Rightarrow 4^{ x }\left( 1-\frac { 1 }{ 4 } \right) =24\Rightarrow (2^{ 2 })=\frac { 24\times 4 }{ 3 } =32\)
\(\Rightarrow \)22x=25
On comparing, we get 2x = 5\(\Rightarrow \)x =\(\frac { 5 }{ 2 } \)
Now, (2x)x = (5)5/2
(d) We have, 4x+1 = 256
\(\Rightarrow \)4x+1 = (4)4
On comparing, we get x + 1 = 4 \(\Rightarrow \) x = 3