Tag: maths

Questions Related to maths

Simplify $| |-5| - 16 + |-1| | $.

absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

  1. $-12$

  2. $-10$

  3. $10$

  4. $12$

  5. $22$

Show answer
Correct Option: C
Explanation:

Rule: $|x|=x$ when $x$ is positive integer and $|x|=-x$ when $x$ is negative integer. So

$||-5|-16+|-1||$
$\Rightarrow |-(-5)-16+[-(-1)]|$
$\Rightarrow |+5-16+1|$
$\Rightarrow |-10|=10$

The value of $|7|+|5|-|7|-|1-3|$= ___________.

absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: D
Explanation:

$|7|+|5|-|7|-|1-3|$

$|7|+|5|-|7|-|-2|$
$=7+5-7-2$  ............  $\because|-2|=2$
$=3$
Hence the correct answer is option D.

If $x$ be real and positive, then the value of
$y = x + \frac{1}{x}$ satisfies

absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

  1. $0 < y \leq 0.5$

  2. $0.5 < y \leq 1$

  3. $1 < y < 2$

  4. $y \geq 2$

Show answer
Correct Option: D
Explanation:

$ Given\quad y\quad =\quad x+\frac { 1 }{ x } \ \quad =(\sqrt { x } )^{ 2 }+\left( \frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }\ \quad =\left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }+2\sqrt { x } \frac { 1 }{ \sqrt { x }  } \ \quad =\left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }+2\ First\quad term\quad is\quad a\quad squared\quad term\quad so\quad it\quad is\quad positive.\ \therefore \quad \left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }+2\quad >2 \quad when\quad \left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }\quad has\quad a\quad finite\quad value\ and\quad \left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }+2\quad =\quad 2\quad \quad  when\quad \left( \sqrt { x } -\frac { 1 }{ \sqrt { x }  }  \right) ^{ 2 }\quad is\quad zero.\ \therefore \quad y\ge 2\quad \quad (Ans) $

The largest number of the three consecutive number is $(x+2)$, then the smallest number is

absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

  1. x+2

  2. x+1

  3. x

  4. x-1

Show answer
Correct Option: C

The absolute value of $\dfrac { \displaystyle\int _{ 0 }^{ \pi /2 }{ \left( x\cos { x+1 }  \right) { e }^{ \sin { x }  }dx }  }{ \displaystyle\int _{ 0 }^{ \pi /2 }{ \left( x\sin { x-1 }  \right) { e }^{ \cos { x }  }dx }  } $ is equal to 

absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

  1. $e$

  2. $\pi e$

  3. $\dfrac{e}{2}$

  4. $\dfrac{\pi}{e}$

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} We\, have \ I=\dfrac { { \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ { e^{ \sin  x } }\left( { x\cos  x+1 } \right) dx }  } }{ { \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ { e^{ \cos  x } }\left( { x\sin  x-1 } \right) dx }  } }  \ =\dfrac { { \left[ { x{ e^{ \sin  x } } } \right] _{ 0 }^{ \frac { \pi  }{ 2 }  } } }{ { \int _{ 0 }^{ \frac { \pi  }{ 2 }  }{ { e^{ \cos  x } }\left( { 1-x\sin  x } \right) dx }  } } =\dfrac { { \frac { \pi  }{ 2 } \times e } }{ { \left[ { { e^{ \cos  x } }x } \right] _{ \frac { \pi  }{ 2 }  }^{ 0 } } }  \ =\dfrac { { \frac { \pi  }{ 2 } e } }{ { 0-\frac { \pi  }{ 2 }  } } =-e \ Hence,\, absolute\, value\, =e \ Hence,\, option\, A\; is\, the\, correct\, answer. \end{array}$

Absolute value of $+16$ is:

absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

  1. $-16$

  2. $0$

  3. $+16$

  4. $16$

Show answer
Correct Option: D
Explanation:

Absolute value is always a positive value. 

So absolute value of $+16$ is $16$. 
Because a simple digit always shows the positive value and there is no need to show the positive sign over there.
Hence, the answer is $16$.

The value of x on simplifying $x -2 |x|=-3$ is

absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

  1. -1 or 3

  2. 1 or -3

  3. -1 or -3

  4. 1, 3

Show answer
Correct Option: A
Explanation:

$x-2|x| = -3$

There two cases possible either $x$ is positive or$x$ is negative
We have to solve both cases
Firstly taking x as positive
$x-2x=-3$
$x=3$
Now taking $x$ as negative
$x-2(-x)=-3$
$3x=-3$
$x=-1$
So we get two values of $x$ that is -1 and 3
 So correct answer will be option A

Simplification of $-|-48|$ is

absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

  1. $48$

  2. $-48$

  3. $0$

  4. $-47$

Show answer
Correct Option: B
Explanation:
Since $|-x| = x$
$-|-48| = -(48) = -48$
So, option $B$ is correct.

If $a$ and $b$ are any real numbers, then which of the following expressions is always positive?

absolute value real numbers (rational and irrational numbers) real numbers basic algebra maths

  1. $\left| a \right| $

  2. $\left| a+b \right| $

  3. $\left| a-b \right| +1/2$

  4. ${ a }^{ 2 }+{ b }^{ 2 }$

  5. ${ \left( a+b \right) }^{ 2 }$

Show answer
Correct Option: C
Explanation:

Lets check each option one by one.

A. |a| cab be 0 if a=0 
B. |a+b| can be 0 if a+b=0
C. |a-b|+ 1/2  will always positive because |a+b| is either 0 or positive 
D. ${a}^{2}+{b}^{2}$ can be 0 if both $a$ and $b$ becomes 0
E ${(a+b)}^{2}$ can equal to 0 if $a+b$ = 0
So correct answer will be option C 

A square field is 40m long. The perimeter of the square field is (in cm)

maths perimeter and area of rectilinear figures unit of area units of area units of area and volume converting between units of areas the metric system metric system

  1. $14000 \, cm$

  2. $16000 \, cm$

  3. $12000 \, cm$

  4. $8000 \, cm$

Show answer
Correct Option: B