Mathematics · Quantitative Aptitude

Algebra and Arithmetic

406 Questions

Algebra and arithmetic questions cover fundamental mathematical operations, inequalities, and binomial products. They assess core quantitative reasoning skills required for various aptitude tests. Solving these problems strengthens the understanding of number systems and algebraic identities.

Binomial productsLinear inequalitiesLeast common multipleQuadratic equationsInteger properties

Algebra and Arithmetic Questions

Multiple choice maths applications of matrices and determinants elementary transformations of a matrix multiplicative inverse of a matrix inverse of a matrix

Let A=$\left( {\begin{array}{{20}{c}}{ - 5}&{ - 8}&{ - 7}\3&5&4\2&3&3\end{array}} \right),B = \left( {\begin{array}{{20}{c}}x\y\z\end{array}} \right)$. If AB is scalar $\left( { \ne 0} \right)$ multiple of B, then x+y=

  1. $z$
  2. $-z$
  3. $0$
  4. $2z$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$A=\begin{pmatrix} -5 & -8 & -7 \\ 3 & 5 & 4 \\ 2 & 3 & 3 \end{pmatrix}\quad B=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
Given $AB=k\ B$
$AB\Rightarrow \begin{pmatrix} -5 & -8 & -7 \\ 3 & 5 & 4 \\ 2 & 3 & 3 \end{pmatrix}\begin{pmatrix} x \\ y \\ z \end{pmatrix}\quad k=\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
$\Rightarrow \begin{pmatrix} -5x & -8y & -7z \\ 3x & +5y & +4z \\ x2 & +3y & +3z \end{pmatrix}=\quad k\begin{pmatrix} x \\ y \\ z \end{pmatrix}$
On adding all the elements on left and right side 
$\Rightarrow \ (-5x-8y-7z)+(3x+5y+47)+(2x+3y+3z)$
$=k(x+y+z)$
$\Rightarrow \ D=k(x+y+z)$
$k\neq 0$
$\Rightarrow \ x+y+z=0$
$\Rightarrow \ x+y=-z$

Multiple choice maths fun with numbers some special sequences triangular numbers properties and patterns of perfect squares

For real number $a,b,c$ and $d$ , if $a^2+b^2=4$ and $c^2+d^2=1$, then possible value of $ac+bd$ is / are 

  1. $2$
  2. $3$
  3. $1$
  4. $\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
A,C,D Correct answer
Explanation

According to Cauchy-Schwarz inequality:


$(ac+bd)^2\le(a^2+b^2)(c^2+d^2)$      $(\forall a, b, c, d \in \mathbb{R})$

Substituting the values here gives:

$(ac+bd)^2\le4$

$(ac+bd)^\le2$

Therefore, it can take all values given in the options except 3.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $x=1+a+{ a }^{ 2 }+{ a }^{ 3 }+....$ to $\infty \left( \left| a \right| <1 \right) $ and 
$y=1+b+{ b }^{ 2 }+{ b }^{ 3 }+...$ to $\infty \left( \left| b \right| <1 \right) $ then
$1+ab+{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 3 }{ b }^{ 3 }+...$ to $\infty =\cfrac { xy }{ x+y-1 } $

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ x }=\dfrac { 1 }{ 1-a } $ ${ y }=\dfrac { 1 }{ 1-b } $ [summing infinite $G.P's$].
$\therefore a=\dfrac { x-1 }{ x } $, $b=\dfrac { y-1 }{ y } $
$\therefore 1+ab+{ a }^{ 2 }{ b }^{ 2 }+...\infty $
$=\dfrac { 1 }{ 1-ab } =\dfrac { 1 }{ 1-\dfrac { (x-1)(y-1) }{ xy }  } =\dfrac { xy }{ x+y-1 }. $

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $x=1+a+a^2+...\infty$ where $|a| <1 $ and $y=1+b+b^2+...\infty$, where $|b| < 1$, then $1+ab+a^2b^2+...\infty =\dfrac{xy}{x+y-1}$.

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is the same identity derived in question 466317. The sum of the series 1 + ab + a^2b^2 + ... is indeed xy / (x + y - 1) given the definitions of x and y.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

${x}^{\cfrac{1}{2}}.{x}^{\cfrac{1}{4}}.{x}^{\cfrac{1}{8}}.{x}^{\cfrac{1}{16}}.....$ to $\infty$

  1. $0$
  2. $1$
  3. $x$
  4. $\infty$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^{1/2}.x^{1/4}......\infty =x^{\left ( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16} \right )}$ 


$=x\left ( 1/2+1/2^{2}+1/2^{3}+1/2^{4}+.......\infty  \right )$

$=x^{\frac{1}{2}\left ( \frac{1-(1/2)\infty }{1-1/2} \right )}$ 

$=x^{1}$ 

$=x$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $p$ is positive, then the sum to infinity of the series, ${1 \over {1 + p}} - {{1 - p} \over {{{(1 + p)}^2}}} + {{{{(1 - p)}^2}} \over {{{(1 + p)}^3}}} - ......$ is

  1. $1/2$
  2. $3/4$
  3. $1$

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$a=\dfrac { 1 }{ 1+P } \\ r=-\dfrac { 1-P }{ 1+P } $
Sum to infinity $=\dfrac { a }{ 1-r } \\ =\dfrac { \dfrac { 1 }{ 1+P }  }{ 1+\dfrac { 1-P }{ 1+P }  } \\ =\dfrac { \dfrac { 1 }{ 1+P }  }{ \dfrac { 1+P+1-P }{ 1+P }  } =\dfrac { 1 }{ 2 } $
Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $0<x,y,a,b<1$,then the sum of infinite terms of the series $\sqrt x (\sqrt a  + \sqrt x ) + \sqrt x (\sqrt {ab}  + \sqrt {xy} ) + \sqrt x (b\sqrt a  + y\sqrt x ) + .......$ is

  1. $\dfrac{{\sqrt {ax} }}{{1 + \sqrt b }} + \dfrac{x}{{1 + \sqrt y }}$
  2. $\dfrac{{\sqrt x }}{{1 + \sqrt b }} + \dfrac{{\sqrt x }}{{1 + \sqrt y }}$
  3. $\dfrac{{\sqrt x }}{{1 - \sqrt b }} + \dfrac{{\sqrt x }}{{1 - \sqrt y }}$
  4. $\dfrac{{\sqrt {ax} }}{{1 - \sqrt b }} + \dfrac{x}{{1 - \sqrt y }}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$0<x,y,a,b<1\ S=\sqrt { x } (\sqrt { a } +\sqrt { x } )+\sqrt { x } (\sqrt { ab } +\sqrt { xy } )+\sqrt { x } (b\sqrt { a } +y\sqrt { x } )\ =\sqrt { x } \left[ (\sqrt { a } +\sqrt { ab } +b\sqrt { a } ....) \right] +(\sqrt { x } +\sqrt { xy } +y\sqrt { x } )\ =\sqrt { x } \left[ (\cfrac { \sqrt { a }  }{ 1-\sqrt { b }  } )+\cfrac { \sqrt { x }  }{ 1-\sqrt { y }  }  \right] \ S=\cfrac { \sqrt { ax }  }{ 1-\sqrt { b }  } +\cfrac { x }{ 1-\sqrt { y }  } $

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

 If  $0<\phi < \pi /2,$   and
 $x= \sum _{n=0}^{\infty} \cos ^{2n} \phi$, $ y=\sum _{n=0}^{\infty } \sin ^{2n} \phi$                     
and $z=\sum _{n=0}^{\infty} \cos ^{2n} \phi \sin ^{2n} \phi $ 
then

  1. xyz $=$xz+y
  2. xyz$=$xy+z
  3. xyz$=$x+y+z
  4. xy$=$yz+z
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

The series x converges to a value $\dfrac{1}{1-{\cos^2 \phi}}\;=\;{cosec^2 \phi}$

Similarly y converges to  a value ${sec^2 \phi} $

In a similar fashion.. z converges to $\dfrac{{cosec^2 \phi}{sec^2 \phi}}{{cosec^2 \phi}{sec^2 \phi} - 1}$
 $ z = \dfrac{xy}{xy - 1} $

$xy + z = xyz&gt;$; Option B.
And, also
$x + y = xy$
$\therefore xyz = x+ y+ z&gt;$; Option C.

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $a=\sum _{ n=0 } ^{\infty  }{x^n } ,b=\sum _{n=0  }^{ \infty  }{ y^n } , c=\sum _{n=0  }^{ \infty  }{ (xy)^n } $ where $|x| ,| y| < 1$ ; then

  1. $abc = a + b + c$
  2. $ab + bc = ac + b$
  3. $ac + bc = ab + c$
  4. $ab + ac = bc + a$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Clearly every summation is infinite series,
$a=\cfrac{1}{1-x}, b = \cfrac{1}{1-y}$ and $c=\cfrac{1}{1-xy}$
or $x=1-\cfrac{1}{a}, y=1-\cfrac{1}{b}$
Simplifying above equation we get, $ac+bc=ab+c$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If $|x| > 1$, then
$\left(1-\dfrac{1}{x}\right)+\left(1-\dfrac{1}{x}\right)^2+\left(1-\dfrac{1}{x}\right)^3+.....=$

  1. $x-1$
  2. $x+1$
  3. $x$
  4. $\dfrac{1}{x-1}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$S=\left( 1-\cfrac { 1 }{ x }  \right) +{ \left( 1-\cfrac { 1 }{ x }  \right)  }^{ 2 }+{ \left( 1-\cfrac { 1 }{ x }  \right)  }^{ 3 }+........$
This is a G.P whose first term, $a=1-\cfrac { 1 }{ x } $ and common ration,  $r=1-\cfrac { 1 }{ x } $
${ S } _{ \infty  }=\cfrac { a }{ 1-r } $
$=\cfrac { 1-\cfrac { 1 }{ x }  }{ 1-\left( 1-\cfrac { 1 }{ x }  \right)  } $
$=\cfrac { { \left( x-1 \right)  }/{ x } }{ { \left[ x-\left( x-1 \right)  \right]  }/{ x } } $
$=\cfrac { x-1 }{ x-x+1 } $
$=x-1$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

For $0 < \phi < \pi/2$ if $x=\sum _{n=0}^{\infty }\cos ^{2n} \phi, y=\sum _{n=0}^{\infty }\sin ^{2n} \phi, z=\sum _{n=0}^{\infty }\cos ^{2n} \phi \sin^{2n}\phi$, then 

  1. $xyz=xz+y$
  2. $xyz=xy+z$
  3. $xyz=x+y+z$
  4. $xyz=yz+x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given $\displaystyle x=\sum _{ n=0 }^{ \infty  }{ { \cos }^{ 2n }\phi ,\quad y=\sum _{ n=0 }^{ \infty  }{ { \sin }^{ 2n }\phi  }  } $ and $\displaystyle z=\sum _{ n=0 }^{ \infty  }{ { \cos }^{ 2n }\phi  } { \sin }^{ 2n }\phi $


\since $\displaystyle 0<\phi <\frac { \pi  }{ 2 } $, so each series is geometric series with common ratio $\displaystyle r<1$. Therefore, the series are convergent.


Now, $\displaystyle x=\frac { 1 }{ 1-{ \cos }^{ 2 }\phi  } $


$\displaystyle =\frac { 1 }{ { \sin }^{ 2 }\phi  } $      $(\because S _{ \infty  }=\dfrac { a }{ 1-r } )$


$\displaystyle y=\frac { 1 }{ 1-{ \sin }^{ 2 }\phi  } $      $(\because S _{ \infty  }=\dfrac { a }{ 1-r } )$


$\displaystyle =\frac { 1 }{ { \cos }^{ 2 }\phi  } $


$\displaystyle z=\frac { 1 }{ 1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi  } $ 

   

$(\because S _{ \infty  }=\dfrac { a }{ 1-r } )$


Consider, $\displaystyle xyz=\frac { 1 }{ { \sin }^{ 2 }\phi { \cos }^{ 2 }\phi (1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi ) } $      $(1)$


Also, $\displaystyle =\frac { 1 }{ { \sin }^{ 2 }\phi { \cos }^{ 2 }\phi  } +\frac { 1 }{ 1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi  } $


$\displaystyle xy+z=\frac { 1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi +{ \sin }^{ 2 }\phi { \cos }^{ 2 }\phi  }{ { \sin }^{ 2 }\phi { \cos }^{ 2 }\phi (1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi ) } $


$\displaystyle =\frac { 1 }{ { \sin }^{ 2 }\phi { \cos }^{ 2 }\phi (1-{ \sin }^{ 2 }{ \phi \cos }^{ 2 }\phi ) } $

 

$\displaystyle =xyz$      $[From (1)]$

 

Multiple choice maths concept of directed numbers and number line subtraction of directed numbers subtraction of integers subtraction of integers on number line

Addition of integer and its additive inverse is

  1. Zero

  2. double the number

  3. half the number

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Additive inverse of an integer is a number which is of opposite sign and when added to the integer, gives zero.

Thus, Addition of integer and its additive inverse is Zero.