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 applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

If the zeroes of the rational expression $ (ax+b)(3x+2)$ are $-\dfrac{2}{3}$ and $ \dfrac{1}{2}$, then $ a+b=$

  1. $4$
  2. $0$
  3. $-b$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$(ax+b)(3x+2)=0$

$\Rightarrow$  $3ax^2+2ax+3bx+2b$
$\Rightarrow$  $3ax^2+(2a+3b)x+2b=0$
It is given that $\dfrac{-2}{3}$ and $\dfrac{1}{2}$ are zeros of the equation $3ax^2+(2a+3b)x+2b=0$
Here, $A=3a,\,B=2a+3b,\,C=2b$
Let $\alpha=\dfrac{-2}{3}$ and $\beta=\dfrac{1}{2}$
We know,
$\Rightarrow$  $\alpha+\beta=\dfrac{-B}{A}$
$\Rightarrow$  $\dfrac{-2}{3}+\dfrac{1}{2}=\dfrac{-(2a+3b)}{3a}$

$\Rightarrow$  $\dfrac{-4+3}{6}\times 3a=-2a-3b$

$\Rightarrow$  $\dfrac{-a}{2}=-2a-3b$
$\Rightarrow$  $-a=-4a-6b$
$\Rightarrow$  $3a+6b=0$              ----- ( 1 )
Now,
$\Rightarrow$  $\alpha.\beta=\dfrac{CA}{A}$

$\Rightarrow$  $\dfrac{-2}{3}\times\dfrac{1}{2}=\dfrac{2b}{3a}$

$\Rightarrow$  $\dfrac{-1}{3}\times 3a=2b$

$\Rightarrow$  $-a+2b=0$          ----- ( 2 )
Adding equation ( 1 ) and ( 2 ) we get,
$b=0$
Put $b=0$ in ( 2 ) we get,
$a=0$
$\therefore$  $a+b=0+0=0$

Multiple choice the nth roots of unity complex numbers maths

For positive integers ${ n } _{ 1 },{ n } _{ 2 }$ the value of the expression; ${ (1+i) }^{ { n } _{ 1 } }+{ (1+i) }^{ { n } _{ 1 } }+{ (1+i) }^{ { n } _{ 2 } }+{ (1+i) }^{ { n } _{ 2 } }$, where $i=\sqrt { -1 } $, is a real number if :

  1. ${ n } _{ 1 }={ n } _{ 2 }+1$
  2. ${ n } _{ 1 }={ n } _{ 2 }-1$
  3. ${ n } _{ 1 }={ n } _{ 2 }$
  4. ${ n } _{ 1 }>0,{ n } _{ 2 }>0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression (1+i)^n1 + (1+i)^n1 + (1+i)^n2 + (1+i)^n2 simplifies to 2 * ((1+i)^n1 + (1+i)^n2). For this to be real, the imaginary parts must cancel or be zero, which occurs when n1 = n2 + 1.

Multiple choice the nth roots of unity complex numbers maths

If $r$ is non-real and $r=\sqrt [ 5 ]{ 1 } $, then the value of $ { 2 }^{ \left| 1+r+{ r }^{ 2 }+{ r }^{ -2 }-{ r }^{ -1 } \right|  }$ is equal to

  1. $2$
  2. $4$
  3. $8$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\left| 1+r+{ r }^{ 2 }+{ r }^{ -2 }-{ r }^{ -1 } \right| =\left| 1+r+{ r }^{ 2 }+{ r }^{ 3 }-{ r }^{ 4 } \right| $


$[\because { r }^{ 5 }=1\Rightarrow { r }^{ 3 }.{ r }^{ 2 }=1\Rightarrow { r }^{ -2 }={ r }^{ 3 }$ and ${ r }^{ 4 }.r=1\Rightarrow { r }^{ -1 }={ r }^{ 4 }]$


$\displaystyle=\left| 1+r+{ r }^{ 2 }+{ r }^{ 3 }+{ r }^{ 4 }-2{ r }^{ 4 } \right| =\left| \frac { 1-{ r }^{ 5 } }{ 1-r } -2{ r }^{ 4 } \right| =\left| 0-2{ r }^{ 4 } \right| \quad \quad \quad \left[ \because { r }^{ 5 }=1 \right] $

$=2{ \left| r \right|  }^{ 4 }=2\left( 1 \right) =2\quad \quad [\because \left| r \right| =1$ as ${ r }^{ 5 }=1]$

$\therefore { 2 }^{ \left| 1+r+{ r }^{ 2 }+{ r }^{ -2 }-{ r }^{ -1 } \right|  }={ 2 }^{ 2 }=4$

Multiple choice maths powers and exponents scientific notation use of exponents power of 10

If $a^{b} = 4  -ab$ and $b^{a} = 1$, where $a$ and $b$ are positive integers, find $a$.

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Plug in real numbers for $a$ and $b$. Since it isn’t clear what numbers to plug in to satisfy the first equation, look at the second equation instead. First, realize that  $a$  cannot be  $0$  since a is a positive integer.
Since, $a\neq0$, so the only way to get $b^a=1$ is if $b=1 $(As $1$ to any power is $1$).
Plugging $b=1$ in to the first equation,
$a^b=4-ab$
$a^1=4-a\times1$
$a=4-a$
$a=2$
Hence, option C is correct.
Multiple choice maths powers and exponents scientific notation use of exponents power of 10

If $x$ is a positive integer satisfying $x^7=k$ and $x^9=m$, which of the following must be equal to $x^{11}$?

  1. $\cfrac{m^2}{k}$
  2. $m^2-k$
  3. $m^2-7$
  4. $2k-\cfrac{m}{3}$
  5. $k+4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $x^7=k, x^9=m$
$x^{11}=\dfrac{x^{18}}{x^{7}}$ 
Put the given values, we get
$x^{11}=\dfrac{m^{2}}{k}$
Multiple choice terms related to matrices matrices and determinants matrices algebra maths

If the traces of $A, B$ are $20$ and $-8$, then the trace of $A+B$ is:

  1. $12$
  2. $-12$
  3. $28$
  4. $-28$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

 the trace of an $n\times n$ square matrix A is defined to be the sum of the elements on the main diagonal (the diagonal from the upper left to the lower right) of A

the traces of A,B are 20 and −8, then the trace of $A+B$ is $trac(A+B)=trace(A)+trac(B)=20+-8=12$

Multiple choice logical equivalence mathematical logic discrete mathematics business maths maths

The contrapositive of the statement "if  $2 ^ { 2 } = 5 ,$  then  $1$  get first class" is

  1. If I do not get a first class, then $2 ^ { 2 } = 5$
  2. If I do not get a first class, then $2 ^ { 2 } \neq 5$
  3. If I get a first class, then $2 ^ { 2 } = 5$
  4. If I get a first class, then $2 ^ { 3 } = 5$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$P:{ 2 }^{ 2 }=5$

$q:I$ get first class
the contrapositive of $p\rightarrow q$ is $\sim q\rightarrow \sim p$. Hence the answer is if $I$ do not get a first class, then ${ 2 }^{ 2 }\neq 5$
Correct Answer : Option B.

Multiple choice reciprocal equations theory of equations maths

The range of reciprocal equation is:

  1. $R$
  2. $R-{0}$
  3. $R^+$
  4. $Q$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The range is defined as the set of all output values or the set of all those possible values which appear on Y axis in the graph of given function. 

The domain and range of a reciprocal function are all the real number except for zero. 
This is because reciprocal of $0$ i.e. is undefined.
The range of reciprocal equation is $R-0$

Multiple choice reciprocal equations theory of equations maths

If the coefficients from one end of an equation are equal in magnitude and sign to the coefficients from the other end, then the equation is said to be 

  1. reciprocal equation of second type

  2. reciprocal equation of first type

  3. reciprocal equation

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation


Multiple choice reciprocal equations theory of equations maths

If the coefficients from one end of an equation are equal in magnitude and opposite in sign to the coefficients from the other end, then the equation is said to be 

  1. reciprocal equation of second type

  2. reciprocal equation of first type

  3. reciprocal equation

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Consider a general equation:

$a _{n}x^n+a _{n-1}x^{n-1}+a _{n-2}x^{n-2}+....................+a _1x+a _0=0$

Now if  $a _n-i=-a _i ,$        $  i=0,1,2,3,...........n$

Then this type of equation is called as reciprocal equation of second type.

Ex- $4x^4-9x^3+9x-4=0$
Multiple choice maths set language different sets de morgan's law de morgan's law for set theory

If $A=\left {x\epsilon C: x^2=1\right }$ and $B=\left {x\epsilon C: x^4=1\right }$, then $A\Delta B$ is equal to

  1. $\left \{-1, 1\right \}$
  2. $\left \{-1, 1, i, -i\right \}$
  3. $\left \{-i, i\right \}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$x^2=1\Rightarrow x=-1, 1.\therefore A=\left {-1, 1\right }$
$x^4=1\Rightarrow x^2=-1, 1$
$\Rightarrow x=-i, i, -1, 1.\therefore B=\left {-i, i, -1, 1\right }$
$\therefore A\Delta B=(A-B)\cup (B-A)=\phi \cup \left {-i, i\right }=\left {-i, i\right }$.

Multiple choice maths square and square root scientific notation use of exponents power of 10

If $x$ and $y$ are any two positive real numbers, then $x > y$ implies

  1. $- x > - y$
  2. $- x < - y$
  3. $\frac{1}{x} > \frac{1}{y}$
  4. $-\frac{1}{x} > \frac{1}{y}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$ Given\quad that\quad x\quad and\quad y\quad are\quad positive\quad number\quad and\quad x>y.\ We\quad may\quad calculate\ (a)\quad on\quad the\quad positive\quad side\quad of\quad number\quad line\quad x\quad will\quad be\ \quad \quad to\quad the\quad right\quad of\quad y\quad and\quad \ (b)\quad on\quad the\quad negative\quad side\quad of\quad the\quad number\quad line\quad x\quad will\ \quad \quad be\quad to\quad the\quad left\quad of\quad y.\ Statement\quad A\longrightarrow -x>-y\quad i.e\quad -x\quad to\quad the\quad right\quad side\quad of\quad -y\ which\quad is\quad not\quad complying\quad with\quad (b).\quad So\quad it\quad is\quad not\quad true.\ Statement\quad B\longrightarrow -x<-y\quad i.e\quad -x\quad lies\quad to\quad the\quad left\quad of\quad -y.\ This\quad complies\quad with\quad (b).\quad So\quad the\quad statement\quad is\quad true.\ Statement\quad C\longrightarrow \frac { 1 }{ x } >\frac { 1 }{ y } .\quad Here\quad the\quad numerators\quad of\quad the\ two\quad fractions\quad are\quad equal.\quad But\quad the\quad denominator\quad of\quad \frac { 1 }{ x } \ is\quad greater\quad than\quad that\quad of\quad \frac { 1 }{ y } .\ So\quad this\quad statement\quad is\quad not\quad true.\ Statement\quad D\longrightarrow -\frac { 1 }{ x } >\frac { 1 }{ y } ,\ -\frac { 1 }{ x } is\quad a\quad negative\quad number\quad so\quad it\quad will\quad always\quad be\quad smaller\ than\quad a\quad positive\quad number.\ \therefore \quad -\frac { 1 }{ x } \ngtr \frac { 1 }{ y } .\quad So\quad this\quad statement\quad is\quad not\quad true.\ Answer-\quad Statement\quad B\quad is\quad correct.\  $