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 complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

Which  of the following is correct ?

  1. $2 + 3i > 1 + 4i$
  2. $2 + 2i > 3 + 3i$
  3. $5 + 8i > 5 + 7i$

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

Option $A$

$2+3i>1+4i$
$2^2+3^2>1^2+4^2$
$4+9>1+16$
$13>17$

It is not correct.

Option $B$
$2+2i>3+3i$

$2^2+2^2>3^2+3^2$
$4+4>9+9$
$8>18$

It is not correct.


Option $C$
$5+8i>5+7i$

$5^2+^82>5^2+7^2$
$25+64>25+49$
$89>74$

It is correct.

Hence, this is the answer.

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

If $a ^ { 2 } + b ^ { 2 } = 1$, then $\dfrac { 1 + b + i a } { 1 + b - i a } = ?$

  1. 1

  2. 2
  3. $b + i a$
  4. $a + i b$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$\dfrac{1+b+ia}{1+b-ia}\times\dfrac{b+ia}{b+ia}$

$=\dfrac{\left(1+b+ia\right)\times\left(b+ia\right)}{b+{b}^{2}-iab+ia+iab+{a}^{2}}$

$=\dfrac{\left(1+b+ia\right)\times\left(b+ia\right)}{1+b+ia}$ since ${a}^{2}+{b}^{2}=1$

$=b+ia$
Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

For positive integers $n _1, n _2$ the value of the expression $(1 + i)^{n _1} + (1 + i^3)^{n _1} + (1 + i^5)^{n _2} + (1 + i^7)^{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
D Correct answer
Explanation

$(1+i)^{n _{1}}+(1+i^{2})^{n _{2}}+(1+i^{5})^{n _{2}}+(1+i^{7})^{n _{2}}$
$=(1+i)^{n _{1}}+(1-i)^{n _{2}}+(1+i)^{n _{2}}+(1-i)^{n _{2}}$
$=2[1+:^{n _{1}}C _{2}i^{2}+:^{n _{1}}C _{4}i^{4}...]+2[1+:^{n _{2}}C _{2}i^{2}+:^{n _{2}}C _{4}i^{4}...]$
$=2[1+-:^{n _{1}}C _{2}+:^{n _{1}}C _{4}-...]+2[1-:^{n _{2}}C _{2}+:^{n _{2}}C _{4}-...]$
Hence
For all $n _{1}>0$ and $n _{2}>0$ the above expression yields real integral number.
Where $n _{1},n _{2}\epsilon N$.
Hence, option 'D' is correct.

Multiple choice business maths linear programming problems structure of linear programming model linear programming problem operations research

Let $a _1,a _2....,a _n$ be a non negative real number such that $a _1+a _2....+a _n=m$ and let $S=\underset{i<j}\sum a _ia _j$, then

  1. $S\leq \dfrac {m^2}2$
  2. $S> \dfrac {m^2}4$
  3. $S< \dfrac {m}2$
  4. $S> \dfrac {m^2}2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For non-negative real numbers, the sum of products S = sum(a_i * a_j) for i < j is related to the square of the sum. Since (sum a_i)^2 = sum(a_i^2) + 2*S, and sum(a_i^2) >= 0, then 2*S <= m^2, so S <= m^2/2.

Multiple choice economics consumption and investment functions keynesian law of consumption and propensity to consume ex ante and ex post concept of consumption function, saving function and investment function

 If MPC =1, the value of multiplier is ________.

  1. 0

  2. 1

  3. Between 0 and 1

  4. Infinity
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Investment multiplier refers to the number of time by which the increase in output or income exceeds the increase in investment. It is measured as the ratio between change in income and change in investment and it is denoted as 'k'.

Multiplier(k) => Change in income / change in investment = 1/ {1-MPC(c)} where c is the marginal propensity to consume.

If MPC = 1, then 

Multiplier(k)= 1/(1-1)= 1/0 = Infinity.

Therefore, the value of the multiplier is infinity. 

Multiple choice maths congruence and inequalities of triangles inequalities of a triangle triangle inequality inequalities in triangle

In a triangle $ABC$, $(a+b+c)(b+c-a)=k$$bc$ if:

  1. $k< 0$
  2. $k> 6$
  3. $0< k< 4$
  4. $k> 4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The expression (a+b+c)(b+c-a) = (b+c)^2 - a^2 = b^2 + c^2 + 2bc - a^2. By the Law of Cosines, a^2 = b^2 + c^2 - 2bc cos(A). Substituting this, we get 2bc + 2bc cos(A) = 2bc(1+cos(A)). Since 0 < 1+cos(A) < 2, the value k must be between 0 and 4.

Multiple choice maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

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

  1. $\dfrac { \sqrt { a x } } { 1 + \sqrt { b } } + \dfrac { x } { 1 + \sqrt { y } }$
  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. $\dfrac { \sqrt { a x } } { 1 - \sqrt { b } } + \dfrac { x } { 1 - \sqrt { y } }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

The value of $ (\bar { A } +\bar { B } )\times (\bar { A } -\bar { B } )$ is 

  1. $0$
  2. ${A}^{2}-{B}^{2}$
  3. $\bar { B } \times \bar { A }$
  4. $2(\bar { B } \times \bar { A })$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$(\overrightarrow A+\overrightarrow B)\times(\overrightarrow A-\overrightarrow B)=\overrightarrow A\times \overrightarrow A-\overrightarrow A\times \overrightarrow B+\overrightarrow B\times \overrightarrow A-\overrightarrow B\times \overrightarrow B=2(\overrightarrow B\times \overrightarrow A)$


Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

If $  \overrightarrow{A} \times \overrightarrow{B}=0,$ $  \overrightarrow{B} \times \overrightarrow{C}=0, $then $  \overrightarrow{A} \times \overrightarrow{C}= $

  1. $AC$
  2. $\dfrac{AB^2}{C} $
  3. $Zero$
  4. $None of these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$  \overrightarrow{A} \times \overrightarrow{B}=0   \overrightarrow{A}  \ and\  \overrightarrow{B} $ are parallel.
$  \overrightarrow{B} \times \overrightarrow{C}=0  \Rightarrow  \overrightarrow{B}  \ and \  \overrightarrow{C} $ are parallel 
 $\Rightarrow \vec{A} \parallel \vec{C}$

Multiple choice business maths inverse of a matrix and linear equations non singular matrix singular & non-singular matrix determinants

If $\left[ {\begin{array}{*{20}{c}}1&{ - 1}&x\1&x&1\x&{ - 1}&1\end{array}} \right]$ has no inverse, then the real value of $x$ is 

  1. $2$
  2. $3$
  3. $0$
  4. $1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A=\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}has\quad no\quad inverse\quad x$

$if\,\left| A \right| = 0$

$\begin{vmatrix} 1 & -1 & x \\ 1 & x & 1 \\ x & -1 & 1 \end{vmatrix}=0$

$1\left( {x + 1} \right) - 1\left( { - 1 + x} \right) + x\left( { - 1 - {x^2}} \right) = 0$

$x + 1 + 1 - x - x - {x^3} = 0$

$ - {x^3} - x + 2 = 0$

${x^3} + x - 2 = 0$

$\left( {x - 1} \right)\left( {{x^2} + x + 2} \right) = 0$

$x = 1\,is\,real\,value\,$