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 negative numbers and integers odd and even numbers operations on integers different types of numbers

If k is a positive integer, then $k(k+1)(k+3)$ is

  1. even only when k is even

  2. even only when k is odd

  3. Always odd

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

Suppose $k$ is even

Then $k(k+1)(k+3)$ is even as $k$ will be divisible by $2$
Now if $k$ is odd.
Then $(k+1)$ and $(k+3)$ are even
$\Rightarrow k(k+1)(k+3)$ is also even.
So $k(k+1)(k+3)$ is always even.
So option $D$ is correct.

Multiple choice mathematical modelling proof by contradiction similar triangles

To prove " If  $x,x\in N$  is even then $x^2$ is even". By direct method, we must start with the assumption:

  1. Let $x ^2 $ not even
  2. Let $x$ not even
  3. Let $x$ be even natural number.
  4. Let $x\notin N$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

To prove an implication by straight forward approach or direct method, we have to take the hypothesis of the statement as an assumption and then reach at the conclusion. 

In the statement, "If $p$ then $q$",  $p$ is hypothesis and $q$  is conclusion.
So, here $C$ is correct

Multiple choice decimal representation of rational numbers rational and irrational numbers maths

If a and b are any two such real numbers that ab $ = 0 $ , then

  1. $a = 0, b \leq 0$
  2. $b = 0, a \leq 0$

  3. a = 0 or b = 0 or both
  4. $a = b$ and $b = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

if both number are real the either a or b or both should be zero.
then only ab will be 0.
if any real number is multiplied by 0 then result will be zero.
So, answer is
C a = 0 or b = 0 or both

Multiple choice business mathematics and statistics moving averages introduction to time series introduction to time series and forecasting time series and forecasting uses of average in day-to-day life

If $A : B =3 : 4$ and $B : C =5 : 6$, then $A : C$ is

  1. $7:3$
  2. $3:4$
  3. $1:2$
  4. $5:9$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$A:B=3:4$
 $B:C=5:6$
 $\therefore \frac { A }{ B } =\frac { 3 }{ 4 } & \frac { B }{ C } =\frac { 5 }{ 6 } $
 $\therefore \frac { A }{ B } \times \frac { B }{ C } =\frac { A }{ C } $
$\therefore \frac { 3 }{ 4 } \times \frac { 5 }{ 6 } =A:C$
 $=\frac { 15 }{ 24 } =\frac { 5 }{ 8 } $
A:C=5:8

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

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

  1. $0 < y \leq 0.5$
  2. $0.5 < y \leq 1$
  3. $1 < y < 2$
  4. $y \geq 2$
Reveal answer Fill a bubble to check yourself
D Correct answer
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) $

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

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

  1. $\left| a \right| $
  2. $\left| a+b \right| $
  3. $\left| a-b \right| +1/2$
  4. ${ a }^{ 2 }+{ b }^{ 2 }$
  5. ${ \left( a+b \right) }^{ 2 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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