Mathematics · Quantitative Aptitude

Algebra and Arithmetic

333 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 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 

Multiple choice maths hcf-lcm introduction to multiples multiples lcm

If A, B and C are three numbers such that L.C.M. of A and B is B and the L.C.M. of B and C is C then the L.C.M. of A, B and C is

  1. A

  2. B

  3. C
  4. $\displaystyle \frac{A+B+C}{3}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

LCM of A and B is B it means that B is multiple of A. LCM of B and C is C it means C is multiple of B or we can say that C is multiple of A also.

So LCM of A,B ,C is C so correct answer is option C

Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

If $a +2b +3c = 0$, then $a \times b + b\times c + c\times a = ka\times b,$
Where $k$ is equal to ?

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

Given,

$\begin{array}{l} a+2b+3c=0 \ a=-\left( { 2b+3c } \right)  \ a\times a=-\left( { 2\left( { b\times a } \right) +3\left( { c\times a } \right)  } \right)  \ \Rightarrow 0=-\left( { 2b\times a+3\left( { c\times a } \right)  } \right) ..........\left( i \right)  \end{array}$
Again,
$\begin{array}{l} 2b=-\left( { a+3c } \right)  \ 2b\times b=-\left( { a\times b+3c\times b } \right)  \ \Rightarrow 0=-\left( { a\times b+3c\times b } \right) ..........\left( { ii } \right)  \end{array}$
equation (i) and (ii)
$\begin{array}{l} -\left( { 2b\times a+3c\times a } \right) +\left( { a\times b } \right) +3c\times b=0 \ \Rightarrow 3a\times b-3c\times a-3b\times c=0 \end{array}$
We know that,
$a \times b =  - b \times a$
$a \times b = c \times a + b \times c...............\left( {iii} \right)$
Now,$ATQ;$
$a \times b + b \times c + c \times a = K\,\,\,a \times b$
$1 = 2a \times b = k\,a \times b$         {from equation (iii)}
$\therefore k=2$
Option $C$ is correct answer.

Multiple choice maths direct proportion and inverse proportion rule of three types of proportions direct proportion

If $a:b=c:d$ then how many of the following statements are true?

  1. $c(a+b)=a(c+d)$
  2. $d(a-b)=b(c-d)$
  3. $(a^{2}+b^{2})(ac-bd)=(a^{2}-b^{2})(ac+bd)$
  4. $(a^{2}-b^{2})(ad-bc)=(a^{2}+b^{2})(ac-bd)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\dfrac{a}{b}=\dfrac{c}{d}$
$\dfrac{b}{a}=\dfrac{d}{c}$

$\left(1+\dfrac{b}{a}\right)=\left(1+\dfrac{d}{c}\right)$
$\dfrac{(a+b)}{a}=\dfrac{(c+d)}{c}$
$c(a+b)=a(c+d)$

Multiple choice maths direct proportion and inverse proportion rule of three types of proportions direct proportion

If $a : b = 5 : 9$ and $b : c = 4 : 7$ find $a : b : c$

  1. $5:9:\dfrac{63}{4}$
  2. $20:63:36$
  3. $4:36:63$
  4. $20:36:63$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Explanation

Given, $a : b = 5 : 9 $ and $b : c = 4 : 7$ $=$ $\displaystyle \left ( 4\times

\frac{9}{4} \right ):\left ( 7\times \frac{9}{4} \right

)=9:\frac{63}{4}$
$\displaystyle \Rightarrow a:b:c=5:9:\frac{63}{4}=20:36:63$

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

a, b, c are even numbers and x, y, z are odd numbers. Which of the following relationships can't be justified at any cost?
(a) $\dfrac{a\times b}{c} = x\times y$ (b) $\dfrac{a\times b}{x}=yz$ (c) $\dfrac{xy}{z} = ab$

  1. Only B

  2. Only C

  3. All the three

  4. Only B and C
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

In option B a×b is always even & xyz is always odd therefore equality not holds.

In option C ab is always even therefore abz is also even & xy is always odd hence equality not holds.
In option A xy is always odd but (a×b)/c can be odd or even therefore equality can hold in this case.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

If $a$ and $b$ are odd numbers, then which of the following is even?

  1. $a+b$
  2. $a+b+1$
  3. $ab$
  4. $ab+2$

  5. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know the following rule :

odd + odd = even,

even + even = even,

odd + even = odd,

even + odd = odd,

odd × odd = odd.

(A) The given expression is

a + b = odd + odd = even.

(B) The given expression is

a + b + 1 = odd + odd + odd = even + odd = odd.

(C) The given expression is

ab = odd × odd = odd.

(D) The given expression is

ab + 2 = odd × odd + 2= odd + even = odd.

Thus, the correct option is (A) a+b.
Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

Multiplication of one odd and one even integer is always :

  1. Even

  2. Odd

  3. Can't be determined

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Even integer is an integer having unit digit as a multiple of $2$

So on multiplication with odd integer, unit digit will still remain a multiple of $2$, hence, multiplication of odd and even integer gives even integer.

Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

$a, b, c$ are even numbers and $x, y, z$ are odd numbers. Which of the following relationships can't be justified at any cost?
(a) $\dfrac{a \times b}{c} = x \times y$  (b)  $\dfrac{a \times b}{x} = yz$  (c) $\dfrac{xy}{z} = ab$

  1. Only a

  2. Only c

  3. All the three

  4. Only b and c
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Use the rules that 
(a) product of two odd or even numbers are odd and even respectively.
(b) the quotient of the odd number $ \div $ even number or vice-versa may or may not be strictly odd or even.

a. $ ab $ must be a even number and $ \dfrac{ab}{c} $ may be even number or a fraction or odd number and $ xy $ must be a odd number
therefore it can be justified in some case.

b. $ \dfrac { ab }{ x } $ is always a fraction or even number and $ yz $ is always a odd number
therefore it cannot be justified in any case

c.$ \dfrac { xy }{ z } $ maybe odd number or fraction but $ ab $ will always be even.
therefore it cannot be justified.

Relationships given in b and c can't be justified at any cost.
Multiple choice maths negative numbers and integers even and odd numbers sum of numbers odd and even numbers

If $f(x)=x^{2}+6x+c$, where $'c'$ is an integer, then $f(0)+f(-1)$ is

  1. an even integer
  2. an odd integer always disable by $3$
  3. an odd integer not divisible by $3$
  4. an odd integer may or not be divisible by $3$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$f(0) = c$

$f(-1) = c-5$

$f(0)+f(-1) = 2c-5 = 2(c-3) + 1$

As the above is of the form $2k+1$, it is always odd.

For $c=3$, the above is not divisible by 3 but for $c=4$, it is. Therefore, it may or may not be divisible by 3.