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 power and exponent power of powers laws of exponents and powers law of indices

If $a^x=\sqrt{b},b^y = \sqrt [3]{c}$ and $c^z = \sqrt {a}$ then the value of $xyz$

  1. $\displaystyle \frac {1}{2}$
  2. $\displaystyle \frac {1}{3}$
  3. $\displaystyle \frac {1}{6}$
  4. $\displaystyle \frac {1}{12}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$a^x=\sqrt{b}=b^{1/2}$
$a=b^{1/2x}$
$b^y=\sqrt[3]{c}$
$b^y=c^{1/3}$
$b=c^{1/3y}$
$c^z=a^{1/2}$
$c=a^{1/2z}=b^{1/4xz}$
$c^1=c^{1/12xyz}$
$\displaystyle 1= \frac {1}{12xyz}$
$\displaystyle xyz = \frac {1}{12}$

Multiple choice maths power and exponent power of powers laws of exponents and powers law of indices

If $2^a\,>\,4^c\;and\;3^b\,>\,9^a\;and\;a,\,b,\,c$ all positive, then

  1. $c\,<\,a\,<\,b$
  2. $b\,<\,c\,<\,a$
  3. $c\,<\,b\,<\,a$
  4. $a\,<\,b\,<\,c$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$2^a\,>\,4^c\;\;\;\;\;\;3^b\,>\,9^a$
$2^a\,>\,2^{2c}\;\;\;\;\;3^b\,>\,3^{2a}$
$a\,>\,2c\;\;\;\;\;\;\;b\,>\,2a$
$\therefore\;a\,>\,c-(i)\;\;\therefore\;b\,>\,a-(ii)$
From (1) & (2), we have
$c\,<\,a\,<\,b$

Multiple choice maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

If $a+b=30$ and $ab=176$, find $a^{3}+b^{3}$

  1. $10160$
  2. $11060$
  3. $11160$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$ a+b=30,ab=176 \ a+b=30 \ \Rightarrow { \left( { a+b } \right) ^{ 2 } }={ \left( { 30 } \right) ^{ 2 } } $


$\Rightarrow { a^{ 2 } }+{ b^{ 2 } }+2ab=900 \ \Rightarrow { a^{ 2 } }+{ b^{ 2 } }+2\times 176=900 \ \Rightarrow { a^{ 2 } }+{ b^{ 2 } }=900-352 \ \Rightarrow { a^{ 2 } }+{ b^{ 2 } }=548$

$Now \ { a^{ 3 } }+{ b^{ 3 } }=\left( { a+b } \right) \left( { { a^{ 2 } }+{ b^{ 2 } }-ab } \right)  \ =30\left( { 548-176 } \right)  \ =30\times 372 \ =11160 $

Multiple choice maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

If a and b are two whole numbers, then commutative law is applicable to subtraction if and only if

  1. $a = b$
  2. a $\neq $ b
  3. $a > b$
  4. $a < b$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Commutative property : The subtraction of whole numbers is not commutative, that is, if $a$ and $b$ are two whole numbers, then in general $a – b$ is not equal to $(b – a)$.

Verification:

We know that $9 – 5 = 4$ but $5 – 9=-4$ which is not a whole number. Thus, for two whole numbers $a$ and $b$ if $a > b$, then $a – b$ is a whole number but $b – a$ is not possible and if $b > a$, then $b – a$ is a whole number but $a – b$ is not possible.

Now, if $a=b=3$ then, $a-b=3-3=0$ which is also a whole number.

Hence, whole numbers are commutative under subtraction if and only if $a=b$.
Multiple choice maths operations adding and subtracting numbers using place value addition & subtraction mental additions and subtractions

Find the value of A and B in the following sum:
$6 A 3$
$\underline {+2 2 B}$
$\underline {B B 1}$

  1. $A=5,B=8$
  2. $A=6,B=8$
  3. $A=5,B=7$
  4. $A=5,B=9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since 3 + B given us 1 in the answer, so B has to be 8 as $3+8=11$.
Now 1 is carried over so, $A+1+2=B$
i.e. $A+1+2=8$. So, A is 5.
Also $6+2=B$ in hundred's place to confirm that B is 8.
The value of A is 5 and B is 8.

Multiple choice maths average arithmetic mean of ap introduction to averages means

The A.M. of a + 2, a, 2-a is

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

Given values are : $a+2, a, 2-a$


Arithmetic Mean, $AM = \dfrac{a+2+a+2-a}{3}$

$AM = \dfrac{a+4}{3}$

Hence, option B is corret

Multiple choice maths average arithmetic mean of ap introduction to averages means

The arithmetic mean of $1 + \sqrt { 2 }$ and $7 + 5 \sqrt { 2 }$ is $\sqrt { a } + \sqrt { b }$ . Then $a - b =$

  1. -1

  2. 1

  3. 2

  4. -2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The arithmetic mean of $1+\sqrt{2}$ and $7+5\sqrt{2}$ is $\dfrac{1+\sqrt{2}+7+5\sqrt{2}}{2}=\dfrac{8+6\sqrt{2}}{2}=4+3\sqrt{2}=\sqrt{16}+\sqrt{18}$

$\implies \sqrt{a}+\sqrt{b}=\sqrt{16}+\sqrt{18}$
$\implies a=16,b=18$
$a-b=16-18=-2$

Multiple choice maths average arithmetic mean of ap introduction to averages means

The airthmatic mean of $1 + \sqrt { 2 }$ and $7 + 5 \sqrt { 2 }$ is $\sqrt { a } + \sqrt { b }$ . Then a $- b =$

  1. -1

  2. 1

  3. 2

  4. -2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The arithmetic mean of $1+\sqrt{2}$ and $7+5\sqrt{2}$ is $\dfrac{1+\sqrt{2}+7+5\sqrt{2}}{2}=\dfrac{8+6\sqrt{2}}{2}=4+3\sqrt{2}=\sqrt{16}+\sqrt{18}$

$\implies \sqrt{a}+\sqrt{b}=\sqrt{16}+\sqrt{18}$
$\implies a=16,b=18$
$a-b=16-18=-2$

Multiple choice maths average arithmetic mean of ap introduction to averages means

The arithmetic mean between $2+\sqrt {(2)}$ and $2-\sqrt {(2)}$ is

  1. $2$
  2. $\sqrt {(2)}$
  3. $0$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since, the arithmetic mean between $a$ and $b$ is $\displaystyle  \frac {a+b}{2}$
$\therefore $the arithmetic mean between $2+\sqrt 2$ and $2-\sqrt 2$  $=\displaystyle \frac {2+\sqrt 2+2-\sqrt 2}{2}$

$=\dfrac {4}{2}$

$=2$
Option A is correct.

Multiple choice maths average arithmetic mean of ap introduction to averages means

The arithmetic mean between $\cfrac { x+a }{ x } $ and $\cfrac { x-a }{ x } $ when $x\ne 0$, is (the symbol $\ne$ means "not equal to"):

  1. $2$, if $a\ne 0$
  2. $1$
  3. $1$, only if $a=0$
  4. $\dfrac {a}{x}$
  5. $x$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The arithmetic mean of two numbers a and b is (a + b) / 2. Here, ((x+a)/x + (x-a)/x) / 2 = (2x/x) / 2 = 2 / 2 = 1.

Multiple choice maths average arithmetic mean of ap introduction to averages means

If $\cfrac {a^n+b^n}{a^{n-1}+b^{n-1}}$ is the AM between a and b, then the value of n is 

  1. 0

  2. 1

  3. -1

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

The A.M. of a and b is $\dfrac{a+b}{2}$.


Therefore, $\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}$ will be the A.M. of a and b, if 

$\dfrac{a^{n+1}+b^{n+1}}{a^n+b^n}=\dfrac{a+b}{2}$

$\Rightarrow 2(a^{n+1}+b^{n+1})=(a^n+b^n)(a+b)$

$\Rightarrow 2a^{n+1}+2b^{n+1}=a^{n+1}+a^nb+b^na+b^{n+1}$

$\Rightarrow a^{n+1}+b^{n+1}=a^nb+b^na$

$\Rightarrow a^n(a-b)=b^n(a-b)$

$\Rightarrow a^n=b^n$

$\Rightarrow \dfrac{a^n}{b^n}=1$

$\Rightarrow (\dfrac{a}{b})^n=1$

$\Rightarrow (\dfrac{a}{b})^n=(\dfrac{a}{b})^0$

$\Rightarrow n=0$

Multiple choice maths average arithmetic mean of ap introduction to averages means

If the arithmetic mean of the numbers $x _{1}, x _{2}, x _{3}.......,x _{3}$ is $\overline{X}$, then the arithmetic mean of numbers $ax _{1}+b, ax _{2}+b, ax _{3}+b,........, ax _{n}+b$, where  $a, b$ are two constants would be 

  1. $\overline{X}$
  2. $na\overline{X}+nb$
  3. $a\overline{X}$
  4. $a\overline{X}+b$
Reveal answer Fill a bubble to check yourself
A Correct answer