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 mathematics and statistics relations cartesian product of sets cartesian product of two sets cartesian product

If ${ y }^{ 2 }={ x }^{ 2 }-x+1$ and $\quad { I } _{ n }=\int { \cfrac { { x }^{ n } }{ y }  } dx$ and $A{ I } _{ 3 }+B{ I } _{ 2 }+C{ I } _{ 1 }={ x }^{ 2 }y$ then ordered triplet $A,B,C$ is

  1. $\quad \left( \cfrac { 1 }{ 2 } ,-\cfrac { 1 }{ 2 } ,1 \right) $
  2. $\left( 3,1,0 \right) $
  3. $\left( 1,-1,2 \right) $
  4. $\left( 3,-\cfrac { 5 }{ 2 } ,2 \right) $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We have given


${y^2} = {x^2} - x + 1$

${I _n} =\displaystyle \int {\dfrac{{{x^n}}}{y}dx} $
And,

$A{I _3} + B{I _2} + C{I _1} = {x^2}y$

Order triplet $A,\, B,\,C$ $ = \left( {\dfrac{1}{2},\,\dfrac{{ - 1}}{2},1} \right)$

Hence, the option $(A)$ is correct.

Multiple choice chemistry further aspects of equilibria dissociation constants ionisation of weak acids and weak bases ionization constants of weak acids and weak bases

At $100^\circ C$, value of $K _{w}$ is 

  1. $1.0\times 10^{-14}\quad m^{2}$
  2. less than $1.0\times 10^{-14}\quad m^{2}$
  3. greater than $1.0\times 10^{-14}\quad m^{2}$

  4. Zero
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

At higher temperature the value of $kw$ increases.This is in according with le-chatelier principle.

At $100^o kw=51.3\times 106{-14}$
C is the correct answer.

Multiple choice logarithm and its uses basic mathematical concepts physics

If $x=500,y=100$ and $z=5050$, then the value of $(\log _{ xyz }{ { x }^{ z } } )(1+\log _{ x }{ yz } )$ is equal to.

  1. 500

  2. 100

  3. 5050

  4. 10
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given,

$\left(\log _{xyz}\left(x^z\right)\right)\left(1+\log _x\left(yz\right)\right)$

$\left(\log _{xyz}\left(x^z\right)\right)\left(1+\log _x\left(yz\right)\right)$

$=z\log _{xzy}\left(x\right)\left(\log _x\left(zy\right)+1\right)$

from given w have,

$=5050\log _{(500 \times 5050 \times 100)}\left(500\right)\left(\log _{500}\left((5050 \times 100)\right)+1\right)$

$=\dfrac{5050\log _e \left(505000\right)}{\log _e \left(252500000\right)}+5050\log _{252500000}\left(500\right)$

$=5050$

Multiple choice logarithm and its uses basic mathematical concepts physics

If there are $n$ zeros after the decimal point, then the characteristic of that number will be

  1. $n+1$
  2. $-n+1$
  3. $-(n+1)$
  4. $n-1$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For number less than $1,$ if there are $n$ zeroes after the decimal point, then the characteristics of that number will be $-(n+1).$

and For number  greater than $1,$ if  there are $n$ number of zeroes are on the left sides of the digits then characteristics will be $(n+1).$
Hence, C is the correct option.

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

G.C.D. of $(a + b -c)^6$ and $(a + b -c)^4$ is

  1. $(a+b-c)^6$
  2. $(a+b-c)^{10}$
  3. $(a+b-c)^2$
  4. $(a+b-c)^4$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since, $(a + b -c)^6 = (a + b -c)^4 \times (a + b -c)^2 $
$\therefore$  G.C.D. of $(a + b -c)^6$ and $(a + b -c)^4$ = $(a + b -c)^4$
$\because (a + b -c)^4$ is greatest common in $(a + b -c)^4$ and $(a + b -c)^6$.
Option D is correct.

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

The G.C.D of $x^3+x^2-x-1$ and $x^2-1$ is

  1. $x^2-1$
  2. $x+1$
  3. $x^3-1$
  4. $x-1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let $p(x) = x^3+x^2-x-1$ and $ q(x) = x^2-1$
$p(x) = x^3+x^2-x-1$
         $ = x^2(x+1)-1(x+1) $
         $ = (x^2-1) (x+1) $
         $ = (x-1)(x+1)(x+1) $
and
$ q(x) = x^2-1$
         $= (x+1)(x-1) $
$\therefore $ G.C.D of $p(x)$ and $q(x)$ =$ (x+1)(x-1) = x^2 - 1 $
Option A is correct.

Multiple choice maths hcf-lcm common factors and hcf hcf highest common factor (h.c.f.)

If $G.C.D\ (a , b) = 1$ then $G.C.D\ ( a+b , a-b )$=?

  1. $1$ or $2$
  2. $a$ or $b$
  3. $a+b$ or $a-b$
  4. $4$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
It is given that GCD$\left(a,b\right)=1$

Let GCD$\left(a-b,a+b\right)=d$

$\Rightarrow\,d$ divides $a-b$ and $a+b$

there exists integers $m$ and $n$ such that 

$a+b=m\times d$        ..........$(1)$

and $a-b=n\times d$        ..........$(2)$

Upon adding and subtracting equation $(1)$ and $(2)$ we get

$2a=\left(m+n\right)\times d$         ..........$(3)$

and $2b=\left(m-n\right)\times d$         ..........$(4)$

Since, GCD$\left(a,b\right)=1$(given)

$\therefore\,2\times GCD\left(a,b\right)=2$

$\therefore\,GCD\left(2a,2b\right)=2$ since $GCD\left(ka,kb\right)=kGCD\left(a,b\right)$

Upon substituting  value of $2a$ and $2b$ from equations $(3)$ and $(4)$ we get

$\therefore\,gcd\left(\left(m+n\right)\times d,\left(m-n\right)\times d\right)=2$

$\therefore\,d\times gcd\left(\left(m+n\right),\left(m-n\right)\right)=2$

$\therefore\,d\times$ some integer$=2$

$\therefore\,d$ divides $2$

$\therefore\,d\le 2$ if $x$ divides $y,$ then $\left|x\right|\le \left|y\right|$

$\therefore\,d=1$ or $2$ since, gcd is always a positive integer.
Multiple choice maths constructions mid-point formula midpoints division of a line segment

The midpoint of the interval in which $x^{2}-2(\sqrt{-x})^{2}-3<0$ is satisfied, is

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

The expression sqrt(-x)^2 is defined for x <= 0 and equals -x. The inequality is x^2 - 2(-x) - 3 < 0 => x^2 + 2x - 3 < 0 => (x+3)(x-1) < 0. Since x <= 0, the interval is [-3, 0]. The midpoint is (-3+0)/2 = -3/2.

Multiple choice maths constructions mid-point formula midpoints division of a line segment

$P and Q$ are points on the line joining $A(-2,5) and (3,1)$ such that $AP=PQ=QB$ then the mid point of $PQ$ is ?

  1. $\dfrac{1}{23}$
  2. $\dfrac{-1}{24}$
  3. $\dfrac{-24}{1}$
  4. $(1,4)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since $AP = PQ = BQ$


hence mid point of PQ will be the mid point of AB.



Thus mid point of AB = [( –2+3)/2 , (5+1)/2 ]





mid point of AB = [1/2 , 3] = mid point of PQ



 

Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

The product of a rational number and its reciprocal is

  1. $0$
  2. $1$
  3. $-1$

  4. none
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The product of rational number with its reciprocal is always equal to $1$.


Let's take an example,
Rational number $= \dfrac{2}{3}$
Its reciprocal $= \dfrac{3}{2}$

Product $= \dfrac{2\times3}{3\times2} = 1$

Multiple choice maths parts and whole multiplication of a fraction multiplication of a fractions multiplication of fraction finding the whole when a fraction is given

The product of a fractional number and its multiplicative inverse is

  1. 0

  2. 1

  3. number itself

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

If the product of two numbers is 1, then each number is known as multiplicative inverse or the reciprocal of one another. ... Hence, the product of a fractional number and its multiplicative inverse is 1