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 multiplication and division of integers division of integers and its properties multiplying and dividing integers multiplication of integers

If the dividend and divisor have unlike signs then the quotient will be _____

  1. positive

  2. negative

  3. zero

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

If the dividend and divisor have unlike signs then the quotient will be Negative.


Division of Integers is similar to the division of whole numbers (both positive) except the sign of the quotient needs to be determined.

If both the dividend and divisor are positive, the quotient will be positive.
(+16) ÷ (+4) = +4

If both the dividend and divisor are negative, the quotient will be positive.
(-16) ÷ (-4) = +4

If only one of the dividend or divisor is negative, the quotient will be negative.
(+16) ÷ (-4) = -4     or      (-16) ÷ (+4) = -4

In other words, if the signs are the same the quotient will be positive, if they are different, the quotient will be negative
Hence Option B

Multiple choice maths multiplication and division of integers division of integers and its properties multiplying and dividing integers multiplication of integers

If ${ B }^{ 3 }A< 0$ and $A> 0$, which of the following must be negative?

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

If $A$ is positive ${B}^{3}$ must be negative. Therefore, $B$ must be negative. If $A$ is positive and $B$ is negative, the product $AB$ must be negative.

Multiple choice maths multiplication and division of integers division of integers and its properties multiplying and dividing integers multiplication of integers

Which of the following statements is true?

  1. The product of a positive and a negative integer is negative

  2. The product of a negative and a positive integer may be zero
  3. For all non-zero integers a and b, $a\times b$ is always greater than either a or b

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

a) The product of a positive and a negative integer is negative = True

b) The product of a negative and a positive integer may be zero = False
c) For all non-zero integers $a$ and $b$, $a\times b$ is always greater than either $a$ or $b$ = False
Hence option A is correct answer.

Multiple choice maths multiplication and division of integers division of integers and its properties multiplying and dividing integers multiplication of integers

The sign of the product of two unlike integers is __________.

  1. Positive

  2. Negative

  3. Positive or negative

  4. Cannot be determined
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The sign of the product of two unlike integers i.e.positive integer and negative integer is always negative.
Hence the correct answer is option B.

Multiple choice maths multiplication and division of integers division of integers and its properties multiplying and dividing integers multiplication of integers

Which of the following statement is CORRECT?

  1. The product of a positive and a negative integer is always negative.

  2. The addition of a negative and a positive integer is always zero.

  3. For all non-zero integers a and b, a X b is always greater than either a or b.

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

In option (A)       $-x\times x=-x^{2}$ which is always negative. option is correct.

In option (B)        $ -x+y$ not equals to $0$ it zero only when $x=y$  option is incorrect.
In option (C)         let $a=-3$ and $b=4$ then $a\times b=-3\times 4=-12$  which is less than both a and b. hence this option is also incorrect.
hence only option $A$ is correct.

Multiple choice maths multiplication and division of integers division of integers and its properties multiplying and dividing integers multiplication of integers

If the dividend and divisor have like signs then the quotient will be .......... .

  1. positive

  2. negative

  3. zero

  4. none
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
If both the dividend and divisor are positive, the quotient will be positive. For example:

$(+16)\div (+4) = +4$

If both the dividend and divisor are negative, the quotient will be positive. For example:

$(-16)\div (-4) = +4$

If only one of the dividend or divisor is negative, the quotient will be negative. For example:

$(-16)\div (+4) = -4$     or     $(+16)\div (-4) = -4$

Therefore, we conclude that if the signs are the same/like, the quotient will be positive, if they are different/unlike, the quotient will be negative.

Hence, if the dividend and divisor have like signs then the quotient will be positive.
Multiple choice maths multiplication and division of integers division of integers and its properties multiplying and dividing integers multiplication of integers

Sign of the product of 231 negative integer and 9 positive integer is

  1. Negative

  2. Positive

  3. 0

  4. None of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
We know that the product of two integers with unlike signs (two integers, one with positive sign and one with negative sign) is always negative.

The given two integers are $-231$ and $+9$ and the product of these integers is:

$(-231)\times (+9)=-(231\times 9)=-2079$ which is a negative integer.

Hence, sign of the product of the given integers is negative.
Multiple choice maths multiplication and division of integers division of integers and its properties multiplying and dividing integers multiplication of integers

The product of each positive integer with $-1$ is always ______.

  1. Positive

  2. Negative

  3. 0

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

Let us take a positive integer $+2$ and multiply it with the given negative integer $-1$.


We know that the  product of two integers with unlike signs is always negative.

Therefore, the product of the integers with unlike signs $+2$ and $-1$ is:

$(+2)\times (-1)=-(2\times 1)=-2$ which is a negative integer.

Hence, the positive integer whose product with $-1$ is always negative.

Multiple choice maths multiplication and division of integers division of integers and its properties multiplying and dividing integers multiplication of integers

$(-1)^{11}$ value is

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

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

We know that if the power $n$ of any negative integer $x$ is even then the resulting integer will always be positive and if the power $n$ of any negative integer $x$ is odd then the resulting integer will always be negative. For example, if the negative integer is $x=-2$, then


Odd power:
$(-2)^3=-2\times -2\times -2=-8$ which is a negative integer.

Even power:

$(-2)^2=-2\times -2=4$ which is a positive integer.

Similarly, $(-1)^{11}=-1$ because $-1$ is a negative integer and $11$ is an odd number, so the result will be a negative integer.

Hence, $(-1)^{11}=-1$

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

Let $a,b,c$ be in AP and $k\neq 0$ be a real number. WHich of trhe following are correct ?
1.$ka,kb,kc$ are in Ap
2. $k-a,k-b,k-c$ are in AP
3. $\dfrac{a}{k},\dfrac{b}{k},\dfrac{c}{k}$ are in AP
Select the correct answer using the code given below :

  1. 1 and 2 only

  2. 2 and 3 only

  3. 1 and 3 only

  4. 1, 2 and 3 only
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$a, b, c \in A.P. \quad \left( \text{Given} \right)$

Therefore,
$(1) \quad\text{when a constant term is multiplied to each term of an A.P. then the resultant is also an A.P.}$
$ka, kb, kc \in A.P., \quad k \ne 0$

$(2)\quad\text{when a constant is subtracted from each term of an A.P. then the resultant is also an A.P.}$
$\Rightarrow k - a, k-b, k - c \in A.P.$

$(3)\quad\text{when a constant is divided from each term of an A.P. then the resultant is also an A.P.}$
$\Rightarrow \cfrac{a}{k}, \cfrac{b}{k}, \cfrac{c}{k} \in A.P.$
Hence all statements are correct.

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If $x \in R,$ the numbers ${2^{1 + x}} + {2^{1 - x}},b/2,{36^x} + {36^{ - x}}$ form an A.P. , then $b$ may lie in the interval

  1. $\left[ {16,\infty } \right)$
  2. $\left[ {6,\infty } \right)$
  3. $\left[ {\infty , - 6} \right)$
  4. $\left[ {6,12} \right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given 


$2^{1+x}+2^{1-x}, \dfrac b2 ,36^x+36^{-x}$ form an AP

The condition to be in AP is

$2\times \left(\dfrac b2\right)=2^{1+x}+2^{1-x}+36^x+36^{-x}$

$b=2.2^x+2.\dfrac1{2^x}+36^x+\dfrac 1{36^x}$

$b=2\left(2^x+\dfrac 1{2^x}\right)+\left( 36^x+\dfrac 1{36^x}\right)$

Let $2^x=y \quad 36^x=k$

$b=2\left(y+\dfrac 1y\right)+\left(k+\dfrac 1k\right)$

The min value of $f(x)+\dfrac 1{f(x)}$ is $2$

The max value is $\infty$

$\implies 2(2)+2 \leq b\leq 2(\infty)+(\infty)$

$\implies 6\leq b\leq\infty$

$\implies b\in [6,\infty)$

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If a, b, c are in A.P., then  $a ^ { 3 } + c ^ { 3 } - 8 b ^ { 3 }$ is equal to: 

  1. $2 a b c$
  2. -$6 a b c$
  3. $4 a b c$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$a,b,c$ are in A.P
$\Rightarrow\,b-a=c-b$
$\Rightarrow\,2b=a+c$
${a}^{3}+{c}^{3}-8{b}^{3}$
$={a}^{3}+{c}^{3}-{\left(2b\right)}^{3}$
$={a}^{3}+{c}^{3}-{\left(a+c\right)}^{3}$
$={a}^{3}+{c}^{3}-{a}^{3}-{c}^{3}-3ac\left(a+c\right)$
$=-3ac\left(2b\right)=-6abc$

Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

If $ab + bc + ca =0$  , then the value of  $\frac{1}{{{a^2} - bc}} + \frac{1}{{{b^2} - ca}} + \frac{1}{{{c^2} - ab}}$ will be 

  1. -1

  2. a+b+c

  3. 0

  4. ab
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\dfrac{1}{a^2-bc}+\dfrac{1}{b^2-ca}+\dfrac{1}{c^2-ab}$
Given $ab+bc+ca=0$
now $-bc=ab+ca$
$-ca=ab+bc$
$-ab=bc+ca$
$\dfrac{1}{a^2+(ab+ca)}+\dfrac{1}{b^2+(ab+bc)}+\dfrac{1}{c^2+(bc+ca)}$
$=\dfrac{1}{a(a+b+c)}+\dfrac{1}{b(a+b+c)}+\dfrac{1}{c(a+b+c)}$
$=\dfrac{bc+ca+ab}{abc(a+b+c)}=0$.