Mathematics · Reasoning

Algebraic Expressions

183 Questions

Algebraic expressions form the foundation of mathematics, involving variables, constants, and arithmetic operations. This topic includes evaluating coefficients, solving proportional relationships, and understanding boolean logic. It is widely tested across various competitive exams to assess analytical skills.

Variable evaluationExpression coefficientsBoolean logic variablesDirect proportionalityProgramming expressions

Algebraic Expressions Questions

Multiple choice

(ax)y

  1. ax+y

  2. axy
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The power of a power rule states that (a^x)^y = a^(x*y). Multiplying the exponents gives a^(xy).

Multiple choice

axay

  1. ax+y

  2. axy
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The product rule for exponents states that a^x * a^y = a^(x+y). Adding the exponents is the correct operation.

Multiple choice physics measurements and experimentation vernier calliper and screw gauge least count of vernier calliper and screw gauge measurement of length

If $x=10.0 \pm 0.1$ and $y=10.0 \pm 0.1$, then $2x-2y$ is equal to

  1. $(0.0 \pm 0.1)$
  2. $Zero$
  3. $(0.0 \pm 0.4)$
  4. $(20 \pm 0.2)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Apply formula

$(A\pm \Delta A)-(B\pm \Delta B)=(A-B)\pm (\Delta A-\Delta B)$

Similarly

  $ 2x-2y=2\left( 10.0\pm 0.1 \right)-2\left( 10\pm 0.1 \right) $

 $ =\left( 2\times 10-2\times 10 \right)\pm (2\times 0.1-2\times 0.1) $

 $ =0 $

Hence, $2x-2y=ZERO$ 

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

If $x = {y^{\frac{1}{a}}},\,y = {z^{\frac{1}{b}}}\,\,{\text{and}}\,\,z = {x^{\frac{1}{c}}}\,{\text{where}}\,x \ne 1,y \ne 1,\,z \ne 1$, then what is the value of $abc$?

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

Given $x=y^{\dfrac{1}{a}}.....(1)$

$y=z^{\dfrac{1}{b}}.....(2)$
$z=x^{\dfrac{1}{c}}.....(3)$
Putting the value of y from equation (2) in equation (1)
$x=[(z)^{\dfrac{1}{b}}]^{\dfrac{1}{a}}\Rightarrow x=z^{\dfrac{1}{ab}}$
Putting the value of z from equation (3) in the above equation
$x=[(x)^{\dfrac{1}{c}}]^{\dfrac{1}{ab}}\Rightarrow x^1=x^{\dfrac{1}{abc}}$
$\therefore\dfrac{1}{abc}=1\Rightarrow abc=1$

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

The mean of $x, y, z$ is $y$, then $x + z = .............$

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

Mean of 3 numbers $=\dfrac {\mbox {Sum of three numbers}}{3}$

$\Rightarrow y = \dfrac {x+y+z}{3}$

$\Rightarrow 3y=x+y+z$

$\Rightarrow x+z = 2y$

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

$-32\times x= 160$, $-23\times y= -115$
What is the value of $x\div y$?

  1. $-1$
  2. $1$
  3. $-5$
  4. $5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given, $-32\times x=160, -23\times y=-115$
We have $x=\dfrac{160}{-32}$

and $ y=\dfrac{-115}{-23}=\dfrac{115}{23}$

Thus $\dfrac{x}{y}=\dfrac{160\times 23}{-32\times 115}=-\dfrac{32\times23}{32\times 23}=-1$

Multiple choice

What is the range of the relation (R = {(x, y) | x + y = 5)?

  1. {x | x \in \mathbb{R}}

  2. {y | y \in \mathbb{R}}

  3. {(x, y) | x \in \mathbb{R}, y \in \mathbb{R}}

  4. {(x, y) | x + y = 5}
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The range of a relation is the set of all possible values of the dependent variable. In this case, the dependent variable is (y), so the range of the relation (R = {(x, y) | x + y = 5) is the set of all real numbers, (\mathbb{R}).

Multiple choice

If x + y = 10 and xy = 21, what is the value of x^2 + y^2?

  1. 29

  2. 49

  3. 69

  4. 89
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the identity (x + y)^2 = x^2 + y^2 + 2xy, we can solve for x^2 + y^2: (x + y)^2 = x^2 + y^2 + 2xy 10^2 = x^2 + y^2 + 2(21) 100 = x^2 + y^2 + 42 x^2 + y^2 = 100 - 42 = 58.

Multiple choice

If (x^2 + y^2 = 25) and (xy = 6), find the value of (x + y).

  1. 5

  2. 7

  3. 9

  4. 11
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the identity ((x + y)^2 = x^2 + y^2 + 2xy), we have ((x + y)^2 = 25 + 2(6) = 37). Therefore, (x + y = \sqrt{37} = 7).

Multiple choice

If 3x + 2y = 12 and 2x - 3y = 5, find the value of x and y.

  1. x = 2, y = 3

  2. x = 3, y = 2

  3. x = 4, y = 1

  4. x = 1, y = 4
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Solving the given equations simultaneously, we get x = 2 and y = 3.

Multiple choice

If (x^2 + y^2 = 25) and (x - y = 3), find the value of (x + y).

  1. 5

  2. 7

  3. 9

  4. 11
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

From (x - y = 3), we can express (x) as (x = y + 3). Substituting this into (x^2 + y^2 = 25), we get ((y + 3)^2 + y^2 = 25). Expanding and simplifying, we get (2y^2 + 6y + 9 = 25). Subtracting 9 from both sides, we get (2y^2 + 6y - 16 = 0). Factoring, we get ((2y - 4)(y + 4) = 0). Therefore, (y = 2) or (y = -4). If (y = 2), then (x = y + 3 = 2 + 3 = 5). Therefore, (x + y = 5 + 2 = 7). If (y = -4), then (x = y + 3 = -4 + 3 = -1). Therefore, (x + y = -1 + (-4) = -5). Since (x + y) cannot be negative, the correct answer is (7).

Multiple choice

If (x^2 + y^2 = 25) and (x - y = 3), find the value of (x + y).

  1. 4

  2. 6

  3. 8

  4. 10
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Squaring both sides of (x - y = 3), we get ((x - y)^2 = 3^2). Expanding the square, we get (x^2 + y^2 - 2xy = 9). Substituting (x^2 + y^2 = 25) in this equation, we get (25 - 2xy = 9). Rearranging the equation, we get (2xy = 16). Dividing both sides by 2, we get (xy = 8). Now, adding (x + y) and (x - y), we get (2x = x + y + x - y = 10). Therefore, (x + y = 10).

Multiple choice

If (x^3 - y^3 = 27) and (x - y = 3), find the value of (x^2 + y^2).

  1. 18

  2. 24

  3. 30

  4. 36
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Using the identity (a^3 - b^3 = (a - b)(a^2 + ab + b^2)), we can write (x^3 - y^3 = (x - y)(x^2 + xy + y^2)). Substituting (x - y = 3) in this equation, we get (27 = 3(x^2 + xy + y^2)). Dividing both sides by 3, we get (x^2 + xy + y^2 = 9). Now, squaring both sides of (x - y = 3), we get ((x - y)^2 = 3^2). Expanding the square, we get (x^2 + y^2 - 2xy = 9). Adding (x^2 + xy + y^2) and (x^2 + y^2 - 2xy), we get (2(x^2 + y^2) = 18). Therefore, (x^2 + y^2 = 9).