Tag: order of operations

Questions Related to order of operations

$24[x+1]-[x^2-24+x]-[2x^2]\div [x^2]$ using BODMAS rule to reduce the expression.

  1. $x^2+23x+46$

  2. $-x^2+23x+46$

  3. $-x^2-23x+46$

  4. $-x^2+23x-46$


Correct Option: B
Explanation:

$24[x+1]-[x^2-24+x]-[2x^2]\div [x^2]$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $24[x+1]-[x^2-24+x]-[2x^2]\div [x^2]$
$=$ $24x+24-x^2+24-x-2$
$=$ $-x^2+46+23x$
$=$ $-x^2+23x+46$

Solve the expression using BODMAS rule: $3x(x-2)+x(x^2\times 2x)-12x$

  1. $2x^5-3x^2-18x$

  2. $2x^5+3x^2-18x$

  3. $2x^5+3x^2+18x$

  4. $-2x^5-3x^2-18x$


Correct Option: B
Explanation:

$3x(x-2)+x(x^2\times 2x)-12x$
We need to follow BODMAS rule.
=> Brackets (parts of a calculation inside brackets always come first).
=> Orders (numbers involving powers or square roots).
=> Division.
=> Multiplication.
=> Addition.
=> Subtraction.
$=$ $3x(x-2)+x(x^2\times 2x)-12x$
$=$ $3x^2-6x+x^3\times2x^2-12x$
$=$ $3x^2-6x+2x^5-12x$
$=$ $2x^5+3x^2-18x$

If $1\le a\le 2$, then $\sqrt { a-2\sqrt { a-1 }  } -\sqrt { a+2\sqrt { a-1 }  } =$.......

  1. $2$

  2. $2\sqrt{a-1}$

  3. $-2$

  4. $1$


Correct Option: C
Explanation:

$\sqrt{a - 2 \sqrt{a - 1}} - \sqrt{a + 2 \sqrt{a - 1}}$

$\Rightarrow \sqrt{(\sqrt{a - 1})^2 - 2 (1) \sqrt{a - 1} + 1^2} - \sqrt{(\sqrt{(a - 1)})^2 + 2 \sqrt{a - 1} + 1^2}$
$\Rightarrow \sqrt{(\sqrt{a - 1} - 1)^2} - \sqrt{(\sqrt{a - 1} + 1)^2}$
$\Rightarrow (\sqrt{a - 1} - 1) - ( \sqrt{a - 1} + 1)$
$= -2$

A number x is decreased by m% and some other number y is increased by m%. If both the results are equal, find m in terms of x and y. Also, find m if $\displaystyle 2x=3y$.


  1. $\displaystyle m=\frac{100\left ( x+y \right )}{x+y};m=10$

  2. $\displaystyle m=\frac{100\left ( x-y \right )}{x-y};m=20$

  3. $\displaystyle m=\frac{100\left ( x+y \right )}{x+y};m=30$

  4. $\displaystyle m=\frac{100\left ( x-y \right )}{x+y};m=20$


Correct Option: D
Explanation:

Given $ \frac {100 - m}{100} \times x = \frac {100 + m}{100} \times y $
$ => 100x - mx = 100y + my $
$ => 100x - 100y = mx + my $
$ => 100 (x-y) = m (x + y) $
$ => m = \frac {100(x-y)}{x+y} $

When , $ 2x = 3y => x = \frac {3y}{2} $
We have, $ m = \frac {100(\frac {3y}{2} -y)}{\frac {3y}{2} +y} $
$ => m =\frac {100(\frac {y}{2})}{\frac {5y}{2}} $
$ => m =\frac {100}{5} = 20 $

 Make b the subject of formula : $\displaystyle a=\frac{1+b^2}{1-b^2}$. 

  1. $\displaystyle b=\sqrt{\frac{a+1}{a-1}}$

  2. $\displaystyle b=\sqrt{\frac{a-1}{a+1}}$

  3. $\displaystyle b=2\sqrt{\frac{a-1}{a+1}}$

  4. $\displaystyle b=2\sqrt{\frac{a+1}{a-1}}$


Correct Option: B
Explanation:

Given, $ a = \frac {1 +{b}^{2}}{1 - {b}^{2}} $
$ => a -a{b}^{2} = 1 +{b}^{2} $
$ => a- 1 = {b}^{2} (1 + a) $
$ =>{b}^{2} = \frac {a-1}{1+a} $


$ x + y -(z - x -[y + z - (x + y - \left { z + x - (y + z + x) \right})])$ is equal to

  1. 3x

  2. 2y

  3. x

  4. 0


Correct Option: C
Explanation:

On simplifying, we get
$x + y -(z - x -[y + z - (x + y - \left { z + x - (y + z + x) \right})])\$ 
$=  x + y -(z - x -[y + z - (x + y - \left { z + x - y - z - x \right})])\$ 
$=  x + y -(z - x -[y + z - (x + y - \left { - y \right})])\$ 
$=  x + y -(z - x -[y + z - (x + y + y)])\$ 
$= x + y -(z - x -[y + z - x - y - y)])\$ 
$= x + y -(z - x -[ z - x - y])\$ 
$= x + y -(z - x - z + x + y)\$ 
$= x + y -(y)\$ 
$= x.$

$\frac{1}{3}(-2p+6q-9r)-\frac{1}{6}(-4p -18q +24r) = $

  1. $-\frac{4}{3}p$

  2. 5q

  3. -7r

  4. 5q-7r


Correct Option: D
Explanation:

$\frac{1}{3}(-2p+6q-9r)-\frac{1}{6}(-4p -18q +24r) = \frac{1}{3}(-2p+6q-9r)-\frac{1}{3}(-2p -9q +12r) $
$=\frac{1}{3} [\left (-2p+6q-9r)-(-2p -9q +12r) \right]$
$=\frac{1}{3} (-2p + 6q - 9r + 2p + 9q -12r)$
$=\frac{1}{3} (15q - 21r)$
$=(5q - 7r)$

$ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left ( y + a -\frac{1}{4}(a+y) \right )\right ]$

  1. $(a+y)^{2}$

  2. $\frac{3a}{16}$

  3. $\frac{9}{16}(a+y)^{2}$

  4. 1


Correct Option: C
Explanation:

$ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left ( y + a -\frac{1}{4}(a+y) \right )\right ]$
= $ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{3} \left (\frac{3}{4}(a+y) \right) \right ]$
= $ \frac{3}{4}(a+y) \left [ y + a - \frac{1}{4}(a+y)\right ]$
= $ \frac{3}{4}(a+y) \left [ \frac{3}{4}(a+y)\right ]$
= $ \frac{9}{16}(a+y)^2$

$-84\times 29+365=$?

  1. $2436$

  2. $2801$

  3. $-2801$

  4. $-2071$

  5. None of these


Correct Option: D
Explanation:

Given Exp. $=-84\times (30-1)+365$
$=-(84\times 30)+84+365$
$=-2520+449$
$=-2071$

In a class, there are 18 boys who are over 160 em tall. If these constitute three-fourths of the boys and the total number of boys is two-third of the total number of students in the class, what is the number of girls in the class?

  1. 6

  2. 12

  3. 18

  4. 24


Correct Option: B
Explanation:

Total number of boys $\displaystyle=18\div\frac{3}{4}=24$

Total number of students $\displaystyle=24\times\frac{3}{2}=36$

$\therefore$ Number of girls $=36-24=12$