Arithmetic aptitude :: Simplification :: General Questions

• Simplification - General Questions
• Simplification - Important Formulas and Facts

BODMAS Rule

B -> Bracket
O -> Of (orders i.e. Powers and Square Roots, Cube Roots, etc)
DM -> Division and Multiplication (start from left to right)
AS -> Addition and Subtraction (start from left to right)

Thus to simplify an expression, solve the expression inside the brackets in the order (),{},[]
then apply i) Of ii) Division and Multiplication (start from left to right) iii) addition and subtraction (start from left to right)

e.g $(5+4)\times 6$

First solve inside ‘brackets’ i.e. $(5+4) = 9$, then $9\times 6 = 54$

Modulus of a real number

Modulus of a real number n is defined as
$|n|=\begin{cases}n, & if &n > 0\\-n, &if & n < 0\end{cases}$
Thus |5| = 5 and |-5| = - |-5| = 5

Virnaculum or Bar

When a question or an expression contains Virnaculum or bar, we first simplify the expression under the Virnaculum and then apply the BODMAS rule.

If a, b, c are integers; $a^{2} + b^{2}= 45$ and $b^{2} + c^{2} = 40$, then the values of a,b and c respectively are :

Answer:

Explanation:

$a^{2}+b^{2}=45$----(1)
$b^{2}+c^{2}=40$----(2)
Subtracting we get, $a^{2}-c^{2}=5$
$\Rightarrow (a+c)(a-c)=5$
$\therefore$ $(a+c)=5$----(3) and $(a-c)=1$----(4)
Sovling, we get : 2a = 6 $\therefore$ a =3
Put value of a in equ.(3), we get c = 2
Put value of a in equ.(1), we get $b^{2}=36$
$\therefore$ b = 6.
So values of a, b, c are (3, 6, 2) respectively

If 0 < a < 1, then the value of $a + \frac{1}{a}$ is :

Answer:

Explanation:

a is possitive and a < 1
$\Rightarrow\frac{1}{a}>1$
$\therefore\left[a+\frac{1}{a}\right]>2$

If (a - b) is 6 more than (c + d) and (a + b) is 3 less than (c - d), then (a - c) is :

Answer:

Explanation:

(a-b) - (c+d) = 6
(c-d)-(a+b) = 3
$\Rightarrow$ (a-c) - (b+d) = 6 and (c-a) - (b+d) = 3
(b+d) = (a-c) - 6 and (b+d) = (c-a)-3
(a-c)-6 = (c-a) - 3 $\Rightarrow$ 2(a-c) = 3
$\Rightarrow$ (a-c) = $\frac{3}{2} = 1.5$

If $\frac{a}{b}=\frac{4}{5}$ and $\frac{b}{c}=\frac{15}{6}$, then $\frac{c^{2}-a^{2}}{c^{2}+a^{2}}$ is

Answer:

Explanation:

$\frac{a}{b}=\frac{4}{5}$ and $\frac{b}{c}=\frac{15}{6}$
$\Rightarrow\left(\frac{a}{b} \times \frac{b}{c}\right)-\left(\frac{4}{5} \times \frac{15}{6}\right)$
$\Rightarrow\frac{a}{c}=\frac{3}{4}$
$\therefore \frac{c^{2}-a^{2}}{c^{2}+a^{2}}=\frac{1-\left(\frac{a^2}{c^2}\right)}{1+\left(\frac{a^2}{c^2}\right)}$
$=\frac{1-\left(\frac{a}{c}\right)^2}{1+\left(\frac{a}{c}\right)^2}$
$=\frac{1-\frac{9}{16}}{1+\frac{9}{16}}$
$=\frac{7/16}{25/16}=\frac{7}{25}$

What is the values of $\frac{\left(P+Q\right)}{\left(P-Q\right)}$ if $\frac{P}{Q}=7$

Answer:

Explanation:

Given  $\frac{P}{Q}=7 $ 

from that P = 7Q,substitute in $\frac{P+Q}{P-Q} $

So,we will get $\frac{7Q+Q}{7Q-Q} =\frac{8Q}{6Q}=\frac{4}{3}$

• Simplification - General Questions