Quantitative Aptitude

Number System and Simplification

548 Questions

Number system and simplification questions assess mathematical proficiency in handling fractions, decimals, and complex algebraic expressions. Test takers must simplify surds, calculate reciprocals, and solve intricate number pattern equations. This foundational quantitative aptitude topic is critical for achieving high scores in SSC and banking exams.

Fraction simplificationDecimal operationsSurds and indicesPercentage calculationsComplex number algebra

Number System and Simplification Questions

Multiple choice maths numbers and place value many forms of ten thousand standard form of numbers number names, numerals and place values

Simplify $\displaystyle (27)^{\frac{-2}{3}} \div \displaystyle (64)^{\frac{-2}{3}}$ is---

  1. $\displaystyle \frac{9}{16}$
  2. 16

  3. $\displaystyle \frac{16}{9}$
  4. 9

Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle \frac{(27)^{-2/3}}{ ( 64)^{-2/3}} = \frac{\displaystyle \frac{1}{9}}{ \displaystyle \frac{14}{16}} = \frac{16}{9}$

Multiple choice maths numbers and place value many forms of ten thousand standard form of numbers number names, numerals and place values

The value of $\dfrac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)}$ is?

  1. $324$
  2. $400$
  3. $373$
  4. $1024$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This expression uses the Sophie Germain identity: a^4 + 4b^4 = (a^2 + 2b^2 + 2ab)(a^2 + 2b^2 - 2ab). Here, 324 = 4 * 81 = 4 * 3^4, so b=3. Applying this to each term allows for cancellation of factors between the numerator and denominator, leaving only the ratio of the remaining terms.

Multiple choice maths numbers and place value many forms of ten thousand standard form of numbers number names, numerals and place values

Solve:

$ \displaystyle 9^{\dfrac{3}{2}\div (243)^{-\dfrac{2}{3}}} $  simplifies to

  1. $ \displaystyle 3^{\dfrac{10}{3}} $
  2. <p><span lang="EN-US">${{3}^{{{3}^{\dfrac{13}{3}}}}} $</p>
  3. $ \displaystyle 3^{\dfrac{1}{3}} $
  4. $ \displaystyle 3^{19} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Consider the given expression,


  $ \Rightarrow {{9}^{\dfrac{3}{2}\div {{\left( 243 \right)}^{-\,\dfrac{2}{3}}}}}={{9}^{\dfrac{3}{2}\times {{\left( 243 \right)}^{\dfrac{2}{3}}}}} $

 $ ={{9}^{\dfrac{3}{2}\times {{\left( {{3}^{5}} \right)}^{\dfrac{2}{3}}}}}={{9}^{\dfrac{3}{2}\times {{3}^{\dfrac{10}{3}}}}} $

 $ ={{\left( {{3}^{2}} \right)}^{\dfrac{3}{2}\times {{3}^{\dfrac{10}{3}}}}}={{\left( 3 \right)}^{3\times {{3}^{\dfrac{10}{3}}}}} $

 $ ={{3}^{{{3}^{1+\dfrac{10}{3}}}}}={{3}^{{{3}^{\dfrac{13}{3}}}}} $


Hence, this is the answer. 

Multiple choice maths numbers and place value many forms of ten thousand standard form of numbers number names, numerals and place values

If $0.00044=$$\displaystyle 4\cdot 4\times 10^{n}$ then, find the value of $ n$.

  1. $4$
  2. $5$
  3. $-4$
  4. $-5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\displaystyle 0\cdot 00044= \frac{44}{100000}= \frac{4\cdot 4\times10^{1}}{10^{5}}$
$=\displaystyle 4\cdot 4\times 10^{-4}$
Now
$\displaystyle 4\cdot 4\times 10^{-4}=4\cdot 4\times 10^{n}$
$\displaystyle \therefore n=-4$

Multiple choice maths numbers and place value many forms of ten thousand standard form of numbers number names, numerals and place values

The distance of the sun from the earth is $1,49,60,00,00,000$ m. Express it in standard form.

  1. $\displaystyle 1\cdot 496\times 10^{-10}$
  2. $\displaystyle 1\cdot 496\times 10^{-11}$
  3. $\displaystyle 1\cdot 496\times 10^{10}$
  4. $\displaystyle 1\cdot 496\times 10^{11}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$1,49,60,00,00,000=$$\displaystyle 1\cdot 496\times 10^{11}$ m

Multiple choice applications of quadratic equations solving (simple) problems word problems based on quadratic equations quadratic equation maths

What is the value of $x$ in the equation $\displaystyle \sqrt{1+\sqrt{1-\frac{2176}{2401}}}=1+\frac{x}{7}$?

  1. 0

  2. 1

  3. 2

  4. 3

Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$\displaystyle \sqrt{1+\sqrt{1-\cfrac{2176}{2401}}}=1+\cfrac{x}{7}$
$\Rightarrow \sqrt{1+\sqrt{\cfrac{2401-2176}{2401}}}=1+\cfrac{x}{7}$
$\Rightarrow \sqrt{1+\sqrt{\cfrac{225}{2401}}}=1+\cfrac{x}{7}$
$\displaystyle \Rightarrow \sqrt{1+\cfrac{\sqrt{225}}{\sqrt{2401}}}=1+\cfrac{x}{7}$                                                                   
$\displaystyle \Rightarrow \sqrt{1+\cfrac{15}{49}}=1+\cfrac{x}{7}$
$\Rightarrow \sqrt{\cfrac{64}{49}}=1+\cfrac{x}{7}$
$\Rightarrow \cfrac{8}{7}=1+\cfrac{x}{7}$
$\Rightarrow \cfrac{x}{7}=\cfrac{8}{7}-1$
$\Rightarrow \cfrac{x}{7}=\cfrac{1}{7}$
$\Rightarrow x=\cfrac{1}{7}\times 7$
$\Rightarrow x=1$