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Questions Related to maths

Multiple choice maths 5-digit numbers expanded form introduction to numbers and number systems numbers in general form

Expanded form of 78.059 is

  1. $78\, +\, \displaystyle \frac {5}{10}\, +\, \displaystyle \frac {9}{100}$
  2. $70\, +\,8\, +\,0\, +\, \displaystyle \frac {5}{100}\, +\, \displaystyle \frac {9}{1000}$
  3. $70\, +\,8\, +\, \displaystyle \frac {5}{10}\, +\, \displaystyle \frac {9}{100}$

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

$\Rightarrow$  The given number is $78.059$

$\Rightarrow$  Expanded form = $70+8+0+\dfrac{5}{100}+\dfrac{9}{1000}$

Multiple choice maths 5-digit numbers expanded form introduction to numbers and number systems numbers in general form

If $(2t - 3)(? + 6 + 9) = 8t$$^3$ $- 27$, then ? will be replaced by 

  1. $-4t$$^2$
  2. $5t$$^2$
  3. $4t$$^2$

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

$8t^3$ $- 27 = (2t)$$^3$ $- (3)$$^3$
$= (2t - 3)$$[(2t)$$^2$ $+ 2t \times 3 + 3$$^2$]
$= (2t - 3)(4t$^2$ + 6t + 9)$
$\because$ $(2t - 3)(4t$$^2$ $+ 6t + 9)$
$= (2t - 3)(a + 6t + 9)$
$\Rightarrow$ $4t$$^2$ $+ 6t + 9 = (a + 6t + 9)$
$\therefore$ $a = 4t$$^2$

Multiple choice maths 5-digit numbers expanded form introduction to numbers and number systems numbers in general form

Eighteen lakh nineteen thousand eight hundred eighteen is _________.

  1. $1,81,81,818$
  2. $18,19,818$
  3. $18,17,818$

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

There are three special numbers in the Indian numbering system – lakh, crore, and arab. A lakh is 1,00,000, 1 , 00 , 000 , a crore is equal to a hundred lakhs and is expressed as 1,00,00,000. An arab is 100 crores and is expressed as 1,00,00,00,000.

So, eighteen lakh nineteen thousand eight hundred eighteen is,
18,19,818
Hence option B is the correct answer

Multiple choice maths 5-digit numbers expanded form introduction to numbers and number systems numbers in general form

The sum of the reciprocals of $\dfrac {x+3}{x^2+1}$ and $\dfrac {x^2-9}{x^2+3}$ is

  1. $\dfrac {x^3+2x^2-x}{x^2-9}$
  2. $\dfrac {x^3-2x^2+x}{x^2-9}$

  3. 1

  4. 0
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Consider the sum of the reciprocals of ,

$\dfrac{x+3}{{{x}^{2}}+1}$ and $\dfrac{{{x}^{2}}-9}{{{x}^{2}}+3}$


  $ \Rightarrow \dfrac{{{x}^{2}}+1}{x+3}+\dfrac{{{x}^{2}}+3}{{{x}^{2}}-9}=\dfrac{{{x}^{2}}+1}{x+3}+\dfrac{{{x}^{2}}+3}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{\left( {{x}^{2}}+1 \right)\left( x-3 \right)+{{x}^{2}}+3}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{{{x}^{3}}-3{{x}^{2}}+x-3+{{x}^{2}}+3}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{{{x}^{3}}-2{{x}^{2}}+x}{\left( x-3 \right)\left( x+3 \right)} $

 $ \Rightarrow \dfrac{{{x}^{3}}-2{{x}^{2}}+x}{\left( x^2-9 \right)} $