Tag: numbers and place value

Questions Related to numbers and place value

The distance of the earth from the sun is 149,000,000 km. In scientific notation the distance is:

  1. $\displaystyle 149\times 10^{6}$ km

  2. $\displaystyle 14.9\times 10^{7}$ km

  3. $\displaystyle 1.49\times 10^{8}$ km

  4. $\displaystyle 0.149\times 10^{9}$ km


Correct Option: C
Explanation:

$\displaystyle 149,000,000 km=149\times1000000=1.49\times100000000=1.49\times 10^{8}km$

Solve:

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

  1. $ \displaystyle 3^{\dfrac{10}{3}} $

  2. ${{3}^{{{3}^{\dfrac{13}{3}}}}} $

  3. $ \displaystyle 3^{\dfrac{1}{3}} $

  4. $ \displaystyle 3^{19} $


Correct Option: B
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. 

The standard form of $0.000000000000487$ is ______

  1. $\displaystyle 4\cdot 87\times 10^{-13}$

  2. $\displaystyle 4\cdot 87\times 10^{-14}$

  3. $\displaystyle 4\cdot 87\times 10^{-15}$

  4. $\displaystyle 4\cdot 87\times 10^{-12}$


Correct Option: A
Explanation:

$0.000000000000487=\displaystyle \frac{487}{1000000000000000}$

=$ \dfrac{487}{10^{15}}$ =$\displaystyle \frac{4\cdot 87\times 10^{2}}{10^{15}}$

=$\displaystyle 4\cdot 87\times 10^{-13}$

The standard form of $8,60,00,00,00,00,000$ is

  1. $\displaystyle 8\cdot 6\times 10^{13}$

  2. $\displaystyle 8\cdot 6\times 10^{-13}$

  3. $\displaystyle 8\cdot 6\times 10^{12}$

  4. $\displaystyle 8\cdot 6\times 10^{14}$


Correct Option: A
Explanation:

$8,60,00,00,00,00,000=$$\displaystyle 8\cdot 6\times 10^{13}$

A number is said to be in the standard form when it is written as $\displaystyle k\times 10^{n}$ where $n$ is an integer and:

  1. $\displaystyle 1< k< 10$

  2. $\displaystyle 1< k\leqslant 10$

  3. $\displaystyle 1\leqslant k< 10$

  4. $\displaystyle 1\leqslant k\leqslant 10$


Correct Option: C
Explanation:

$k\times10^n$


If $n$ is an integer, then $k$ should lie between $1$ and $10$.

Lets consider the equalties, $k$ cannot be equal to $10$, if it is then we can update $k = 1$ and increment $n$ by $1$.

So, $1\leq k <10$

The usual form of $\displaystyle 5\times 10^{-8}$ is

  1. $0.000005$

  2. $0.00000005$

  3. $0.0000005$

  4. $0.000000005$


Correct Option: B
Explanation:

$\displaystyle 5\times 10^{-8}=0\cdot 00000005$

The usual form of $\displaystyle 4\cdot 56\times 10^{-5}$ is:

  1. $0.0000456$

  2. $0.00000456$

  3. $0.000456$

  4. $456000$


Correct Option: A
Explanation:

$\displaystyle 4\cdot 56\times 10^{-5}=0\cdot 00000456$

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

  1. $4$

  2. $5$

  3. $-4$

  4. $-5$


Correct Option: C
Explanation:

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

Find the value of $n$ such that $502000000=$$\displaystyle 5\cdot 02\times 10^{n}$.

  1. $5$

  2. $7$

  3. $-8$

  4. $8$


Correct Option: D
Explanation:

$\displaystyle 50,20,00,000=5\cdot 02\times 10^{8}$
Now $\displaystyle 5\cdot 02\times 10^{8}= 5\cdot 02\times 10^{n}$
$\displaystyle \therefore n= 8$

The standard form of $\displaystyle \frac{1}{10000000}$ is:

  1. $\displaystyle 1\times 10^{-7}$

  2. $\displaystyle 0\cdot 1\times 10^{-7}$

  3. $\displaystyle 1\times 10^{7}$

  4. $\displaystyle 1\times 10^{-6}$


Correct Option: A
Explanation:

$\displaystyle \frac{1}{10000000}= \frac{1}{10^{7}}= 1\times 10^{-7}$