Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

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

Find the unit digit of ${3^{46}} + 125 \times 436 + 256 \times {7^{345}}$

  1. $1$
  2. $3$
  3. $7$
  4. $9$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Find the unit digit of each term: 3^46 has a cycle of 4 for powers (46 mod 4 = 2, so 3^2 ends in 9). The middle term ends in 0 because 125 * 436 is an even number times 5. The last term ends in 6 * 7^345 (345 mod 4 = 1, so 7^1 ends in 7, and 6 * 7 ends in 2). Summing the unit digits: 9 + 0 + 2 = 11, which ends in 1.

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

Find the last two digits of $3^{1997}$.

  1. $67$
  2. $63$
  3. $80$
  4. $56$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

This is same as asking what is remainder when $3^{1997}\div 100$
$3^{4}\equiv 81  mod  100$
$3^{8}\equiv 61  mod  100$
$3^{12}\equiv 41  mod  100$
$3^{16}\equiv 21  mod  100$
$3^{20}\equiv 1  mod  100$


Now, $3^{40}, 3^{60}, 3^{80}, 3^{100}, ...., 3^{1980}$ all are $\equiv 1  mod  100$

We know $3^{16}\equiv 21  mod  100$

$3^{17}\equiv 21\times 3  mod  100$

$3^{17}\equiv 63  mod  100$

$\therefore 3^{1997}\equiv 3^{1980}\times 3^{17}$

since, $3^{1980}\equiv 1  mod  100$

and $3^{17}\equiv 63  mod  100$

$\therefore 3^{1997}\equiv 63  mod  100$

$\therefore $ Last two digit is 63

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 ${\left( {{{27}^{\tfrac{{ - 2}}{3}}}} \right)^{\tfrac{1}{2}}} \times {\left( {{{64}^{\tfrac{1}{3}}}} \right)^2} \times {\left( {{{81}^{\tfrac{{ - 3}}{2}}}} \right)^{\tfrac{1}{6}}}$

  1. $\dfrac{1}{9}$
  2. $\dfrac{16}{9}$
  3. $\dfrac{2}{9}$
  4. $-\dfrac{16}{9}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
${\left( {{{27}^{\tfrac{{ - 2}}{3}}}} \right)^{\tfrac{1}{2}}} \times {\left( {{{64}^{\tfrac{1}{3}}}} \right)^2} \times {\left( {{{81}^{\tfrac{{ - 3}}{2}}}} \right)^{\tfrac{1}{6}}}$

$\displaystyle =\left(\dfrac{1}{(27)^{\tfrac{2}{3}}}\right)^{\tfrac{1}{2}}\times (4^{3\times \tfrac{1}{3}})^{2}\times \left(\dfrac{1}{(81)^{\frac{3}{2}}}\right)^{\tfrac{1}{6}}$

$\displaystyle =\left(\dfrac{1}{27}\right)^{\tfrac{2}{3}\times \tfrac{1}{2}}\times (4)^{2}\times \left(\dfrac{1}{81}\right)^{\tfrac{3}{2}\times \tfrac{1}{6}}$

$\displaystyle =\left(\dfrac{1}{3^{3}}\right)^{\tfrac{1}{3}}\times (4)^{2}\times \left(\dfrac{1}{3^{4}}\right)^{\tfrac{1}{4}}$

$\displaystyle =\frac{1}{3}\times 16\times \frac{1}{3}$

$=\dfrac{16}{9}$
Multiple choice maths numbers and place value many forms of ten thousand standard form of numbers number names, numerals and place values

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

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

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

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. $\displaystyle 4\cdot 87\times 10^{-12}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

=$ \dfrac{487}{10^{15}}$ =$\displaystyle \frac{4\cdot 87\times 10^{2}}{10^{15}}$
=$\displaystyle 4\cdot 87\times 10^{-13}$

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

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. $\displaystyle 1\leqslant k\leqslant 10$
Reveal answer Fill a bubble to check yourself
C Correct answer
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, $1\leq k <10$