Tag: maths

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Questions Related to maths

Multiple choice maths numbers in indian and international systems indian system of numeration numbers to 1 million formation of large numbers

Unit's digit of the number ${36^{1001}} \times$ ${7^{1002}} \times$${13^{1003}}$ is 

  1. $1$
  2. $3$
  3. $7$
  4. $8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Unit's digit of $36^{1001}$ will be $6$

$7^{1002}=(7^2)^{501}=(49^4)^{125}\times 49$
Unit's digit of $(49^4)$ is $1$. So, unit's digit of $(49^4)^{125}$ will be $1$ and thus, unit's digit of $7^{1002}$ will be $9$.
$13^{1003}=(13^2)^{501}\times 13=(169^4)^{125}\times 169\times 13$
Unit's digit of $169^4$ is $1$. So, unit's digit of $13^{1003}$ will be $7$.
Hence, unit's digit of $36^{1001}\times 7^{1002}\times 13^{1003}$ will be $8$.

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

Which  of the following is correct ?

  1. $2 + 3i > 1 + 4i$
  2. $2 + 2i > 3 + 3i$
  3. $5 + 8i > 5 + 7i$

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

Option $A$

$2+3i>1+4i$
$2^2+3^2>1^2+4^2$
$4+9>1+16$
$13>17$

It is not correct.

Option $B$
$2+2i>3+3i$

$2^2+2^2>3^2+3^2$
$4+4>9+9$
$8>18$

It is not correct.


Option $C$
$5+8i>5+7i$

$5^2+^82>5^2+7^2$
$25+64>25+49$
$89>74$

It is correct.

Hence, this is the answer.

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

Find the value of $\begin{vmatrix} 2+i & 2-i \ 1+i & 1-i \end{vmatrix}$ if $i^2=-1$.

  1. A complex quantity

  2. real quantity
  3. $0$

  4. cannot be determined
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\left| \begin{matrix} 2+i & 2-i \ 1+i & 1-i \end{matrix} \right| \ =(2+i)(1-i)-(2+i)(1+i)\ =2-2i+i-{ i }^{ 2 }-2-2i-i-{ i }^{ 2 }\ =-4i+2$

So it is complex quantity

Multiple choice maths complex numbers and linear inequations identities of complex numbers powers of imaginary unit i algebra of complex numbers

$\displaystyle i+\frac{1}{i}=$

  1. $1$
  2. $-1$
  3. $0$
  4. $2i$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $Z= i+\dfrac{1}{i}$


Mutiplying numerator and denominator by i. We get,

          $=i+\dfrac{i}{i^{2}}$

          $=i+\dfrac{i}{-1} \quad \dots (i^2=-1)$

          $=i-i$

      $Z=0$

Hence, 

$i+\dfrac{1}{i}=0$.