Tag: maths

Quizzes (518) Questions (9324)

Questions Related to maths

Multiple choice maths numbers and place value face value of digit large numbers general form of number

How many times does the digit $1$ appear in numbers from $1$ to $100$?

  1. $18$
  2. $19$
  3. $20$
  4. $21$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\Rightarrow$  1–10 = 2 times

$\Rightarrow$  11–20 = 10 times
$\Rightarrow$  21–30 = 1 time
$\Rightarrow$  31–40 = 1 time
$\Rightarrow$  41–50 = 1 time
$\Rightarrow$  51–60 = 1 time
$\Rightarrow$  61–70 = 1 time
$\Rightarrow$  71–80 = 1 time
$\Rightarrow$  81–90 = 1 time
$\Rightarrow$  91–100 = 2 times
$\Rightarrow$  $Total = 2+10+1+1+1+1+1+1+1+2=21$
$\therefore$   The digit 1 appear in number from 1 to 100 is $21$.

Multiple choice maths numbers and place value face value of digit large numbers general form of number

What is the sum of all integers between $50$ and $350$ which have $1$ as the units digit?

  1. $5880$
  2. $5985$
  3. $6230$
  4. $6800$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sequence $51+61+71...+341$ is an arithmetic progression.

$\Rightarrow$  To find the sum of n terms of an AP we use the formula.
$\Rightarrow$  Here, $n=30,\,a=51$ and $d=10$.
$\therefore$   $S _n=\dfrac{n}{2}[2a+(n-1)d]$

$\therefore$   $S _n=\dfrac{30}{2}[2\times 51+(30-1)10]$

$\therefore$   $S _n=15[102+290]$
$\therefore$   $S _n=15\times 392$
$\therefore$   $S _n=5880$

Multiple choice maths numbers and place value face value of digit large numbers general form of number

A number consists of two digits whose sum is $11$. If $27$ is added to the number, then the digits change their places. What is the number?

  1. $47$
  2. $65$
  3. $83$
  4. $92$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the ten's digit be $x$. Then, unit's digit $= \left(11 - x\right)$.
So, number $= 10x + \left(11 - x\right) = 9x + 11$.
Therefore $\left(9x + 11\right) + 27 = 10 \left(11 - x\right) + x  \Leftrightarrow  9x + 38 = 110 - 9x \Leftrightarrow   18x = 72 \Leftrightarrow   x = 4$.
Thus, ten's digit $= 4$ and unit's digit $= 7$.
Hence, required number $= 47$.

Multiple choice maths numbers and place value face value of digit large numbers general form of number

For $Z _1=\displaystyle \sqrt[6]{\frac{1-i}{1+i\sqrt{3}}}; Z _2=\sqrt[6]{\frac{1-i}{\sqrt{3}+i}}; Z _3=\sqrt[6]{\frac{1+i}{\sqrt{3}-i}}$ which of the following holds good?

  1. $\displaystyle\sum|Z _1|^2=\frac{3}{2}$
  2. $\displaystyle|Z _1|^4+|Z _2|^4=|Z _3|^{-8}$
  3. $\displaystyle\sum|Z _1|^3+|Z _2|^3=|Z _3|^{-6}$
  4. $|Z _1|^4+|Z _2|^4=|Z _3|^8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
$z _1 =\sqrt {\dfrac {1-i}{1+i\sqrt 3}}, z _2=\sqrt {\dfrac {1-i}{\sqrt 3 +i}}, z _3=\sqrt {\dfrac {1+i}{\sqrt 3-1}}$
$z _1 =\sqrt [6]{\dfrac {1-i}{1+\sqrt 3}}=\sqrt [6]{\dfrac {(1-i)(1-i\sqrt 3)}{1+3}}=\sqrt [6]{\dfrac {1(1-\sqrt 3)-i(1+\sqrt 3)}{4}}$
$|z _1|^2 =z _1 \bar {z} _1 =\sqrt [6]{\dfrac {(1-\sqrt 3)}{4}}\times \sqrt [6]{\dfrac {(1-\sqrt 3)+i(1+\sqrt 3)}{4}}$
$|z _1|^2 =\sqrt [6]{\dfrac {(1-\sqrt 3)^2 +(1+\sqrt 3)^2}{16}}$
$|z _1|^2 =\sqrt [6]{\dfrac {8}{16}}=\dfrac {1}{(2) 1/6}$
$z _2 =\sqrt [6]{\dfrac {1-i}{\sqrt 3+i}}=\sqrt [6]{\dfrac {(1-i) (\sqrt 3 -i)}{(3+1)}}=\sqrt [6]{\dfrac {(\sqrt 3-1)-i (1+\sqrt 3)}{4}}$
$|z _2|^2 =z _2 \bar {z} _2=\sqrt [6]{\dfrac {(\sqrt 3-1)-i (1+\sqrt 3)+i(1+\sqrt 3)}{4}}$
$|z _2|^2 =\sqrt [6]{\dfrac {(\sqrt 3-1)^2 +(1+\sqrt 3)^2}{16}}=\sqrt {\dfrac {8}{16}}=\dfrac {1}{(2) 1/6}$
$z _3 =z _3 \bar {z} _3 =\sqrt [6]{\dfrac {(\sqrt 3-1)+(1+\sqrt 3)}{4}\times \dfrac {(\sqrt 3-1)-i (1+\sqrt 3)}{4}}$
$=\sqrt [6]{\dfrac {(\sqrt 3-1)^2 +(1+\sqrt 3)^2}{16}}=\sqrt {\dfrac {8}{16}}=\dfrac {1}{(2)1/6}$
$|z _1|^4 =\dfrac {1}{2^{2/6}}\quad |z _2|^4 =\dfrac {1}{2^{2/6}}$
$|z _3|^8 =\dfrac {1}{2^{4/6}}\ \Rightarrow \ |z _3|^{-8}=2^{4/6}$
$\Rightarrow \ |z _1|^4 +|z _2|^4 =\dfrac {1}{2^{2/6}}+\dfrac {1}{2^{2/6}}=\dfrac {2}{2^{2/6}}=2^{4/6}$
$=|z _3|^{-8}$
so, $\boxed {|z _1|^4 +|z _2|^4 =|z _3|^{-8}}$ as
so, option $(B)$ is right.