Tag: face value of digit

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Questions Related to face value of digit

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

The digit in the units place of the number $2015!+3^{7886}$ is

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

$2015! + 3^{7886}$

$2015! \Rightarrow$ Unit digit $\Rightarrow \underline { 0 }$

$3^{7886}=3^{4n+2}$, So unit digit $\Rightarrow 9$

$\therefore 0+9 \Rightarrow \underline{9}$.

$\therefore$ Digit at unit place $\Rightarrow 9$ 

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

A man has 480 rupees in the denominations of one-rupee five rupee notes and ten-rupee notes The number of notes of each denomination is equal. What is the total number of notes that he has?

  1. 90

  2. 75

  3. 45

  4. 60
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Let the numbers Of all kind is $x$

Then amount of one rupee note=$1\times x=x$
And  amount of Five rupee note=$5\times x=5x$
And  amount of Ten  rupee note=$10\times x=10x$

Then $x+5x+10x=480$
Or $16x=480$  or $x=30$

Then total numbers of note =$3\times 30=90$

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.