MATHS M.C.Q. [1 Marks Each]

We know that, $1$ decimeter $=10\ cm,$
$1\text{cm}=\frac{1}{10}\text{dm}$
$ 745\text{cm}=\frac{1}{10}\times745\text{dm}$
$=\frac{745}{10}\text{dm}$
$= 74.5$ decimeter $= 74\ dm \ 5\ cm$

$10.5$ litres can be written as $10.5 \times 1000 = 10500\ ml.$

We know that-If a number has more than $2$ factors then it is a composite number. $11$ is a prime number
$\therefore$ it has only $2$ factors $12$ is a composite number
$\therefore$ it has more than $2$ factors $10$ is a composite number
$\therefore$ it has more than $2$ factors $16$ is a composite number
$\therefore$ it has more than $2$ factors $10, 12,16$
$\therefore$ are composite numbers.The least composite number is $10$

 $\Rightarrow $ The given numbers are $9,6,0,5,8$
$\Rightarrow $ To form smallest $6-$digit number,
we have to start number with smallest digit and end with largest digit.
$\Rightarrow $ Here, we can not use 0 as first digit because then number will becomes $5$ digit.
The smallest $6-$digit number $= 1,05,689$

  The given number is $90,08,00,700$
$90,08,00,700 = 90 \times 1,00,00,000 + 08 \times 1,00,000 + 7 \times 100$
So, the name is "Ninety crore eight lakh seven hundred".

We have, $27\ km\ 241\ m\ 319\ mm$
We know that $1\ km = 1000\ m\ 1\ mm={10}-^{3}\text{m}$
$⇒ 27 \times{10}^{3}\ m + 241\ m + 319 × {10}-^{3}\text{m}$
$⇒ 27000\ m + 241\ m + 0.319\ m$
$⇒ 27241.319\ m$

 Number reads Ten thousand eight hundred six. $1$ is at ten thousands place, $0$ at thousands place,
$8$ at hundreds place, $0$ at tens place and $6$ at ones place.
This written in numeral is $10,806$
Missing digit is $0$ in thousands place

Three digit numbers that can be formed from $0, 1$ and $2$ are $102$ and $201.$

$X$ can be subtracted from $L$ and $C$ only.
i.e., $XC = (100 - 10) = 90$

In International System commas are introduced after every $3$ digits starting from right to left side
In Indian System commas are introduced after $3$ digits for one time and then after $2$ digits starting from right to left side.
given the number $648340021$
International System:
$648340021 = 648,340,021$
Indian System :
​​​​​​​$648340021 = 64,83,40,021$

Million Period consists of places: Million, Hundred Millions, Thousand Millions
$→ M, HM, TM$

All numbers from $6,500$ to $7,499$ are rounded off to the nearest thousand as $7,000.$ Therefore, the greatest number is $7,499.$

$1000 - 400 = DC$

$ 31505 < 31550$
$\therefore$ at tens places number $31550$ has $5$ where as same place number $31505$ has $0$ and $0 < 5.$

We know that $1km = 1000m \rightarrow 10 km = 10 \times 1000m = 10000m$
$\therefore \frac{1}{5}$​ of $10km$
$\frac{1}{5}$ of $10000m = \frac{1}{5}\times 10000 = 2000m$ Hence,

By using $10$ dots, we can make a line, a triangle and a rectangle.

Ninety thousand nine hundred nineThis written in numerals gives $90,909$
$9$ is at ten thousands place, $0$ at thousands place, $9$ at hundreds place and so on.

Greatest n digits number is a number with all of its digits equal to $9.$
Because the next greater number will have $′n+1′$ digits. So, greatest 8 digits number is $99999999$

We know $1$ trillion $= 1000$ billions
Hence, the correct answer is option $(c).$

For greatest number, we write the digits in descending order.
So, the greatest $5-$digit number having three different digits is $99987.$

$6,749$ is the greatest number that when rounded off to the nearest hundred will become $6,700.$
$6,650$ is the least number that when rounded off to the nearest hundred will become $6,700.$
$\therefore$ Difference $= 6,749 - 6,650 = 99$

Prime factorization of $10,15$ and $20$
$\begin{array}{c|c}2&10,15,20\\\hline2&5,15,10\\\hline3&5,15,5\\\hline5&5,5,5\\\hline&1,1,1,\end{array}$
$LCM$ of $10, 15$ and $20 = 22 \times 3 \times 5 = 60$

We know that, if $HCF$ of a pair of numbers is $1,$ then they are called coprimes.
So, $8$ and $10$ are not coprimes.

I can be subtracted from $V$ and $X$ only. Thus, $IC$ is wrong.

The greatest $7-$digit number $= 99,99,999$
$\therefore 1 $ added to $99,99,999 = 99,99,999 + 1 = 1,00,00,000$

The given number is $9798745995.$
As per the international place value system, the number can be written as: $9,798,745,995$
Hence, $9798745995$ is written with periods in international system as $9,798,745,995.$

Given digits are $5, 9, 2$ and $6.$
Descending order of the given digits is $9, 6, 5, 2.$
Since, we can use any one digit twice, so the required largest $4-$digit number is $9965$

 Vol. of cylinder $=$ Vol. of cone
$\Rightarrow \pi \times{10}^{2}\times\text{h}=\frac{1}{3}\pi{14}^{2}\times{36}$
$\Rightarrow \text{h}=23.52\text{cm}$

 To form the greatest $5$ digit number we write the given numbers in descending order.
The number we form is $97520.$

 Let $x$ and $y$ be two numbers such that $x > y$
Then $x + y = 10......$
$(i)$ and $x - y = 2......$
$(ii)$ Solving them $2 = 12$
$\Rightarrow x = 6 and 4 + y = 0$
$\Rightarrow y = 4$
Thus, the bigger number is $6.$

$ 85642$ rounded off to nearest thousands $= 86000$

 We know that, if a number is divisible by two coprime numbers, then it is divisible by their product also.
If $7254 * 98$ is divisible by $22,$ then it must be divisible by $2$ and $11$ both.
Since, the last digit is $8,$ so it is divisible by $2.$ Now, for divisibility by $11,$ the difference of sum of digits on even and odd places must be either zero or divisible by $11.$
Sum of digits at even places $= 9 + 4 + 2 = 15$
Sum of digits at odd places $= 8 + * + 5 + 7 = 20 + *$
Required difference $= (20 + *) - 15 = 5 + *$
For divisibility by $11, 5 + * = 11 * = 6$
Hence, the digit at $*$ is $6.$

$\frac{3}{4}=0.75, \frac{5}{6}=0.833,$
$\frac{1}{2}=0.5,$$\frac{2}{3}=0.66,$
$\frac{4}{5}=0.8,\frac{9}{10}=0.9.$
Clearly, $0.8$ lies between $0.75$ and $0.833$
$\frac{4}{5}$ lies between$\frac{3}{4}$and $\frac{5}{6}$

The smallest digit is 0, but the smallest counting number is $1.$

$7$ is at lakhs place followed by $8$ at ten thousands place, $8$ at thousands place, $7$ at hundreds place, $8$ at tens place and $9$ at ones place.
so$ 7, 88,789$ is read as seven lakh eighty eight thousand seven hundred eighty nine.

$1$ Feet $= 0.3048m$

We know that, $1$ centimeter $= 10\ mm, 1$ meter $= 100\ cm\ 1$ meter $= 1000$
mm or $1\ mm =\frac{1}{10}\text{cm} \text { or}1\text{ mm}=\frac{1}{1000}\text{m}$
To convert $4000\ mm\ 400\ cm$ to meters
$4000\ mm \ 400\ cm = 4000\ mm + 400\ cm$
Now, $4000\ mm =\frac{1}{1000}\times{4000}\text{m}=4\text{m }{ 400}\text{cm}=\frac{1}{100}\times{400}\text{m} =4\text{m}$
$4000\ mm\ 400\ cm = 4000\ mm + 400 \ cm = (4+4)\ m = 8\ m$

In $1827$ the last two digits i.e., $27$ is less than $50.$
Hence, the $3rd$ last digit will remain same and the last two digits become zero
Hence, $1827$ when rounded off to the nearest hundred is $1800$
Hence, the correct answer is option $(d).$