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

The smallest counting number is $1.$

On rounding off the given numbers to nearest thousands, we get
$5001 \rightarrow 5000$
$5559 \rightarrow 6000$
$5999 \rightarrow 6000$
$5499 \rightarrow 5000$
Out of $5001$ and $5499,$ the greatest number is $5499,$
which gives $5000$ on rounding off to nearest thousands.

$123,080,603 -$ We can write as One hundred twenty three million eighty thousand six hundred three

Successor of $99999 = 99999 + 1 = 100000$
Predecessor of $99999 = 99999 - 1 = 99998$
Required difference $= 100000 - 99998 = 2$
Hence, the correct answer is option $(c).$

Million $= 1,000,000 = (6$ zeros$)$
$10$ million $= 10,000,000$
Hence the number of zeros after $1$ in $10$ million is $7.$

$128$ can be written using the general form $abc = 100 \times a + 10 \times b + c$
By changing the alphabetic orders of $a, b, c,$
we get $3$ more numbers from $3$ digit number.
Here, $281, 821$ and $182$ are formed from the $3$ digit number $128$

$1$ million $(1,000,000) = 10$ lakh $(10 \times 1,00,000)$

We know that, $HCF$ of an even and an odd number is always an odd number, e.g. $HCF (8, 7) = 1 ($odd$)$

 The descending order of the given numbers $7, 6, 3$ is: $7 > 6 > 3$
We observe that the smallest digit is $3$ and the largest digit is $7$
so the number should start with $7$ and end with $3.$
Thus, the largest number formed is $763$
Hence, the greatest three digit number formed is $763$

 The face value of a digit remains as it is irrespective of the place it occupies in the place value chart.
Thus, the face value of $4$ is always $4$ irrespective of where it may be.

$L.C.M.$ of $8, 16, 40$ and $80 = 80$
$\frac{7}{8}=\frac{70}{80};\frac{13}{16}=\frac{65}{80};\frac{31}{40}=\frac{62}{80}$
Since,$\frac{70}{80}>\frac{65}{80}>\frac{63}{80}>\frac{62}{80}$
So $\frac{7}{8}>\frac{13}{16}>\frac{63}{80}>\frac{31}{40}$
$\therefore \frac{7}{8}$ is the largest.

 All numbers from $550$ to $649$ are rounded off to the nearest hundred as $600.$ Therefore, the smallest number is $550.$

 Prime factorization of $75, 60$ and $105.$
$\begin{array}{c|c}3&75 \\\hline5&25 \\\hline5&5 \\\hline&1\end{array}$
$\begin{array}{c|c}2&60 \\\hline2&30 \\\hline3&15 \\\hline5&5 \\\hline&1\end{array}$
$75 = 3 \times 5 \times 5$
$60 = 3 \times 5 \times 2 \times 2$
$105 = 3 \times 5 \times 1 \times 7$
Common factors of $75, 60$ and $105$ are $3$ and $5.$
Hence, the number of common prime factors of $75, 60$ and $105$ is $2.$

 $10$ crores $= 10,00,00,000$
$1$ billion $= 1,000,000,000$
So, $1$ billion $= 100 \times 10,00,00,000 = 100$ crore

 The $3$ digit numbers formed by $9, 8$ and $7$ are $987, 978, 897, 879, 798,$ and $789.$
The greatest number is $987.$

 We know that, $1$ kilometer $= 1000\ m$
$1$ meter $= 100\ cm$ or $1$ kilometer $= 100,000\ cm$
To convert $3\ km\ 4\ m\ 350\ cm$ to centimeter
$3\ km\ 4\ m\ 350\ cm = 3\ km + 4 \ m + 350\ cm$
$3\ km = 100000 \times 3\ cm = 300000$
$4\ m = 100 \times 4\ cm = 400\ cm$
$\therefore 3\ km\ 4\ m\ 350\ cm = 3\ km + 4\ m + 350\ cm = (300000 + 400 + 350)\ cm$
$= 300750\ cm$

\Whole numbers between $38$ and $68$ are $39, 40, 41,42, 43, 44, 45, 46, 47, 48, 49, 50, 51,52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66$ and $67.$
Total whole numbers between $38$ and $68 = 29$
Alternate Method
Let $a = 38$ and $b = 68$
Then, number of whole numbers between $a$ and $b = b - a -1 [$when $b > a]$
Number of whole numbers between $38$ and $68 = 68 - 38 - 1 = 29$

$ 61000285 < 610000285$ as the first number has $8$ digits while the second number has $9$ digits.

 We know $1$ million $= 10$ lakhs
Hence, the correct answer is option $(b).$

 Any number divisible by $5$ and $6$ will be either $30$ or multiple of $30,$ but $30$ is not divisible by $60.$

 Numeral for five hundred three million eight thousand seven hundred two is $503, 008, 702$

 A number is divisible by $11,$ if the difference of the sum of the digits on even places and odd places is either $0$ or divisible by $11.$
From option $(a),$ we get
Sum of the digits at even places $= 1 + 1 + 0 = 2$
Sum of the digits at odd places $= 1 + 0 +1 + 1 = 3$
Difference $= 3 - 2 = 1$
It is not divisible by $11.$
From option $(b),$ we get
Sum of the digits at even places $= 1 + 1 + 1 = 3$
Sum of the digits at odd places $= 1 + 1 + 1 + 1 = 4$
Difference $= 4 - 3 = 1$
It is not divisible by $11.$
From option $(c),$ we get
Sum of the digits at even places $= 2 + 2 + 2 + 2 = 8$
Sum of the digits at odd places $= 2 + 2 + 2 + 2 = 8$
Difference $= 8 - 8 = 0$
It is divisible by $11.$
From option $(d),$ we get
Sum of the digits at even places $= 3 + 3 + 3 = 9$
Sum of the digits at odd places $= 3 + 3 + 3 + 3 = 12$
Difference $= 12 - 9 = 3$
It is not divisible by $11.$
Hence, $22222222$ is divisible by $11.$

$ 70x$ $1,00,00,000 + 1x$ $1,000 = 70,00,01,000$

The largest three distinct digits are $9, 8$ and $7.$ So, the largest number using these digits can be obtained by arranging the digits in descending order.

The largest three-digit number $= 999$
The largest three-digit number with distinct digits $= 987$
$\therefore$ Difference $= 999 - 987 = 12$

Seventy crore one thousand $= 70 \times 1,00,00,000 + 1 \times 1,000 = 70,00,01,000$

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 $9,798,745,995.$

 We know that, $1$ meter $= 100\ cm, 1$ decimeter $= 10$ centimeter
To convert $23\ dm\ 9\ cm$ to centimeter
now, $23\ dm = 23 \times 10\ cm = 230$ centimeter
$23\ dm\ 9\ cm = 23\ dm + 9\ cm$
$= 230\ cm + 9\ cm$
$= 239\ cm$

We know that
$1$ meter $=100\ cm$
$1$ centimeter $=10$ millimeter
$1$ meter $=1000\ mm$
Given That, we have to convert $3\ m\ 40\ cm$ to millimeter
$40\ cm = 10 × 40\ mm$
$= 400$ millimeter And
$3\ m = 1000 × 3\ mm$
$= 3000$ millimeter
$3\ m\ 40\ cm = 3\ m + 40\ cm$
$= 3000\ mm + 400\ mm$
$= 3400 \ mm$

As per the rules of writing Roman numbers, Only $I, X, C,$ and $M$ can be repeated; $V, L,$ and $D$ cannot be repeated.
Hence, Numerals that can be repeated in Roman system are $I, X$ and $C$

 Numeral for five hundred three million eight thousand seven hundred two, using the international place value system, is $503,008,702$

To get successor of a number, we add $1$ to the given number.
So, the successor of $1$ million $= 1000000 + 1 = 1000001$

$(C)$ Smallest $7-$digit number
$= 1000000$
also $1 + 9999991$
$= 1000000$

given that jessica walks $2\ km$ a day.
Total distance travelled by jessica in two days is $2 \times 2 = 4\ km$
We know that
$1$ kilometer =$1000\ m$
$1\text{m}=\frac{1}{100}$
Given that, we have to convert $4\ km$ into $m$
​​​​​​​$4\ km = 4 \times 1000\ m$
$= 4000\ m$

$\therefore 1$ in $= 0.0254m$
$\therefore 118.1103$ in
$= 0.0254 \times 118.11.3m$
$= 3m$

$ 76, 987$ reads as Seventy six thousand nine hundred eighty seven.
$7$ is at ten thousands place, $6$ at thousands place, $9$ at hundreds place and so on.

Place value of $8 = 8 \times 100 = 800$
Face value of $8 = 8$
Required difference $= 800 - 8 = 792$
Hence, the correct answer is option $(c).$

 In international numbering system
The $1st$ period consists of - ones, tens and hundred.
The $2nd$ period consists of - thousand, $10$ thousand and $100$ thousand.
The $3rd$ period consists of - million, $10$ million and $100$ million.
The $4th$ period consists of - billion, $10$ billion and $100$ billion.
$\therefore 9849475825$ is written with commas as (International System) $= 9,849,475,825$

The digits in the given number $6350947$ are $6, 3, 5, 0, 9, 4$ and $7.$
Keeping the digit $6$ at ten lakh’s place, the rest of the digits fill other places like lakh, ten thousands, thousand, hundreds, tens and ones place by decreasing order of remaining number, i.e. $0, 3, 4, 5, 7, 9.$
Hence, the required smallest number is $6034579.$

The smallest pair of consecutive odd numbers is $1$ and $3.$
Their sum $= 1 + 3 = 4,$ which is divisible by $4.$

$4$ tens $12$ tenths and $25$ hundredths
$=4\times10+\frac{12}{10}+\frac{25}{100}$
$= 40 + 1.2 + 0.25 = 41.45$
$3$ tens $14$ tenths and $45$ thousandths
$=3\times{10}+\frac{14}{10}+\frac{45}{1000}$
$= 30 + 1.4 + 0.045 = 31.445$
Required difference
$= 41.45 - 31.445 = 10.005.$

Greatest $7-$digit number $= 9999999$
On adding $1$ to greatest $7-$digit number,
we get $9999999 + 1 = 10000000 , = 1$ crore

 We know that $1$ meter $=100\ cm$
$1\text{cm}=\frac{1}{100}\text{m}$
Given that we have to convert $1436\ cm$ into $m$
$1436\text{cm}=\frac{1}{100}\times1436\text{m}$
$=1436\text{m}$