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

$3$ is not a positive multiple of $12$ as it is smaller than $12.$
Rest others are multiples of $12.$

$100 = 1 \times 2 \times 2 \times 5 \times 5$
$101 = 1 \times 101$
Since, $100$ is a composite number and $101$ is a prime number.
Thus, their $LCM = 100 \times 101 = 10100$
Hence, the correct answer is option $(a).$

Required number $= H.C.F.$ of $245 - 5 = 240$ and $1029 - 5 = 1024$
$H.C.F.$ of $240 = 2 \times 2 \times 2 \times 2 \times 3 \times 5$
$1024 = 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2$
$= 2 \times 2 \times 2 \times 2 = 16$
Therefore,$16$ is the largest number which divides $245$ and $1024$ leaving remainder $5$ in each

Let proper fraction $=\frac{3}{2}$
$\therefore$ Resulting fraction $=\frac{2+1}{3+1}=\frac{3}{4}$
Hence $\frac{2}{3}<\frac{3}{4}$
$\frac{3}{5}<\frac{3+1}{5+1}$
$\Rightarrow\frac{3}{5}<\frac{4}{6}$ etc.

$2 \times 11 \times 13 = 286$

$2$ is the least prime number. It is the only even prime number.

First six multiples of $17 = 17, 34, 51, 68, 85$ and $102.$

$12 \times 1 = 12$
$12 \times 3 = 36$
$\therefore 12$ and $36$ are multiples,
while $4$ is a factor of $12$
So, options $A$ and $B$ are both correct.

The given expression can be simplified as follows:
$:\frac{(0.0036) (2.8)}{(0.04) (0.1) (0.003)}$ $=\frac{0.0036\times2.8}{0.04\times0.1\times0.003}=\frac{0.01008}{0.000012}=840$
Hence, the value of $:\frac{(0.0036) (2.8)}{(0.04) (0.1) (0.003)}$ is $840$

No. of factors for $50$
$50 =5\times 5\times 2=5^2\times 2^1$
Total no. of factors are $=(2+1)\times (1+1)=6$

$36 = 2 \times 2 \times 3 \times 3$
$72 = 2 \times 2 \times 2 \times 3 \times 3$
$\therefore L.C.M$ of $36$ and $72 = 2 \times 2 \times 2 \times 3 \times 3 = 72$

A prime number (or a prime) is a natural number greater than
$1$ that has no positive divisors other than $1$ and itself.
A natural number greater than $1$ that is not a prime
number is called a composite number.

Multiples of $8 = 8, 16, 24, 32, ..$
Multiples of $12 = 12, 24, 36, 48...$
The first common multiple will be $24$
And the next common multiples will be multiples of $24$
Hence, first four common multiples of $8, 12$ are $24, 48, 72, 96$

$p = 4, 4 ÷ 2 = 2$
$p + 2 = 4 + 2, 6 ÷ 2 = 3$
$p + 4 = 4 + 4, 8 ÷ 2 = 4$
Product is divisible by $2.$
Therefore, $C$ is the correct answer.

Since on dividing by $3$ the remainder is $1,$ the sum of digits of the number must add upto a number, dividing which by $3$
we get remainder $1$ Since on dividing by $5$ remainder is $3,$ unit place digit has to be either $3$ or $8$ Since on dividing by $9$
remainder is $7,$ sum of digits should give remainder $7$ when divided by $9$
Only options $(A), (B)$ satisfy these criteria Since $(A)$ is bigger, we divide it by $7$ and find remainder which turns out to be $5$.

In the fraction, numerator means the upper part of fraction.
So, the numerator of the fraction $\frac{4}{7}$ is $4.$
Hence, the answer is $4.$

$42 = 1 \times 2 \times 3 \times 7.$
The factors are $1, 2, 3, 6, 7, 14, 21, 42.$

$36 = 2 \times 2 \times 3 \times 3$
$72 = 2 \times 2 \times 2 \times 3 \times 3$
$\therefore L.C.M$ of 36 and $72 = 2 \times 2 \times 2 \times 3 \times 3 = 72$

$(a)\ 4th$ multiple of $52 = 52 \times 4 = 208$
$(b)\ 8th$ multiple of $37 = 37 \times 8 = 296$
$(c)\ 5th$ multiple of $25 = 25 \times 5 = 125$
$(d)\ 7th$ multiple of $50 = 50 \times 7 = 350$
So, $350$ is the greatest value amongst all.

Two fractions are called as unlike fractions, if the denominators of those fractions are different.
For example: Consider $\frac{1}{5}$ and $\frac{3}{6}$ here both the fractions have different denominators, so they are unlike fractions.
Hence, fractions with different denominators are called unlike fractions.

Multiples of $3 = 3, 6, 9, 12, 15, 18..$
Multiples of $4 = 4, 8, 12, 16, 20..$
Multiples of $6 = 6, 12, 18, 36..$
The first common multiple will be $12$
And the next common multiples will be multiples of $12$
Hence, first four common multiples of $3, 4, 6$ are $12, 24, 36, 48$

 We know that the common factor of two consecutive odd numbers is $1.$
Thus, $HCF$ of two consecutive odd numbers is $1.$

$2, 3$ are primes.
$\therefore $ Each number has no factor other than $11$ and itself.
$\therefore $ Their $LCM$ is the product of the numbers.
$\therefore LCM$ of $1, 2, 3 = 2 \times 3 = 6.$
Also here $1 + 2 + 3 = 6.$
Answer- Option A and Option $C.$

$12 \times 1 = 12$
$12 \times 3 = 36$
$\therefore 12$ and $36$ are multiples, while $4$ is a factor of $12$
So, options $A$ and $B$ are both correct.

Fractions that are greater than $0$ but less than $1$ are called proper fractions.
In proper fractions, the numerator is less than the denominator.
When a fraction has a numerator that is greater than or equal to the denominator, then the fraction is an improper fraction.
An improper fraction is always $1$ or greater than $1.$
Now looking at options
$\frac{4}{3}=1.33>1$
$\frac{3}{2}=1.5>1$
$\frac{5}{3}=1.66>1$
$\frac{7}{11}=0.63>1$
So $\frac{7}{11}$is Not a Improper fraction.

Since we need $2$ as the remainder, we will subtract $2$ from each of the numbers.
$167 - 2 = 165$
$134 - 2 = 132$
Now, any of the common factors of $165$ and $132$ will be the required divisor.
On factorising:
$165 = 3 \times 5 \times 11$
$132 = 2 \times 2 \times 3 \times 11$
Their common factors are $11$ and $3.$
So, $3 \times 11 = 33$ is the required divisor.

Two numbers are $3 \times HCF$ and $4 \times HCF$
i.e. $3 \times 4 = 12$ and $4 \times 4 = 16$
$LCM$ of $12$ and $16 = 48$

Multiple of $6$ like $6, 12, 18,24, 30$ and they can all be divided evenly by $3.$
So option $C$ is correct.

$6th$ multiple of $7 = 42$
$10th$ multiple of $7 = 70$
Now, $x = 2 × (70 − 42) = 2 × 28 = 56$

A unit fraction is a rational number written as a fraction in which numerator is $1$ and denominator is a positive integer.
Example- $\frac{1}{2},\frac{1}{3},\frac{1}{5}$​ etc.

$LCM$ of $9,10,129 = 3 \times 310 = 2 \times 512 = 2 = 2 \times 2 \times 3$
$LCM (9,10,12) = 2 \times 2 \times 3 \times 3 \times 5 = 180$
$180$ is also multiply of $18,36,45.$
If $v$ is $18,36$ and $45$ than $LCM$ of all the number would be $180$ but its $540$

$(a) 4th$ multiple of $52 = 52 \times 4 = 208$
$(b) 8th$ multiple of $37=37 \times 8 = 296$
$(c) 5th$ multiple of $25=25 \times 5 = 125$
$(d) 7th$ multiple of $50 = 50 \times 7 = 350$
So, $350$ is the greatest value amongst all.

Taking out the factors of the given numbers, $12 = 2 \times 2 \times 3$
$24 = 2 \times 2 \times 2 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
$\therefore LCM$ of $12, 24$ and $36 = 2 \times 2 \times 2 \times 3 \times 3 = 72$

 We know:
$HCF \times LCM =$ Product of two numbers
$\because 16 \times LCM = 3,072$
$\therefore LCM = 3,07216=192$

The decimal expansion of a rational number always either terminates after a finite number of digits or begins to repeat the same finite sequence of digits over and over.
Moreover, any repeating or terminating decimal represents a rational number.
So,Decimal expansion of a rational number cannot be
Answer $(A)$ Non-terminating and non-recrring

Factors of the given numbers are, $4 = 2 \times 2$
$9 = 3 \times 3$
$\therefore LCM$ of $4$ and $9 = 2 \times 2 \times 3 \times 3 = 36$

Let the divisor be $a$ and remainder be $r$
Let $31513 = am + r$ and $34369 = an + r$
Then, $a (n - m) = 2856 = 24 \times 119 = 12 \times 238 = 8 \times 357 = 6 \times 476 = 4 \times 714$
All three digit numbers give same remainder $= 97$

Factors of $24$ are $1, 2, 3, 4, 6, 8, 12, 24$

Required smallest number $= LCM$ of $(11, 28, 36, 45) + 3 = 13,860 + 3 = 13,863$

Factor of $1955 = 5 × 17 × 23$
$1, 5, 17, 23$ are not even, so there are no even factors.

A decimal is a fractional number and is indicated by digits after a period which is called a decimal point.
Tenths have one digit after the decimal point. The decimal $0.8$ is pronounced as eight tenths.
Hundredths have two digits after the decimal point.
The decimal $0.06$ is pronounced as six hundredths.
Thousandths follow a similar pattern.
They have three digits after the decimal point.
The decimal $0.005$ is pronounced as five thousandths.
Hence, five thousandths is $0.005.$

$75\%$ can be written in regular fraction as $75\%=\frac{75}{100}$
On reducing the fraction, we get
$\frac{75}{100}=\frac{3}{4}$

Denominator is the number at the bottomOut of the given options, only $\frac{9}{4}$ has denominator as $4.$
Hence, the answer is $\frac{9}{4}.$

A prime number is a whole number greater than $1$ whose only factors are $1$ and itself.
A factor is a whole numbers that can be divided evenly into another number.
The first few prime numbers are $2, 3, 5, 7, 11, 13, 17, 19, 23$ and $29.$
The next prime number after $43$ is $47,$So option C is the correct answer.

 Consider the given expression.
$\Bigg[\bigg(-2\frac{3}{4}\bigg)-\bigg(-1\frac{3}{4}\bigg)\Bigg]+\Bigg[\bigg(-2\frac{3}{4}\bigg)-\bigg(-1\frac{3}{4}\bigg)\Bigg]+ ....$ upto $30$ times
Sum $=\Bigg[\bigg(\frac{-11}{4}\bigg)-\bigg(\frac{-7}{4}\bigg)\Bigg]+\Bigg[\bigg(\frac{-11}{4}\bigg)-\bigg(\frac{-7}{4}\bigg)\Bigg]+ ....$ upto $30$ times
$=\big[-1\big]+\big[-1\big]+....$ upto $30$ times
$=-30$
Hence, the value of the expression is $-30.$

 Factors of $1955 = 5 \times 17 \times 23$
Number of factors are $3.$

Sum of the given digits $= 1 + 5 + 4 + 8 = 18$
Since $18$ is a multiple of $3,$ the required digit is $0.$

 The place value of every one-digit number is the same as and equal to its face value.
$(i)$ Place value and face value of $1, 2, 3, 4, 5, 6, 7, 8$ and $9$ are $1, 2, 3, 4, 5, 6, 7, 8$ and $9$ respectively.
$(ii)$ The place value of zero $(0)$ is always $0.$ As, in $105, 350, 42017, 90218$ the place value of $0$ in each number is $0.$
So option $A$ is the correct answer.

Improper fraction is a fraction in which the numerator is greater than the denominator, such as $\frac{3}{2}$
Hence, $\frac{15}{14}$ is an improper fraction.

 Given, $0.8\times\frac{\frac{7}{12}}{\frac{5}{24}}=0.8\times\frac{7}{12}\times\frac{24}{5}=0.8\times\frac{14}{5}$
$=\frac{8}{10}\times\frac{14}{5}$
$=\frac{56}{25}$
$=2\frac{6}{25}$

Sum of the given digits $= 1 + 5 + 4 + 8 = 18$
Since $18$ is a multiple of $3,$ the required digit is $0.$

Numerator of a fraction is the upper number.
So, the numerator of the fraction $\frac{5}{6}$ is $5.$
Hence, the answer is $5.$

Taking out the factors of the given numbers,$12 = 2 \times 2 \times 3$
$24 = 2 \times 2 \times 2 \times 3$
$36 = 2 \times 2 \times 3 \times 3$
$\therefore LCM$ of 12, 24 and $36 = 2 \times 2 \times 2 \times 3 \times 3 = 72$

The least number divisible by $15, 20, 24, 32,$ and $36$ can be found by taking their $LCM$ as:

$\therefore LCM$ of $15, 20, 24, 32$ and $36 = 2 \times 2 \times 2 \times 2 \times 2 \times 3 \times 3 \times 5 = 1,440$
Hence, $1,440$ is the least number that is divisible by $15, 20, 24, 32$ and $36.$

 Given that
we have to find the value of given expression
$​​​​\frac{1}{1+\frac{1}{3}}-\frac{1}{1+\frac{1}{2}}$
$=\frac{1}{\frac{1+3}{3}}-\frac{1}{\frac{1+2}{2}}$
$=\frac{3}{4}-\frac{2}{3} $
$=\frac{3\times3-2\times4}{12}$
$=\frac{1}{12}$

 Decimals having the same number of decimal places are called like decimals i.e.
decimals having the same number of digits on the right of the decimal point are known as like decimals.
For example, $16.37$ and $18.97$ are like decimals as both of these decimal numbers are written up to $2$ places of decimal.
Hence, $16.37$ and $18.97$ are like decimal fractions.

$ 16 = 2 \times 2 \times 2 \times 2$ The factors are $1, 2, 4, 8, 16.$

 Factors of the given numbers are,$4 = 2 \times 2$
$9 = 3 \times 3$
$\therefore LCM$ of $4$ and $9 = 2 \times 2 \times 3 \times 3 = 36$

By getting factors of $24 = 2 \times 2 \times 2 \times 3$
The mean is the average of the numbers. It is easy to calculate: add up all the numbers,
then divide by how many numbers there are.
Mean of the factors will be = $\frac{2 + 2 + 2 + 3}{4} = \frac{9}{4}$
So, the correct answer is option $B.$

 Example:
$HCF$ of $8$ and $21$ is $1.$
$HCF$ of $6$ and $9$ is $3.$
$HCF$ of $9$ and $36$ is $9.$
So there is no fixed number that can be the $HCF$ of an even number and an odd number.

The $LCM$ of two co-prime numbers is their product.

$6th$ multiple of $7 = 42$
$10th$ multiple of $7 = 70$
Now, $x = 2 \times (70 − 42) = 2 \times 28 = 56$

 first five multiple of $6$ are $6 × 1 = 66 × 2 = 126 × 3 = 186 × 4 = 246 × 5 = 30$
Their sum will be $6 + 12 + 18 + 24 + 30 = 90$

Prime numbers out of $17, 8, 21, 13, 41, 2, 27, 31, 51$ are $17, 13, 41, 2, 31.$
Sum of prime numbers $= 17 + 13 + 41 + 2 + 31 = 104.$

$LCM$ of $A$ and $B$ is $B$ it means that $B$ is multiple of $A.\ LCM$ of $B$ and $C$ is $C$ it means $C$ is multiple of $B$ or we can say that $C$ is multiple of $A$ also.
So $LCM$ of $A, B, C$ is $C$

 Factors are $17 = 1 \times 17$
$5 = 1 \times 5$
$\therefore LCM$ of $17$ and $5 = 1 \times 17 \times 5 = 85$

$59 = 1 \times 59$
$1$ and $59$ are the factors.So, options $A$ and $B$ are correct.

 A number is divisible by $5$ if its last digit is either $0$ or $5$ out of which $5$ is maxim
$\therefore b = 5$
A number is divisible by $3$ if the sum of its digits is divisible by $3$
$2 + 3 + 4 + 5 + 6 + 0 + 5 = 25$
So, we can add maximum $8$ to $25$ which will give us $33$ which is divisible by $3$
$\therefore a = 8$
Now, $a + b = 8 + 5 = 13$
Hence, the correct answer is option $(b).$

 The time after which all the three will be together will be $LCM$ of
$15\frac{1}{6}, 16\frac{1}{4}, 18\frac{2}{3}$
$LCM$ of $\frac{91}{6}, \frac{65}{4}, \frac{56}{3}=\frac{\text{LCM of 91,65,56}}{\text{HCF of 6,4,3}}=3640$ second $=1$ hour $40$ minits

 Factor $\times $ factor $=$ product

Prime numbers between $60$ and $75$ are $61, 67, 71,$ and $73.$
Their sum is given by:
$61 + 67 + 71 + 73 = 272$

$0.35=\frac{35}{100}=\frac{7}{20}$

 Every number is a factor and a multiple of itself.
For example, $10$ has a factor $10$ as well as a multiple $10.$

Here $1\%$ can be written as $\frac{1}{100}$
So, $44\%\Rightarrow(44\times1)\%$
$=44\times\frac{1}{100}$
$=\frac{44}{100}$
$=\frac{4\times11}{4\times25}$
$=\frac{11}{25}$
$44\%\Rightarrow\frac{11}{25}$ (Fraction)

There is only one prime number between $90$ and $100,$ i.e. $97.$

Here, $HCF = 12$
Product of two number $= 2160$
We know:
$LCM \times HCF =$ Product of the two numbers
$LCM =\frac{2160}{\text{HCF}}$
$=\frac{2160}{12}$$= 180$
$LCM = 180$

Lets consider prime and composite numbers separately and prove $1$ is a factor in each case.
Prime:
Example: $7$
Factors of $7$ are $1,7.$

Composite:
Example: $10$
Factors of $10$ are $1, 10, 2, 5.$

$6 + 7 + 3 + 0 + 1 + * + 2 = 19 + *$
$8$ is the least number that should be added to $19$ such that number will be divisible by $9.$
Sum of the digits:
$6 + 7 + 3 + 0 + 1 + 8 + 2 = 27$
$27$ is divisible by $9.$

$2\frac{1}{3}=\frac{3\times2+1}{3}=\frac{7}{3}$

Pairs of prime numbers that differ by $2$ are called twin primes.
The difference between $17$ and $23$ is $6.$
Hence, $17$ and $23$ are not twin primes.

$\begin{array}{c|c}17&289\\\hline17&17\\\hline&1\end{array}$
$\begin{array}{c|c}17&391\\\hline23&23\\\hline&1\end{array}$
$289 = 17 × 17$
$391 = 17 × 23$
The $HCF$ of $289$ and $391$ is $17.$
Dividing both the numerator and the denominator by $17:$
$\frac{289\div17}{391\div17}=\frac{17}{23}$

Multiple of $6$ like $6, 12, 18, 24, 30$ and they can all be divided evenly by $3.$
So option $C$ is correct.

$p = 4, 4 ÷ 2 = 2$
$p + 2 = 4 + 2, 6 ÷ 2 = 3$
$p + 4 = 4 + 4, 8 ÷ 2 = 4$
Product is divisible by $2.$
Therefore, $C$ is the correct answer.

Sum of the digits at odd places $= 6 + 3 + 4 = 13$
Sum of the digits at even places $= 5 + * + 7 = 12 + *$
Difference $= 13 - [12 + *] = 1 − *$
If $6,53,*47$ is divisible by $11,$ then $1 - *$ must be zero or multiple of $11.$
$1 - * = 0 or 11$
$* = 1$ or $- 10$
But $*$ is a digit, so $*$ must be $1.$

$LCM$ of two co -prime numbers is their product.
Example: Consider $6$ and $7,$
Multiple of $6 = 6, 12, 18, 24, 30, 36, 42, 48$
Multiple of $7 = 7, 14, 21, 28, 35, 42$
$L.C.M$ of $6$ and $7 = 42$
The product of $6$ and $7 = 6 \times 7 = 42$

The $HCF$ of any two consecutive natural numbers is $1$ because two consecutive natural numbers are always co-prime.

$\frac{\sqrt3+\sqrt2}{\sqrt3-\sqrt2}+\frac{\sqrt3-\sqrt2}{\sqrt3+\sqrt2}$
$=\frac{\big(\sqrt3+\sqrt2\big)^2+\big(\sqrt3-\sqrt2\big)^2}{3-2}$
$=\frac{3+2+3+2}{1}=10$

$GCD$ of two numbers is $17$
So, the numbers can be $17a$ and $17b.$
Now, $17a \times 17b = 17 \times 765$
$\Rightarrow ab = 45$
So, we can get two pairs
$a = 5$ and $b = 9$ or $a = 9$ and $b = 5$
Thus, the numbers are $17 \times 5 = 85$ and $17 \times 9 = 153.$
Also, we can get the other pair $17 \times 1 = 17$ and $765.$
Hence, the correct answer is option $(b).$

$\frac{7}{4}=1\frac{3}{4}$

Factors of $512 = 1, 2, 4, 8, 16, 32, 64, 128, 256, 512$
Therefore, number of factors of $512 = 10.$

Two fractions are called as unlike fractions, if the denominators of those fractions are different.
For example: Consider $\frac {1}{5}$ and $\frac {3}{6},$ here both the fractions have different denominators, so they are unlike fractions.
Hence, fractions with different denominators are called unlike fractions.

$LCM \times HCF =$ product of numbers
$HCF$ of co-prime numbers $=1$
So, $LCM = LCM =$ product of numbers
Therefore, $D$ is the correct answer.

When an object is divided into a number of equal parts then each part is called a fraction.
For example, in a fraction $\frac{2}{5},$ the numerator is $2$ and the denominator is $5,$
where the numerator represents how many parts is there in the fraction and the denominator represents how many equal parts in the whole object.
Hence, numerator of the fraction $\frac{5}{6}$​ is $5.$

$HCF = 17$
Dividing both the numerator and the denominator by the $HCF$ of $289$ and $391:$
$\begin{array}{c|c}17&289\\\hline17&17\\\hline&1\end{array}$
$\begin{array}{c|c}17&391\\\hline23&23\\\hline&1\end{array}$
$\frac{289\div17}{391\div17}=\frac{17}{23}$

Every counting number has an infinite number of multiples.
If $p$ is a counting number, its multiples are $1p, 2p, 3p....$

Factors of $78$ are $1, 2, 3, 6, 13, 26, 39$ and $78.$
Required sum $= 1 + 2 + 3 + 6 + 13 + 26 + 39 + 78 = 168.$

$3$ is not a positive multiple of $12$ as it is smaller than $12.$
Rest others are multiples of $12.$

To convert an improper fraction to a mixed fraction, we divide the numerator by the denominator, then write down the whole number answer.
Finally we write down any remainder above the denominator.
$39 ÷ 12 = 3$ leaving remainder $3$
Therefore, the answer will be, $3$ whole $\frac{3}{12}$
Hence, the mixed fraction of $\frac{39}{12}$ is $3\frac{3}{12}$.

To find : How many three-quarters part of $24$ pancakes
$\frac{3}{4}\times24=18$
Three - quarters part of $24$ pancakes $= 18$

Required sum $= (2 + 3 + 5 + 7 + 11) = 28$
Note: $11$ is not a prime number.

$ 9 = 3 \times 3 \times 1$
$10 = 2 \times 5 \times 1$
Though both $9$ and $10$ are composite numbers, the only factor common to them is $1.$
Therefore, $9$ and $10$ are co-primes.

$ 176 = 2 \times 2 \times 2 \times 2 \times 11$
$182 = 2 \times 7 \times 13$
$99 = 3 \times 11 \times 3$
$101$ is a prime so has only $1$ and itself as factors.
Hence, clearly, $176176$ has the greatest number of divisors.

Every number is a factor and a multiple of itself. For example,
$10$ has a factor $10$ as well as a multiple $10.$

$100 = 1 \times 2 \times 2 \times 5 \times 5$
$101 = 1 \times 101$
Since, $100$ is a composite number and $101$ is a prime number.
Thus, their $HCF$ is $1.$
Hence, the correct answer is option $(a).$

  Multiples of $18 = 18, 36, 54, 72.....$
Multiples of $6 = 6, 12, 18, 24....$
The first common multiple will be $18$
And the next common multiples will be multiples of $18$
Hence, the common multiples of $18, 6$ are $18, 36, 54, 72...$

$HCF$ of two consecutive even numbers is always $2.$
For example:
$HCF$ of $4$ and $6$ is $2.$
$HCF$ of $10$ and $12$ is $2$ and so on.

$2, 3$ are primes.
$\therefore $ Each number has no factor other than $1$ and itself.
$\therefore $ Their $LCM$ is the product of the numbers.
$\therefore LCM$ of $1, 2, 3 = 2 \times 3 = 6.$
Also here $1 + 2 + 3 = 6.$
Answer- Option $A$ and Option $C.$

The $LCM$ of two co-prime numbers is equal to their product.
Thus, $LCM$ of $'x'$ and $'y'$ will be $xy.$

$5005 = 5 \times 7 \times 11 \times 13$

A number which is a factor of every number is $1.$

 A fraction equivalent to $\frac{2}{7}$ is $\frac{4}{14}$.
Equivalent fractions are got by multiplying the numerator and denominator with the same number.
In this case, it is.$2.$

 A number which is a factor of every number is $1.$

 $\frac{7\sqrt3}{\sqrt10+\sqrt3}-\frac{2\sqrt5}{\sqrt6+\sqrt5}-\frac{3\sqrt2}{\sqrt15+3\sqrt2}$
$=\frac{7\sqrt3\big(\sqrt10-\sqrt3\big)}{\big(\sqrt10+\sqrt3\big)\big(\sqrt10-\sqrt3\big)}-\frac{2\sqrt5\big(\sqrt6-\sqrt5\big)}{\big(\sqrt6+\sqrt5\big)\big(\sqrt6-\sqrt5\big)}-\frac{3\sqrt2\big(\sqrt15-3\sqrt2\big)}{\big(\sqrt15-3\sqrt2\big)\big(\sqrt15+3\sqrt2\big)}$
$=\frac{7\sqrt3\big(\sqrt10-\sqrt3\big)}{10-3}-\frac{2\sqrt5\big(\sqrt6-\sqrt5\big)}{6-5}-\frac{3\sqrt2\big(\sqrt15-3\sqrt2\big)}{15-18}$
$=\frac{7\sqrt3\big(\sqrt10-\sqrt3\big)}{7}-\frac{2\sqrt5\big(\sqrt6-\sqrt5\big)}{1}-\frac{3\sqrt2\big(\sqrt15-3\sqrt2\big)}{3}$
$=\frac{21\sqrt30-63425\sqrt30+210+21\sqrt30-18*7}{21}$
$=\frac{21}{21}=1$

 A number for which the sum of all its factors is equal to twice the number is called a perfect number.
Factors of $6$ are $1, 2, 3,$ and $6.$
Sum of the factors of $6 = 1 + 2 + 3 + 6 = 12 = 2 \times 6$
Hence, $6$ is a perfect number.

 $47$ is a prime number.
So, the divisors of $47$ is $1,47.$

 Since acc. to given ques.
$\frac{\text{a}{+2}}{2}=\frac{\text{b}{+4}}{4}=\frac{\text{c}{+6}}{6}=12 $
$\Rightarrow\text{a}=22, \text{b}=44,\text{c}=66 $
$\Rightarrow\text{abc}=(22)(44)(66)$
$=63888=11^3×2^4×3^6$
Total no. of factors of composite numbers $N=p^m q^n$
where $N$ is composite number and p and q are prime numbers Then,
Total factors of $N = (m+1)(n+1)$
Number of factors
$= (3+1)(4+1)(1+1) = 40$
$= (3+1)(4+1)(1+1) = 40$

$22352$
$21776$
$2588$
$2294$
$3147$
$749$
$7$
$L.C.M$ of $ 2352 =2^2\times 2^2\times 7^2\times 3$
To make $23522352 $ a perfect square it must be multiplied by $3$

The given number is divisible by $8$ if the number formed by its last three digits is divisible by $8.$
Hence, $63,57,6*2$ is divisible by $8$ if $ 6*2$ is divisible by $8.$
Thus, the least value of $*$ will be $3.$

 Let $\frac{1}{2}$ is original fraction.$\Rightarrow\frac{1}{2}=0.5$
$\Rightarrow $ Numerator and denominator increased by $5=\frac{1+5}{2+5}=\frac{5}{7}=0.71$
$\therefore\frac{1}{2}<\frac{5}{7}$
$\therefore$ If the numerator and denominator of a proper fraction are increased by the same quantity, then the resulting fraction is Always greater than the original fraction.

$36 = 2 \times 2 \times 3 \times 3$
$72 = 2 \times 2 \times 2 \times 3 \times 3$
$\therefore L.C.M$ of $36$ and $72 = 2 \times 2 \times 2 \times 3 \times 3 = 72$

 $0.25\times0.4=\frac{25}{100}\times\frac{4}{10}$
$=\frac{100}{100 \times10}=\frac{1}{10}$

 Let the number be $x$
So, $\frac{36}{48}=\frac{\text{x}}{4}\Rightarrow\text{x}=\frac{4\times36}{48}$
now, $\text{x}=\frac{4.12.3}{12.4}=3$
$\Rightarrow $ Equivalent Fraction $=\frac{3}{4}$

 Multiples of $14$ are $14 \times 1, 14 \times 2 ....$
And $7$ and $1$ are the factors of $14.$
So, option $C$ is correct.

$ HCF$ of two co-primes is $1.$
This is because two co-prime numbers do not have any common factor.
For example, $15$ and $16$ are co-primes.Their.
$HCF$ is $1.$

 Number when divides $70$ and $125$ leaves remainders $5$ and
$8,$ then
$70 - 5 = 65$
$125 − 8 = 117$
then $HCF$ of $65$ and $117$ is
$65 = 5 \times 13117 = 3 \times 3 \times 13$
Hence, $HCF$ of $65$ and $117$ is $13.$
$13$ is the largest number which divides $70$ and $125$ and leaves remainders $5$ and $8.$

 Factor of $4500 = 2 \times 2250$
$\Rightarrow 2 \times 2 \times 1125$
$\therefore$ odd numbered factor of $4500$ is $1125.$

A prime number (or a prime) is a natural number greater than
$1$ that has no positive divisors other than $1$ and itself.
By Euclids theorem, there is an infinite number of prime numbers.
The prime number which comes just after $43$ is $47.$
So option $C$ is the correct answer.

$LCM$ of two co -prime numbers is their product.
Example: Consider $6$ and $7,$
Multiple of 6$ = 6, 12, 18, 24, 30, 36, 42, 48$
Multiple of $7 = 7, 14, 21, 28, 35, 42$
$L.C.M$ of $66$ and $7 = 42$
The product of $6$ and $7 = 6 × 7 = 42$

 Sum of digits at odd places $= 4 + 5 + 2 = 11$
Sum of digits at even places $= 1 + 6 + 4 = 11$
Difference of these two sums $= 11 - 11 = 0$
Therefore, $4,15,624 $ is divisible by $11.$

$ LCM$ of $9$ and $13 = 9 × 13 = 117$
Largest 5-digit number is $99999$
Now, if we divide $99999$ by $117,$ we will get $854.69$ as quotient.
The integer just less than $854.69$ is $854$
$\therefore$ Required number $= 117 × 854 = 99918$
Hence, the correct answer is option $(b).$