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

$LCM \times HCF =$ 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.