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

$36 = 2 \times 2 \times 3 \times 3$
$72 = 2 \times 2 \times 2 \times 3 \times 3$
$\therefore L.C.M$ 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 $(b).$

 The two numbers which have only $1$ as their common factor are called co-primes.
For example, Factors of $5$ are $1, 5$
Factors of $3$ are $1, 3$
Common factors is $1.$
So they are co-prime numbers.
To find their $LCM,$ we
then choose each prime number with the greatest power and multiply them to get the $LCM.$
$\Rightarrow LCM = 3 \times 5 = 15$
Hence, $LCM$ of two co-prime numbers is their product.

 The positive integers factor of $8080$ are $1, 2, 4, 5, 8, 10, 20, 40$
and $80$ Then in given option the option $C$ is $12$ not
the factor a factor of $80$. Answer is $12.$

$I.$ If $x$ is divisible by $y$ and $y$ is divisible by $z$ then $x$ must be divisible by $z$
Let $x $be $12, y $ be $4$ and $z$ be $2.$
So, here we can see that $12$ is divisible by $2.$ Hence true.

$II.$ If $x$ is a factor of $y$ and $z$ then $x$ must be a factor of $y + z$
Let $x$ be $3, y$ be $6$ and $z$ be $15.$
So, here we can see that $21$ is divisible by $3.$ Hence true.

$III$ If $x$ is a factor of $y$ and $z$ then $x$ must be a factor of$\frac{\text{y}}{\text{z}}$
Let$ x$ be $2, y$ be $4$ and $z$ be $6.$
So, here we can see that $\frac{4}{6}$ is not divisible by $2.$ Hence untrue.

Therefore, option $C$ is correct.

 $\frac{5718\times5718-4135\times4135}{5718+4135}$
$=\frac{(5718)^2-(4135)^2}{5718 + 4135}$
$\therefore(\text{a}^2-\text{b}^2)=(\text{a}+\text{b})+(\text{a}-\text{b})$
$=\frac{5718\times5718-4135\times4135}{5718+4135}$
$=5718-4135$
$=1583$

 $\therefore5\frac{4}{7}=\frac{(7\times5)+4}{7}=\frac{35+4}{7}$
$=\frac{39}{5}$

 In $90,76,185:$
Sum of the digits $= 9 + 0 + 7 + 6 + 1 + 8 + 5 = 36$
Since $36$ is divisible by $9, 9076185$ is divisible by $9.$

$263 = 1 \times 263$
The number $263$ has only two factors, $1$ and $263.$
Hence, it is a prime number.

To write a number as product of its primes, we divide it by various prime numbers $2, 3, 5, 7$ etc one by one and check by which prime numbers it is divisible with and how many times.
Hence, $20 = 2 \times 10 = 2 \times 2 \times 5$

Factor of $860 = 2 \times 430 = 2 \times 2 \times 215215$ is an odd number,
so the largest odd number factor of $860860$ is $215215.$

Co-prime number is a set of numbers or integers which have only $1$ as their common factor i.e. their $(HCF)$ will be $1.$
The factors of prime number is $11$ and number itself, so $HCF$ of such numbers is $1.$
Therefore, $C$ is the correct answer.

The prime numbers between $0$ and $20$ are $2, 3, 5, 7, 11, 13, 172,3,5,7,11,13,17$ and $19$
$\therefore $ number of prime numbers between $0$ and $20$ is $8.$

$8\frac{5}{9}=\frac{9\times8+5}{9}$
$=\frac{72+5}{9}=\frac{77}{9}$

A number is divisible by $9$ if the sum of its digits is a multiple of $9.$
Sum of the given digits $= 9 + 1 + 5 + 2 + 6 = 23$
We know that multiple of $9$ greater than $23$ is $27.$
$\therefore 27 - 23 = 4$
Hence, the smallest required digit is $4.$

$132 = 4 × 3 × 11$
So, if the number divisible by all the three number $4, 3, 114, 3, 11,$ then the number is divisible by $132$ also.
$264 \rightarrow 11, 3, 4264 \rightarrow 11, 3, 4 (/)$
$396 \rightarrow 11, 3, 4396 \rightarrow 11, 3, 4 (/)$
$462 \rightarrow 11, 3, 4462 \rightarrow 11, 3, 4 (X)$
$792 \rightarrow 11, 3, 4792 \rightarrow 11, 3, 4 (/)$
$968 \rightarrow 11, 3, 4968 \rightarrow 11, 3, 4 (X)$
$2178 \rightarrow 11, 3, 42178 \rightarrow 11, 3, 4 (X)$
$5184 \rightarrow 11, 3, 45184 \rightarrow 11, 3, 4 (X)$
$6336\rightarrow 11,3,46336 \rightarrow 11, 3, 4 (/)$
Therefore the following numbers are divisible by $132 : 264, 396, 792, 6336$
Required number of number $= 4$

Three numbers are $1\times HCF, 2 \times HCF,$ and $3 \times HCF,$
i.e. $1 \times 6 = 6, 2 \times 6 = 12,$ and $3 \times 6 = 18.$
Thus, the numbers are $6, 12, 18.$

Let the given number be $x=2.525252....$
multiplying with $100$ on both sides
$\Rightarrow 100x = 252.525252.$
$\Rightarrow 100x = 250 + 2.5252...$
$\Rightarrow 99x = 250$
$\Rightarrow \text{x}=\frac{250}{99}$
$\therefore$ Sum of numerator and denominator $=25099 = 349$

One of the numbers is $725.$
$HCF = 145$
$LCM = 2175$
We know:
$LCM \times HCF =$ Product of the two numbers
$\therefore$ Product of the two numbers $= 145 \times 2175$
$= 315375$
$\therefore$ Other number $=\frac{315375}{725}$
$=435$

In $2,22,22,222,$ the difference of the sum of alternate digits $2 + 2 + 2 + 2 = 8$ and $2 + 2 + 2 +2 = 8$ is zero.
Hence, the number is divisible by $11.$

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

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

The two numbers which have only $1$ as their common factor are called co-primes.
For example, Factors of $5$ are $1,5$
Factors of $3$ are $1, 3$
Common factors is $1.$
So they are co-prime numbers.
To find their $LCM,$ we
then choose each prime number with the greatest power and multiply them to get the $LCM.$
$\Rightarrow LCM = 3 \times 5 = 15$
Hence, $LCM$ of two co-prime numbers is their product.