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 $16 = 48$