M.C.Q (1 Marks)

Download App

Questions · Page 5 of 5

MCQ 2011 Mark

The least number which when divided by $18,24,30$ and $42$ will leave in each case the same remainder $1$ , would be

A
$2520$
B
$2519$
✓
$2521$
D
None of these

Answer

Correct option: C.

$2521$

We have, $18=2 \times 3^2, 24=2^3 \times 3$
$30=2 \times 3 \times 5,42=2 \times 3 \times 7$
$\text{LCM} =2^3 \times 3^2 \times 5 \times 7=2520$
So, required least number is $2520+1=2521$.

View full question & answer→

MCQ 2021 Mark

If $p$ and $q$ are positive integers such that $p=a b^2$ and $q=a^3 b$, where $a, b$ are prime numbers, then $\operatorname{HCF}(p, q)=$

✓
$a b$
B
$a^2 b^2$
C
$a^3 b^2$
D
$a^3 b^3$

Answer

Correct option: A.

$a b$

(a) : $\operatorname{HCF}(p, q)=\operatorname{HCF}\left(a b^2, a^3 b\right)=a b$.

View full question & answer→

MCQ 2031 Mark

The $\text{HCF}$ of two numbers is $23$ and their $\text{LCM}$ is $1449$. If one of the numbers is $161,$ then the other number is

✓
$207$
B
$307$
C
$1449$
D
None of these

Answer

Correct option: A.

$207$

Let the other number be $x$.
Product of two numbers $= \text{HCF} \times \text{LCM}$ of two numbers
$\therefore x \times 161=23 \times 1449$
$\Rightarrow x=\frac{23 \times 1449}{161}$
$ \Rightarrow x=207$

View full question & answer→

MCQ 2041 Mark

If LCM of $a$ and 18 is 36 and HCF of $a$ and 18 is 2 , then $a=$