MCQ
Statement-1 $(A)$: If 11 divides 627264, then 11 divides 792.
Statement-2 $(R)$ : Let $p$ be a prime number and a be a positive integer, if $p$ divides $a^2$, then pdivides $a$.
  • Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
  • B
    Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.
  • C
    Statement-1 is true, Statement-2 is false.
  • D
    Statement- 1 is false, Statement-2 is true.

Answer

Correct option: A.
Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
(A) Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.
Using Theorem 2 (page 1.3 of main book), we find that the statement- 2 is true. We observe that 11 is prime and it divides 627264 . Therefore, by using statement-2, 11 divides $\sqrt{627264}=792$. So, statement-1 is also true. Also, statement-2 is a correct explanation for statement-1.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

Statement-1 (A): The value of the product $P=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ}, \ldots \ldots, \tan 89^{\circ}$ is 1 .
Statement-2 (R): For $0 < \theta \leq 90^{\circ}, \tan \left(90^{\circ}-\theta\right)=\cot \theta$ and $\tan 45^{\circ}=1$.
Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : If we number is added to its own $- \text{ve}$ number then the result is zero is known as additive inverse.
Reason : $3 + (-3) = 0$ is additive inverse.
Statement A (Assertion) : $\sqrt{2}$ is an irrational number.
Statement $R$ (Reason) : If $p$ be a prime, then $\sqrt{p}$ is an irrational number.
Statement $A\ ($Assertion$)$ : If $-1$ is the zero of the polynomial $p(x)=x^2-3 a x+3 a-7,$ then value of $a$ is $3$ .
Statement $R\ ($Reason$)$ : The zeroes of polynomial $a x^2+b x+c, a \neq 0$ are the $x-$ coordinate of the points where the parabola representing $y$
$=a x^2+b x+c$ intersects the $x-$ axis.
Statement-1 (A): If the system of equations $3 x+6 y=10$ and $2 x-k y+5=0$ is inconsistent, then $k=-4$.
Statement-2 $(R)$ : The system of equations $a_1 x+b_1 y+c_1=0$ and $a_2 x+b_2 y+c_2=0$ is inconsistent iff $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$.
Statement-1 (A): The system of equations $2 x+y+9=0$ and $x+3 y+7=0$ is consistent having unique solution.
Statement-2 (R): The system of equations $a x+b y+c=0$ and $p x+q y+r=0$ is always consistent with unique solution, if $a q \neq b p$.
Statement A (Assertion): The polynomial $p(x)=x^3+x$ has one real zero.
Statement R (Reason) : A polynomial of $n^{\text {th }}$ degree has at most $n-1$ zeroes.
Statement-1 (A) : If roots of the equation $(2 k-1) x^2+4 x-3=0$ are reciprocal of each other, then $k=-1$.
Statement- 2(R) : If $a=c$, then roots of $a x^2+b x+c=0$ are reciprocal of each other.
Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following:
Assertion : A pair of linear equations has no solution $(s)$ if it is represented by intersecting lines graphically.
Reason : If the pair of lines are intersecting, then the pair has unique solution and is called consistent pair of equations.
Statement $A ($Assertion$)$ : In the given figure, $\text{AOB}$ is a diameter of a circle with centre $O$ and $A C$ is a tangent to the circle at $A$. If $\angle B O C=125^{\circ}$, then $\angle A C O=35^{\circ}$.
Image
Statement $R ($Reason$)$ : $\angle A C O$
and $\angle B O C$ form a linear pair.