MCQ
$\text{H.C.F}.$ of $26$ and $91$ is :
- ✓$13$
- B$2366$
- C$91$
- D$182$
✓
Answer
Correct option: A.
$13$
Prime factors of $26 = 2 \times 13$
Prime factors of $91 = 7 \times 13$
As $13$ is the only common factor.
Therefore, $= \text{HCF}\ (26, 91) = 13z$
Prime factors of $91 = 7 \times 13$
As $13$ is the only common factor.
Therefore, $= \text{HCF}\ (26, 91) = 13z$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The capacity of the cylindrical vessel with the hemispherical bottom portion raised upwards as shown in Fig., is
Which of the following is a quadratic equation?
If the centroid of the triangle formed by $(7, x), (y, -6)$ and $(9, 10)$ is at $ (6, 3),$ then $ (x, y) =$
→$\frac{2\tan30^\circ}{1+\tan^230^\circ}$ is equal to:
→The area of a circle is $38.5 \ cm^2$. The circumference of the circle is
→If the point $(x, 4)$ lies on a circle whose centre is at the origin and radius is 5 , then $x=$
→If $p$ is a root of the quadratic equation $x^2-(p+q) x+k=0, $ then the value of $k$ is
→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 : Sum of first hundred even natural numbers divisible by $5$ is $500.$
Reason : Sum of first $n-$ terms of an $A.P$. is given by $\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a+}\ell] \text{ where }\ell =$ last term.
→Assertion : Sum of first hundred even natural numbers divisible by $5$ is $500.$
Reason : Sum of first $n-$ terms of an $A.P$. is given by $\text{s}_\text{n}=\frac{\text{n}}{2}[\text{a+}\ell] \text{ where }\ell =$ last term.
A single letter is selected at random from the word 'PROBABILITY'. The probability that it is a vowel is
In Fig. $Q R$ is a common tangent to the given circles touching externally at the point $T$. The tangent at $T$ meets $Q R$ at $P$. If $P T=3.8 cm$, then the length of $Q R( in cm )$ is
→