Select the correct statement from the following. - Vidyadip

Question

Select the correct statement from the following.

✓

Answer

$\frac{-2}{3}<\frac{-4}{5}$
Solution:
$\frac{-2}{3}<\frac{-4}{5}$
Taking LCM of 3 and 5,
LCM = 15,
So, $\frac{-2\times5}{3\times5},\frac{-4\times3}{5\times3}$
$\Rightarrow\frac{-10}{15},\frac{-12}{15}$
Now, since both the denominator is equal so, we compare its numerator,
and since, -10 < -12
So, $\frac{-10}{15}>\frac{-12}{15}$
thus, $\frac{-2}{3}>\frac{-4}{5}$

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

Explore more

More "M.C.Q" from Number systems→Number systems — all question types→MATHS STD 9 question papers→Rajasthan Board STD 9 question papers→Rajasthan Board question paper generator→

Similar questions

An equilateral triangle of side 9cm is inscribed in a circle. The radius of the circle is:

Write the correct answer in the following:
The perimeter of an equilateral triangle is 60m. The area is:

$10\sqrt{3}\text{m}^2$
$15\sqrt{3}\text{m}^2$
$20\sqrt{3}\text{m}^2$
$100\sqrt{3}\text{m}^2$

In the following, write the correct answer.
It is given that $\triangle\text{ABC}=\triangle\text{FDE}$ and AB = 5 cm, $∠\text{B} = 40°$and $∠\text{A} = 80°$then which of the following is true?

The value of $\sqrt{20}\times\sqrt{5}$ is :

If $\frac{\text{a}}{\text{b}}+\frac{\text{b}}{\text{a}}=1,$ then $a^3 + b^3 =$

Write the correct answer of the following:
The median of a triangle divides it into two:

Triangles of equal area.
Congruent triangles.
Right triangles.
Isosceles triangles.

In a square ABCD, the diagonals AC and BD bisects at O. Then $\triangle\text{AOB}$ is:

Three cubes of metal with edges $3\ cm, 4\ cm$ and $5\ cm$ respectively are melted to form a single cube. The lateral surface area of the new cube formed is.

If x = 3 and y = -2 satisfies 5x - y = k, then the value of k is:

If bisectors of $\angle\text{A}$ and $\angle\text{B}$ of a quadrilateral ABCD intersect each other at P, of $\angle\text{B}$ and $\angle\text{C}$ at Q, of $\angle\text{C}$ and $\angle\text{D}$ at R and of $\angle\text{D}$ and $\angle\text{A}$ at S, then PQRS is a: