(C) Let the missing value be $x$. Then,the equation is $\frac{250}{\sqrt{x}} = 10$. By cross-multiplying,we get $\sqrt{x} = \frac{250}{10}$. $\sqrt{x}

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

(C) Let the missing value be $x$. Then,the equation is $\frac{250}{\sqrt{x}} = 10$. By cross-multiplying,we get $\sqrt{x} = \frac{250}{10}$. $\sqrt{x} = 25$. Squaring both sides,we get $x = (25)^2$. Therefore,$x = 625$.

Competitive Quantitative Aptitude Roots of Numbers Squares and Square Roots

Medium

$\frac{250}{\sqrt{?}}=10$

A
$25$
B
$250$
C
$625$
D
$2500$

Explore More

Browse All

Question Bank

All Subjects

Competitive Questions

Competitive

Quantitative Aptitude Questions

Chapter

Roots of Numbers

Subtopic

Squares and Square Roots

If $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3},$ then

Medium

View Solution

$(15.01)^{2} \times \sqrt{730} = ?$

Easy

View Solution

The value of $\sqrt{\frac{(0.03)^{2}+(0.21)^{2}+(0.065)^{2}}{(0.003)^{2}+(0.021)^{2}+(0.0065)^{2}}}$ is

Easy

View Solution

The value of $\frac{\sqrt{80}-\sqrt{112}}{\sqrt{45}-\sqrt{63}}$ is

Easy

View Solution

Find the square root of $4356$.

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$\frac{250}{\sqrt{?}}=10$

Similar Questions

If $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a+b \sqrt{3},$ then

$(15.01)^{2} \times \sqrt{730} = ?$

The value of $\sqrt{\frac{(0.03)^{2}+(0.21)^{2}+(0.065)^{2}}{(0.003)^{2}+(0.021)^{2}+(0.0065)^{2}}}$ is

The value of $\frac{\sqrt{80}-\sqrt{112}}{\sqrt{45}-\sqrt{63}}$ is

Find the square root of $4356$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module