frac{frac{64}{121}-frac{9}{64}}{frac{8}{11}+f | vedclass

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

Question

(B) Given expression: $\frac{\frac{64}{121} - \frac{9}{64}}{\frac{8}{11} + \frac{3}{8}}$Numerator: $\frac{64}{121} - \frac{9}{64} = \frac{64^2 - 9 	i

Vedclass · Accepted Answer

$\frac{31}{88}$

Vedclass · Answer

(B) Given expression: $\frac{\frac{64}{121} - \frac{9}{64}}{\frac{8}{11} + \frac{3}{8}}$Numerator: $\frac{64}{121} - \frac{9}{64} = \frac{64^2 - 9 	imes 121}{121 	imes 64} = \frac{4096 - 1089}{7744} = \frac{3007}{7744}$Denominator: $\frac{8}{11} + \frac{3}{8} = \frac{8 	imes 8 + 3 	imes 11}{11 	imes 8} = \frac{64 + 33}{88} = \frac{97}{88}$Now,divide the numerator by the denominator:$\frac{3007}{7744} \div \frac{97}{88} = \frac{3007}{7744} 	imes \frac{88}{97}$Since $3007 = 31 	imes 97$ and $7744 = 88 	imes 88$:$= \frac{31 	imes 97}{88 	imes 88} 	imes \frac{88}{97} = \frac{31}{88}$

$\frac{\frac{64}{121}-\frac{9}{64}}{\frac{8}{11}+\frac{3}{8}} = ?$

Similar Questions

The value of $3+\frac{3}{3+\frac{1}{3+\frac{1}{3}}}$ is:

$A$ person was to multiply a fraction by $\frac{6}{7}$. Instead,he divided the fraction by $\frac{6}{7}$ and got an answer which exceeds the correct answer by $\frac{1}{7}$. The correct answer is:

Which of the following statement$(s)$ is/are $TRUE$?
$I.$ $\sqrt{12} > \sqrt[3]{16} > \sqrt[4]{24}$
$II.$ $\sqrt[3]{25} > \sqrt[4]{32} > \sqrt[6]{48}$
$III.$ $\sqrt[4]{9} > \sqrt[3]{15} > \sqrt[6]{24}$

The value of $\frac{(81)^{3.6} \times (9)^{2.7}}{(81)^{4.2} \times 3}$ is

If $x=2$,then find the value of $x^{3}+27x^{2}+243x+631$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module