What will come in the place of the question m | vedclass

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

Question

(D) To find the value of $?$,we rearrange the equation:$? = 2 \frac{1}{8} - 1 \frac{1}{16} - 1 \frac{1}{32} + 1 \frac{9}{64}$Separate the whole number

Vedclass · Accepted Answer

$1 \frac{11}{64}$

Vedclass · Answer

(D) To find the value of $?$,we rearrange the equation:$? = 2 \frac{1}{8} - 1 \frac{1}{16} - 1 \frac{1}{32} + 1 \frac{9}{64}$Separate the whole numbers and the fractions:$? = (2 - 1 - 1 + 1) + (\frac{1}{8} - \frac{1}{16} - \frac{1}{32} + \frac{9}{64})$Calculate the whole number part:$2 - 1 - 1 + 1 = 1$Calculate the fractional part by finding the common denominator,which is $64$:$\frac{1}{8} - \frac{1}{16} - \frac{1}{32} + \frac{9}{64} = \frac{8}{64} - \frac{4}{64} - \frac{2}{64} + \frac{9}{64}$$= \frac{8 - 4 - 2 + 9}{64} = \frac{11}{64}$Combining both parts,we get:$? = 1 + \frac{11}{64} = 1 \frac{11}{64}$

What will come in the place of the question mark $(?)$ in the following equation?
$2 \frac{1}{8} - 1 \frac{1}{16} = ? + 1 \frac{1}{32} - 1 \frac{9}{64}$

Similar Questions

If $x + \frac{1}{x} = \sqrt{13}$,then $\frac{3x}{x^2 - 1}$ is equal to:

The value of: $\sqrt{\frac{0.324 \times 0.081 \times 4.624}{1.5625 \times 0.0289 \times 72.9 \times 64}}$

$3889 + 12.952 - ? = 3854.002$

The least number to be subtracted from $16800$ to make it a perfect square is

If $p = 99$,then the value of $p(p^{2} + 3p + 3)$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

What will come in the place of the question mark $(?)$ in the following equation?$2 \frac{1}{8} - 1 \frac{1}{16} = ? + 1 \frac{1}{32} - 1 \frac{9}{64}$