Approximation Test-1 (Rounding off to Integers)

Answer: Option D

Explanation

Given,

$(\sqrt{\text{746}}\times \sqrt{\text{93}})\times \sqrt{25}=?$

It can be approximated as

=$\begin{array}{l}(\sqrt{\text{729}}\times \sqrt{100})\times \sqrt{25}=?\end{array}$

= 27×10×5=?

= 1350 ≈ 1390

$\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}[\because \sqrt{746}=27and\sqrt{100}\text{ = 25]}$

Hence, Option D is correct.

Answer: Option D

Explanation

Given

$$\begin{gathered} \text{4096}\times \frac{2}{7}\times \frac{3}{4}=? \\ \Rightarrow \text{ 4096}\times \frac{6}{28}=? \end{gathered}$$

It can be approximated as

$$\begin{gathered} \begin{array}{l}\Rightarrow \text{ 4096}\times \frac{6}{30}=?\end{array} \\ \begin{array}{l}\Rightarrow \text{4096}\times \frac{1}{5}=?\\ \end{array} \\ \Rightarrow \text{\hspace{0.17em}?=}\frac{4096}{5}=819.2\approx 820 \end{gathered}$$

Hence, Option D is correct.

Answer: Option D

Explanation

Given,

4563÷63×25=?

It can be approximated as

= 4500÷60×3=?

= 75×3=?

= 225 ≈200

Hence, Option D is correct.

Answer: Option C

Explanation

Given,

3942÷64÷3=?

It can be approximated as

= 3900÷60÷3=?

$$\begin{gathered} \text{=3900}\times \frac{1}{60}\times \frac{1}{3}= ? \\ \text{=}\frac{3900}{180}=21.66\approx 21 \end{gathered}$$

Hence, Option C is correct.

Answer: Option D

Explanation

Given,

$$\begin{gathered} 2\frac{3}{10}\times 4\frac{6}{7}\times 7\frac{1}{2}=? \\ =\frac{23}{10}\times \frac{34}{7}\times \frac{15}{2} \end{gathered}$$

It can be approximated as

$\begin{array}{l}\Rightarrow \frac{20}{10}\times \frac{35}{7}\times \frac{16}{2}=?\end{array}$

= 2×5×8= ?

= 80 ≈ 84

Hence, Option D is correct.