☰ अधिक माहितीसाठी येथे क्लीक करा

#### Question 2:

Show the following fractions on the number line.

(1)

(2)

(1)

(2)

#### Question 1:

Multiply.

(i)

(ii) $\frac{6}{7}×\frac{2}{5}$

(iii) $\frac{5}{9}×\frac{4}{9}$

(iv)

(v) $\frac{1}{5}×\frac{7}{2}$

(vi) $\frac{9}{7}×\frac{7}{8}$

(vii) $\frac{5}{6}×\frac{6}{5}$

(viii) $\frac{6}{17}×\frac{3}{2}$

(i)

(ii)
$\frac{6}{7}×\frac{2}{5}=\frac{6×2}{7×5}\phantom{\rule{0ex}{0ex}}=\frac{12}{35}$

(iii)
$\frac{5}{9}×\frac{4}{9}=\frac{5×4}{9×9}\phantom{\rule{0ex}{0ex}}=\frac{20}{81}$

(iv)
$\frac{4}{11}×\frac{2}{7}=\frac{4×2}{11×7}\phantom{\rule{0ex}{0ex}}=\frac{8}{77}$

(v)
$\frac{1}{5}×\frac{7}{2}=\frac{1×7}{5×2}\phantom{\rule{0ex}{0ex}}=\frac{7}{10}$

(vi)
$\frac{9}{7}×\frac{7}{8}=\frac{9}{\overline{)7}}×\frac{\overline{)7}}{8}\phantom{\rule{0ex}{0ex}}=\frac{9}{8}$

(vii)
$\frac{5}{6}×\frac{6}{5}=\frac{\overline{)5}}{\overline{)6}}×\frac{\overline{)6}}{\overline{)5}}\phantom{\rule{0ex}{0ex}}=1$

(viii)
$\frac{6}{17}×\frac{3}{2}=\frac{3\overline{)6}}{17}×\frac{3}{\overline{)2}}\phantom{\rule{0ex}{0ex}}=\frac{3×3}{17}\phantom{\rule{0ex}{0ex}}=\frac{9}{17}$

#### Question 2:

Ashokrao planted bananas on $\frac{2}{7}$ of his field of 21 acres. What is the area of the banana plantation?

Ashokrao planted bananas on $\frac{2}{7}$ of his field of 21 acres.
The area of the banana plantation = $\frac{2}{7}×21$

Hence, the area of the banana plantation is 6 acres.

#### Question 3:

Of the total number of soldiers in our army, $\frac{4}{9}$ are posted on the northern border and one - third of them on the north - eastern border. If the number of soldiers in the north is 540000, how many are posted in the north - east?

Let the total number of the soldiers be x.
Number of soldiers posted on the northern border = 540000
$⇒\frac{4}{9}x=540000\phantom{\rule{0ex}{0ex}}⇒x=\frac{540000×9}{4}\phantom{\rule{0ex}{0ex}}=1215000$
Now, the number of soldiers posted on the northern border = $\frac{1}{3}×1215000$
= 405000
Hence, 405000 soldiers are posted in the north - east.

#### Question 1:

Write the reciprocals of the following numbers.
(i) 7

(ii) $\frac{11}{3}$

(iii) $\frac{5}{13}$

(iv) 2

(v) $\frac{6}{7}$

(i) The reciprocal of 7 is $\frac{1}{7}$.

(ii) The reciprocal of $\frac{11}{3}$ is $\frac{3}{11}$.

(iii) The reciprocal of $\frac{5}{13}$ is $\frac{13}{5}$.

(iv) The reciprocal of 2 is $\frac{1}{2}$.

(v) The reciprocal of $\frac{6}{7}$ is $\frac{7}{6}$.

#### Question 2:

Carry out the following divisions.
(i) $\frac{2}{3}÷\frac{1}{4}$

(ii) $\frac{5}{9}÷\frac{3}{2}$

(iii)

(iv) $\frac{11}{12}÷\frac{4}{7}$

(i)
$\frac{2}{3}÷\frac{1}{4}=\frac{2}{3}×\frac{4}{1}\phantom{\rule{0ex}{0ex}}=\frac{2×4}{3×1}\phantom{\rule{0ex}{0ex}}=\frac{8}{3}$

(ii)
$\frac{5}{9}÷\frac{3}{2}=\frac{5}{9}×\frac{2}{3}\phantom{\rule{0ex}{0ex}}=\frac{5×2}{9×3}\phantom{\rule{0ex}{0ex}}=\frac{10}{27}$

(iii)

(iv)
$\frac{11}{12}÷\frac{4}{7}=\frac{11}{12}×\frac{7}{4}\phantom{\rule{0ex}{0ex}}=\frac{11×7}{12×4}\phantom{\rule{0ex}{0ex}}=\frac{77}{48}$

#### Question 3:

There were 420 students participating in the Swachh Bharat campaign. They cleaned $\frac{42}{75}$ part of the town, Sevagram. What part of Sevagram did each student clean if the work was equally shared by all?

420 students cleaned $\frac{42}{75}$ part of the Sevagram town.
1 student cleaned the part of the Sevagram town equal to $\frac{42}{75}÷420$
$=\frac{42}{75}×\frac{1}{420}\phantom{\rule{0ex}{0ex}}=\frac{42}{75×420}\phantom{\rule{0ex}{0ex}}=\frac{1}{75×10}\phantom{\rule{0ex}{0ex}}=\frac{1}{750}\phantom{\rule{0ex}{0ex}}$
Hence, each student cleaned $\frac{1}{750}$ part of the Sevagram town.