☰ अधिक माहितीसाठी येथे क्लीक करा

#### Question 1:

Convert into improper fractions.

(i)
$7\frac{2}{5}$

(ii)
$5\frac{1}{6}$

(iii)
$4\frac{3}{4}$

(iv)
$2\frac{5}{9}$

(v)
$1\frac{5}{7}$

#### Answer 1:

(i)

$7\frac{2}{5}=\frac{5×7+2}{5}\phantom{\rule{0ex}{0ex}}=\frac{35+2}{5}\phantom{\rule{0ex}{0ex}}=\frac{37}{5}$

(ii)

$5\frac{1}{6}=\frac{6×5+1}{6}\phantom{\rule{0ex}{0ex}}=\frac{30+1}{6}\phantom{\rule{0ex}{0ex}}=\frac{31}{6}$

(iii)

$4\frac{3}{4}=\frac{4×4+2}{4}\phantom{\rule{0ex}{0ex}}=\frac{16+2}{4}\phantom{\rule{0ex}{0ex}}=\frac{18}{4}$

(iv)

$2\frac{5}{9}=\frac{9×2+5}{9}\phantom{\rule{0ex}{0ex}}=\frac{18+5}{9}\phantom{\rule{0ex}{0ex}}=\frac{23}{9}$

(v)

$1\frac{5}{7}=\frac{7×1+5}{7}\phantom{\rule{0ex}{0ex}}=\frac{7+5}{7}\phantom{\rule{0ex}{0ex}}=\frac{12}{7}$

#### Question 2:

Convert into mixed numbers.

(i)
$\frac{30}{7}$

(ii)
$\frac{7}{4}$

(iii)
$\frac{15}{12}$

(iv)
$\frac{11}{8}$

(v)
$\frac{21}{4}$

(vi)
$\frac{20}{7}$

#### Answer 2:

(i)

$\frac{30}{7}=\frac{28+2}{7}\phantom{\rule{0ex}{0ex}}=\frac{28}{7}+\frac{2}{7}\phantom{\rule{0ex}{0ex}}=4+\frac{2}{7}\phantom{\rule{0ex}{0ex}}=4\frac{2}{7}$

(ii)

$\frac{7}{4}=\frac{4+3}{4}\phantom{\rule{0ex}{0ex}}=\frac{4}{4}+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=1+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=1\frac{3}{4}$

(iii)

$\frac{15}{12}=\frac{12+3}{12}\phantom{\rule{0ex}{0ex}}=\frac{12}{12}+\frac{3}{12}\phantom{\rule{0ex}{0ex}}=1+\frac{3}{12}\phantom{\rule{0ex}{0ex}}=1\frac{3}{12}$

(iv)

$\frac{11}{8}=\frac{8+3}{8}\phantom{\rule{0ex}{0ex}}=\frac{8}{8}+\frac{3}{8}\phantom{\rule{0ex}{0ex}}=1+\frac{3}{8}\phantom{\rule{0ex}{0ex}}=1\frac{3}{8}$

(v)

$\frac{21}{4}=\frac{20+1}{4}\phantom{\rule{0ex}{0ex}}=\frac{20}{4}+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=5+\frac{1}{4}\phantom{\rule{0ex}{0ex}}=5\frac{1}{4}$

(vi)

$\frac{20}{7}=\frac{14+6}{7}\phantom{\rule{0ex}{0ex}}=\frac{14}{7}+\frac{6}{7}\phantom{\rule{0ex}{0ex}}=2+\frac{6}{7}\phantom{\rule{0ex}{0ex}}=2\frac{6}{7}\phantom{\rule{0ex}{0ex}}$

#### Question 3:

Write the following examples using fractions.
(i) If 9 kg rice is shared amongst 5 people, how many kilograms of rice does each person get?
( ii) To make 5 shirts of the same size, 11 metres of cloth is needed. How much cloth is needed for one shirt?

#### Answer 3:

(i) If 9 kg rice is shared amongst 5 people, then each person will get
$\frac{9}{5}$
kilograms of rice.

( ii) If 11 metres of cloth is needed to make 5 shirts of the same size, then one shirt will need
$\frac{11}{5}$
metres of cloth.

#### Question 1:

Add.

(i)
$6\frac{1}{3}+2\frac{1}{3}$

(ii)
$1\frac{1}{4}+3\frac{1}{2}$

(iii)
$5\frac{1}{5}+2\frac{1}{7}$

(iv)
$3\frac{1}{5}+2\frac{1}{3}$

#### Answer 1:

(i)

$6\frac{1}{3}+2\frac{1}{3}=\frac{6×3+1}{3}+\frac{2×3+1}{3}\phantom{\rule{0ex}{0ex}}=\frac{18+1}{3}+\frac{6+1}{3}\phantom{\rule{0ex}{0ex}}=\frac{19}{3}+\frac{7}{3}$

$=\frac{19+7}{3}\phantom{\rule{0ex}{0ex}}=\frac{26}{3}\phantom{\rule{0ex}{0ex}}=\frac{24+2}{3}$

$=\frac{24}{3}+\frac{2}{3}\phantom{\rule{0ex}{0ex}}=8+\frac{2}{3}\phantom{\rule{0ex}{0ex}}=8\frac{2}{3}$

(ii)

$1\frac{1}{4}+3\frac{1}{2}=\frac{1×4+1}{4}+\frac{3×2+1}{2}\phantom{\rule{0ex}{0ex}}=\frac{5}{4}+\frac{7}{2}\phantom{\rule{0ex}{0ex}}=\frac{5}{4}+\frac{7×2}{2×2}$

$=\frac{5}{4}+\frac{14}{4}\phantom{\rule{0ex}{0ex}}=\frac{5+14}{4}\phantom{\rule{0ex}{0ex}}=\frac{19}{4}\phantom{\rule{0ex}{0ex}}=\frac{16+3}{4}$

$=\frac{16}{4}+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=4+\frac{3}{4}\phantom{\rule{0ex}{0ex}}=4\frac{3}{4}$

(iii)

$5\frac{1}{5}+2\frac{1}{7}=\frac{5×5+1}{5}+\frac{2×7+1}{7}\phantom{\rule{0ex}{0ex}}=\frac{26}{5}+\frac{15}{7}\phantom{\rule{0ex}{0ex}}=\frac{26×7}{5×7}+\frac{15×5}{7×5}\phantom{\rule{0ex}{0ex}}$

$=\frac{182}{35}+\frac{75}{35}\phantom{\rule{0ex}{0ex}}=\frac{182+75}{35}\phantom{\rule{0ex}{0ex}}=\frac{257}{35}$

$=\frac{245+12}{35}\phantom{\rule{0ex}{0ex}}=\frac{245}{35}+\frac{12}{35}\phantom{\rule{0ex}{0ex}}=7+\frac{12}{35}\phantom{\rule{0ex}{0ex}}=7\frac{12}{35}$

(iv)
$3\frac{1}{5}+2\frac{1}{3}$

$3\frac{1}{5}+2\frac{1}{3}=\frac{3×5+1}{5}+\frac{2×3+1}{3}\phantom{\rule{0ex}{0ex}}=\frac{16}{5}+\frac{7}{3}\phantom{\rule{0ex}{0ex}}=\frac{16×3}{5×3}+\frac{7×5}{3×5}\phantom{\rule{0ex}{0ex}}$

$=\frac{48}{15}+\frac{35}{15}\phantom{\rule{0ex}{0ex}}=\frac{48+35}{15}\phantom{\rule{0ex}{0ex}}=\frac{83}{15}$

$=\frac{75+8}{15}\phantom{\rule{0ex}{0ex}}=\frac{75}{15}+\frac{8}{15}\phantom{\rule{0ex}{0ex}}=5+\frac{8}{15}\phantom{\rule{0ex}{0ex}}=5\frac{8}{15}$

#### Question 2:

Subtract.

(i)
$3\frac{1}{3}-1\frac{1}{4}$

(ii)

(iii)

(iv)

#### Answer 2:

(i)

$3\frac{1}{3}-1\frac{1}{4}=\frac{10}{3}-\frac{5}{4}\phantom{\rule{0ex}{0ex}}=\frac{10×4}{3×4}-\frac{5×3}{4×3}\phantom{\rule{0ex}{0ex}}=\frac{40}{12}-\frac{15}{12}\phantom{\rule{0ex}{0ex}}=\frac{40-15}{12}$

$=\frac{25}{12}\phantom{\rule{0ex}{0ex}}=\frac{24+1}{12}\phantom{\rule{0ex}{0ex}}=\frac{24}{12}+\frac{1}{12}\phantom{\rule{0ex}{0ex}}=2+\frac{1}{12}\phantom{\rule{0ex}{0ex}}=2\frac{1}{12}$

(ii)

$=\frac{33-20}{6}\phantom{\rule{0ex}{0ex}}=\frac{13}{6}\phantom{\rule{0ex}{0ex}}=2\frac{1}{6}$

(iii)

$=\frac{285-244}{40}\phantom{\rule{0ex}{0ex}}=\frac{41}{40}\phantom{\rule{0ex}{0ex}}=1\frac{1}{40}$

(iv)

$=\frac{75-32}{10}\phantom{\rule{0ex}{0ex}}=\frac{43}{10}\phantom{\rule{0ex}{0ex}}=4\frac{3}{10}$

#### Question 3:

Solve.

(1) Suyash bought
$2\frac{1}{2}$
kg of sugar and Ashish bought
$3\frac{1}{2}$
kg. How much sugar did they buy altogether? If sugar costs 32 rupees per kg, how much did they spend on the sugar they bought?

(2) Aradhana grows potatoes in
$\frac{2}{5}$
part of her garden, greens in
$\frac{1}{3}$
part and brinjals in the remaining part. On how much of her plot did she plant brinjals?

(3) Sandeep filled water in
$\frac{4}{7}$
of an empty tank. After that, Ramakant filled
$\frac{1}{4}$
part more of the same tank. Then Umesh used
$\frac{3}{14}$
part of the tank to water the garden. If the tank has a maximum capacity of 560 litres, how many litres of water will be left in the tank?

#### Answer 3:

(1) The amount of sugar they bought altogether =
$2\frac{1}{2}+3\frac{1}{2}$

$=\frac{5}{2}+\frac{7}{2}\phantom{\rule{0ex}{0ex}}=\frac{5+7}{2}\phantom{\rule{0ex}{0ex}}=\frac{12}{2}$
= 6 kg
Now, cost of 1 kg of sugar = Rs 32
Therefore, the cost of 6 kg of sugar is = 6 × 32
= Rs 192
Hence, they spend Rs 192 on the sugar they bought.

(2) The part of the garden in which Aradhana grew brinjals is given by
$1-\frac{2}{5}-\frac{1}{3}$

$=\frac{1×15}{1×15}-\frac{2×3}{5×3}-\frac{1×5}{3×5}\phantom{\rule{0ex}{0ex}}=\frac{15}{15}-\frac{6}{15}-\frac{5}{15}\phantom{\rule{0ex}{0ex}}=\frac{15-6-5}{15}\phantom{\rule{0ex}{0ex}}=\frac{4}{15}$

Hence, Aradhana grew brinjals in
$\frac{4}{15}$
part of her garden.

(3) The amount of water will be left in the tank is given by
$\frac{4}{7}\left(560\right)+\frac{1}{4}\left(560\right)-\frac{3}{14}\left(560\right)$

Hence, 340 l of water will be left in the tank.
Std 6 th Mathematics- 4. Operations on Fractions page 1 Reviewed by Amol Uge on March 02, 2019 Rating: 5
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