Std 6 th Mathematics- 4. Operations on Fractions page 1

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)

$7\frac{2}{5}$

$=\frac{5\times 7+2}{5}=\frac{35+2}{5}=\frac{37}{5}$

(ii)

$5\frac{1}{6}$

$=\frac{6\times 5+1}{6}=\frac{30+1}{6}=\frac{31}{6}$

(iii)

$4\frac{3}{4}$

$=\frac{4\times 4+2}{4}=\frac{16+2}{4}=\frac{18}{4}$

(iv)

$2\frac{5}{9}$

$=\frac{9\times 2+5}{9}=\frac{18+5}{9}=\frac{23}{9}$

(v)

$1\frac{5}{7}$

$=\frac{7\times 1+5}{7}=\frac{7+5}{7}=\frac{12}{7}$

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)

$\frac{30}{7}$

$=\frac{28+2}{7}=\frac{28}{7}+\frac{2}{7}=4+\frac{2}{7}=4\frac{2}{7}$

(ii)

$\frac{7}{4}$

$=\frac{4+3}{4}=\frac{4}{4}+\frac{3}{4}=1+\frac{3}{4}=1\frac{3}{4}$

(iii)

$\frac{15}{12}$

$=\frac{12+3}{12}=\frac{12}{12}+\frac{3}{12}=1+\frac{3}{12}=1\frac{3}{12}$

(iv)

$\frac{11}{8}$

$=\frac{8+3}{8}=\frac{8}{8}+\frac{3}{8}=1+\frac{3}{8}=1\frac{3}{8}$

(v)

$\frac{21}{4}$

$=\frac{20+1}{4}=\frac{20}{4}+\frac{1}{4}=5+\frac{1}{4}=5\frac{1}{4}$

(vi)

 

$\frac{20}{7}=\frac{14+6}{7}=\frac{14}{7}+\frac{6}{7}=2+\frac{6}{7}=2\frac{6}{7}$

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)

$6\frac{1}{3}$

$+2\frac{1}{3}=\frac{6\times 3+1}{3}+\frac{2\times 3+1}{3}=\frac{18+1}{3}+\frac{6+1}{3}=\frac{19}{3}+\frac{7}{3}$
$=\frac{19+7}{3}$

$=\frac{26}{3}=\frac{24+2}{3}$
$=\frac{24}{3}$

$+\frac{2}{3}=8+\frac{2}{3}=8\frac{2}{3}$

(ii)

$1\frac{1}{4}$

$+3\frac{1}{2}=\frac{1\times 4+1}{4}+\frac{3\times 2+1}{2}=\frac{5}{4}+\frac{7}{2}=\frac{5}{4}+\frac{7\times 2}{2\times 2}$
$=\frac{5}{4}$

$+\frac{14}{4}=\frac{5+14}{4}=\frac{19}{4}=\frac{16+3}{4}$
$=\frac{16}{4}$

$+\frac{3}{4}=4+\frac{3}{4}=4\frac{3}{4}$

(iii)

$5\frac{1}{5}$

$+2\frac{1}{7}=\frac{5\times 5+1}{5}+\frac{2\times 7+1}{7}=\frac{26}{5}+\frac{15}{7}=\frac{26\times 7}{5\times 7}+\frac{15\times 5}{7\times 5}$
$=\frac{182}{35}$

$+\frac{75}{35}=\frac{182+75}{35}=\frac{257}{35}$
$=\frac{245+12}{35}$

$=\frac{245}{35}+\frac{12}{35}=7+\frac{12}{35}=7\frac{12}{35}$

(iv) 

$3\frac{1}{5}+2\frac{1}{3}$
$3\frac{1}{5}$

$+2\frac{1}{3}=\frac{3\times 5+1}{5}+\frac{2\times 3+1}{3}=\frac{16}{5}+\frac{7}{3}=\frac{16\times 3}{5\times 3}+\frac{7\times 5}{3\times 5}$
$=\frac{48}{15}$

$+\frac{35}{15}=\frac{48+35}{15}=\frac{83}{15}$
$=\frac{75+8}{15}$

$=\frac{75}{15}+\frac{8}{15}=5+\frac{8}{15}=5\frac{8}{15}$

Question 2:

Subtract.

(i) 

$3\frac{1}{3}-1\frac{1}{4}$

(ii) 

$5\frac{1}{2}-3\frac{1}{3}$

(iii) 

$7\frac{1}{8}-6\frac{1}{10}$

(iv) 

$7\frac{1}{2}-3\frac{1}{5}$

Answer 2:

(i)

$3\frac{1}{3}$

$-1\frac{1}{4}=\frac{10}{3}-\frac{5}{4}=\frac{10\times 4}{3\times 4}-\frac{5\times 3}{4\times 3}=\frac{40}{12}-\frac{15}{12}=\frac{40-15}{12}$
$=\frac{25}{12}$

$=\frac{24+1}{12}=\frac{24}{12}+\frac{1}{12}=2+\frac{1}{12}=2\frac{1}{12}$

(ii)

$5\frac{1}{2}$

$-3\frac{1}{3}=\frac{11}{2}-\frac{10}{3}=\frac{11\times 3}{2\times 3}-\frac{10\times 2}{3\times 2}=\frac{33}{6}-\frac{20}{6}$
$=\frac{33-20}{6}$

$=\frac{13}{6}=2\frac{1}{6}$

(iii)

$7\frac{1}{8}$

$-6\frac{1}{10}=\frac{57}{8}-\frac{61}{10}=\frac{57\times 5}{8\times 5}-\frac{61\times 4}{10\times 4}=\frac{285}{40}-\frac{244}{40}$
$=\frac{285-244}{40}$

$=\frac{41}{40}=1\frac{1}{40}$

(iv)

$7\frac{1}{2}$

$-3\frac{1}{5}=\frac{15}{2}-\frac{16}{5}=\frac{15\times 5}{2\times 5}-\frac{16\times 2}{5\times 2}=\frac{75}{10}-\frac{32}{10}$
$=\frac{75-32}{10}$

$=\frac{43}{10}=4\frac{3}{10}$

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 = 

$2\frac{1}{2}+3\frac{1}{2}$
$=\frac{5}{2}$

$+\frac{7}{2}=\frac{5+7}{2}=\frac{12}{2}$

= 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 

$1-\frac{2}{5}-\frac{1}{3}$

$=\frac{1\times 15}{1\times 15}$

$-\frac{2\times 3}{5\times 3}-\frac{1\times 5}{3\times 5}=\frac{15}{15}-\frac{6}{15}-\frac{5}{15}=\frac{15-6-5}{15}=\frac{4}{15}$

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 

$\frac{4}{7}\left(560\right)+\frac{1}{4}\left(560\right)-\frac{3}{14}\left(560\right)$

 

$=320+140-120=340l$

Hence, 340 l of water will be left in the tank.