Class 7 Maths Chapter 8 Working with Fractions Questions

Final answer: $\frac{7}{2}$ glasses in a week and $\frac{31}{2}$ glasses in January.

1. A week has $7$ days, so the weekly milk is $7\times\frac{1}{2}=\frac{7}{2}$ glasses.
2. January has $31$ days, so the monthly milk is $31\times\frac{1}{2}=\frac{31}{2}$ glasses.
3. The answers may also be read as $3\frac{1}{2}$ glasses and $15\frac{1}{2}$ glasses.

Repeated daily use is multiplication of the daily amount by the number of days.

Final answer: $\frac{7}{2}$ glasses in a week and $\frac{31}{2}$ glasses in January.

Final answer: $\frac{1}{8}$ km in one day and $\frac{5}{8}$ km in five days.

1. In $8$ days the work is $1$ km.
2. In $1$ day the work is $1\div8=\frac{1}{8}$ km.
3. In $5$ days the work is $5\times\frac{1}{8}=\frac{5}{8}$ km.

If a total quantity is shared equally across days, one day's work is found by division.

Final answer: $\frac{1}{8}$ km in one day and $\frac{5}{8}$ km in five days.

Final answer: $\frac{5}{3}$ litres in one week and $\frac{20}{3}$ litres in four weeks.

1. There are $3$ families in all.
2. Oil per family per week is $5\div3=\frac{5}{3}$ litres.
3. For $4$ weeks, multiply by $4$: $4\times\frac{5}{3}=\frac{20}{3}$ litres.

Equal sharing among three families means division by $3$.

Final answer: $\frac{5}{3}$ litres in one week and $\frac{20}{3}$ litres in four weeks.

Final answer: $2\frac{1}{2}$ hours after 10 pm, that is 12:30 am.

1. Tuesday is one shift after Monday, Wednesday is two shifts, and Thursday is three shifts.
2. Total delay is $3\times\frac{5}{6}=\frac{15}{6}=\frac{5}{2}$ hours.
3. $\frac{5}{2}$ hours is $2\frac{1}{2}$ hours, so the time is 12:30 am.

From Monday to Thursday there are three daily shifts.

Final answer: $2\frac{1}{2}$ hours after 10 pm, that is 12:30 am.

Final answer: $4\frac{1}{5}$, $1\frac{1}{3}$, $7\frac{5}{7}$ and $7\frac{1}{11}$.

1. $7\times\frac{3}{5}=\frac{21}{5}=4\frac{1}{5}$.
2. $4\times\frac{1}{3}=\frac{4}{3}=1\frac{1}{3}$.
3. $\frac{9}{7}\times6=\frac{54}{7}=7\frac{5}{7}$.
4. $\frac{13}{11}\times6=\frac{78}{11}=7\frac{1}{11}$.

A whole number can be written as a fraction with denominator $1$.

Final answer: $4\frac{1}{5}$, $1\frac{1}{3}$, $7\frac{5}{7}$ and $7\frac{1}{11}$.

Final answer: $\frac{1}{15}$, $\frac{1}{12}$, $\frac{1}{10}$, $\frac{1}{30}$ and $\frac{1}{216}$.

1. $\frac{1}{3}\times\frac{1}{5}=\frac{1}{15}$.
2. $\frac{1}{4}\times\frac{1}{3}=\frac{1}{12}$.
3. $\frac{1}{5}\times\frac{1}{2}=\frac{1}{10}$.
4. $\frac{1}{6}\times\frac{1}{5}=\frac{1}{30}$.
5. $\frac{1}{12}\times\frac{1}{18}=\frac{1}{216}$.

For unit fractions, multiply the denominators and keep numerator $1$.

Final answer: $\frac{1}{15}$, $\frac{1}{12}$, $\frac{1}{10}$, $\frac{1}{30}$ and $\frac{1}{216}$.

Final answer: $\frac{8}{15}$, $\frac{1}{6}$, $\frac{3}{10}$ and $\frac{2}{5}$.

1. $\frac{2}{3}\times\frac{4}{5}=\frac{8}{15}$.
2. $\frac{1}{4}\times\frac{2}{3}=\frac{2}{12}=\frac{1}{6}$.
3. $\frac{3}{5}\times\frac{1}{2}=\frac{3}{10}$.
4. $\frac{4}{6}\times\frac{3}{5}=\frac{12}{30}=\frac{2}{5}$.

Multiply numerators together and denominators together, then reduce if possible.

Final answer: $\frac{8}{15}$, $\frac{1}{6}$, $\frac{3}{10}$ and $\frac{2}{5}$.

Final answer: $\frac{7}{30}$, $\frac{14}{30}$, $\frac{21}{40}$, $\frac{49}{100}$ and $\frac{10}{7}$ hours.

1. $\frac{1}{3}$ hour fills $\frac{1}{3}\times\frac{7}{10}=\frac{7}{30}$.
2. $\frac{2}{3}$ hour fills $\frac{2}{3}\times\frac{7}{10}=\frac{14}{30}$.
3. $\frac{3}{4}$ hour fills $\frac{3}{4}\times\frac{7}{10}=\frac{21}{40}$.
4. $\frac{7}{10}$ hour fills $\frac{7}{10}\times\frac{7}{10}=\frac{49}{100}$.
5. For one full tank, time is $1\div\frac{7}{10}=\frac{10}{7}$ hours.

Tank filling is rate multiplied by time. Full-tank time is found by division.

Final answer: $\frac{7}{30}$, $\frac{14}{30}$, $\frac{21}{40}$, $\frac{49}{100}$ and $\frac{10}{7}$ hours.

Final answer: Krishna gets $\frac{5}{12}$, Bora gets $\frac{5}{18}$, and Somu keeps $\frac{5}{36}$.

1. Remaining land after the road is $1-\frac{1}{6}=\frac{5}{6}$.
2. Krishna gets half of the remaining land: $\frac{1}{2}\times\frac{5}{6}=\frac{5}{12}$.
3. Bora gets one-third of the remaining land: $\frac{1}{3}\times\frac{5}{6}=\frac{5}{18}$.
4. Somu keeps $\frac{5}{6}-\frac{5}{12}-\frac{5}{18}=\frac{30-15-10}{36}=\frac{5}{36}$.

First find the remaining land, then take fractions of that remaining part.

Final answer: Krishna gets $\frac{5}{12}$, Bora gets $\frac{5}{18}$, and Somu keeps $\frac{5}{36}$.

Final answer: $36$ square feet.

1. Convert mixed fractions: $3\frac{3}{4}=\frac{15}{4}$ and $9\frac{3}{5}=\frac{48}{5}$.
2. Area is $\frac{15}{4}\times\frac{48}{5}$.
3. Cancel $15\div5=3$ and $48\div4=12$.
4. The product is $3\times12=36$ square feet.

Area of a rectangle equals length multiplied by breadth.

Final answer: $36$ square feet.

Final answer: $2\frac{1}{4}$ m.

1. From first to last sapling, there are $4-1=3$ gaps.
2. Each gap is $\frac{3}{4}$ m.
3. Total distance is $3\times\frac{3}{4}=\frac{9}{4}=2\frac{1}{4}$ m.

Four saplings in a row create three equal gaps.

Final answer: $2\frac{1}{4}$ m.

Final answer: $\frac{3}{20}$ of $4$ kg is heavier.

1. $\frac{12}{15}$ of $500$ grams is $\frac{12}{15}\times500=400$ grams.
2. $4$ kg is $4000$ grams.
3. $\frac{3}{20}$ of $4000$ grams is $600$ grams.
4. $600$ grams is heavier than $400$ grams.

Compare both quantities in the same unit.

Final answer: $\frac{3}{20}$ of $4$ kg is heavier.

Final answer: $\frac{27}{7}$, $\frac{7}{4}$, $1$, $1$, $\frac{16}{9}$, $\frac{49}{4}$, $15$, $\frac{9}{5}$, $\frac{2}{11}$ and $\frac{8}{3}$.

1. $3\div\frac{7}{9}=3\times\frac{9}{7}=\frac{27}{7}$.
2. $\frac{14}{4}\div2=\frac{14}{4}\times\frac{1}{2}=\frac{7}{4}$.
3. $\frac{2}{3}\div\frac{2}{3}=1$ and $\frac{14}{6}\div\frac{7}{3}=1$.
4. $\frac{4}{3}\div\frac{3}{4}=\frac{16}{9}$, $\frac{7}{4}\div\frac{1}{7}=\frac{49}{4}$, and $\frac{8}{2}\div\frac{4}{15}=15$.
5. $\frac{1}{5}\div\frac{1}{9}=\frac{9}{5}$, $\frac{1}{6}\div\frac{11}{12}=\frac{2}{11}$, and $3\frac{2}{3}\div1\frac{3}{8}=\frac{8}{3}$.

To divide by a fraction, multiply by its reciprocal.

Final answer: $\frac{27}{7}$, $\frac{7}{4}$, $1$, $1$, $\frac{16}{9}$, $\frac{49}{4}$, $15$, $\frac{9}{5}$, $\frac{2}{11}$ and $\frac{8}{3}$.

Final answer: The correct choices are (iii), (iv), (iii), giving $32$, $\frac{1}{16}$ m and $30$.

1. For $8$ m lace with $\frac{1}{4}$ m per bag, use $8\div\frac{1}{4}=32$ bags.
2. For $\frac{1}{2}$ m ribbon shared among $8$ badges, use $\frac{1}{2}\div8=\frac{1}{16}$ m per badge.
3. For $5$ kg flour with $\frac{1}{6}$ kg per loaf, use $5\div\frac{1}{6}=30$ loaves.

Choose division when a total quantity is split into equal small amounts.

Final answer: The correct choices are (iii), (iv), (iii), giving $32$, $\frac{1}{16}$ m and $30$.

Final answer: $\frac{1}{8}$ kg.

1. The flour for $12$ rotis is $\frac{1}{4}$ kg.
2. $6$ rotis are half of $12$ rotis.
3. So flour needed is $\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}$ kg.

Six rotis are half of twelve rotis.

Final answer: $\frac{1}{8}$ kg.

Final answer: $40$.

1. $1\div\frac{1}{6}=6$.
2. $1\div\frac{1}{10}=10$, $1\div\frac{1}{13}=13$, $1\div\frac{1}{9}=9$, and $1\div\frac{1}{2}=2$.
3. Add them: $6+10+13+9+2=40$.

Dividing $1$ by a unit fraction gives its denominator.

Final answer: $40$.

Final answer: $200$ pages.

1. Yesterday she read $\frac{1}{5}\times400=80$ pages.
2. Today she read $\frac{3}{10}\times400=120$ pages.
3. Total read is $80+120=200$ pages.
4. Pages left are $400-200=200$ pages.

The pages read are found by multiplying total pages by the fraction read.

Final answer: $200$ pages.

Final answer: $44$ km.

1. Convert $2\frac{3}{4}$ to $\frac{11}{4}$.
2. Distance is $16\times\frac{11}{4}$.
3. Since $16\div4=4$, the product is $4\times11=44$ km.

Distance equals mileage multiplied by litres used.

Final answer: $44$ km.

Final answer: $4\frac{2}{3}$ hours.

1. $5\frac{1}{6}=\frac{31}{6}$ hours.
2. $\frac{1}{2}=\frac{3}{6}$ hours.
3. Difference is $\frac{31}{6}-\frac{3}{6}=\frac{28}{6}=\frac{14}{3}=4\frac{2}{3}$ hours.

Time saved equals longer time minus shorter time.

Final answer: $4\frac{2}{3}$ hours.

Final answer: $\frac{1}{15}$ of the whole cake.

1. Cake remaining is $1-\frac{4}{5}=\frac{1}{5}$.
2. This remaining cake is shared by $3$ friends.
3. Each friend gets $\frac{1}{5}\div3=\frac{1}{5}\times\frac{1}{3}=\frac{1}{15}$.

Find the remaining fraction first, then divide it equally.

Final answer: $\frac{1}{15}$ of the whole cake.

Final answer: Options (a), (c) and (e).

1. $\frac{565}{465}>1$ because $565>465$.
2. $\frac{707}{676}>1$ because $707>676$.
3. Multiplying by a number greater than $1$ makes the other positive factor larger.
4. So the product is greater than $\frac{565}{465}$, greater than $\frac{707}{676}$, and greater than $1$.

Both factors are greater than $1$, so the product is greater than each factor and greater than $1$.

Final answer: Options (a), (c) and (e).

Final answer: $\frac{3}{32}$ of the whole square.

1. The top-right small square is $\frac{1}{4}$ of the whole square.
2. The triangle inside that square is half of it, so its area is $\frac{1}{2}\times\frac{1}{4}=\frac{1}{8}$.
3. The shaded part is $\frac{3}{4}$ of this triangle.
4. So shaded area is $\frac{3}{4}\times\frac{1}{8}=\frac{3}{32}$ of the whole square.

The shaded part is three-fourths of a triangle that is half of one-fourth of the big square.

Final answer: $\frac{3}{32}$ of the whole square.

Final answer: Mango tree: $\frac{29}{32}$; sugarcane field: $\frac{3}{32}$.

1. Trace the branches that end near the mango tree and add their shares.
2. The mango side receives $\frac{29}{32}$ of the original group.
3. The sugarcane side receives $\frac{3}{32}$ of the original group.
4. The fractions add to $\frac{29}{32}+\frac{3}{32}=1$, so the whole colony is accounted for.

At each split, the group divides equally, so each branch carries half of the incoming group.

Final answer: Mango tree: $\frac{29}{32}$; sugarcane field: $\frac{3}{32}$.

Final answer: $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{5}$, $\frac{1}{10}$, and the general product is $\frac{1}{n}$.

1. $1-\frac{1}{2}=\frac{1}{2}$.
2. $(1-\frac{1}{2})(1-\frac{1}{3})=\frac{1}{2}\times\frac{2}{3}=\frac{1}{3}$.
3. The product up to $(1-\frac{1}{5})$ is $\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times\frac{4}{5}=\frac{1}{5}$.
4. The product up to $(1-\frac{1}{10})$ is $\frac{1}{10}$.
5. In general, $(1-\frac{1}{2})(1-\frac{1}{3})\cdots(1-\frac{1}{n})=\frac{1}{n}$.

Each factor $1-\frac{1}{k}$ equals $\frac{k-1}{k}$, so consecutive factors cancel.

Final answer: $\frac{1}{2}$, $\frac{1}{3}$, $\frac{1}{5}$, $\frac{1}{10}$, and the general product is $\frac{1}{n}$.