Year 7 fractions worksheet answers to download as PDF

These Year 7 fractions worksheet answers give a full worked solution to all 22 questions of the fractions worksheet. Try each question yourself first, then check your method step by step.

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Answer to Question 1
1. \(12\div3=4\), so multiply the top by 4: \(2\times4=8\)
2. \(20\div5=4\), so multiply the bottom by 4: \(8\times4=32\)
3. \(48\div4=12\), so divide the top by 12: \(36\div12=3\)
4. Multiply the top and bottom by the same number: \(\dfrac{3\times2}{5\times2}=\dfrac{6}{10}\) and \(\dfrac{3\times3}{5\times3}=\dfrac{9}{15}\). Any correct pair is accepted.
Answer to Question 2
To simplify fully, divide the numerator and the denominator by their highest common factor. If you divide by a smaller common factor, keep going until no common factor is left.
1. HCF = 6, so \(\dfrac{3}{4}\)
2. HCF = 15, so \(\dfrac{3}{4}\)
3. HCF = 42, so \(\dfrac{2}{3}\)
4. HCF = 50, so \(\dfrac{3}{4}\)
\(\dfrac{18}{24}\), \(\dfrac{45}{60}\) and \(\dfrac{150}{200}\) all equal \(\dfrac{3}{4}\), so (a), (b) and (d). Part (c) is \(\dfrac{2}{3}\).
Answer to Question 3
To change an improper fraction into a mixed number, divide the numerator by the denominator: the quotient is the whole number and the remainder is the new numerator. To go back, multiply the whole number by the denominator and add the numerator.
1. \(17\div5=3\) remainder 2: \(3\dfrac{2}{5}\)
2. \(29\div8=3\) remainder 5: \(3\dfrac{5}{8}\)
3. \(3\times4+3=15\): \(\dfrac{15}{4}\)
4. \(5\times9+2=47\): \(\dfrac{47}{9}\)
5. \(\dfrac{23}{6}=3\dfrac{5}{6}\) because \(23\div6=3\) remainder 5. They are equal.
Answer to Question 4
1. Common denominator 24: \(\dfrac{16}{24}\), \(\dfrac{18}{24}\), \(\dfrac{15}{24}\), \(\dfrac{14}{24}\). Order: \(\dfrac{7}{12}\), \(\dfrac{5}{8}\), \(\dfrac{2}{3}\), \(\dfrac{3}{4}\).
2. Common denominator 20: \(\dfrac{14}{20}\), \(\dfrac{15}{20}\), \(\dfrac{13}{20}\), \(\dfrac{16}{20}\). Largest first: \(\dfrac{4}{5}\), \(\dfrac{3}{4}\), \(\dfrac{7}{10}\), \(\dfrac{13}{20}\).
3. Using twelfths: \(\dfrac{6}{12}\) and \(\dfrac{8}{12}\), so \(\dfrac{7}{12}\) works (many answers are possible).
Writing every fraction with the same denominator turns the comparison into a comparison of whole numbers, which is much easier than guessing from the size of the pieces.
Answer to Question 5
1. The denominators are already the same, so add the numerators: \(2+1=3\), giving \(\dfrac{3}{5}\).
2. The lowest common multiple of 3 and 4 is 12: \(\dfrac{4}{12}+\dfrac{3}{12}=\)\(\dfrac{7}{12}\).
3. LCM of 6 and 4 is 12: \(\dfrac{10}{12}-\dfrac{3}{12}=\)\(\dfrac{7}{12}\).
4. LCM of 3 and 5 is 15: \(\dfrac{10}{15}+\dfrac{9}{15}=\dfrac{19}{15}\), which is \(1\dfrac{4}{15}\) as a mixed number.
5. Change to improper fractions: \(\dfrac{5}{2}+\dfrac{7}{4}=\dfrac{10}{4}+\dfrac{7}{4}=\dfrac{17}{4}=\)\(4\dfrac{1}{4}\).
6. \(\dfrac{16}{5}-\dfrac{17}{10}=\dfrac{32}{10}-\dfrac{17}{10}=\dfrac{15}{10}=\)\(1\dfrac{1}{2}\) after simplifying.
In every part, check that the answer is fully simplified and that improper fractions have been written as mixed numbers.
Answer to Question 6
1. Divide by the denominator, then multiply by the numerator: \(36\div3=12\), \(12\times2=\)24.
2. One fifth of 45 is \(45\div5=9\), so three fifths is \(9\times3=\)27.
3. One eighth of 64 kg is 8 kg, so seven eighths is \(8\times7=\)56 kg.
4. One twelfth of £84 is £7, so five twelfths is \(7\times5=\)£35.
Check: each answer is less than the whole amount, as it should be for a fraction smaller than 1.
Answer to Question 7
1. Common denominator 24: \(\dfrac{15}{24}\) and \(\dfrac{14}{24}\). Pizza A has had more eaten.
2. \(1-\dfrac{5}{8}=\)\(\dfrac{3}{8}\)
3. Pizza A: 3 slices. Pizza B: \(12-7=5\) slices. Total 8 slices.
4. Dev added the tops and the bottoms. You must use a common denominator: \(\dfrac{15}{24}+\dfrac{14}{24}=\dfrac{29}{24}=\)\(1\dfrac{5}{24}\) pizzas.
Answer to Question 8
1. \(48\div4=\)12
2. \(48\div3=\)16
3. \(1-\dfrac{1}{4}-\dfrac{1}{3}=\dfrac{12-3-4}{12}=\)\(\dfrac{5}{12}\)
4. Sold: \(12+16=28\). Money: \(28\times75=2100\)p = £21
5. \(\dfrac{5}{6}\text{ of }48=40\) cupcakes, \(40\times0.75=30\). She raises exactly £30, so she is not right (it is not more than £30).
Check: \(12+16=28\) cupcakes sold and \(48-28=20\) left, and \(\dfrac{20}{48}=\dfrac{5}{12}\), which agrees with part (c).
Answer to Question 9
1. \(360\div5=72,\ 72\times2=144\): £144
2. \(360\div4=\)£90
3. \(1-\dfrac{2}{5}-\dfrac{1}{4}=\dfrac{20-8-5}{20}=\)\(\dfrac{7}{20}\)
4. Pupils pay \(\dfrac{7}{20}\times360=126\), so each pays \(126\div24=5.25\), that is £5.25.
Check: \(144+90+126=360\), so the three shares add up to the whole cost of the trip. Each pupil pays £5.25.
Answer to Question 10
1. Change to improper fractions with denominator 4: \(\dfrac{5}{2}+\dfrac{7}{4}=\dfrac{10}{4}+\dfrac{7}{4}=\dfrac{17}{4}=\)\(4\dfrac{1}{4}\) litres of green paint.
2. She needs \(3\dfrac{1}{5}\) litres and has \(1\dfrac{7}{10}\). In tenths: \(\dfrac{32}{10}-\dfrac{17}{10}=\dfrac{15}{10}=\)\(1\dfrac{1}{2}\) litres more.
3. Dividing by 5 is the same as multiplying by \(\dfrac{1}{5}\): \(\dfrac{17}{4}\times\dfrac{1}{5}=\)\(\dfrac{17}{20}\) of a litre in each jar, a little less than 1 litre.
Answer to Question 11
First write the route as an improper fraction: \(5\dfrac{1}{2}=\dfrac{11}{2}\) km. A fraction “of” the route means multiply.
1. \(\dfrac{3}{4}\times\dfrac{11}{2}=\dfrac{33}{8}=\)\(4\dfrac{1}{8}\) km
2. Wednesday: 11 km. Friday: \(\dfrac{2}{3}\times\dfrac{11}{2}=\dfrac{11}{3}\) km. In 24ths: Monday \(\dfrac{99}{24}\), Wednesday \(\dfrac{264}{24}\), Friday \(\dfrac{88}{24}\). Total \(\dfrac{451}{24}=\)\(18\dfrac{19}{24}\) km
3. \(20-18\dfrac{19}{24}=\)\(1\dfrac{5}{24}\) km
She has run just under 19 km, so she still needs a little over 1 km to reach the coach’s target.
Answer to Question 12
1. \(\dfrac{1}{5}\) of 60 is 12, so \(\dfrac{2}{5}\) is \(12\times2=\)24 m² of potatoes.
2. Common denominator 20: \(1-\dfrac{8}{20}-\dfrac{5}{20}=\)\(\dfrac{7}{20}\) is grass and path.
3. Carrots use \(\dfrac{1}{4}\) of 60, which is 15 m². Shared between 3 rows: \(15\div3=\)5 m² per row.
4. \(\dfrac{1}{4}\) of 24 is 6, so \(\dfrac{3}{4}\) is \(6\times3=\)18 m² of new potatoes.
Answer to Question 13
1. \(\dfrac{1}{8}\) of 240 is 30 litres, so \(\dfrac{3}{8}\) is \(30\times3=\)90 litres.
2. The level rises from \(\dfrac{3}{8}\) to \(\dfrac{7}{8}\): \(\dfrac{7}{8}-\dfrac{3}{8}=\dfrac{4}{8}=\)\(\dfrac{1}{2}\) of the butt.
3. \(\dfrac{7}{8}\) of 240 is \(30\times7=\)210 litres (90 litres plus the 120 litres he added).
4. The plants use \(\dfrac{1}{4}\) of 210, which is 52.5 litres. Left: \(210-52.5=\)157.5 litres.
Answer to Question 14
1. How many \(\dfrac{3}{20}\) fit into \(\dfrac{9}{10}\) is a division. Flip and multiply: \(\dfrac{9}{10}\times\dfrac{20}{3}=\dfrac{180}{30}=\)6 pieces.
2. \(1\dfrac{1}{2}=\dfrac{3}{2}\). \(\dfrac{3}{2}\div\dfrac{3}{20}=\dfrac{3}{2}\times\dfrac{20}{3}=\dfrac{60}{6}=\)10 pieces.
3. In tenths: \(\dfrac{9}{10}+\dfrac{15}{10}=\dfrac{24}{10}=\)\(2\dfrac{2}{5}\) m, enough for \(6+10=16\) pieces.
Answer to Question 15
1. Common denominator 24: \(\dfrac{8}{24}+\dfrac{9}{24}+\dfrac{2}{24}=\dfrac{19}{24}\), so prizes are \(\dfrac{5}{24}\)
2. Music: £80. Food: £90. Decorations: £20. Prizes: \(\dfrac{5}{24}\times240=\)£50. (Check: \(80+90+20+50=240\).)
3. Prizes get £50, which is more than £30. Yes, there is enough.
4. \(240\div2.50=96\) tickets: 96
The fractions in the table must add up to 1 (the whole budget), which is why we subtract them from 1 to find the share for prizes.
Answer to Question 16
1. Week 1: 2.5 km. Week 2: 4 km. Week 3: 6 km. Week 4: 9 km.
2. \(\dfrac{1}{4}+\dfrac{2}{5}=\dfrac{5}{20}+\dfrac{8}{20}=\)\(\dfrac{13}{20}\)
3. Twice week 1 is 5 km, plus 0.5 km is 5.5 km. No week has 5.5 km, so the answer is no.
4. Three times week 1 is 7.5 km. Week 4 is 9 km, which is bigger, so yes.
5. He needs 10 km in week 5, so 1 km more than week 4: 1 km (\(\dfrac{1}{10}\) of 10 km).
Answer to Question 17
Work in stages: the afternoon fraction is a fraction of what is left, not of the original 4 kg.
1. \(4\times\dfrac{3}{8}=\dfrac{12}{8}=\)1.5 kg
2. \(4-1.5=\)2.5 kg
3. \(\dfrac{2}{5}\times2.5=\)1 kg
4. Sold: 2.5 kg = 25 lots of 100 g, so \(25\times1.20=\)£30
5. Left: 1.5 kg. Fraction: \(\dfrac{1.5}{4}=\dfrac{3}{8}\): \(\dfrac{3}{8}\)
Check: \(1.5+1+1.5=4\) kg, so the amounts sold and left add up to the original stock.
Answer to Question 18
1. Common denominator 24: \(\dfrac{9}{24}+\dfrac{6}{24}+\dfrac{4}{24}+\dfrac{2}{24}=\dfrac{21}{24}\), so train is \(\dfrac{3}{24}=\dfrac{1}{8}\)
2. Walk 45, bus 30, car 20, cycle 10, train 15. (Total 120.)
3. \(1-\dfrac{3}{8}=\)\(\dfrac{5}{8}\)
4. Car + bus = \(\dfrac{1}{6}+\dfrac{1}{4}=\dfrac{5}{12}\), which is less than \(\dfrac{1}{2}=\dfrac{6}{12}\). She is not correct (50 pupils out of 120).
Answer to Question 19
1. 36 is 3 times 12. Flour \(4\dfrac{1}{2}\) cups, butter \(2\dfrac{1}{4}\) cups, sugar 1 cup, milk 2 cups.
2. Half the recipe: flour \(\dfrac{3}{4}\) cup, butter \(\dfrac{3}{8}\) cup, sugar \(\dfrac{1}{6}\) cup, milk \(\dfrac{1}{3}\) cup.
3. Flour for one scone is \(1\dfrac{1}{2}\div12=\dfrac{1}{8}\) cup. \(2\div\dfrac{1}{8}=16\), so she can make 16 scones.
4. No. \(\dfrac{2}{3}-\dfrac{1}{2}=\dfrac{4}{6}-\dfrac{3}{6}=\dfrac{1}{6}\), so she is short by \(\dfrac{1}{6}\) cup.
Answer to Question 20
1. Ali added tops and bottoms. Correct: \(\dfrac{3}{6}+\dfrac{2}{6}=\)\(\dfrac{5}{6}\)
2. Beth multiplied by the fraction instead of flipping it. Correct: \(\dfrac{3}{4}\times\dfrac{2}{1}=\dfrac{6}{4}=\)\(1\dfrac{1}{2}\)
3. Carl multiplied the whole numbers and the fractions separately. Correct: \(\dfrac{3}{2}\times\dfrac{7}{3}=\dfrac{21}{6}=\)\(3\dfrac{1}{2}\)
4. Ali: \(\dfrac{2}{5}\) is smaller than \(\dfrac{1}{2}\), but we added something to \(\dfrac{1}{2}\), so the answer must be bigger than \(\dfrac{1}{2}\).
Answer to Question 21
1. \(\dfrac{1}{5}\) of £20 is £4, so \(\dfrac{2}{5}\) is £8 saved.
2. \(\dfrac{1}{4}\) of £20 is £5 on snacks.
3. \(20-8-5=\)£7 is left after saving and snacks.
4. The magazine is \(\dfrac{1}{5}\) of the £7 that remains: \(7\div5=\)£1.40.
5. Bus fares: \(7-1.40=£5.60\). As a fraction of £20: \(\dfrac{5.60}{20}=\dfrac{56}{200}=\)\(\dfrac{7}{25}\) after dividing by 8.
Answer to Question 22
1. \(45\div3\times5=\)75 litres
2. \(2.5-0.5=\)2 litres per minute
3. Needed: \(75-45=30\) litres. \(30\div2=\)15 minutes
4. Net rate is \(2.5-0.75=1.75\). \(30\div1.75=17\dfrac{1}{7}\) minutes, so it takes \(2\dfrac{1}{7}\) minutes longer (about 2 min 9 s).
The tank fills more slowly with the bigger leak because the net rate falls from 2 to 1.75 litres per minute, so each litre takes longer to add.

How to use these Year 7 fractions worksheet answers

First, try every question on the worksheet without looking. Then compare your working with the solution line by line, not only the final answer. When your answer is different, find the first step that changes, because that is where the mistake is. After that, read the matching part of the fractions lesson again. Finally, check that the method is secure with the quiz or the test.

Also, the Year 7 fractions worksheet answers are available as a PDF, so you can mark printed work quickly. Moreover, all the pages of this chapter work together, so you can move between them at any time. For example, many pupils read the lesson on Monday, practise on Wednesday and test themselves on Friday. In short, these Year 7 fractions worksheet answers show every step, so nothing is left unexplained.