Adding and subtracting fractions · applications

Applications: Adding and Subtracting Fractions

13 problems

PSLE · GCSE Higher

01

Two Unlike Fractions of One Whole Used

heuristicCommon Denominator / Add the Parts Used, Then Subtract from One Whole

Mr Lee used 12 of a tin of paint on a door and 13 of the same tin on a gate. (a) What fraction of the tin of paint did he use altogether? (b) What fraction of the tin of paint is left?

Tin1/61/61/61/61/61/6
6 is a multiple of 2 and of 3, so the tin is cut into 6 equal parts.
Draw the tin as one bar. The denominators are 2 and 3, and 6 is the smallest number that is a multiple of both, so cut the bar into 6 equal parts.
step 1 of 4

Both fractions are parts of the same tin, so draw the tin as one bar. Halves and thirds are parts of different sizes, so cut the bar into sixths, which can show both.

All steps
  1. Draw the tin as one bar. The denominators are 2 and 3, and 6 is the smallest number that is a multiple of both, so cut the bar into 6 equal parts.
  2. Write each fraction in sixths: 12 = 36 and 13 = 26.
  3. (a) He used 36 + 26 = 56 of the tin.
  4. (b) The whole tin is 66, so 66 − 56 = 16 of the tin is left. Check: 3 + 2 + 1 = 6 parts.

answer(a) 56; (b) 16

techniqueAdding Fractions with Unlike Denominators · A Common Denominator · Subtracting Fractions

examsPSLE · GCSE Higher

Common pitfalls

  • Adding the numerators and the denominators to get 1 + 12 + 3 = 25. That is less than 12, the amount used on the door alone, so it cannot be the total. The parts must be the same size before they are counted together.
  • Giving 56 as the fraction that is left. 56 is the fraction that was used, and the fraction left is what remains of the whole tin after it is taken away.
02

Three Shares of One Job Compared and Added

heuristicCompare with a Common Denominator / Then Add the Shares

Three friends painted a fence. Aisha painted 25 of the fence, Ben painted 14 of it and Chen painted 310 of it. (a) Who painted the largest part of the fence? (b) What fraction of the fence is still not painted?

Aisha1/51/51/51/51/52/5Ben1/41/41/41/41/4Chen1/101/101/101/101/101/101/101/101/101/103/10
Fifths, quarters and tenths are parts of different sizes, so the three shares cannot be compared yet.
To compare the shares they need the same denominator. 20 is the smallest number that is a multiple of 5, 4 and 10, so cut the fence into 20 equal parts.
step 1 of 5

Draw the fence as one bar cut into twentieths, because fifths, quarters and tenths can all be written as twentieths. Each share is then a number of equal parts.

All steps
  1. To compare the shares they need the same denominator. 20 is the smallest number that is a multiple of 5, 4 and 10, so cut the fence into 20 equal parts.
  2. Write each share in twentieths: 25 = 820, 14 = 520 and 310 = 620.
  3. (a) 820 is the greatest of the three, so Aisha painted the largest part of the fence.
  4. Add the three shares: 820 + 520 + 620 = 1920 of the fence is painted.
  5. (b) The whole fence is 2020, so 2020 − 1920 = 120 of the fence is not painted.

answer(a) Aisha, who painted 820 of the fence; (b) 120

techniqueA Common Denominator · Ordering Fractions · Adding Fractions with Unlike Denominators

examsPSLE

Common pitfalls

  • Choosing Chen because 310 has the largest numerator and the largest denominator. Tenths are smaller parts than fifths, so the shares can be compared only after they are written with the same denominator.
  • Adding the numerators and the denominators to get 2 + 1 + 35 + 4 + 10 = 619. That is less than Aisha's share alone. Only the numerators are added, and only after the denominators are the same.
03

A Count Written as a Fraction in Simplest Form

heuristicEquivalent Fractions / Divide or Multiply the Numerator and the Denominator by the Same Number

In Class 4A, 18 of the 24 pupils walk to school. (a) What fraction of the pupils in Class 4A walk to school? Give the answer in its simplest form. (b) Class 4B has 28 pupils, and the same fraction of them walk to school. How many pupils in Class 4B walk to school?

Class 4A18/24
18 of the 24 equal parts are shaded: 1824 of Class 4A walk to school.
Write the count as a fraction: 18 of 24 pupils is 1824. Draw a strip of 24 equal parts and shade 18 of them.
step 1 of 5

Draw Class 4A as a strip of 24 equal parts. Grouping the parts makes a strip of the same length with fewer and larger parts, which is the same fraction in a simpler form.

All steps
  1. Write the count as a fraction: 18 of 24 pupils is 1824. Draw a strip of 24 equal parts and shade 18 of them.
  2. 18 and 24 can both be divided by 6, so put the parts in groups of 6: 18 ÷ 6 = 3 and 24 ÷ 6 = 4. The strip is now 4 larger parts with 3 of them shaded.
  3. (a) 1824 = 34 of the pupils in Class 4A walk to school. It is in its simplest form because 3 and 4 have no common factor other than 1.
  4. For Class 4B, write 34 as an equivalent fraction with the denominator 28. Since 4 × 7 = 28, multiply the numerator by 7 as well: 3 × 7 = 21.
  5. (b) 34 = 2128, so 21 pupils in Class 4B walk to school. Check: 21 ÷ 7 = 3 and 28 ÷ 7 = 4.

answer(a) 34; (b) 21 pupils

techniqueSimplest Form · Equivalent Fractions

examsPSLE

Common pitfalls

  • Dividing by 2 once and stopping at 912. 9 and 12 can still both be divided by 3, so the fraction is not yet in its simplest form.
  • Adding 4 to both numbers because Class 4B has 4 more pupils, which gives 2228. Adding the same number to the numerator and the denominator changes the fraction. Equivalent fractions are made by multiplying or dividing both by the same number.
04

Two Strips Glued with an Overlap

heuristicAdd the Two Lengths / Subtract the Overlap Once

A paper strip 12 m long and a paper strip 45 m long are glued together to make one long strip. The two strips overlap by 110 m where they are glued. (a) What is the total length of the two strips before they are glued? Give the answer as a mixed number. (b) How long is the glued strip? Give the answer as a mixed number in its simplest form.

First1/2 = 5/10 mSecond4/5 = 8/10 m
Each part is one tenth of a metre: 12 = 510 and 45 = 810.
10 is a multiple of 2 and of 5, so write both lengths in tenths: 12 = 510 and 45 = 810.
step 1 of 4

Draw both strips in tenths of a metre. Placed end to end they show the total length. Sliding the second strip over the end of the first hides one tenth, and the glued strip is what can still be seen.

All steps
  1. 10 is a multiple of 2 and of 5, so write both lengths in tenths: 12 = 510 and 45 = 810.
  2. (a) Before gluing, the strips are 510 + 810 = 1310 = 1310 m long altogether.
  3. In the overlap the two strips lie one on top of the other, so that 110 m is counted twice in the total. Subtract it once: 1310 − 110 = 1210.
  4. (b) Simplify: 1210 = 65 = 115 m. The glued strip is 115 m long.

answer(a) 1310 m; (b) 115 m

techniqueAdding Fractions with Unlike Denominators · Subtracting Fractions · Simplest Form

examsPSLE · GCSE Higher

Common pitfalls

  • Subtracting the overlap twice, once for each strip, to get 1110 m. The overlap is a part of both strips, but only one of the two layers is hidden, so it is subtracted once.
  • Adding the overlap to the total to get 1410 m. An overlap makes the glued strip shorter than the two strips placed end to end, not longer.
05

How Much More One Distance Is Than Another

heuristicComparison Bars / Subtract with a Common Denominator, Then Simplify

Mei ran 34 km and Siti ran 512 km. (a) How much farther did Mei run than Siti? (b) Devi ran 14 km farther than Siti. How much farther did Mei run than Devi? Give each answer in its simplest form.

Mei9/12 kmSiti5/12 km
Each part is 112 km. Mei ran 34 = 912 km and Siti ran 512 km.
12 is a multiple of 4, so write Mei's distance in twelfths: 34 = 912. Mei's bar is 9 parts long and Siti's bar is 5 parts long.
step 1 of 5

Draw one bar for each girl, all on the same scale, with one kilometre cut into twelfths. The difference between two distances is the part of the longer bar that goes past the end of the shorter bar.

All steps
  1. 12 is a multiple of 4, so write Mei's distance in twelfths: 34 = 912. Mei's bar is 9 parts long and Siti's bar is 5 parts long.
  2. Mei's bar is 9 − 5 = 4 parts longer, so the difference is 912 − 512 = 412 km.
  3. (a) Divide the numerator and the denominator by 4: 412 = 13. Mei ran 13 km farther than Siti.
  4. Devi ran 14 = 312 km farther than Siti, so Devi ran 512 + 312 = 812 km.
  5. (b) Mei ran 912 − 812 = 112 km farther than Devi. Check: 13 − 14 = 412 − 312 = 112.

answer(a) 13 km; (b) 112 km

techniqueSubtracting Fractions · A Common Denominator · Simplest Form

examsPSLE · GCSE Higher

Common pitfalls

  • Subtracting the numerators and the denominators to get 5 − 312 − 4 = 28. The denominator names the size of the parts. It is made the same for both fractions first, and then only the numerators are subtracted.
  • Leaving the answer to part (a) as 412. 4 and 12 can both be divided by 4, so the simplest form is 13.
06

Fractions of an Hour Added and Changed to Minutes

heuristicAdd the Fractions First / One Hour Is 60 Minutes

Ravi practiced the piano for 23 hour and then read a book for 15 hour. (a) What fraction of an hour did the two activities take altogether? (b) How many minutes is that?

One hourpiano 10/15reading 3/15
The hour is cut into 15 equal parts: 23 = 1015 and 15 = 315.
15 is the smallest number that is a multiple of 3 and 5, so cut the hour into 15 equal parts: 23 = 1015 and 15 = 315.
step 1 of 4

Draw one hour as a bar. Thirds and fifths can both be written as fifteenths, so cut the hour into 15 equal parts. Each part is the same number of minutes.

All steps
  1. 15 is the smallest number that is a multiple of 3 and 5, so cut the hour into 15 equal parts: 23 = 1015 and 15 = 315.
  2. (a) Altogether the two activities took 1015 + 315 = 1315 hour.
  3. One hour is 60 minutes, so each of the 15 parts is 60 ÷ 15 = 4 minutes.
  4. (b) 13 parts are 13 × 4 = 52 minutes. Check: 23 hour is 40 minutes and 15 hour is 12 minutes, and 40 + 12 = 52.

answer(a) 1315 hour; (b) 52 minutes

techniqueAdding Fractions with Unlike Denominators · A Common Denominator

examsPSLE · GCSE Higher

Common pitfalls

  • Reading 15 hour as 5 minutes. One fifth of an hour is 60 ÷ 5 = 12 minutes.
  • Adding the numerators and the denominators to get 38 hour. That is less than the 23 hour spent on the piano alone. Write both fractions in fifteenths before adding.
07

Cups Combined and Compared with the Size of a Bowl

heuristicAdd Three Fractions / Write the Improper Fraction as a Mixed Number

A recipe mixes 34 cup of flour, 58 cup of sugar and 12 cup of cocoa in a bowl that holds 2 cups. (a) How many cups of ingredients go into the bowl altogether? Give the answer as a mixed number. (b) How much space is left in the bowl?

Bowl1 cup1 cupflour 6/8sugar 5/8cocoa 4/8
The bowl is 2 cups, each cut into eighths: 34 = 68 and 12 = 48.
8 is a multiple of 4 and of 2, so write every amount in eighths: 34 = 68 and 12 = 48. The sugar is already 58.
step 1 of 4

Draw the bowl as two cups, each cut into eighths, because quarters and halves can both be written as eighths. Fill the parts one ingredient at a time.

All steps
  1. 8 is a multiple of 4 and of 2, so write every amount in eighths: 34 = 68 and 12 = 48. The sugar is already 58.
  2. Add the three amounts: 68 + 58 + 48 = 158 cups.
  3. (a) 8 eighths make 1 whole cup, so 158 = 88 + 78 = 178 cups.
  4. (b) The bowl holds 2 cups, which is 168. The space left is 168 − 158 = 18 cup.

answer(a) 178 cups; (b) 18 cup

techniqueAdding Fractions with Unlike Denominators · Adding Fractions with the Same Denominator · Mixed Numbers

examsPSLE · GCSE Higher

Common pitfalls

  • Adding the numerators and the denominators to get 914 cup. That is less than the 34 cup of flour alone, so it cannot be the total.
  • Deciding that the ingredients do not fit because the numerator 15 is larger than 2. Compare like with like: the bowl holds 2 cups, which is 168, and 158 is less than 168.
08

A Missing Part That Completes One Whole

heuristicThe Parts Add Up to One Whole / Subtract the Known Parts

A relay race is exactly 1 km long and is run in three legs. The first leg is 512 km long and the third leg is 13 km long. (a) How long is the second leg? (b) How far from the start is the end of the second leg? Give each answer in its simplest form.

Racesecond ?1 kmfirst 5/12third 4/12
The race is 12 equal parts. The first leg is 5 of them and the third leg is 13 = 412, the last 4.
12 is a multiple of 3, so cut the 1 km into 12 equal parts. The first leg is 512 km and the third leg is 13 = 412 km.
step 1 of 5

Draw the race as one bar 1 km long, cut into twelfths. Mark the first leg from the start and the third leg from the finish. The parts between them are the second leg.

All steps
  1. 12 is a multiple of 3, so cut the 1 km into 12 equal parts. The first leg is 512 km and the third leg is 13 = 412 km.
  2. The two known legs take 512 + 412 = 912 km of the race.
  3. (a) The whole race is 1212 km, so the second leg is 1212 − 912 = 312 = 14 km.
  4. The end of the second leg comes after the first two legs: 512 + 312 = 812 km from the start.
  5. (b) Divide the numerator and the denominator by 4: 812 = 23 km. Check: 1 − 13 = 23.

answer(a) 14 km; (b) 23 km

techniqueSubtracting Fractions · Adding Fractions with Unlike Denominators · Simplest Form

examsPSLE · GCSE Higher

Common pitfalls

  • Subtracting only the first leg from 1 km and giving 712 km. That is the second and third legs together, so the third leg must be subtracted as well.
  • Writing 1 − 912 as 1 − 912, which cannot be worked out. Write the whole as 1212 first, then subtract the numerators.
09

Forward and Back Along a Number Line

heuristicNumber Line in Twelfths / Forward Adds and Back Subtracts

A toy robot starts at 0 on a number line marked in metres. It moves 56 m forward, then 12 m back, then 14 m forward. (a) Where is the robot after the second move? (b) Where is the robot after the third move? Give each answer in its simplest form.

01Each mark is 1/12 m.
From 0 to 1 m there are 12 marks. The moves are 1012 forward, 612 back and 312 forward.
12 is the smallest number that is a multiple of 6, 2 and 4. In twelfths the moves are 56 = 1012, 12 = 612 and 14 = 312.
step 1 of 5

Draw a number line from 0 to 1 and cut it into twelfths, because sixths, halves and quarters can all be written as twelfths. Each move is then a whole number of marks.

All steps
  1. 12 is the smallest number that is a multiple of 6, 2 and 4. In twelfths the moves are 56 = 1012, 12 = 612 and 14 = 312.
  2. The first move takes the robot 10 marks forward from 0, to 1012.
  3. The second move takes it 6 marks back: 1012 − 612 = 412.
  4. (a) 412 = 13, so after the second move the robot is at 13 m.
  5. (b) The third move takes it 3 marks forward: 412 + 312 = 712. The robot ends at 712 m.

answer(a) 13 m; (b) 712 m

techniqueSubtracting Fractions · Adding Fractions with Unlike Denominators · Simplest Form

examsPSLE · GCSE Higher

Common pitfalls

  • Adding all three moves to get 1912 m. That is the distance the robot travelled. Its position is found by subtracting the move back.
  • Changing 412 to 13 and then adding 13 + 14 as 27. Thirds and quarters are parts of different sizes. Keep the position as 412 until the last move is added.
10

Fractions of Two Different Wholes

heuristicFractions Add Only over the Same Whole / Change Each Share to Grams First

A small pizza has a mass of 300 g and a large pizza has a mass of 600 g. Ali eats 12 of the small pizza and Ben eats 13 of the large pizza. (a) How many grams of pizza does each boy eat? (b) What fraction of the large pizza is the amount the two boys eat altogether? Give the answer in its simplest form.

SmallAli 150 g300 gLarge600 g
The small pizza is cut into 2 equal parts: 300 ÷ 2 = 150 g for Ali.
Ali eats 12 of the small pizza. Cut the small bar into 2 equal parts: 300 ÷ 2 = 150 g.
step 1 of 5

Draw the two pizzas as bars on the same scale, so that the large bar is twice as long as the small bar. A half of the short bar and a third of the long bar can then be compared in grams.

All steps
  1. Ali eats 12 of the small pizza. Cut the small bar into 2 equal parts: 300 ÷ 2 = 150 g.
  2. Ben eats 13 of the large pizza. Cut the large bar into 3 equal parts: 600 ÷ 3 = 200 g.
  3. (a) Ali eats 150 g and Ben eats 200 g. Ben eats more, although 13 is less than 12, because his fraction is of a larger whole.
  4. Altogether they eat 150 + 200 = 350 g. As a fraction of the large pizza this is 350600.
  5. (b) Divide the numerator and the denominator by 50: 350600 = 712 of the large pizza.

answer(a) Ali: 150 g, Ben: 200 g; (b) 712

techniqueSimplest Form · Adding Fractions with Unlike Denominators · A Fraction of an Amount

examsPSLE

Common pitfalls

  • Adding 12 + 13 = 56 and calling it 56 of a pizza. The two fractions are of different wholes, so they cannot be added until both are written as fractions of the same pizza.
  • Deciding that Ali eats more because 12 is greater than 13. Two fractions can be compared in that way only when they are fractions of the same whole.
11

Resin in Three Containers Linked by More and Less

heuristicMixed Numbers on Comparison Bars / Wholes with Wholes, Parts with Parts

A workshop keeps epoxy resin in three containers. Container A holds 316 litres of resin. Container B holds 134 litres more than Container A. Container C holds 223 litres less than Containers A and B together. During a project, a technician uses 458 litres of resin from Container B and 156 litres from Container C. (a) How much resin do Containers A and B hold together at first? (b) How much resin is left in the three containers altogether at the end of the project? Give each answer as a mixed number in its simplest form.

A3 1/6 LB3 1/61 3/44 11/12 L
In twelfths, 316 = 3212 and 134 = 1912, so B holds 3212 + 1912 = 41112 litres.
12 is a multiple of 6 and of 4, so write the parts in twelfths: 316 = 3212 and 134 = 1912. Container B holds 3212 + 1912 = 41112 litres.
step 1 of 6

Draw each container as a bar: B is as long as A with a piece added, and C is as long as A and B together with a piece taken off. Add and subtract the whole litres and the parts separately, changing twelve twelfths into one litre, or one litre into twelve twelfths, where the parts need it.

All steps
  1. 12 is a multiple of 6 and of 4, so write the parts in twelfths: 316 = 3212 and 134 = 1912. Container B holds 3212 + 1912 = 41112 litres.
  2. (a) Containers A and B hold 3212 + 41112 = 71312 litres together. Twelve twelfths make one litre, so this is 8112 litres.
  3. Container C holds 223 = 2812 litres less. 112 is too small to take 812 from, so change one of the 8 litres into twelfths: 8112 = 71312, and C holds 71312 − 2812 = 5512 litres.
  4. At first the three containers hold 8112 + 5512 = 13612 = 1312 litres.
  5. 24 is a multiple of 8 and of 6, so the resin used is 41524 + 12024 = 53524 = 61124 litres.
  6. (b) Write 1312 as 131224: 131224 − 61124 = 7124 litres are left. Check, container by container: A still holds 3424, B keeps 42224 − 41524 = 724 and C keeps 51024 − 12024 = 31424, and 3424 + 724 + 31424 = 62524 = 7124.

answer(a) 8112 litres; (b) 7124 litres

techniqueAdding Fractions with Unlike Denominators · Subtracting Fractions · Adding and Subtracting Mixed Numbers

examsPSLE · GCSE Higher

Common pitfalls

  • Working out 8112 − 2812 as 6712 by taking the smaller numerator from the larger one. The first fraction is the smaller, so one whole litre is changed into 1212 first: 8112 = 71312, which gives 5512.
  • Taking Container C as 223 litres less than Container B alone, which gives 214 litres. The question compares C with Containers A and B together, which is 8112 litres.
12

Trays of Lasagna Eaten in Portions, and the Portions Left

heuristicCount Portions as Twelfths of a Tray / Improper Fraction to Mixed Number and Back

A caterer prepares 18 identical trays of lasagna for a charity dinner and cuts each tray into 12 equal portions. By the middle of the dinner, the guests have eaten 14712 trays of lasagna. During the second half of the dinner, the caterer serves another 334 trays. (a) How many trays of lasagna have the guests eaten by the middle of the dinner? Give the answer as a mixed number in its simplest form. (b) How many portions of lasagna are left at the end of the dinner?

Trays 1-6Trays 7-12Trays 13-18A portion is 1/12 of a tray: 147/12 trays is 147 portions.
Each tray is 12 portions, so a portion is 112 of a tray and 14712 trays is 147 portions.
Each tray is cut into 12 equal portions, so one portion is 112 of a tray, and 14712 trays is 147 portions.
step 1 of 6

Draw the 18 trays as bars of 12 portions each. One portion is 112 of a tray, so a number of twelfths is a number of portions. Filling whole trays twelve portions at a time turns the improper fraction into a mixed number.

All steps
  1. Each tray is cut into 12 equal portions, so one portion is 112 of a tray, and 14712 trays is 147 portions.
  2. Take the portions twelve at a time, one tray after another: 12 trays take 12 × 12 = 144 portions, and the other 147 − 144 = 3 portions come from the thirteenth tray. So 14712 = 12312.
  3. (a) Divide the numerator and the denominator of 312 by 3: the guests have eaten 1214 trays by the middle of the dinner.
  4. Change the second half into portions: 34 = 912, so 334 trays is 3 × 12 + 9 = 45 portions. They are the other 9 portions of the thirteenth tray and three more whole trays.
  5. The caterer began with 18 × 12 = 216 portions, and 147 + 45 = 192 portions have been eaten.
  6. (b) 216 − 192 = 24 portions are left. Check: 1214 + 334 = 16 trays are eaten, and the 18 − 16 = 2 trays left are 2 × 12 = 24 portions.

answer(a) 1214 trays; (b) 24 portions

techniqueEquivalent Fractions · Simplest Form · Mixed Numbers and Improper Fractions

examsPSLE

Common pitfalls

  • Writing 14712 as 14712 by splitting the digits of 147. The whole number is how many twelves go into 147, which is 12, and the 3 portions over are 312 of a tray.
  • Giving (b) as 2, the number of trays left. The question asks for portions, and each of the 2 trays is 12 portions, so 24 portions are left.
13

A Dog Walked Distances in Fifths of a Kilometre over Three Days

heuristicSame Denominator: Add the Numerators, Keep the Denominator / Five Fifths Make One Whole

Mia walks her dog every day. On Monday they walk 25 km, on Tuesday 45 km and on Wednesday 35 km. (a) How far do they walk on Monday and Tuesday together? Give the answer as a mixed number. (b) How far do they walk in the three days, and how much farther would they need to walk to make 2 km?

Monday2/5 kmTuesday4/5 kmWednesday3/5 kmeach part is 1/5 km, and 5 parts make 1 km
Each kilometre is a bar of 5 equal parts, and each part is 15 km.
Draw each kilometre as a bar cut into 5 equal parts. Each part is 15 km.
step 1 of 5

The three distances are all counted in fifths of a kilometre, so every piece is the same size. Draw each kilometre as a bar cut into 5 equal parts. Adding counts the parts, and 5 fifths fill one whole kilometre.

All steps
  1. Draw each kilometre as a bar cut into 5 equal parts. Each part is 15 km.
  2. Monday and Tuesday: 25 + 45. Count the parts, 2 + 4 = 6, and keep the denominator: 65 km. The parts are still fifths.
  3. (a) 5 fifths fill one whole kilometre, and 6 − 5 = 1 fifth is left over: 65 = 115 km.
  4. Add Wednesday: 65 + 35 = 95 km. 5 fifths make 1 km, and 4 fifths are left over, so 95 = 145 km.
  5. (b) 2 km is 10 fifths, and 10 − 9 = 1. They walk 145 km and need 15 km more to make 2 km. Check: 25 + 45 + 35 + 15 = 105 = 2.

answer(a) 115 km; (b) 145 km, and 15 km more to make 2 km

techniqueAdding Fractions with the Same Denominator

examsPSLE

Common pitfalls

  • Adding the denominators as well: 25 + 45 = 610. That is less than the 45 km walked on Tuesday alone, so it cannot be the total. Fifths added to fifths are still fifths: the denominator stays 5.
  • Thinking that 95 km is less than 1 km because it is written as a fraction. 5 fifths already make a whole kilometre, so 95 km is more than 1 km: it is 145 km.