Sum and Difference of Two Lengths
A red ribbon and a blue ribbon are 7.4 m long altogether. The red ribbon is 1.8 m longer than the blue ribbon. (a) How long is the blue ribbon? (b) How long is the red ribbon?
Draw one bar for each ribbon. The red bar is the blue bar with 1.8 m more on the end, so taking that piece away leaves two equal bars.
All steps
- Draw the blue ribbon as one bar. The red ribbon is the same bar with an extra 1.8 m on the end.
- Take the extra 1.8 m away from the total: 7.4 − 1.8 = 5.6 m.
- What is left is two equal bars, so one bar is 5.6 ÷ 2 = 2.8 m.
- (a) The blue ribbon is 2.8 m long.
- (b) The red ribbon is 2.8 + 1.8 = 4.6 m long. Check: 2.8 + 4.6 = 7.4.
answer(a) 2.8 m; (b) 4.6 m
techniqueSubtracting in Columns · Dividing a Decimal
examsPSLE · GCSE Higher
Common pitfalls
- Halving the total first (7.4 ÷ 2 = 3.7) and then adding or subtracting 1.8. The difference must be taken off before the total is shared equally.
- Forgetting to regroup when subtracting, for example writing 7.4 − 1.8 = 6.6. Four tenths is less than eight tenths, so one of the ones must be exchanged for ten tenths.
A Decimal Point Moved One Place
When the decimal point of a number is moved one place to the right, the new number is 23.85 more than the original number. (a) What is the original number? (b) What is the new number?
Draw the original number as 1 unit. The new number is 10 times as large, so it is 10 units, and the difference between them is 9 units.
All steps
- Moving the decimal point one place to the right multiplies a number by 10. Draw the original number as 1 unit and the new number as 10 units.
- The new number is 10 − 1 = 9 units more than the original, so 9 units = 23.85.
- 1 unit = 23.85 ÷ 9 = 2.65.
- (a) The original number is 2.65.
- (b) The new number is 2.65 × 10 = 26.5. Check: 26.5 − 2.65 = 23.85.
answer(a) 2.65; (b) 26.5
techniqueMultiplying and Dividing Decimals by 10, 100, 1,000 · Dividing a Decimal
examsPSLE
Common pitfalls
- Dividing 23.85 by 10. The 23.85 is the difference between 10 units and 1 unit, which is 9 units.
- Treating the move of the decimal point as adding 10 to the number. It multiplies the number by 10.
Two Purchases with a Shared Part
Three pens and two notebooks cost $8.10. Three pens and five notebooks of the same kinds cost $14.85. (a) How much does one notebook cost? (b) How much does one pen cost?
Draw the two purchases one under the other. The three pens are the same in both, so the second bar is longer only because of its three extra notebooks.
All steps
- Line up the two purchases. Both have three pens, and the second has three more notebooks than the first.
- The extra cost comes only from the three extra notebooks: 14.85 − 8.10 = $6.75.
- (a) One notebook costs 6.75 ÷ 3 = $2.25.
- Two notebooks cost 2 × 2.25 = $4.50, so the three pens cost 8.10 − 4.50 = $3.60.
- (b) One pen costs 3.60 ÷ 3 = $1.20.
answer(a) $2.25; (b) $1.20
techniqueSubtracting in Columns · Dividing a Decimal · Money as Decimals
examsPSLE · GCSE Higher
Common pitfalls
- Dividing the difference of $6.75 by 5, the number of notebooks in the second purchase. The difference pays for only the 3 extra notebooks.
- Stopping after part (a). The pens are found by taking the cost of two notebooks away from the first purchase.
A Container Weighed Full and Half Full
A jar filled completely with rice has a mass of 2.35 kg. After half of the rice is used, the jar and the remaining rice have a mass of 1.3 kg. (a) What was the mass of the rice when the jar was full? (b) What is the mass of the empty jar?
Draw the full jar as the jar followed by two equal halves of rice. The second weighing is the jar followed by one half, so the two bars differ by one half of the rice.
All steps
- Draw the full jar as the jar plus two equal halves of rice. The second weighing is the jar plus one half.
- The difference between the two weighings is one half of the rice: 2.35 − 1.3 = 1.05 kg.
- (a) All the rice has a mass of 2 × 1.05 = 2.1 kg.
- (b) The empty jar has a mass of 2.35 − 2.1 = 0.25 kg. Check: 0.25 + 1.05 = 1.3.
answer(a) 2.1 kg; (b) 0.25 kg
techniqueSubtracting in Columns · Counting the Decimals
examsPSLE · GCSE Higher
Common pitfalls
- Halving 2.35 kg. Only the rice is halved. The mass of the jar is the same in both weighings.
- Writing 2.35 − 1.3 = 2.22 by subtracting 13 hundredths. 1.3 is the same as 1.30, so line up the decimal points before subtracting.
Cutting Equal Pieces with a Remainder
A rope 9.2 m long is cut into pieces that are each 0.75 m long. (a) What is the greatest number of complete pieces that can be cut? (b) What length of rope is left over?
Divide the length of the rope by the length of one piece. The whole-number part of the answer is the number of pieces, and the rope those pieces do not use is the leftover.
All steps
- Divide the rope by the length of one piece: 9.2 ÷ 0.75. Multiply both numbers by 100 so that the divisor is a whole number: 920 ÷ 75.
- 75 × 12 = 900 and 75 × 13 = 975, which is more than 920. So only 12 complete pieces fit.
- (a) The greatest number of complete pieces is 12.
- The 12 pieces use 12 × 0.75 = 9 m of rope.
- (b) The length left over is 9.2 − 9 = 0.2 m.
answer(a) 12 pieces; (b) 0.2 m
techniqueDividing by a Decimal · Counting the Decimals
examsGCSE Higher
Common pitfalls
- Rounding 12.26… up to 13 pieces. The thirteenth piece would be shorter than 0.75 m, so it is not a complete piece.
- Reading the remainder of 920 ÷ 75, which is 20, as 20 m or 2 m. Both numbers were multiplied by 100, so the remainder is 20 hundredths of a metre, which is 0.2 m.
Smallest and Largest Value Before Rounding
The mass of a parcel is measured in kilograms to 2 decimal places. Rounded to 1 decimal place, the mass is 4.7 kg. (a) What is the smallest possible mass of the parcel? (b) What is the largest possible mass of the parcel?
Draw a number line in hundredths from 4.6 to 4.8 and mark the numbers that are nearer to 4.7 than to 4.6 or 4.8.
All steps
- Mark 4.7 on a number line in hundredths, with 4.6 and 4.8 on either side.
- A number rounds up to 4.7 when its hundredths digit is 5 or more. The smallest such number is 4.65.
- (a) The smallest possible mass is 4.65 kg.
- A number rounds down to 4.7 when its hundredths digit is 4 or less. The largest such number with 2 decimal places is 4.74.
- (b) The largest possible mass is 4.74 kg, because 4.75 rounds up to 4.8.
answer(a) 4.65 kg; (b) 4.74 kg
techniqueRounding Decimals to One and Two Decimal Places · Comparing Decimals
examsPSLE · O-Level · GCSE Higher
Common pitfalls
- Giving 4.75 kg as the largest mass. 4.75 rounds up to 4.8, so the largest mass with 2 decimal places is 4.74 kg.
- Giving 4.6 kg as the smallest mass. 4.6 rounds to 4.6. The smallest number that rounds up to 4.7 is 4.65.
Comparing Two Pack Sizes by Unit Price
Pack A holds 0.75 kg of almonds and costs $4.20. Pack B holds 1.2 kg of the same almonds and costs $6.60. (a) What is the price of 1 kg of almonds in each pack? (b) Mrs Lim needs exactly 6 kg of almonds. How much does she save by buying only the cheaper kind of pack?
Two packs of different sizes cannot be compared until both prices are for the same mass. Find the price of 1 kg for each, then the cost of 6 kg made from each kind of pack.
All steps
- Find the price of 1 kg in Pack A: 4.20 ÷ 0.75 = 420 ÷ 75 = $5.60.
- Find the price of 1 kg in Pack B: 6.60 ÷ 1.2 = 66 ÷ 12 = $5.50.
- (a) Pack A costs $5.60 per kilogram and Pack B costs $5.50 per kilogram, so Pack B is cheaper.
- For 6 kg she needs 6 ÷ 0.75 = 8 of Pack A, costing 8 × 4.20 = $33.60, or 6 ÷ 1.2 = 5 of Pack B, costing 5 × 6.60 = $33.
- (b) She saves 33.60 − 33 = $0.60.
answer(a) Pack A: $5.60 per kg, Pack B: $5.50 per kg; (b) $0.60
techniqueDividing by a Decimal · Counting the Decimals
examsGCSE Higher
Common pitfalls
- Choosing Pack A because $4.20 is less than $6.60. The packs hold different masses, so compare the price of 1 kg.
- Dividing the mass by the cost (0.75 ÷ 4.20). That gives the kilograms bought with one dollar, not the price of one kilogram.
Equal Spacing Along a Line
Twelve lamp posts stand in a straight line along a road, equally spaced. The distance from the first lamp post to the last is 46.2 m. (a) What is the distance between two neighbouring lamp posts? (b) What is the distance from the 3rd lamp post to the 9th?
Draw the posts as marks on a line. The distance is made of the gaps between the marks, and there is always one gap fewer than there are posts.
All steps
- Count the gaps, not the posts. Twelve posts in a line have 12 − 1 = 11 gaps between them.
- The 11 equal gaps make 46.2 m, so one gap is 46.2 ÷ 11 = 4.2 m.
- (a) Neighbouring lamp posts are 4.2 m apart.
- From the 3rd post to the 9th there are 9 − 3 = 6 gaps.
- (b) The distance is 6 × 4.2 = 25.2 m.
answer(a) 4.2 m; (b) 25.2 m
techniqueDividing a Decimal · Counting the Decimals
examsPSLE · GCSE Higher
Common pitfalls
- Dividing 46.2 by 12, the number of posts. The distance is made of 11 gaps.
- Counting 7 gaps from the 3rd post to the 9th because there are 7 posts in that stretch. Seven posts have 6 gaps between them.
Two Kinds of Coins with a Known Total
Mei has 40 coins in her coin box. Some are 20-cent coins and the rest are 50-cent coins. The total value of the coins is $13.10. (a) How many 50-cent coins does she have? (b) What is the total value of her 20-cent coins?
Suppose every coin is a 20-cent coin and compare the supposed total with the real one. Each 50-cent coin that was counted as a 20-cent coin explains $0.30 of the shortfall.
All steps
- Suppose all 40 coins were 20-cent coins. Their value would be 40 × 0.20 = $8.
- The real total is $13.10, which is 13.10 − 8 = $5.10 more than the supposed total.
- Changing one 20-cent coin into a 50-cent coin adds 0.50 − 0.20 = $0.30.
- (a) The number of 50-cent coins is 5.10 ÷ 0.30 = 17.
- (b) There are 40 − 17 = 23 20-cent coins, worth 23 × 0.20 = $4.60. Check: 17 × 0.50 + 4.60 = $13.10.
answer(a) 17 coins; (b) $4.60
techniqueCounting the Decimals · Dividing by a Decimal
examsPSLE · GCSE Higher
Common pitfalls
- Dividing the shortfall of $5.10 by 0.50. Each exchange adds only the difference between the two coins, which is $0.30.
- Giving 17 as the number of 20-cent coins. Supposing every coin is a 20-cent coin finds the number of the other kind, the 50-cent coins.
Spending in Stages with the Final Amount Known
Ali spent $3.75 on lunch. He then spent half of his remaining money on a book and had $6.40 left. (a) How much money did Ali have just before he bought the book? (b) How much money did Ali have at first?
Start from the money left at the end and undo each step in reverse order: first the book, then the lunch.
All steps
- Work backwards from the $6.40 he had left. The book cost half of his money, so the $6.40 is the other half.
- (a) Just before he bought the book he had 2 × 6.40 = $12.80.
- Before that he spent $3.75 on lunch, so add it back: 12.80 + 3.75 = $16.55.
- (b) Ali had $16.55 at first. Check: 16.55 − 3.75 = 12.80, and half of 12.80 is 6.40.
answer(a) $12.80; (b) $16.55
techniqueMultiplying and Dividing Undo Each Other · Adding in Columns
examsPSLE · GCSE Higher
Common pitfalls
- Undoing the steps in the order they happened: adding $3.75 to $6.40 first and then doubling gives $20.30. The last step, the book, must be undone first.
- Halving $6.40 instead of doubling it. He spent half, so the $6.40 he kept is the other half of what he had.
Luge Times Measured to the Thousandth of a Second
In one run of a luge race, each slider is timed to the thousandth of a second. A reporter copied the four times into her notebook without their zeros at the end: Kim 47.407 s, Lea 47.47 s, Mia 47.4 s and Nora 47.074 s. The shortest time wins. (a) List the four sliders from first place to last place. (b) By how many seconds did the winner beat the slider who came last?
Write the four times in a place-value table with the decimal points lined up, and compare them one place at a time, starting from the left.
All steps
- Line up the decimal points in a place-value table. Write zeros in the empty places so that every time has three decimal places: 47.407, 47.470, 47.400 and 47.074.
- Every time has 47 whole seconds, so compare the tenths. Nora's time has 0 tenths and the other three have 4 tenths, so Nora's time is the shortest.
- Compare the hundredths of the other three times. Lea's time has 7 hundredths and the other two have 0 hundredths, so Lea's time is the longest.
- Compare the thousandths of Mia's and Kim's times. 0 thousandths is less than 7 thousandths, so 47.400 < 47.407. (a) From first place to last place: Nora, Mia, Kim, Lea.
- (b) The winner beat the last slider by 47.470 − 47.074 = 0.396 s. Check: 47.074 + 0.396 = 47.470.
answer(a) Nora, Mia, Kim, Lea; (b) 0.396 s
techniqueThousandths · Comparing Decimals · Subtracting in Columns
examsPSLE · GCSE Higher
Common pitfalls
- Deciding that 47.407 is longer than 47.47 because 407 is more than 47. Compare place by place instead: both have 4 tenths, and 7 hundredths is more than 0 hundredths, so 47.47 is the longer time.
- Lining up the last digits instead of the decimal points when subtracting, so that the hundredths of 47.47 are taken from the thousandths of 47.074. Write 47.47 as 47.470 first.
A Packed Suitcase and the Airline's Weight Limit
Mr Tan weighs everything he packs for a flight. The empty suitcase has a mass of 3.6 kg. He puts in clothes with a mass of 6.85 kg, a pair of boots with a mass of 1.2 kg and books with a mass of 4.375 kg. The airline allows a suitcase of no more than 20 kg. (a) What is the mass of the packed suitcase? (b) He wants to add three gift boxes, all with the same mass. What is the greatest mass each box can have?
Draw the 20 kg limit as one bar. The packed suitcase fills part of it, and the three boxes must share the part that is left.
All steps
- Draw the 20 kg limit as one bar. The suitcase and everything in it fill part of the bar, and the rest is the room left for the boxes.
- Add the four masses in columns with the decimal points lined up. Write zeros so that each mass has three decimal places: 3.600 + 6.850 + 1.200 + 4.375 = 16.025.
- (a) The packed suitcase has a mass of 16.025 kg.
- The room left under the limit is 20 − 16.025 = 3.975 kg.
- (b) The three boxes share that room equally, so each box can have a mass of at most 3.975 ÷ 3 = 1.325 kg. Check: 16.025 + 3 × 1.325 = 20.
answer(a) 16.025 kg; (b) 1.325 kg
techniqueLining Up the Dot · Adding in Columns · Dividing a Decimal
examsPSLE
Common pitfalls
- Lining up the last digits instead of the decimal points, so that the 5 thousandths of 4.375 is added to the 5 hundredths of 6.85. Line up the points, then add tenths to tenths, hundredths to hundredths and thousandths to thousandths.
- Giving 3.975 kg as the answer to (b). That is the mass all three boxes may have together, and one box may have a third of it.
Reading a Height Chart Marked in Hundredths
A height chart on a wall is marked in metres. Its long lines are labelled 1.2, 1.3 and 1.4. Between each pair of long lines there are nine short lines, which divide the space into ten equal parts. In January, the top of Mei's head reached the 7th short line above 1.2. In July of the next year, eighteen months later, it reached the 5th short line above 1.3. (a) What was Mei's height at each reading? (b) How many centimetres did Mei grow in those eighteen months?
Read each mark as a tenth and some hundredths, then count the hundredths along the scale from one mark up to the other.
All steps
- The ten equal parts between two long lines split one tenth of a metre into ten, so each short line marks one hundredth of a metre, 0.01 m.
- January's mark is 7 hundredths above 1.2: 1.2 + 0.07 = 1.27 m. July's mark is 5 hundredths above 1.3: 1.3 + 0.05 = 1.35 m.
- (a) Mei was 1.27 m tall in January and 1.35 m tall in July.
- Count up the scale from 1.27: 3 hundredths reach 1.3, and 5 more reach 1.35. That is 3 + 5 = 8 hundredths of a metre.
- (b) One hundredth of a metre is 1 cm, so Mei grew 8 cm. Check: 1.35 − 1.27 = 0.08 m, which is 8 cm.
answer(a) January 1.27 m, July 1.35 m; (b) 8 cm
techniqueHundredths · Decimals · Subtracting in Columns
examsPSLE · GCSE Higher
Common pitfalls
- Reading the 7th short line above 1.2 as 1.9 by counting each short line as a tenth. The long lines mark the tenths, and a short line marks one tenth of a tenth, which is a hundredth.
- Giving the growth as 0.08 cm. The subtraction gives 0.08 m, and each hundredth of a metre is one centimetre, so the growth is 8 cm.
Flour for a Bake Sale from a Recipe in Quarters
Ms Rao's muffin recipe uses 34 kg of flour for one batch. For a school bake sale she makes 5 batches. Flour is sold in bags of 2.5 kg, and she has no flour at home. (a) How many kilograms of flour does she need? Give your answer as a decimal. (b) How many bags of flour must she buy, and how much flour will be left over?
Change the fraction to a decimal so that it can be multiplied. Then draw the flour needed against whole bags of flour.
All steps
- Three quarters is 0.75, because 34 = 75100. So one batch needs 0.75 kg of flour.
- Five batches need 5 × 0.75 = 3.75 kg of flour.
- (a) She needs 3.75 kg of flour.
- One bag holds only 2.5 kg, which is less than 3.75 kg. Two bags hold 2 × 2.5 = 5 kg, which is enough.
- (b) She must buy 2 bags, and 5 − 3.75 = 1.25 kg of flour is left over. Check: 3.75 + 1.25 = 5.
answer(a) 3.75 kg; (b) 2 bags, with 1.25 kg left over
technique0.5 Is a Half · Counting the Decimals · Decimals
examsPSLE · SAT · GCSE Higher
Common pitfalls
- Writing 34 as 0.34 or 0.3. The digits of a fraction are not the digits of its decimal: 34 = 75100 = 0.75.
- Answering 1.5 bags because 3.75 ÷ 2.5 = 1.5. Flour is sold only in whole bags, and one bag is too little, so she must buy 2.
Rover Lap Times Placed Between Two Known Times
A timer records the lap times of four robot rovers, P, Q, R and S, in seconds. Rover P's time was 14.502 s, and Rover Q was 0.35 s faster than Rover P. (a) What was Rover Q's time? (b) Rover S's time is written with the digits 1, 4, 5 and 2, each used once, and no other digits. It is more than Rover Q's time and less than Rover P's time. What was Rover S's time? (c) Rover R's time has 3 decimal places and is 14.18 s when rounded to 2 decimal places. Rover R was faster than Rover S. What is the greatest possible difference between the times of Rover P and Rover R?
Mark the times of Rovers P and Q on a number line. Rover S's time must fall between them, and Rover R's time must fall in the short stretch of times that round to 14.18.
All steps
- Faster means a shorter time, so subtract. Line up the decimal points, writing 0.35 as 0.350: 14.502 − 0.350 = 14.152.
- (a) Rover Q's time was 14.152 s.
- Every time between 14.152 and 14.502 has 14 whole seconds, which uses the digits 1 and 4. The digits 5 and 2 are left for the tenths and the hundredths, so Rover S's time is 14.25 or 14.52.
- (b) 14.52 is 14.520, which is more than 14.502, so it lies past Rover P's time. Rover S's time was 14.25 s.
- A time with 3 decimal places rounds to 14.18 when it is from 14.175 to 14.184. All of these are less than 14.25, so each of them is faster than Rover S. The difference from Rover P's time is greatest when Rover R's time is shortest, 14.175 s.
- (c) The greatest possible difference is 14.502 − 14.175 = 0.327 s. Check: 14.175 + 0.327 = 14.502.
answer(a) 14.152 s; (b) 14.25 s; (c) 0.327 s
techniqueComparing Decimals · Thousandths · Rounding Decimals to One and Two Decimal Places
examsPSLE · GCSE Higher
Common pitfalls
- Choosing 14.52 for Rover S because it has fewer digits than 14.502. Compare place by place: both have 5 tenths, and 2 hundredths is more than 0 hundredths, so 14.52 is the longer time and lies past Rover P.
- Taking 14.18 s or 14.184 s as Rover R's time in (c). The difference from Rover P is greatest when Rover R's time is as short as possible, and the shortest time with 3 decimal places that rounds to 14.18 is 14.175 s.
Solvent in Three Tanks Given as a Mixed Number, a Decimal and a Fraction
A factory stores solvent in three tanks. Tank A holds 4725 litres of solvent. Tank B holds 0.485 litres more than Tank A. Tank C holds 38 of the amount in Tanks A and B together. (a) Write the volume of solvent in Tank A as a decimal. (b) What volume of solvent is in Tank B? (c) What is the total volume of solvent in the three tanks? Give your answer as a decimal correct to 3 decimal places.
Write Tank A's amount as a decimal first. Then draw Tanks A and B together as 8 units, so that Tank C, three eighths of them, is 3 units.
All steps
- Write the fraction as a decimal. 25 × 4 = 100, so 725 = 28100 = 0.28.
- (a) Tank A holds 4725 = 4.28 litres.
- (b) Line up the decimal points, writing 4.28 as 4.280: Tank B holds 4.280 + 0.485 = 4.765 litres.
- Tanks A and B hold 4.280 + 4.765 = 9.045 litres together. Draw this as 8 units. Tank C holds 38 of it, which is 3 units, and 1 unit is 9.045 ÷ 8 = 1.130625 litres.
- (c) The three tanks hold 8 + 3 = 11 units: 11 × 1.130625 = 12.436875 litres. The fourth decimal place is 8, so round up: 12.437 litres, correct to 3 decimal places. Check: 3 × 1.130625 = 3.391875, and 9.045 + 3.391875 = 12.436875.
answer(a) 4.28 litres; (b) 4.765 litres; (c) 12.437 litres
technique0.5 Is a Half · Adding in Columns · Rounding Decimals to One and Two Decimal Places
examsPSLE · SAT · GCSE Higher
Common pitfalls
- Writing 4725 as 4.725 or 4.7 by copying the digits of the fraction. A fraction becomes a decimal when its denominator is made 100: 725 = 28100 = 0.28.
- Cutting 12.436875 off after the third decimal place to give 12.436. The digit after the third decimal place is 8, which is 5 or more, so the thousandths digit rounds up to 7.
A Cable Cut into Equal Pieces and the Lengths That Round to One Value
A contractor buys a steel cable exactly 89 m long and cuts all of it into 14 pieces of equal length. (a) What is the length of each piece, correct to 2 decimal places? (b) Lengths are measured in metres to 3 decimal places. A support line needs a piece whose length is 6.36 m when rounded to 2 decimal places. (i) What is the shortest length such a piece can have? (ii) What is the greatest length such a piece can have? (c) The contractor has another steel cable, whose mass is a whole number of kilograms. When it is cut into 7 pieces of equal mass, each piece has a mass of 5.4 kg, correct to 1 decimal place. What is the mass of the whole cable?
Draw a number line in thousandths around 6.36 and mark the lengths that are nearer to 6.36 than to 6.35 or 6.37. For the mass, find the stretch for one piece, then stretch it seven times.
All steps
- Divide the length of the cable by the number of pieces: 89 ÷ 14 = 6.357… m.
- (a) The third decimal place is 7, which is 5 or more, so round up: each piece is 6.36 m long, correct to 2 decimal places.
- On a number line in thousandths, the lengths that round to 6.36 start at the halfway mark 6.355 and stop just before the next halfway mark 6.365, which rounds up to 6.37.
- (b) (i) The shortest length is 6.355 m. (ii) With 3 decimal places, the greatest length below 6.365 is 6.364 m.
- In the same way, a mass of 5.4 kg to 1 decimal place is at least 5.35 kg and less than 5.45 kg. Seven such pieces have a mass of at least 7 × 5.35 = 37.45 kg and less than 7 × 5.45 = 38.15 kg.
- (c) The only whole number of kilograms from 37.45 up to 38.15 is 38, so the cable has a mass of 38 kg. Check: 38 ÷ 7 = 5.428…, which is 5.4 correct to 1 decimal place.
answer(a) 6.36 m; (b) (i) 6.355 m, (ii) 6.364 m; (c) 38 kg
techniqueRounding Decimals to One and Two Decimal Places · Dividing a Decimal · Thousandths
examsPSLE · O-Level · GCSE Higher
Common pitfalls
- Giving 6.365 m as the greatest length. 6.365 rounds up to 6.37, so the greatest length with 3 decimal places that rounds to 6.36 is 6.364 m.
- Multiplying the rounded mass, 7 × 5.4 = 37.8, and rounding that to 38 kg. The 5.4 kg is already rounded, so the whole cable can be any mass from 37.45 kg up to 38.15 kg; the answer comes from the whole numbers in that stretch, and a longer stretch could hold more than one.
Two Kinds of Bolts Weighed by the Hundred and by the Thousand
A packing machine handles two types of bolt. 1000 Type A bolts have a mass of 24.8 kg altogether, and 100 Type B bolts have a mass of 4.15 kg altogether. (a) What is the mass of one Type A bolt in grams? (b) What is the mass of one Type B bolt in grams? (c) A container holds 60 bolts, some of Type A and the rest of Type B. The bolts have a total mass of 1.822 kg. How many Type B bolts are in the container?
Find the mass of one bolt of each type. Then suppose every bolt in the container is the lighter Type A, and compare the supposed mass with the real one.
All steps
- Change the masses to grams: 24.8 kg = 24.8 × 1000 = 24800 g and 4.15 kg = 4.15 × 1000 = 4150 g.
- (a) One Type A bolt has a mass of 24800 ÷ 1000 = 24.8 g.
- (b) One Type B bolt has a mass of 4150 ÷ 100 = 41.5 g.
- Suppose all 60 bolts were Type A. Their mass would be 60 × 24.8 = 1488 g. The real mass is 1.822 kg = 1822 g, which is 1822 − 1488 = 334 g more.
- Changing one Type A bolt for a Type B bolt adds 41.5 − 24.8 = 16.7 g.
- (c) The number of Type B bolts is 334 ÷ 16.7 = 3340 ÷ 167 = 20. Check: 20 × 41.5 + 40 × 24.8 = 830 + 992 = 1822 g.
answer(a) 24.8 g; (b) 41.5 g; (c) 20 Type B bolts
techniqueMultiplying and Dividing Decimals by 10, 100, 1,000 · Counting the Decimals · Dividing by a Decimal
examsPSLE
Common pitfalls
- Giving 0.0248 g as the mass of one Type A bolt. 24.8 ÷ 1000 = 0.0248 is in kilograms, the unit of the 24.8; in grams it is 0.0248 × 1000 = 24.8 g.
- Dividing the extra 334 g by 41.5 g, the mass of one Type B bolt. A Type B bolt that was supposed to be a Type A bolt adds only the difference between them, 16.7 g.
Paper Frog Jumps Measured to the Hundredth of a Metre
In a paper-frog jumping contest, each frog's best jump is measured in metres: Asha's frog 0.4 m, Ben's frog 0.35 m, Chloe's frog 0.42 m and Dev's frog 0.38 m. (a) List the frogs from the longest jump to the shortest. (b) How many centimetres farther did Asha's frog jump than Ben's?
Cut every tenth of a metre into ten hundredths. Then every length is a number of hundredths, and the lengths can be compared by counting.
All steps
- Write Asha's 0.4 m in hundredths. 0.4 is 4 tenths, and each tenth is 10 hundredths, so 0.4 = 0.40, which is 40 hundredths.
- The other lengths are already in hundredths: 0.35 m is 35 hundredths, 0.42 m is 42 hundredths and 0.38 m is 38 hundredths.
- Compare the numbers of hundredths: 42 is the most, then 40, then 38, then 35.
- (a) From the longest jump to the shortest: Chloe (0.42 m), Asha (0.4 m), Dev (0.38 m), Ben (0.35 m).
- (b) Asha's frog jumped 40 − 35 = 5 hundredths of a metre farther than Ben's. One hundredth of a metre is 1 cm, so that is 5 cm. Check: 0.40 − 0.35 = 0.05 m.
answer(a) Chloe, Asha, Dev, Ben; (b) 5 cm
techniqueHundredths · Comparing Decimals · Subtracting in Columns
examsPSLE · GCSE Higher
Common pitfalls
- Putting Ben's 0.35 m ahead of Asha's 0.4 m because 35 is more than 4. A decimal with more digits is not always bigger: 0.4 = 0.40, which is 40 hundredths, and 40 is more than 35.
- Giving the answer to (b) as 0.05 cm. The subtraction 0.40 − 0.35 = 0.05 is in metres, and each hundredth of a metre is 1 cm, so the difference is 5 cm.
Two Paper Chains, Each Made of Two Parts Joined End to End
For a class party, Lily and Tom each make a paper chain in two parts and join the parts end to end. Lily's parts are 2.5 m and 1.75 m long. Tom's parts are 3.2 m and 0.95 m long. (a) How long is Lily's paper chain? (b) Whose paper chain is longer, and by how many centimetres?
Write the lengths in columns with the decimal points lined up, and write a zero in any empty place. Then add hundredths to hundredths, tenths to tenths and ones to ones.
All steps
- Line up the decimal points. 2.5 has one decimal place and 1.75 has two, so write 2.5 as 2.50.
- Add from the right. Hundredths: 0 + 5 = 5. Tenths: 5 + 7 = 12 tenths, so write 2 and carry 1 one. Ones: 2 + 1 + 1 = 4.
- (a) Lily's paper chain is 2.50 + 1.75 = 4.25 m long.
- Add Tom's parts the same way, writing 3.2 as 3.20. The tenths are 2 + 9 = 11, so write 1 and carry 1 one: 3.20 + 0.95 = 4.15 m.
- (b) Lily's paper chain is longer, by 4.25 − 4.15 = 0.10 m. One hundredth of a metre is 1 cm, so 0.10 m is 10 cm. Check: 4.15 + 0.10 = 4.25.
answer(a) 4.25 m; (b) Lily's, by 10 cm
techniqueLining Up the Dot · Adding in Columns · Subtracting in Columns
examsPSLE
Common pitfalls
- Adding 5 and 75 as if they were the same kind of piece, and writing 2.5 + 1.75 = 3.80. The 5 in 2.5 is 5 tenths, which is 50 hundredths. With the points lined up the sum is 4.25, and Lily's chain, not Tom's, is the longer one.
- Leaving out the carried 1 from the tenths and writing 3.25 for Lily's chain. Twelve tenths is 1 one and 2 tenths, so the 1 is carried into the ones.