Whole numbers · applications

Applications: Whole Numbers

16 problems

PSLE · GCSE Higher

01

Trees and Lampposts along a Straight Road

heuristicCount the Gaps, Then Add the Tree at the Start / Shared Spots Are Multiples of the LCM

A straight road is 480 m long. Trees are planted along it every 15 m, from one end of the road to the other, so there is a tree at each end. (a) How many trees are planted along the road? (b) Lampposts are also put up every 20 m along the same road, from one end to the other. At how many places do a tree and a lamppost stand at the same spot?

0 m480 mtrees every 15 m (above the line)
Mark a tree at 0 m, then a tree at the end of every 15 m gap.
Draw the road as a line 480 m long. Mark a tree at 0 m, and then a tree at the end of every 15 m gap.
step 1 of 6

Draw the road as a line and mark the trees along it. The gaps between the trees are counted by dividing, and there is one tree more than there are gaps. Then mark the lampposts and look for the spots the two counts share.

All steps
  1. Draw the road as a line 480 m long. Mark a tree at 0 m, and then a tree at the end of every 15 m gap.
  2. The number of 15 m gaps is 480 ÷ 15 = 32.
  3. (a) The first tree stands at 0 m, before any gap, and each gap ends in one more tree. So there are 32 + 1 = 33 trees.
  4. Now mark a lamppost every 20 m. Count in 15s and in 20s: 15, 30, 45, 60 and 20, 40, 60. The first spot both counts reach is 60 m, the lowest common multiple of 15 and 20.
  5. From 0 m, a tree and a lamppost stand together every 60 m, so there are 480 ÷ 60 = 8 gaps between the shared spots.
  6. (b) Add the shared spot at 0 m: 8 + 1 = 9 places. Check: they are at 0, 60, 120, 180, 240, 300, 360, 420 and 480 m, which is 9 places.

answer(a) 33 trees; (b) 9 places

techniqueMultiples · The Lowest Common Multiple

examsPSLE · GCSE Higher

Common pitfalls

  • Answering 32 trees for part (a). 32 is the number of 15 m gaps. The first tree stands at 0 m, before any gap, so there is one more tree than gaps: 32 + 1 = 33.
  • Using 15 × 20 = 300 m as the distance between shared spots. 300 is a common multiple of 15 and 20, but not the lowest: a tree and a lamppost already meet at 60 m, so the shared spots are 60 m apart, not 300 m.
02

The Strikes of a Grandfather Clock

heuristicCount the Gaps Between the Strikes, Not the Strikes

A grandfather clock strikes once for each hour, so at 4 o'clock it strikes 4 times. The strikes are evenly spaced, and at 4 o'clock it takes 6 seconds from the first strike to the last. (a) How many seconds are there between one strike and the next? (b) How many seconds does the clock take from the first strike to the last at 10 o'clock?

4 strikes6 seconds
The first of the 4 strikes comes at 0 seconds, and the last comes 6 seconds later.
Draw 4 marks in a row for the 4 strikes. The first strike comes at 0 seconds, and the last comes 6 seconds later.
step 1 of 5

Draw the strikes as marks on a time line. The first strike comes at the start, at 0 seconds, and the seconds pass in the gaps between the marks.

All steps
  1. Draw 4 marks in a row for the 4 strikes. The first strike comes at 0 seconds, and the last comes 6 seconds later.
  2. Count the gaps between the marks: 4 marks have 3 gaps between them.
  3. (a) The 3 gaps take 6 seconds, so one gap is 6 ÷ 3 = 2 seconds.
  4. At 10 o'clock there are 10 marks, and 10 marks have 10 − 1 = 9 gaps between them.
  5. (b) The 9 gaps take 9 × 2 = 18 seconds. Check: at 4 o'clock, 3 gaps of 2 seconds make 3 × 2 = 6 seconds, as the question says.

answer(a) 2 seconds; (b) 18 seconds

techniqueTime Durations · Multiples

examsPSLE · GCSE Higher

Common pitfalls

  • Sharing the 6 seconds among the 4 strikes, 6 ÷ 4 = 112 seconds each, and answering 10 × 112 = 15 seconds. The first strike comes at 0 seconds, so the time is spent in the 3 gaps between the strikes, not on the 4 strikes.
  • Using 10 gaps for 10 strikes and answering 10 × 2 = 20 seconds. Between 10 marks in a row there are only 9 gaps, so the time is 9 × 2 = 18 seconds.
03

Microscopes and Scales Sharing 104 Batteries

heuristicSuppose All Are Microscopes / Each Change Adds the Difference in Batteries

A science lab has 36 microscopes and electronic scales altogether. Each microscope needs 2 batteries and each electronic scale needs 4 batteries. The 36 devices use 104 batteries altogether. (a) How many electronic scales are there in the lab? (b) How many microscopes are there?

36 microscopes, 0 scales222222222222222222222222222222222222Batteries36 × 2 = 7272
Suppose all 36 devices are microscopes: 36 × 2 = 72 batteries.
Suppose all 36 devices are microscopes. They would use 36 × 2 = 72 batteries.
step 1 of 5

Suppose every device is a microscope and count the batteries. The batteries that are still not counted show how many of the microscopes must really be scales.

All steps
  1. Suppose all 36 devices are microscopes. They would use 36 × 2 = 72 batteries.
  2. The lab really uses 104 batteries, so 104 − 72 = 32 batteries are not yet counted.
  3. Change one microscope into a scale. There are still 36 devices, and the batteries go up by 4 − 2 = 2.
  4. (a) Each change adds 2 batteries, so it takes 32 ÷ 2 = 16 changes to make up the 32 batteries. There are 16 electronic scales.
  5. (b) The other 36 − 16 = 20 devices are microscopes. Check: 20 × 2 + 16 × 4 = 40 + 64 = 104 batteries.

answer(a) 16 electronic scales; (b) 20 microscopes

techniqueTwo-Step Word Problems · Forming Equations

examsO-Level · SAT · GCSE Higher · H2

Common pitfalls

  • Dividing the 32 missing batteries by 4, the batteries in one scale, to get 8 scales. Each scale takes the place of a microscope whose 2 batteries were already counted, so one change adds only 4 − 2 = 2 batteries: 32 ÷ 2 = 16.
  • Giving 16 as the number of microscopes. The working began by supposing all 36 were microscopes, and each change turned a microscope into a scale, so the 16 changes count the scales. The microscopes are 36 − 16 = 20.
04

A Donation Tin of Three Kinds of Coins

heuristicBundle the Coins That Come in a Fixed Ratio / Suppose All Are Dollar Coins

A donation tin holds 54 coins. They are 20-cent, 50-cent and $1 coins, and there are twice as many 20-cent coins as 50-cent coins. The coins are worth $37.20 altogether. (a) How many $1 coins are in the tin? (b) How many 20-cent coins are in the tin?

2020501 bundle: 20¢ + 20¢ + 50¢ = 3 coins, 90¢
Two 20-cent coins for every 50-cent coin: tie them into bundles of 3 coins worth 90 cents.
There are twice as many 20-cent coins as 50-cent coins, so tie two 20-cent coins to each 50-cent coin. One bundle is 3 coins worth 20 + 20 + 50 = 90 cents.
step 1 of 6

Tie every two 20-cent coins to one 50-cent coin, so the small coins come in bundles of 3. Then suppose every coin is a $1 coin, and trade three $1 coins for one bundle until the value is right.

All steps
  1. There are twice as many 20-cent coins as 50-cent coins, so tie two 20-cent coins to each 50-cent coin. One bundle is 3 coins worth 20 + 20 + 50 = 90 cents.
  2. Suppose all 54 coins are $1 coins. They would be worth $54, which is 5400 cents. That is 5400 − 3720 = 1680 cents more than the coins are worth.
  3. Take out three $1 coins and put in one bundle. There are still 54 coins, and the value falls by 300 − 90 = 210 cents.
  4. The value must fall by 1680 cents, so there are 1680 ÷ 210 = 8 bundles.
  5. (a) The bundles hold 8 × 3 = 24 coins, so the other 54 − 24 = 30 coins are $1 coins.
  6. (b) Each bundle has two 20-cent coins, so there are 8 × 2 = 16 twenty-cent coins. Check: 30 × 100 + 16 × 20 + 8 × 50 = 3000 + 320 + 400 = 3720 cents.

answer(a) 30 one-dollar coins; (b) 16 twenty-cent coins

techniqueForming Equations · Counting Coins

examsO-Level · SAT · GCSE Higher · H2

Common pitfalls

  • Trading one $1 coin at a time for a 20-cent coin or a 50-cent coin. A single trade breaks the rule that there are twice as many 20-cent coins as 50-cent coins. Trading three $1 coins for a bundle of two 20-cent coins and one 50-cent coin keeps that rule every time.
  • Answering 54 − 8 = 46 one-dollar coins. 8 is the number of bundles, and each bundle is 3 coins, so the bundles hold 8 × 3 = 24 coins and the $1 coins are 54 − 24 = 30.
05

Markers Shared Two Ways: Some Left Over, Then Some Short

heuristicLeft Over Plus Short Gives the Extra Markers / Divide by the Extra Each Student Gets

A teacher shares markers equally among the students in an art class. If she gives each student 6 markers, she has 14 markers left over. If she gives each student 8 markers, she is 18 markers short. (a) How many students are in the art class? (b) How many markers does the teacher have?

6 eachgiven out14 left8 eachthe markers she has18 short
With 6 each, 14 are left over. With 8 each, 18 more are needed.
With 6 markers each, the teacher has 14 markers left over. With 8 markers each, she needs 18 more markers than she has.
step 1 of 5

Draw the markers as two bars, one for each way of sharing. Going from 14 left over to 18 short takes 14 + 18 more markers, and those markers are the extra 2 markers each student gets.

All steps
  1. With 6 markers each, the teacher has 14 markers left over. With 8 markers each, she needs 18 more markers than she has.
  2. Giving 8 each instead of 6 uses up the 14 markers left over and needs 18 more, so it takes 14 + 18 = 32 more markers.
  3. Each student gets 8 − 6 = 2 more markers, so the 32 markers are 2 markers for each student.
  4. (a) There are 32 ÷ 2 = 16 students.
  5. (b) With 6 each, the teacher has 16 × 6 + 14 = 96 + 14 = 110 markers. Check: with 8 each she needs 16 × 8 = 128 markers, and 128 − 18 = 110.

answer(a) 16 students; (b) 110 markers

techniqueRemainders · Forming Equations

examsPSLE

Common pitfalls

  • Finding the difference, 18 − 14 = 4, instead of the sum. Going from 14 left over to 18 short, the teacher first gives out the 14 markers she has left and then needs 18 more, so the extra markers are 14 + 18 = 32.
  • Dividing the 32 markers by 8, the markers each student gets the second way, to get 4 students. The 32 markers are only the extra 8 − 6 = 2 markers each student gets, so divide by 2: 32 ÷ 2 = 16 students.
06

Oranges Packed Two Ways, Some Left Over Both Times

heuristicSubtract the Two Leftovers / Divide by the Extra Each Box Takes

Uncle Raymond packs oranges into gift boxes. If he packs 5 oranges into each box, 48 oranges are left over. If he packs 9 oranges into each box, 8 oranges are left over. (a) How many gift boxes does Uncle Raymond have? (b) How many oranges does he have altogether?

5 eachpacked48 left9 eachpacked8
Packing 5 to a box leaves 48 over. Packing 9 to a box leaves 8 over.
Packing 5 oranges into each box leaves 48 oranges over. Packing 9 oranges into each box leaves only 8 over.
step 1 of 5

Draw the oranges as two bars of the same length, one for each way of packing. The second way packs away some of the oranges the first way left over, and those oranges are the extra 4 oranges in each box.

All steps
  1. Packing 5 oranges into each box leaves 48 oranges over. Packing 9 oranges into each box leaves only 8 over.
  2. So the second way packs 48 − 8 = 40 more oranges into the boxes.
  3. Each box takes 9 − 5 = 4 more oranges the second way, so the 40 oranges are 4 oranges for each box.
  4. (a) There are 40 ÷ 4 = 10 gift boxes.
  5. (b) He has 10 × 9 + 8 = 90 + 8 = 98 oranges. Check: 10 × 5 + 48 = 50 + 48 = 98.

answer(a) 10 gift boxes; (b) 98 oranges

techniqueRemainders · Forming Equations

examsPSLE

Common pitfalls

  • Adding the two leftovers, 48 + 8 = 56, as in a problem where the sharing is once left over and once short. Here oranges are left over both times, so the second way packs away 48 − 8 = 40 of the oranges the first way left over, and the boxes are 40 ÷ 4 = 10, not 56 ÷ 4 = 14.
  • Answering 10 for part (b) as well. 10 counts the boxes; the oranges are 10 boxes of 9 and the 8 left over: 90 + 8 = 98.
07

A Drum of Oil Weighed Full and a Quarter Full

heuristicThe Drum Is in Both Masses / The Difference Is the Oil Pumped Out

A metal drum full of oil has a mass of 24 kg 800 g. After 34 of the oil is pumped out, the drum and the oil left in it have a mass of 8 kg 600 g. (a) What is the mass of the empty drum? (b) What is the mass of the oil in a full drum?

Full24 kg 800 g24800 g1/4 left8 kg 600 g8600 g
In grams: 24 kg 800 g is 24800 g, and 8 kg 600 g is 8600 g.
Change the masses to grams: 24 kg 800 g is 24800 g, and 8 kg 600 g is 8600 g.
step 1 of 6

Draw the full drum as the drum and 4 equal units of oil, and the drum after pumping as the same drum and 1 unit of oil. The two bars differ only by the 3 units that were pumped out.

All steps
  1. Change the masses to grams: 24 kg 800 g is 24800 g, and 8 kg 600 g is 8600 g.
  2. Cut the oil into 4 equal units. Full, the drum and 4 units weigh 24800 g. After 34 of the oil is pumped out, 1 unit is left, and the drum and 1 unit weigh 8600 g.
  3. The same drum is in both bars, so the difference is the 3 units pumped out: 24800 − 8600 = 16200 g.
  4. One unit of oil is 16200 ÷ 3 = 5400 g.
  5. (a) Take the 1 unit of oil away from the second mass: the empty drum is 8600 − 5400 = 3200 g, which is 3 kg 200 g.
  6. (b) All the oil is 4 units: 4 × 5400 = 21600 g, which is 21 kg 600 g. Check: 3200 + 21600 = 24800 g.

answer(a) 3 kg 200 g; (b) 21 kg 600 g

techniqueForming Equations · Grams and Kilograms

examsO-Level · SAT · GCSE Higher · H2

Common pitfalls

  • Taking 14 of the whole 24800 g, 24800 ÷ 4 = 6200 g, as if the drum were pumped out too. Only the oil is pumped out, and the drum weighs the same in both masses, so only the oil is cut into quarters.
  • Giving 16200 g as the oil in a full drum. 16200 g is the oil pumped out, which is 3 of the 4 units. All the oil is 4 units: 4 × 5400 = 21600 g.
08

Tokens Taken from a Box by Three Children in Turn

heuristicWork Backwards from the End / Put Back the Extra Tokens, Then Find the Whole

A box contained some game tokens. Sean took half of the tokens and 4 more. Then Terry took 13 of the tokens that were left and 6 more. Then Uma took 14 of the tokens that were left and 3 more. In the end, 15 tokens were left in the box. (a) How many tokens were in the box just before Uma's turn? (b) How many tokens were in the box at first?

End15Before Uma1/4315?15 + 3 = 18: 3 quarters
Put back Uma's 3 tokens: 15 + 3 = 18, the 34 she left.
Undo Uma's turn. Put back her 3 extra tokens: 15 + 3 = 18. She took 14, so these 18 tokens are the other 34 of what she found.
step 1 of 5

Start from the 15 tokens at the end and undo the turns in the opposite order. For each turn, first put back the extra tokens that were taken. Then the tokens on hand are the fraction that turn left, and they give the whole from before the turn.

All steps
  1. Undo Uma's turn. Put back her 3 extra tokens: 15 + 3 = 18. She took 14, so these 18 tokens are the other 34 of what she found.
  2. (a) 3 quarters are 18 tokens, so 1 quarter is 18 ÷ 3 = 6 tokens. Before Uma's turn there were 4 × 6 = 24 tokens.
  3. Undo Terry's turn. Put back his 6 extra tokens: 24 + 6 = 30. He took 13, so 30 is 23 of what he found. One third is 30 ÷ 2 = 15, so he found 3 × 15 = 45 tokens.
  4. Undo Sean's turn. Put back his 4 extra tokens: 45 + 4 = 49. He took half, so 49 is the other half of the box.
  5. (b) The box held 2 × 49 = 98 tokens at first. Check: Sean takes 49 + 4 = 53 and leaves 45. Terry takes 15 + 6 = 21 and leaves 24. Uma takes 6 + 3 = 9 and leaves 15.

answer(a) 24 tokens; (b) 98 tokens

techniqueMultiplying and Dividing Undo Each Other · A Fraction of an Amount

examsPSLE · GCSE Higher

Common pitfalls

  • Taking away the extra tokens when working backwards, as in 15 − 3 = 12. Uma took those 3 tokens out of the box, so undoing her turn puts them back: 15 + 3 = 18.
  • Treating the 18 tokens as 14 of what Uma found, because she took 14, and getting 4 × 18 = 72. The 14 is the part she took; the 18 tokens are the 34 she left, so 1 quarter is 18 ÷ 3 = 6 and she found 24.
09

Files Bought on a Buy 4, Get 1 Free Offer

heuristicGroup the Files in Fives / The Remainder Is Paid at Full Price

A bookstore sells files at $3.50 each. It has an offer: buy 4 files and get 1 more file free. Mrs Lee needs 43 files for her office. (a) How many of the 43 files does she pay for? (b) What is the least amount of money she must spend?

1 group$3.50$3.50$3.50$3.50free$141 group: 4 × $3.50 = $14 for 5 files
One group is 4 files at $3.50 and 1 file free: $14 for 5 files.
One group on the offer is 5 files: she pays for 4 of them and gets 1 free. A group costs 4 × $3.50 = $14.
step 1 of 5

Draw the 43 files in groups of 5, each group 4 paid files and 1 free file. Divide 43 by 5: the quotient counts the groups, and the remainder counts the files she pays for at full price.

All steps
  1. One group on the offer is 5 files: she pays for 4 of them and gets 1 free. A group costs 4 × $3.50 = $14.
  2. Divide the files she needs by 5: 43 ÷ 5 = 8 remainder 3. That is 8 full groups, which make 40 files, and 3 files left over.
  3. The 3 files left over are fewer than 4, so they do not earn a free file, and she pays the full price for them. Buying a fourth file to get a fifth free would cost more.
  4. (a) She pays for 4 files in each group and for the 3 files left over: 8 × 4 + 3 = 35 files.
  5. (b) The groups cost 8 × $14 = $112 and the 3 files cost 3 × $3.50 = $10.50, so she spends $112 + $10.50 = $122.50. Check: 35 × $3.50 = $122.50.

answer(a) 35 files; (b) $122.50

techniqueRemainders · Money as Decimals

examsPSLE

Common pitfalls

  • Dividing 43 by 4 because the offer says to buy 4. Each group on the offer holds 5 files, the free one included, and the free files count toward the 43 she needs, so divide by 5: 43 ÷ 5 = 8 remainder 3.
  • Taking a free file with the 3 files left over as well. A free file comes only with every 4 files bought, and 3 is fewer than 4, so all of the last 3 files cost $3.50 each.
10

Shirts and Shorts Bought in Two Different Sets

heuristicDouble One Set So the Shorts Match / Subtract to Leave Only Shirts

At a shop, 3 shirts and 4 pairs of shorts cost $168 altogether, and 5 shirts and 2 pairs of shorts cost $182 altogether. All the shirts cost the same, and all the pairs of shorts cost the same. (a) How much does 1 pair of shorts cost? (b) How much do 2 shirts and 3 pairs of shorts cost?

First$1683 shirts4 pairsSecond$1825 shirts2 pairs
The first set is 3 shirts and 4 pairs of shorts for $168. The second is 5 shirts and 2 pairs for $182.
Draw the two sets as bars: 3 shirts and 4 pairs of shorts cost $168, and 5 shirts and 2 pairs of shorts cost $182.
step 1 of 6

Draw each set as a bar of shirts and shorts. Double the second set, so that both bars hold 4 pairs of shorts. The bars then differ only in shirts, and the difference in price is the price of those shirts.

All steps
  1. Draw the two sets as bars: 3 shirts and 4 pairs of shorts cost $168, and 5 shirts and 2 pairs of shorts cost $182.
  2. Double the second set: 10 shirts and 4 pairs of shorts cost 2 × $182 = $364. Now both bars hold 4 pairs of shorts.
  3. The 4 pairs of shorts are the same in both bars, so the bars differ by 10 − 3 = 7 shirts, and the 7 shirts cost $364 − $168 = $196.
  4. One shirt costs $196 ÷ 7 = $28.
  5. (a) In the first set, 3 shirts cost 3 × $28 = $84, so the 4 pairs of shorts cost $168 − $84 = $84. One pair costs $84 ÷ 4 = $21.
  6. (b) 2 shirts and 3 pairs of shorts cost 2 × $28 + 3 × $21 = $56 + $63 = $119. Check with the second set: 5 × $28 + 2 × $21 = $140 + $42 = $182.

answer(a) $21; (b) $119

techniqueSimultaneous by Elimination · Solving Simultaneous Equations by Scaling

examsO-Level · SAT · GCSE Higher

Common pitfalls

  • Doubling the clothes in the second set but not the price, so that 10 shirts and 4 pairs of shorts are said to cost $182. Twice as many clothes cost twice as much: 2 × $182 = $364.
  • Subtracting the two sets as they are: the second set has 2 more shirts and 2 fewer pairs of shorts, and costs $14 more. Both kinds of clothes change at once, so $14 is not the price of either. Make the shorts match first, then subtract.
11

A Library Order of Dictionaries and Atlases

heuristicLong Multiplication for the Known Part, Then Part and Whole

A primary school buys 48 dictionaries for its library at $27 each. It also buys some atlases at $15 each. The whole bill is $1776. (a) How much do the 48 dictionaries cost? (b) How many atlases does the school buy?

Bill48 × $27? atlases$1776
The bill of $1776 has two parts: 48 dictionaries at $27, and the atlases at $15.
The bill of $1776 is one bar with two parts: the 48 dictionaries at $27 each, and an unknown number of atlases at $15 each.
step 1 of 6

Draw the bill as one bar in two parts, the dictionaries and the atlases. Long multiplication gives the first part, and the second part is what is left of the bill.

All steps
  1. The bill of $1776 is one bar with two parts: the 48 dictionaries at $27 each, and an unknown number of atlases at $15 each.
  2. Find 48 × 27 by long multiplication. Multiply by the ones digit first: 48 × 7 = 336.
  3. Now multiply by the tens digit. The 2 in 27 stands for 2 tens, so write a 0 in the ones place, then 48 × 2 = 96: the second row is 48 × 20 = 960.
  4. (a) Add the two rows: 336 + 960 = 1296. The dictionaries cost $1296.
  5. The rest of the bill pays for the atlases: $1776 − $1296 = $480.
  6. (b) Each atlas costs $15, so the school buys 480 ÷ 15 = 32 atlases. Check: 1296 + 32 × 15 = 1296 + 480 = 1776.

answer(a) $1296; (b) 32 atlases

techniqueTwo Digits by Two Digits · Two-Step Word Problems

examsPSLE · GCSE Higher

Common pitfalls

  • Writing 48 × 2 = 96 as the second row without the 0 in the ones place. The 2 in 27 is 2 tens, so the row is 960, and 336 + 96 = 432 is far too small for 48 books at $27.
  • Dividing the whole bill by the price of an atlas, 1776 ÷ 15. The bill also pays for the dictionaries, so their $1296 must be taken away first.
12

School Fair Money Shared Equally Among the Classes

heuristicLong Division, Then Share Out the Two Shares No Longer Needed

At a school fair, the parents' association raises $3024. The money is for class outings and is shared equally among the school's 18 classes. (a) How much does each class receive? (b) A local business then offers to pay for two classes' outings, so the whole $3024 is shared equally among the other 16 classes instead. How much more does each of those 16 classes receive than before?

18 classes$30241 share = ?
The $3024 is one bar cut into 18 equal shares, one for each class.
Draw the $3024 as one bar cut into 18 equal shares, one for each class. To find one share, divide 3024 by 18 by long division.
step 1 of 6

Draw the money as one bar cut into 18 equal shares. When two classes no longer need their shares, those two shares are cut into 16 equal pieces, one more piece for each of the other classes.

All steps
  1. Draw the $3024 as one bar cut into 18 equal shares, one for each class. To find one share, divide 3024 by 18 by long division.
  2. Since 3 is less than 18, start with 30: 30 ÷ 18 = 1 remainder 12. Bring down the 2 to make 122.
  3. 18 × 6 = 108, so 122 ÷ 18 = 6 remainder 122 − 108 = 14. Bring down the 4 to make 144.
  4. (a) 18 × 8 = 144, so 144 ÷ 18 = 8 with nothing left over. Each class receives $168.
  5. The two shares no longer needed are 2 × $168 = $336, and they are shared among the other 16 classes.
  6. (b) Long division gives 336 ÷ 16 = 21, so each of the 16 classes receives $21 more. Check: 16 × (168 + 21) = 16 × 189 = 3024.

answer(a) $168; (b) $21 more

techniqueLong Division · Two-Step Word Problems

examsPSLE · GCSE Higher

Common pitfalls

  • Answering $189 for part (b). That is the new share; the question asks how much more it is than the old share of $168.
  • Stopping the long division after two digits and answering $16. Every digit of 3024 must be brought down, so there is a third round, 144 ÷ 18 = 8, and the share is $168.
13

Coaches and Minibuses for a School Trip

heuristicDivide with a Remainder, Then Read the Remainder the Way Each Question Needs

A school is taking 468 students and teachers on a trip to the science centre. A coach seats 44 people. (a) How many coaches must the school hire so that everyone has a seat? (b) Instead, the school hires only as many coaches as it can fill completely, and everyone left over travels in minibuses that seat 12 people each. How many minibuses does it need?

People4444444444444444444428468
The 468 people, cut into coach loads of 44.
Draw the 468 people as one bar and cut it into coach loads of 44. Long division tells us how many loads there are.
step 1 of 6

Draw the 468 people as one bar and cut it into coach loads of 44. The quotient counts the full coaches and the remainder counts the people left over; each part of the question reads that remainder in its own way.

All steps
  1. Draw the 468 people as one bar and cut it into coach loads of 44. Long division tells us how many loads there are.
  2. Since 4 is less than 44, start with 46: 46 ÷ 44 = 1 remainder 2. Bring down the 8 to make 28. Since 28 is less than 44, the next digit of the quotient is 0, so 468 ÷ 44 = 10 remainder 28.
  3. (a) Ten coaches carry 10 × 44 = 440 people, and the 28 left over still need seats. Round up: the school must hire 10 + 1 = 11 coaches.
  4. Part (b) reads the same division another way. Only full coaches go, so the school hires 10 coaches, and the remainder, 28 people, travels in minibuses.
  5. 28 ÷ 12 = 2 remainder 4. Two minibuses carry 24 people, and the last 4 still need seats, so round up again.
  6. (b) The school needs 2 + 1 = 3 minibuses. Check: 10 × 44 + 28 = 440 + 28 = 468, and 3 minibuses have 3 × 12 = 36 seats for the 28 people.

answer(a) 11 coaches; (b) 3 minibuses

techniqueInterpreting the Remainder · Long Division · Remainders

examsPSLE · GCSE Higher

Common pitfalls

  • Answering 10 coaches for part (a) because the quotient is 10. The 28 people in the remainder still have to travel, so one more coach is needed.
  • Starting part (b) from the 11 coaches of part (a), which leaves nobody for the minibuses. Part (b) hires full coaches only, so it keeps the quotient, 10, and the minibuses carry the remainder, 28.
14

Tickets for a School Concert over Three Nights

heuristicMultiply One Digit at a Time / Carry Past Nine into the Next Column

The school hall has 485 seats. The school concert runs for 3 nights, and every seat is sold each night. A ticket costs $7. (a) How many tickets are sold altogether? (b) How much money do the tickets bring in?

(a) 485 seats × 3 nights1485×35
Ones: 5 × 3 = 15. Write the 5 and carry the 1 ten.
Write 485 × 3 in columns and start at the ones: 5 × 3 = 15. Write the 5 and carry the 1 ten.
step 1 of 6

Multiply in columns, starting at the ones. Whenever a product is 10 or more, write its ones digit and carry the tens into the next column, where it is added after that column is multiplied.

All steps
  1. Write 485 × 3 in columns and start at the ones: 5 × 3 = 15. Write the 5 and carry the 1 ten.
  2. Tens: 8 × 3 = 24, and the carried 1 makes 25. Write the 5 and carry the 2 hundreds.
  3. (a) Hundreds: 4 × 3 = 12, and the carried 2 makes 14. So 485 × 3 = 1455 tickets are sold.
  4. Now find 1455 × 7 the same way. Ones: 5 × 7 = 35, so write 5 and carry 3. Tens: 5 × 7 = 35, and the carried 3 makes 38, so write 8 and carry 3.
  5. Hundreds: 4 × 7 = 28, and the carried 3 makes 31, so write 1 and carry 3. Thousands: 1 × 7 = 7, and the carried 3 makes 10, so write 10.
  6. (b) 1455 × 7 = 10185, so the tickets bring in $10185. Check: 1455 is close to 1500, and 1500 × 7 = 10500, which is close to 10185.

answer(a) 1455 tickets; (b) $10185

techniqueMultiplying in Columns

examsPSLE

Common pitfalls

  • Forgetting to add a carried digit. If the 1 carried from 5 × 3 = 15 is left out, the tens digit comes out as 4 instead of 5, and the answer as 1445. The carried 1 is one ten, so the tens are 24 + 1 = 25.
  • Adding the carried digit before multiplying, as (8 + 1) × 3 = 27. The carried 1 ten is not one of the 8 tens being multiplied, so multiply first, 8 × 3 = 24, and then add the 1.
15

Bread Delivered to Two Hotels Every Day of February

heuristicLong Multiplication, a Ones Row and a Tens Row / Compare the Two Totals

A bakery delivers bread to two hotels every day of February, which has 28 days. The Sea View Hotel takes 48 loaves a day, and the Park Hotel takes 52 loaves a day. (a) How many loaves does the Sea View Hotel take in February? (b) How many more loaves does the Park Hotel take than the Sea View Hotel in February?

Sea View48 × 28(a) 48 × 2848×2838448 × 8
Long multiplication, 48 × 28. The ones digit first: 48 × 8 = 384.
Find 48 × 28 by long multiplication. Multiply by the ones digit first: 48 × 8 = 384.
step 1 of 5

Multiply by the ones digit of 28 for the first row. Multiply by the tens digit for the second row, with a 0 in the ones place because the 2 in 28 stands for 2 tens. Add the two rows, then compare the two hotels.

All steps
  1. Find 48 × 28 by long multiplication. Multiply by the ones digit first: 48 × 8 = 384.
  2. The 2 in 28 stands for 2 tens. Write a 0 in the ones place, then 48 × 2 = 96: the second row is 48 × 20 = 960.
  3. (a) Add the two rows: 384 + 960 = 1344. The Sea View Hotel takes 1344 loaves.
  4. Find 52 × 28 the same way: 52 × 8 = 416 and 52 × 20 = 1040, so 52 × 28 = 416 + 1040 = 1456.
  5. (b) The Park Hotel takes 1456 − 1344 = 112 more loaves. Check: it takes 52 − 48 = 4 more loaves each day, and 4 × 28 = 112.

answer(a) 1344 loaves; (b) 112 more loaves

techniqueTwo Digits by Two Digits · Multiplying in Columns

examsPSLE · GCSE Higher

Common pitfalls

  • Leaving out the 0 in the second row and writing 96 under 384. The 2 in 28 is 2 tens, so the row is 960. Without the 0 the total is 384 + 96 = 480, far too few loaves for 28 days.
  • Giving 1456 as the answer to part (b). That is all the bread the Park Hotel takes; the question asks how many more it takes than the Sea View Hotel, which is 1456 − 1344 = 112.
16

Footballs and Whistles Bought with a Fixed Sum

heuristicDivide with a Remainder / Round Down for Whole Items, and the Remainder Is the Money Left

Mr Lim has $500 to spend on sports equipment for the school. A football costs $23. (a) He buys as many footballs as he can. How many footballs does he buy, and how much money is left? (b) He then spends the money that is left on whistles at $3 each, buying as many as he can. How many whistles does he buy, and how much money is left at the end?

Money$500$500, $23 a football2350024640
Divide the money by the price: 500 ÷ 23. Start with 50: 50 ÷ 23 = 2 remainder 4.
Divide the money by the price of one football: 500 ÷ 23. Since 5 is less than 23, start with 50: 50 ÷ 23 = 2 remainder 4. Bring down the 0 to make 40.
step 1 of 5

Divide the money by the price. The quotient counts the items he can pay for, so here the answer rounds down, and the remainder is the money left over.

All steps
  1. Divide the money by the price of one football: 500 ÷ 23. Since 5 is less than 23, start with 50: 50 ÷ 23 = 2 remainder 4. Bring down the 0 to make 40.
  2. 40 ÷ 23 = 1 remainder 17. So 500 ÷ 23 = 21 remainder 17.
  3. (a) The remainder, $17, cannot buy another football, so round down. He buys 21 footballs, and $17 is left. Check: 21 × 23 = 483 and 483 + 17 = 500.
  4. Now divide the money that is left by the price of one whistle: 17 ÷ 3 = 5 remainder 2.
  5. (b) $2 cannot buy another whistle, so round down again. He buys 5 whistles, and $2 is left at the end.

answer(a) 21 footballs, with $17 left; (b) 5 whistles, with $2 left at the end

techniqueInterpreting the Remainder · Long Division · Remainders

examsPSLE · GCSE Higher

Common pitfalls

  • Rounding up to 22 footballs because the remainder, 17, is large. 22 footballs cost 22 × 23 = $506, which is more than the $500 he has. Only the footballs he can pay for count, so round down.
  • Giving the quotient, 21, as the money left. The quotient counts the footballs he buys; the remainder, $17, is the money left.