Measurement · applications

Applications: Measurement

15 problems

PSLE · SAT · GCSE Higher

01

Lengths in Meters and Centimeters Cut from a Roll

heuristicOne Unit Before Any Arithmetic / Change Every Length to Centimeters

A roll of wire is 12 m long. Mr Tan cuts off one piece that is 2 m 85 cm long and another piece that is 75 cm long. He cuts the rest of the wire into 8 equal pieces. (a) How many centimeters of wire are left after the first two pieces are cut off? (b) How long is each of the 8 equal pieces?

12 m = 1200 cm and 2 m 85 cm = 285 cm.Roll285 cm75?1200 cm
Every length is in centimeters: the roll is 1200 cm, and the two pieces cut off are 285 cm and 75 cm.
Change every length to centimeters so that the lengths can be added and subtracted: 12 m = 1200 cm and 2 m 85 cm = 285 cm.
step 1 of 5

Draw the roll as one bar in centimeters. The two pieces that were cut off come first, and the rest of the bar is shared into 8 equal pieces.

All steps
  1. Change every length to centimeters so that the lengths can be added and subtracted: 12 m = 1200 cm and 2 m 85 cm = 285 cm.
  2. Add the two pieces that were cut off: 285 + 75 = 360 cm.
  3. (a) The wire that is left is 1200 − 360 = 840 cm.
  4. The 840 cm is cut into 8 equal pieces, so one piece is 840 ÷ 8 = 105 cm.
  5. (b) Each equal piece is 105 cm long, which is 1 m 5 cm. Check: 8 × 105 + 360 = 1200.

answer(a) 840 cm; (b) 105 cm

techniqueConverting Lengths · Meters, Centimeters and Kilometers

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Working out 12 − 285 − 75 or 12 − 2.85 − 75 with the numbers as they are given. A number of meters and a number of centimeters cannot be added or subtracted until they are in the same unit.
  • Writing 12 m as 120 cm. There are 100 centimeters in a meter, so 12 m is 12 × 100 = 1200 cm.
02

Square Tiles on a Floor Measured in Meters

heuristicConvert the Lengths Before Finding the Area / One Square Meter Is 10 000 Square Centimeters

A rectangular floor is 4 m long and 3 m wide. It is to be covered completely with square tiles of side 25 cm. (a) What is the area of the floor in square centimeters? (b) How many tiles are needed?

4 m = 400 cm3 m= 300 cm
The tile is measured in centimeters, so the floor is too: 400 cm by 300 cm.
Change the length and the width of the floor to centimeters: 4 m = 400 cm and 3 m = 300 cm.
step 1 of 4

The tile is measured in centimeters, so change the length and the width of the floor to centimeters first. Then both areas are in square centimeters and one can be divided by the other.

All steps
  1. Change the length and the width of the floor to centimeters: 4 m = 400 cm and 3 m = 300 cm.
  2. (a) The area of the floor is 400 × 300 = 120 000 cm².
  3. The area of one tile is 25 × 25 = 625 cm².
  4. (b) The number of tiles needed is 120 000 ÷ 625 = 192. Check: 400 ÷ 25 = 16 tiles fit along the length and 300 ÷ 25 = 12 along the width, and 16 × 12 = 192.

answer(a) 120 000 cm²; (b) 192 tiles

techniqueConverting Units of Area · Converting Lengths

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Changing 12 m² to 1200 cm² by multiplying by 100. A square meter is 100 cm long and 100 cm wide, so each square meter is 100 × 100 = 10 000 cm².
  • Dividing the area of the floor by 25, the side of a tile. Each tile covers an area of 25 × 25 = 625 cm², so the area of the floor must be divided by 625.
03

A Box of Identical Tins Weighed Before and After Some Are Removed

heuristicDifference of Two Weighings in Grams / The Difference Is the Tins Removed

A box holding 12 identical tins of beans has a mass of 5.3 kg. After 5 of the tins are taken out, the box and the remaining tins have a mass of 3 kg 550 g. (a) What is the mass of one tin, in grams? (b) What is the mass of the empty box, in kilograms?

At firstboxtintintintintintintintintintintintin5300 g5 tins outboxtintintintintintintin3550 g
In grams the two weighings are 5300 g and 3550 g. The box is the same in both.
Write both masses in grams: 5.3 kg = 5300 g and 3 kg 550 g = 3550 g.
step 1 of 5

Draw each weighing as the box followed by its tins. The box is the same in both bars, so the first bar is heavier only by the 5 tins that were taken out.

All steps
  1. Write both masses in grams: 5.3 kg = 5300 g and 3 kg 550 g = 3550 g.
  2. The box is the same in both weighings, so the difference is the mass of the 5 tins that were taken out: 5300 − 3550 = 1750 g.
  3. (a) One tin has a mass of 1750 ÷ 5 = 350 g.
  4. The 12 tins together have a mass of 12 × 350 = 4200 g.
  5. (b) The empty box has a mass of 5300 − 4200 = 1100 g, which is 1.1 kg. Check: 1100 + 7 × 350 = 3550.

answer(a) 350 g; (b) 1.1 kg

techniqueGrams and Kilograms

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Writing 3 kg 550 g as 3.55 kg and 5.3 kg as 5 kg 3 g. The digit 3 in 5.3 kg is 3 tenths of a kilogram, which is 300 g.
  • Dividing the difference of 1750 g by 7, the number of tins left in the box. The difference is the mass of the 5 tins that were taken out.
04

A Weighing Scale Read Between Its Labeled Marks

heuristicValue of One Small Space / Count the Spaces, Not the Marks

A weighing scale is labeled at every kilogram, and there are 4 small marks between each label and the next. When a basket of mangoes is placed on the scale, the pointer rests on the 3rd small mark after 2 kg. (a) What is the reading on the scale, in grams? (b) The empty basket has a mass of 450 g. What is the mass of the mangoes, in grams?

01 kg2 kg3 kg4 kgpointer1000 g in 5 spaces
From one label to the next is 1000 g, and the 4 small marks make 5 equal spaces.
From one label to the next is 1 kg, which is 1000 g. The 4 small marks divide it into 5 equal spaces.
step 1 of 4

Before reading the pointer, work out what one small space on the scale is worth. Then count the spaces from the label just before the pointer.

All steps
  1. From one label to the next is 1 kg, which is 1000 g. The 4 small marks divide it into 5 equal spaces.
  2. One small space stands for 1000 ÷ 5 = 200 g.
  3. (a) The pointer is 3 spaces after 2 kg, so the reading is 2000 + 3 × 200 = 2600 g.
  4. (b) Take away the basket: the mass of the mangoes is 2600 − 450 = 2150 g, which is 2 kg 150 g.

answer(a) 2600 g; (b) 2150 g

techniqueReading Scales · Grams and Kilograms

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Dividing 1000 g by 4 because there are 4 small marks. Four marks make 5 spaces, so each space is 1000 ÷ 5 = 200 g, not 250 g.
  • Reading the pointer as 2.3 kg because it is on the 3rd mark. Each space is 200 g, not 100 g, so the 3rd mark after 2 kg is 2 kg 600 g.
05

A Drink Mixed in Liters and Poured into Glasses in Milliliters

heuristicChange Liters to Milliliters / Full Glasses and the Amount Left

Mrs Rahim mixes 3 liters 600 ml of water with 0.85 liters of syrup. She pours the drink into glasses that hold 300 ml each. (a) How many glasses can she fill completely? (b) How many milliliters of the drink are left over?

Drinkwater 3600 mlsyrup 850
In milliliters, the water is 3600 ml and the syrup is 0.85 × 1000 = 850 ml.
Write both amounts in milliliters: 3 liters 600 ml = 3600 ml and 0.85 liters = 850 ml.
step 1 of 5

Change both amounts to milliliters and add them. Then divide the total by the capacity of one glass: the whole-number part is the number of full glasses, and the drink they do not use is left over.

All steps
  1. Write both amounts in milliliters: 3 liters 600 ml = 3600 ml and 0.85 liters = 850 ml.
  2. The total amount of drink is 3600 + 850 = 4450 ml.
  3. Divide by the capacity of one glass. 300 × 14 = 4200 and 300 × 15 = 4500, which is more than 4450.
  4. (a) She can fill 14 glasses completely.
  5. (b) The amount left over is 4450 − 4200 = 250 ml.

answer(a) 14 glasses; (b) 250 ml

techniqueMilliliters and Liters

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Writing 0.85 liters as 85 ml. One liter is 1000 ml, so 0.85 liters is 0.85 × 1000 = 850 ml.
  • Giving 15 glasses because 4450 ÷ 300 is nearly 15. The 15th glass would receive only 250 ml, so it is not filled completely.
06

A Journey in Two Legs with a Rest, Crossing Noon

heuristicTimeline / Add Each Duration by Stopping at the Hour

A bus leaves the school at 10.45 a.m. It travels for 1 h 35 min, stops for a rest of 25 min, and then travels for another 50 min to reach the camp. (a) At what time does the bus reach the camp? (b) What is the total traveling time, not counting the rest?

1 h 35 min25 min50 minrest10.45 a.m.12.20 p.m.
From 10.45 a.m., 1 h reaches 11.45 a.m., 15 min more reaches 12 noon, and the last 20 min reaches 12.20 p.m.
Add the first leg to the starting time. One hour after 10.45 a.m. is 11.45 a.m. Of the 35 min, 15 min reaches 12 noon and the other 20 min reaches 12.20 p.m.
step 1 of 5

Draw the journey as a timeline with one block for each leg and one for the rest. Add each duration to the clock time in two moves: first up to the next hour, then the minutes that remain.

All steps
  1. Add the first leg to the starting time. One hour after 10.45 a.m. is 11.45 a.m. Of the 35 min, 15 min reaches 12 noon and the other 20 min reaches 12.20 p.m.
  2. Add the rest of 25 min: 12.20 p.m. becomes 12.45 p.m.
  3. Add the second leg of 50 min. The first 15 min reaches 1.00 p.m. and the other 35 min reaches 1.35 p.m.
  4. (a) The bus reaches the camp at 1.35 p.m.
  5. (b) The rest is not traveling time. The two legs take 1 h 35 min + 50 min = 1 h 85 min = 2 h 25 min. Check: from 10.45 a.m. to 1.35 p.m. is 2 h 50 min, and 2 h 50 min − 25 min = 2 h 25 min.

answer(a) 1.35 p.m.; (b) 2 h 25 min, which is 145 min

techniqueTime Durations

examsPSLE · GCSE Higher

Common pitfalls

  • Adding the times as decimals: 10.45 + 1.35 = 11.80. There are 60 minutes in an hour, not 100, so 45 min + 35 min = 80 min is 1 h 20 min.
  • Giving 2 h 50 min as the traveling time. That is the whole time from leaving to arriving, and it includes the rest of 25 min when the bus was not moving.
07

Prices in Cents and Dollars with Change from a Note

heuristicWrite Every Price in Dollars / Total Cost, Then Change

Mr Lee buys 6 pens at 85 cents each and 4 files at $2.40 each. He pays with a $20 note. (a) How much do the pens and files cost altogether? (b) How much change does he receive?

Pens85¢85¢85¢85¢85¢85¢$5.10
In dollars a pen costs $0.85, so the 6 pens cost 6 × 0.85 = $5.10.
Write the price of a pen in dollars: 85 cents = $0.85. The 6 pens cost 6 × 0.85 = $5.10.
step 1 of 4

Write the price of a pen in dollars so that it can be added to the price of the files. Find the cost of each kind of item, add them, and take the total away from the note.

All steps
  1. Write the price of a pen in dollars: 85 cents = $0.85. The 6 pens cost 6 × 0.85 = $5.10.
  2. The 4 files cost 4 × 2.40 = $9.60.
  3. (a) The total cost is 5.10 + 9.60 = $14.70.
  4. (b) The change from the note is 20 − 14.70 = $5.30. Check: 14.70 + 5.30 = 20.

answer(a) $14.70; (b) $5.30

techniqueMoney as Decimals

examsPSLE · GCSE Higher

Common pitfalls

  • Working out 6 × 85 + 4 × 2.40 = 519.60. The 85 is a number of cents and the 2.40 is a number of dollars, so one of them must be converted before they are added.
  • Writing 85 cents as $8.50 or $0.085. One cent is one hundredth of a dollar, so 85 cents is $0.85.
08

A Sensible Estimate Chosen by Its Unit, Then Used

heuristicCompare with a Length You Know / Multiply, Then Convert

The length of one of Mei's walking steps is one of these: 6 cm, 60 cm or 6 m. (a) Which is the reasonable length for one step? (b) Mei takes 1250 steps to walk from her home to her school. Using the length from part (a), about how far is her school from her home, in meters?

6 cmas long as a finger60 cma little over half a meter6 mas long as a car
The three bars are drawn to the same scale. A walking step is a little more than half a meter.
Compare each length with something familiar. 6 cm is about the length of a finger, and 6 m is about the length of a car. One walking step is a little more than half a meter.
step 1 of 4

Compare each length with something familiar to choose the sensible one. Then multiply by the number of steps in centimeters and change the answer to meters.

All steps
  1. Compare each length with something familiar. 6 cm is about the length of a finger, and 6 m is about the length of a car. One walking step is a little more than half a meter.
  2. (a) The reasonable length for one step is 60 cm.
  3. Multiply by the number of steps: 1250 × 60 = 75 000 cm.
  4. (b) There are 100 cm in 1 m, so the distance is 75 000 ÷ 100 = 750 m.

answer(a) 60 cm; (b) about 750 m

techniqueEstimating with Units · Converting Lengths

examsPSLE · GCSE Higher

Common pitfalls

  • Choosing 6 m because a walk to school is a long distance. The 6 m would be the length of a single step, and nobody can step the length of a car.
  • Dividing 75 000 cm by 1000 to get 75 m. There are 100 centimeters in a meter. It is a kilometer that has 1000 meters.
09

Parcel Charges by Mass Band

heuristicConvert Grams to Kilograms / Find the Band Each Mass Falls In

A courier charges by the mass of a parcel: $2.80 for a parcel up to 0.5 kg, $4.20 for over 0.5 kg and up to 1 kg, $6.50 for over 1 kg and up to 2 kg, and $9.90 for over 2 kg and up to 3 kg. Parcel A has a mass of 850 g and parcel B has a mass of 1 kg 400 g. (a) How much does it cost to send the two parcels separately? (b) How much is saved by packing them together as one parcel?

$2.80$4.20$6.50$9.9000.51 kg2 kg3 kgA 0.85 kgB 1.4 kg
In kilograms, parcel A is 0.85 kg and parcel B is 1.4 kg. Each block is one band of the charges.
The bands are in kilograms, so change each mass: 850 g = 0.85 kg and 1 kg 400 g = 1.4 kg.
step 1 of 5

The charges are given in kilograms, so change each mass to kilograms and mark it on a line divided into the bands. The band a mass falls in gives its charge.

All steps
  1. The bands are in kilograms, so change each mass: 850 g = 0.85 kg and 1 kg 400 g = 1.4 kg.
  2. 0.85 kg is over 0.5 kg and up to 1 kg, so parcel A costs $4.20. 1.4 kg is over 1 kg and up to 2 kg, so parcel B costs $6.50.
  3. (a) Sending them separately costs 4.20 + 6.50 = $10.70.
  4. Packed together, the mass is 0.85 + 1.4 = 2.25 kg. That is over 2 kg and up to 3 kg, so the charge is $9.90.
  5. (b) The saving is 10.70 − 9.90 = $0.80.

answer(a) $10.70; (b) $0.80

techniqueGrams and Kilograms · Money as Decimals

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Writing 850 g as 8.5 kg. There are 1000 g in a kilogram, so 850 g is 850 ÷ 1000 = 0.85 kg, which is less than 1 kg.
  • Charging parcel B at $4.20 because its mass begins with 1 kg. 1.4 kg is over 1 kg, so it falls in the next band and costs $6.50.
10

Two Bottle Sizes Compared by the Cost of 100 Milliliters

heuristicSame Unit, Same Amount / Cents for Each 100 ml

Bottle A holds 600 ml of olive oil and costs $2.70. Bottle B holds 1.5 liters of the same oil and costs $6.30. (a) What is the cost of 100 ml of oil from each bottle, in cents? (b) How much is saved on 3 liters of oil by buying Bottle B instead of Bottle A?

Bottle A100100100100100100$2.70Bottle B100100100100100100100100100100100100100100100$6.30
Each block is 100 ml. Bottle A holds 6 of them, and Bottle B holds 1500 ml, which is 15 of them.
Write both capacities in milliliters: 1.5 liters = 1500 ml. Bottle A holds 6 lots of 100 ml and Bottle B holds 15 lots.
step 1 of 5

The two bottles cannot be compared until both capacities are in the same unit and both prices are for the same amount. Change liters to milliliters, cut each bottle into lots of 100 ml, and find the cost of one lot in cents.

All steps
  1. Write both capacities in milliliters: 1.5 liters = 1500 ml. Bottle A holds 6 lots of 100 ml and Bottle B holds 15 lots.
  2. Bottle A costs $2.70, which is 270 cents. One lot of 100 ml costs 270 ÷ 6 = 45 cents.
  3. Bottle B costs $6.30, which is 630 cents. One lot of 100 ml costs 630 ÷ 15 = 42 cents.
  4. (a) 100 ml costs 45 cents from Bottle A and 42 cents from Bottle B, so Bottle B is the better buy.
  5. (b) 3 liters is 30 lots of 100 ml, and each lot costs 3 cents less from Bottle B: 30 × 3 = 90 cents = $0.90. Check: 5 of Bottle A cost $13.50 and 2 of Bottle B cost $12.60.

answer(a) Bottle A: 45 cents, Bottle B: 42 cents; (b) $0.90

techniqueMilliliters and Liters · Money as Decimals

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Dividing $6.30 by 1.5 and $2.70 by 600, then comparing 4.20 with 0.0045. One price is for a liter and the other for a milliliter. Both bottles must be in the same unit first.
  • Choosing Bottle A because $2.70 is less than $6.30. Bottle B holds more oil, so compare the cost of the same amount from each.
11

One Rod, Two Strips: Undoing the Cuts, and Reading a Rule with No Zero

heuristicUndo the Stages in Reverse Order / A Length to the Nearest Centimeter Lies Within Half a Centimeter of It

A technician cuts a master rod in half. One half is trimmed by 4.6 cm and then welded end to end to an extension piece 14 as long as the trimmed piece; the finished Strip Z is exactly 21.5 cm long. Strip X is cut from the other half. Laid along a workbench rule whose zero mark has broken off, Strip X starts at mark A and ends at mark B, and its length, to the nearest centimeter, is 18 cm. (a) How long was the master rod? (b) Mark A is the 6.7 cm mark on the rule. What are the possible readings of mark B?

Strip Z4.34.34.34.34.321.5 cmtrimmed piece, 4 units1/4
The extension is 14 of the trimmed piece: 4 units and 1 unit make 21.5 cm, so 1 unit is 21.5 ÷ 5 = 4.3 cm.
The extension is 14 of the trimmed piece. Draw the trimmed piece as 4 units and the extension as 1 unit, so Strip Z is 5 units, and 1 unit = 21.5 ÷ 5 = 4.3 cm.
step 1 of 6

Draw Strip Z as the trimmed piece followed by its extension, then undo the stages in reverse order: take away the extension, put back the 4.6 cm, and double. For part (b), find the lengths that round to 18 cm and lay them along the rule from the 6.7 cm mark.

All steps
  1. The extension is 14 of the trimmed piece. Draw the trimmed piece as 4 units and the extension as 1 unit, so Strip Z is 5 units, and 1 unit = 21.5 ÷ 5 = 4.3 cm.
  2. The trimmed piece is 4 × 4.3 = 17.2 cm. Put back the 4.6 cm that was trimmed off: half of the rod was 17.2 + 4.6 = 21.8 cm.
  3. (a) Undo the cut in half: the master rod was 2 × 21.8 = 43.6 cm long.
  4. A length that is 18 cm to the nearest centimeter is at least 17.5 cm and less than 18.5 cm. A length of 18.5 cm would round up to 19 cm.
  5. The zero mark is missing, so the reading at B is the reading at A plus the length of the strip. The least reading is 6.7 + 17.5 = 24.2 cm, and every reading is less than 6.7 + 18.5 = 25.2 cm.
  6. (b) 24.2 cm ≤ B < 25.2 cm. Check: 43.6 ÷ 2 − 4.6 = 17.2, and 17.2 + 14 × 17.2 = 17.2 + 4.3 = 21.5.

answer(a) 43.6 cm; (b) 24.2 cm ≤ B < 25.2 cm

techniqueMeters, Centimeters and Kilometers · Reading Scales · Reading a Ruler

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Taking the extension as 14 of Strip Z and working out 21.5 − 14 × 21.5. The extension is a quarter of the trimmed piece, so the trimmed piece is 4 of the 5 equal units in Strip Z.
  • Giving 25.2 cm as a possible reading of B. A strip 18.5 cm long rounds to 19 cm, not 18 cm, so the length is less than 18.5 cm and B is less than 6.7 + 18.5 = 25.2 cm.
12

Three Freight Containers Weighed in Tonnes, in Grams and on a Dial

heuristicKilograms for Every Mass Before Comparing / Work Out What One Space on the Dial Stands For

A logistics hub weighs three freight containers. Container X has a mass of 2 tonnes 65 kg. Container Y is 480 000 g lighter than Container X. Container Z is weighed on a dial scale that is labeled every 100 kg, with 4 unnumbered marks dividing each interval into equal spaces, and its pointer rests on the 3rd mark past 1800 kg. (a) What is the mass of Container X in kilograms? (b) What is the reading on the dial for Container Z, in kilograms? (c) Arrange the three containers in order of mass, starting with the heaviest. (d) Container Y carries cartons of spare parts, and each carton holds 45 liters. Express this volume in cubic centimeters.

1700180019002000Container Z (kg)X2 tonnes = 2000 kg2065 kg
(a) 1 tonne is 1000 kg, so X is 2 × 1000 + 65 = 2065 kg.
(a) 1 tonne is 1000 kg, so Container X has a mass of 2 × 1000 + 65 = 2065 kg.
step 1 of 6

The three masses are given in three different ways: in tonnes and kilograms, as a difference in grams, and as a pointer on a dial. Change each one to kilograms first. On the dial, work out what one space stands for before counting the spaces.

All steps
  1. (a) 1 tonne is 1000 kg, so Container X has a mass of 2 × 1000 + 65 = 2065 kg.
  2. Change the difference to kilograms: 480 000 g = 480 000 ÷ 1000 = 480 kg. Container Y has a mass of 2065 − 480 = 1585 kg.
  3. On the dial, the 4 unnumbered marks divide each 100 kg into 5 equal spaces, so one space stands for 100 ÷ 5 = 20 kg.
  4. (b) The pointer is 3 spaces past 1800 kg, so the reading is 1800 + 3 × 20 = 1860 kg.
  5. (c) In kilograms the masses are 2065, 1860 and 1585. From the heaviest, the order is Container X, Container Z, Container Y.
  6. (d) 1 liter is 1000 ml, and 1 ml is 1 cm³, so 45 liters = 45 × 1000 = 45 000 cm³.

answer(a) 2065 kg; (b) 1860 kg; (c) Container X (2065 kg), Container Z (1860 kg), Container Y (1585 kg); (d) 45 000 cm³

techniqueGrams and Kilograms · Reading Scales · Milliliters and Liters

examsPSLE · SAT · GCSE Higher

Common pitfalls

  • Dividing 100 kg by 4 because there are 4 unnumbered marks, and reading 1800 + 3 × 25 = 1875 kg. Four marks make 5 spaces, so each space stands for 20 kg.
  • Writing 2 tonnes 65 kg as 2650 kg. The 65 is a number of kilograms, so it is added to 2000 kg to make 2065 kg.
13

Painted Tiles Measured in Square Centimeters, Then Fitted to a Wall in Square Meters

heuristicOne Square Meter Is 100 cm by 100 cm / Divide the Square Centimeters by 10 000

Each of the 28 children in a class paints a square tile that is 50 cm long on each side. The tiles are fixed side by side, with no gaps, on a part of a wall that has an area of 8 m². (a) What is the total area of the 28 tiles, in square centimeters? (b) How many square meters of that part of the wall are left uncovered?

50 cmwall: 8 m²one tile: 50 × 50 = 2500 cm²
One tile is 50 cm by 50 cm, so its area is 50 × 50 = 2500 cm².
One tile is 50 cm by 50 cm, so its area is 50 × 50 = 2500 cm².
step 1 of 4

A square meter is a square 100 cm long on each side. It holds 100 rows of 100 square centimeters, which is 10 000 cm². Find the area of the tiles in square centimeters, then find how many square meters that makes.

All steps
  1. One tile is 50 cm by 50 cm, so its area is 50 × 50 = 2500 cm².
  2. (a) The 28 tiles cover 28 × 2500 cm². 28 × 25 = 700, so 28 × 2500 = 70 000 cm².
  3. A square meter is 100 cm by 100 cm, so 1 m² = 100 × 100 = 10 000 cm². Divide by 10 000: 70 000 ÷ 10 000 = 7 m².
  4. (b) The wall has 8 m², and 8 − 7 = 1, so 1 m² is left uncovered. Check: 4 tiles fill one square meter, 2 along and 2 down, and 28 ÷ 4 = 7 m².

answer(a) 70 000 cm²; (b) 1 m²

techniqueConverting Units of Area

examsPSLE · O-Level · SAT · GCSE Higher

Common pitfalls

  • Dividing 70 000 by 100 and saying that the tiles cover 700 m², as if 1 m² were 100 cm². A square meter is 100 cm long and 100 cm wide, so it holds 100 × 100 = 10 000 cm².
  • Changing the wall's 8 m² to 800 cm² by multiplying by 100, and deciding that the tiles do not fit. 8 m² is 8 × 10 000 = 80 000 cm², which is more than the 70 000 cm² of the tiles.
14

A Hall's Wall Estimated in Doors, Then Checked Against a Measurement with No Unit

heuristicCompare with a Length You Know / Use the Estimate to Choose the Unit

The wall of a school hall is about as tall as 3 doors standing one on top of another. A door is about 2 m tall. (a) About how tall is the wall, in meters? (b) A teacher measures the wall and writes down 640, but forgets to write the unit. Use your estimate to decide whether the unit was meters, centimeters or millimeters. How tall is the wall, in meters?

walldoor2 mdoor2 mdoor2 m?a door is about 2 m tall
A door is about 2 m tall, and the wall is about as tall as 3 doors, one on top of another.
A door is about 2 m tall, and the wall is about as tall as 3 doors.
step 1 of 5

Compare the wall with something whose size you know, the height of a door. Then use the estimate as a check: the right unit gives a height close to it, and a wrong unit gives a height that makes no sense.

All steps
  1. A door is about 2 m tall, and the wall is about as tall as 3 doors.
  2. (a) 3 × 2 = 6, so the wall is about 6 m tall.
  3. Try meters: a wall 640 m tall would be more than 100 times the estimate, taller than almost every building in the world. Try millimeters: 640 mm is 64 cm, about a third of the height of one door.
  4. Try centimeters: 640 cm is 640 ÷ 100 = 6.4 m, which is close to the estimate of 6 m.
  5. (b) The unit was centimeters, and the wall is 6.4 m tall. The estimate was only 6.4 − 6 = 0.4 m short, which is 40 cm.

answer(a) About 6 m; (b) centimeters, so the wall is 6.4 m tall

techniqueEstimating with Units · Converting Lengths

examsPSLE · GCSE Higher

Common pitfalls

  • Choosing meters because a wall is big and 640 looks like a length in meters. A wall 640 m tall would be more than 100 times the estimate of 6 m, so the unit cannot be meters.
  • Changing 640 cm to meters by dividing by 1000 and getting 0.64 m. There are 100 centimeters in a meter, so divide by 100: 6.4 m. It is millimeters that make 1000 to a meter.
15

A Fence Drawn to Scale on a Ruler Whose End Has Broken Off

heuristicA Line Drawn from 0 Ends on the Mark That Names Its Length; from Any Other Mark, Add the Length / Length = End − Start

Priya draws a plan of her garden at a scale of 1 cm to 2 m. The fence along one side is 13 m long. The end of her ruler has broken off, so its first mark is the 3 cm mark. (a) How long is the line she draws for the fence? (b) She starts the line at the 3 cm mark. At which mark does it end? (c) Her brother draws a fence on his own plan, at the same scale, from the 1 cm mark of his ruler to the 6.5 cm mark. How long is the fence his plan shows?

34567891011Priyabroken off1 cm to 2 m: 13 ÷ 2 = 6.5 cm
At 1 cm to 2 m, the 13 m fence is a line 13 ÷ 2 = 6.5 cm long. (a) 6.5 cm.
At 1 cm to 2 m, every 2 m of fence is 1 cm of line: 13 ÷ 2 = 6.5. (a) The line is 6.5 cm long.
step 1 of 4

Turn the fence into centimeters with the scale, then lay the line along the ruler from where it starts. A line from the 0 mark ends on the mark that names its length; from any other mark, add the length to the start.

All steps
  1. At 1 cm to 2 m, every 2 m of fence is 1 cm of line: 13 ÷ 2 = 6.5. (a) The line is 6.5 cm long.
  2. Drawn from the 0 mark, the line would end at the 6.5 cm mark. Priya starts at the 3 cm mark, so the line ends 6.5 cm further on: 3 + 6.5 = 9.5. (b) It ends at the 9.5 cm mark.
  3. Her brother's line ends at the 6.5 cm mark, but it starts at the 1 cm mark, not at 0. Take the start off the end: 6.5 − 1 = 5.5 cm.
  4. (c) His plan shows a fence of 5.5 × 2 = 11 m.

answer(a) 6.5 cm; (b) the 9.5 cm mark; (c) 11 m

techniqueDrawing a Line Segment · Map Scales · Reading a Ruler

examsPSLE

Common pitfalls

  • Ending Priya's line at the 6.5 cm mark: that is where a line drawn from 0 ends. Hers starts at 3, so it ends at 9.5.
  • Reading the brother's line as 6.5 cm because it ends at the 6.5 cm mark, and giving 13 m.