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Either paper 4 marks
A plan of a field is drawn to a scale of 1 cm to 25 m.
(a)On the plan, the track is \(7\) cm long.
Find the real length of the track.
(b)The real fence is \(50\) m long.
Calculate how long the fence should be on the plan.
Show answer
(a) \(175\ \text{m}\)
(b) \(2\ \text{cm}\)Every 1 cm on the plan stands for \(25\) m in the field, so multiply by \(25\).
\(7 \times 25 = 175\)
The track is \(175\) m long.
This one goes the other way, from the real field back to the plan, so divide by \(25\) instead of multiplying.
\(50 \div 25 = 2\)
The fence should be \(2\) cm long on the plan.
- (a) M1\(7 \times 25\)
- (a) A1\(175\) m
- (b) M1\(50 \div 25\)
- (b) A1\(2\) cm
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Either paper 4 marks
The diagram is a scale drawing of a rectangular field. The scale is 1 cm to 10 m.
On the drawing the field is \(9\) cm long and \(7\) cm wide.
(a)Find the real length of the field.
(b)A barn in the field is \(30\) m across.
Work out how wide the barn should be drawn on the scale drawing.
Show answer
(a) \(90\ \text{m}\)
(b) \(3\ \text{cm}\)Each centimetre on the drawing stands for \(10\) m.
\(9 \times 10 = 90\)
The field is \(90\) m long.
Going from the real field to the drawing means dividing by \(10\).
\(30 \div 10 = 3\)
The barn should be drawn \(3\) cm across.
- (a) M1\(9 \times 10\)
- (a) A1\(90\) m
- (b) M1\(30 \div 10\)
- (b) A1\(3\) cm
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Either paper 4 marksPart of a plan of a school is shown in the table. The scale of the plan is 1 cm to 3 m.
Room Length on the plan Real length Corridor 6 cm 18 m Classroom 2 cm Hall 21 m (a)Calculate the real length of the classroom.
(b)Calculate the length of the hall on the plan.
Show answer
(a) \(6\ \text{m}\)
(b) \(7\ \text{cm}\)The corridor row shows the scale at work: \(6\) cm on the plan is \(18\) m in the school, and \(18 \div 6 = 3\).
So multiply the plan length by \(3\).
\(2 \times 3 = 6\)
The classroom is \(6\) m long.
This row gives the real length and asks for the plan length, so divide by \(3\) rather than multiplying.
\(21 \div 3 = 7\)
The hall is \(7\) cm long on the plan.
- (a) M1\(2 \times 3\)
- (a) A1\(6\) m
- (b) M1\(21 \div 3\)
- (b) A1\(7\) cm
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Either paper 3 marks
Here is a plan of a flat with three rectangular rooms.
The plan is not drawn to scale.
Make an accurate scale drawing of the plan on the centimetre grid below, by plotting every corner of every room.
Use a scale of 1 cm represents 3 feet.
The bottom left corner of the Bedroom is at the point \(P\).
You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.
Show answer
Bedroom: 3 cm wide and 4 cm high, from (2, 2) to (5, 6). Kitchen: 4 cm wide and 5 cm high, from (5, 2) to (9, 7). Lounge: 4 cm wide and 6 cm high, from (9, 2) to (13, 8).Divide every real length by the scale, because 1 cm on the drawing represents 3 feet.
Bedroom: \(9\text{ ft } \div 3\text{ ft per cm} = 3\) cm wide and \(12\text{ ft } \div 3\text{ ft per cm} = 4\) cm high.
Kitchen: \(12\text{ ft } \div 3\text{ ft per cm} = 4\) cm wide and \(15\text{ ft } \div 3\text{ ft per cm} = 5\) cm high.
Lounge: \(12\text{ ft } \div 3\text{ ft per cm} = 4\) cm wide and \(18\text{ ft } \div 3\text{ ft per cm} = 6\) cm high.
Draw the Bedroom with its bottom left corner at \(P(2, 2)\), going to \((5, 6)\).
Draw the Kitchen against the right-hand wall of the Bedroom, with its bottom edge level, from \((5, 2)\) to \((9, 7)\).
Draw the Lounge against the right-hand wall of the Kitchen, from \((9, 2)\) to \((13, 8)\).
The corners to plot are the ends of every wall: four along the bottom, counting \(P\), and the top corners of each room.
- M1real lengths divided by 3, shown by at least three corners in the right place: widths 3 cm, 4 cm, 4 cm and heights 4 cm, 5 cm, 6 cm
- B1at least six of the nine corners in the right place
- B1every corner of the three rooms in the right place: 3 by 4 cm, 4 by 5 cm, 4 by 6 cm, side by side from P
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Either paper 3 marks
A plan of a flat is drawn with the scale 1 cm represents 0.3 m.
On the plan, an alcove is \(6.7\) cm wide and \(3\) cm deep.
Noah wants to put a sofa in the alcove. The sofa is \(170\) cm wide and \(85\) cm deep.
Will the sofa fit in the alcove? You must show your working.
Show answer
Yes. The alcove is \(201\) cm by \(90\) cm and the sofa is \(170\) cm by \(85\) cm.1 cm on the plan represents \(0.3\) m, which is \(30\) cm.
Change the plan lengths to real lengths in cm.
Width: \(6.7 \times 30 = 201\) cm
Depth: \(3 \times 30 = 90\) cm
Compare each with the sofa, which is \(170\) cm wide and \(85\) cm deep.
Width: \(201\) cm against \(170\) cm, so the alcove is wide enough.
Depth: \(90\) cm against \(85\) cm, so the alcove is deep enough.
The sofa fits only if the alcove is big enough in both directions, so the answer is Yes.
- M1alcove changed to real lengths: \(201\) cm wide and \(90\) cm deep
- M1both compared with the sofa in cm: \(170\) cm and \(85\) cm
- A1Yes, as the alcove is wide enough and deep enough
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Either paper 5 marks
Here is a map of a town.
Each square of the grid is 1 cm by 1 cm.
Scale: 1 cm represents 136 m
North is at the top of the map.
(a)Which place is exactly south west of the station?
Circle your answer.
library, monument, museum, school, town hall
(b)Circle the three-figure bearing of the town hall from the station.
000° 090° 180° 270°
(c)What is the distance, in metres, from the station to the town hall?
(d)Why might the shortest walking distance from the station to the town hall be greater than your answer to part (c)?
Show answer
(a) Library
(b) 090°
(c) \(544\ \text{m}\)
(d) The answer to part (c) is the straight-line distance, but a walk has to follow roads and paths, which cannot go in a straight line between the two places, so the route is longer.Exactly south west of the station means the same number of centimetres south west on the grid, along a diagonal of the squares.
The library is 4 cm west and 4 cm south of the station, so it is on that diagonal.
The town hall and the station are on the same horizontal grid line, so the direction is due east.
Bearings are measured clockwise from north, so due east is \(090^\circ\).
The two places are on the same horizontal grid line, so count the centimetre gaps: \(4\) cm.
Each centimetre represents 136 m:
\(4 \times 136 = 544\)
The distance is \(544\) m.
The distance in part (c) is the straight line between the two places on the map, which is the shortest distance possible.
People walk along roads and paths, which bend and turn rather than going straight, so the walking route is longer.
- (c) M1counting \(4\) cm between them on the map
- (c) A1\(4 \times 136 = 544\) m
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Either paper 2 marks
Here is a sketch of a shape made from a square and a semicircle. The diameter of the semicircle is one side of the square.
The sketch is not drawn to scale.
The point \(A\) is marked on the centimetre grid.
On the grid, make an accurate drawing of the shape.
Draw it with corner \(A\) at the marked point and facing the same way as the sketch.
You cannot draw on this page. Draw this on paper, then show the answer and check your drawing against the model.
Show answer
Square with corners \((11, 7)\), \((11, 11)\), \((7, 11)\) and \((7, 7)\), and a semicircle outside it on the side from \((7, 11)\) to \((7, 7)\), centre \((7, 9)\), radius \(2\) cmCorner \(A\) is marked at \((11, 11)\). Read the sketch to see which way the sides run from \(A\), then count 4 cm along each side to plot the four corners of the square:
\(D(11, 7)\), \(A(11, 11)\), \(B(7, 11)\), \(C(7, 7)\)
Join them to draw the square.
The semicircle sits on the side \(BC\), so that side is its diameter. The centre of the arc is the midpoint of \(BC\), at \((7, 9)\).
The radius is half of 4 cm, which is \(2\) cm.
Draw the arc from \(B\) to \(C\), bulging left away from the square, keeping every point \(2\) cm from the centre.
- B1the three outer sides of the square drawn, between corners \(C(7, 7)\), \(D(11, 7)\), \(A(11, 11)\) and \(B(7, 11)\)
- B1the square, and the semicircle on BC outside it, with centre \((7, 9)\) and radius \(2\) cm
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Calculator 6 marks
Here is a map showing points \(P\) and \(Q\).
(a)Kemal wants to measure the bearing of \(P\) from \(Q\).
Kemal draws the line \(QP\) and a north line at \(P\), then measures the angle clockwise from the north line to the line \(QP\). The angle is \(121^\circ\).
Kemal says that the bearing of \(P\) from \(Q\) is \(121^\circ\).
Is Kemal's method correct? Give a reason for your answer.
(b)On a different map, the bearing of \(D\) from \(C\) is \(225^\circ\).
Nina says, "\(D\) is south west of \(C\)."
Is Nina correct? Give a reason for your answer.
(c)A map of an island is drawn on a grid of squares of side \(1.5\) cm. An airport, \(E\), is marked on the map.
Scale: \(1\) cm represents \(42\) km.
A plane flies due south from the airport. On the map the coast is \(4\) squares due south of \(E\).
How far does the plane fly until it reaches the sea?
Show answer
(a) No: the angle was measured at P, but a bearing of P from Q is measured at Q
(b) Yes: a bearing of 225° points south west
(c) \(252\ \text{km}\)A bearing of \(P\) from \(Q\) starts at \(Q\), the place the bearing is measured from, and turns clockwise from the north line to the line \(QP\).
Kemal measured clockwise from the north line at \(P\).
Answer: No, because the angle was measured at P, but a bearing of P from Q is measured at Q.
The compass points between north, east, south and west lie at \(045^\circ\), \(135^\circ\), \(225^\circ\) and \(315^\circ\), taken clockwise from north in that order: north east, south east, south west, north west.
A bearing of \(225^\circ\) is south west.
Nina says south west, so the answer is Yes.
Distance on the map: \(4 \times 1.5 = 6\) cm.
Each centimetre is \(42\) km, so \(6 \times 42 = 252\) km.
- (a) A1No
- (a) B1reason: the angle must be measured clockwise from north at Q
- (b) A1Yes
- (b) B1reason: 225° is south west
- (c) M1map distance \(6\) cm
- (c) A1distance \(252\) km
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Either paper 3 marks
A map has a scale of \(1 : 8000\).
On the map, the distance from the school to the library is \(6\) cm.
Is the actual distance from the school to the library less than \(500\) m?
Tick a box. Yes / No
Show working to support your answer.
Show answer
Yes: the actual distance is \(480\) m, which is less than \(500\) mThe scale \(1 : 8000\) means \(1\) cm on the map is \(8000\) cm on the ground.
\(6 \times 8000 = 48000\) cm
There are \(100\) cm in a metre:
\(48000 \div 100 = 480\) m
The actual distance is \(480\) m, which is less than \(500\) m.
The question asks about less than \(500\) m, so the answer is Yes.
- M1multiplying by the scale: \(6 \times 8000 = 48000\) cm
- M1converting to metres: \(480\) m
- A1Yes, comparing 480 m with 500 m
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Either paper 3 marks
A map is drawn to a scale of \(1 : 20000\).
On the map, the distance from Garsdale to Hollowbrook is \(8\) cm.
Calculate the real distance from Garsdale to Hollowbrook, in kilometres.
Show answer
\(1.6\ \text{km}\)A scale written as a ratio has no units in it, so both sides are in whatever unit you choose. Start by finding what 1 cm on the map stands for on the ground.
1 cm on the map is \(20000\) cm on the ground, and there are \(100\) cm in a metre:
\(20000 \div 100 = 200\)
So 1 cm on the map stands for \(200\) m.
\(8 \times 200 = 1600\) m
There are \(1000\) m in a kilometre:
\(1600 \div 1000 = 1.6\)
The real distance is \(1.6\) km.
- M11 cm stands for \(200\) m
- M1\(8 \times 200 = 1600\) m
- A1\(1.6\) km