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Either paper 4 marks
A rainwater barrel has a hole in it. The graph shows the volume of water left in the rainwater barrel as it leaks out at a steady rate.
(a)How much water was in the rainwater barrel at the start?
(b)Work out the rate at which the water is leaking out.
Give your answer in litres per minute.
(c)How long does it take for the rainwater barrel to empty completely?
Show answer
(a) \(22\ \text{litres}\)
(b) \(1\ \text{litre per minute}\)
(c) \(22\ \text{minutes}\)The start is at \(0\) minutes, where the line meets the volume axis.
There were \(22\) litres in the rainwater barrel.
The rate is the gradient of the line. The line falls \(22\) litres in \(22\) minutes:
\(\frac{ 22 }{ 22 } = 1\)
So the rainwater barrel loses \(1\) litre every minute. (As a gradient it is \(-1\), because the volume is going down.)
The rainwater barrel is empty when the volume is \(0\), which is where the line meets the time axis.
That is at \(22\) minutes.
- (b) M1gradient method: \(22\) litres lost over \(22\) minutes
- (b) A1rate \(1\) litre per minute
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Either paper 5 marks
Two taps, \(A\) and \(B\), are used to fill two identical tanks. Each tank holds \(24\) litres.
Both taps are turned on at the same time. The graph shows the volume of water in each tank.
(a)Work out the rate at which tap \(A\) fills its tank.
Give your answer in litres per minute.
(b)Work out the rate at which tap \(B\) fills its tank.
Give your answer in litres per minute.
(c)Which tap fills its tank at the greater rate?
Show answer
(a) \(3\ \text{litres per minute}\)
(b) \(2\ \text{litres per minute}\)
(c) Tap ALine \(A\) reaches \(24\) litres after \(8\) minutes, so its gradient is
\(\frac{ 24 }{ 8 } = 3\) litres per minute
Line \(B\) reaches \(24\) litres after \(12\) minutes:
\(\frac{ 24 }{ 12 } = 2\) litres per minute
Tap \(A\) fills its tank at \(3\) litres per minute, against \(2\) for tap \(B\).
Its line is the steeper of the two, and it reaches the top first.
- (a) M1volume \(24\) litres divided by time \(8\) minutes
- (a) A1rate \(3\) litres per minute
- (b) M1volume \(24\) litres divided by time \(12\) minutes
- (b) A1rate \(2\) litres per minute
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Either paper 5 marks(a)
A stack of pallets stands on the floor. It presses down on the floor with a force of \(4800\) N.
The base of the stack of pallets is a rectangle measuring \(4\) m by \(3\) m.
Find the pressure the stack of pallets exerts on the floor.
Give your answer in N/m\(^2\).
(b)A different stack of pallets presses down with a force of \(2160\) N.
The pressure it exerts on the floor is \(240\) N/m\(^2\).
Find the area of the base of this second stack of pallets.
Show answer
(a) \(400\ \text{N/m}^2\)
(b) \(9\ \text{m}^2\)Pressure is the force spread over the area it acts on.
\(\text{pressure} = \frac{\text{force}}{\text{area}}\)
The base has area \(4 \times 3 = 12\ \text{m}^2\).
\(\frac{ 4800 }{ 12 } = 400\)
So the pressure is \(400\ \text{N/m}^2\).
Rearrange the formula to make the area the subject.
\(\text{pressure} = \frac{\text{force}}{\text{area}}\), so \(\text{area} = \frac{\text{force}}{\text{pressure}}\)
\(\frac{ 2160 }{ 240 } = 9\)
So the base has area \(9\ \text{m}^2\).
- (a) M1area of base \(4 \times 3 = 12\) m²
- (a) M1force divided by area, \(4800 \div 12\)
- (a) A1pressure \(400\) N/m²
- (b) M1area = force divided by pressure, \(2160 \div 240\)
- (b) A1area \(9\) m²
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Either paper 4 marks
A tank can hold up to \(2300\) litres. At a certain moment it contains \(1143\) litres.
Water is then drained from the tank at a constant rate of \(6\) litres per minute.
Work out how much space is left in the tank \(1\) hour later.
Show answer
\(1517\ \text{litres}\)Change the time to minutes so it matches the rate.
\(1 \times 60 = 60\) minutes
Multiply by the rate to find the volume that has moved.
\(6 \times 60 = 360\) litres
Subtract this from the starting volume.
\(1143 - 360 = 783\) litres
Subtract this from the capacity to find the space left.
\(2300 - 783 = 1517\) litres
- M1time in minutes \(60\)
- M1volume moved \(360\) litres
- M1volume in tank \(783\) litres
- A1space left \(1517\) litres
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Either paper 3 marks
The town of Corwen has a population of 130,720 and an area of 16 square miles.
The town of Dunmore has a population density of 3,399 people per square kilometre.
\(\text{population density} = \frac{ \text{population} }{ \text{area} }\)
Use \(1\) square mile \(= 2.5\) square kilometres.
Which town has the greater population density?
You must show your working to support your answer.
Show answer
Dunmore. Corwen has about 3,268 people per square kilometre and Dunmore has 3,399 people per square kilometre.Find the population density of Corwen in people per square mile:
\(130720 \div 16 = 8170\) people per square mile
Dunmore is given per square kilometre, so convert Corwen to the same unit. One square mile is \(2.5\) square kilometres, so the same people are spread over a bigger number of square kilometres. The density per square kilometre is smaller, so divide by \(2.5\):
\(8170 \div 2.5 \approx 3268\) people per square kilometre
Both densities are now in people per square kilometre: \(3268\) for Corwen and \(3399\) for Dunmore.
\(3399\) is greater than \(3268\), so Dunmore has the greater population density.
- M1Corwen per square mile, \(130720 \div 16 = 8170\)
- M1converted to per square kilometre, \(8170 \div 2.5 \approx 3268\)
- A1Dunmore, comparing 3,399 with 3,268
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Either paper 4 marks
Sand is poured from a hopper at a steady rate of 14.5 kg per second for 26 minutes.
Is more than 22 tonnes of sand poured?
Tick a box.
Yes \(\square\) No \(\square\)
You must show your working.
Show answer
Yes. In 26 minutes the mass is 14.5 \(\times\) 1560 = 22620 kg, which is 22.62 tonnes, more than 22 tonnes.The rate is in kg per second but the time is in minutes, so change the time to seconds first.
\(26 \times 60 = 1560\) s
Multiply the rate by the number of seconds to get the mass in kg.
\(14.5 \times 1560 = 22620\) kg
The threshold is in tonnes, and \(1\) tonne is \(1000\) kg, so divide by \(1000\).
\(22620 \div 1000 = 22.62\) tonnes
\(22.62\) tonnes is more than \(22\) tonnes, so the answer is Yes.
- M1time in seconds, \(26 \times 60 = 1560\) s
- M1mass in kg, \(14.5 \times 1560 = 22620\) kg
- M1mass in tonnes, \(22620 \div 1000 = 22.62\) tonnes
- A1Yes, comparing \(22.62\) tonnes with \(22\) tonnes
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Either paper 5 marks(a)
A tank is a cuboid measuring \(3\) m by \(2\) m by \(2\) m.
\(1\) m\(^3\) holds \(1000\) litres of water.
Work out the capacity of the tank in litres.
(b)Water flows into the tank at a rate of \(50\) litres per minute.
Work out how long the tank takes to fill.
Give your answer in hours.
Show answer
(a) \(12000\ \text{litres}\)
(b) \(4\ \text{hours}\)The volume of a cuboid is length times width times height.
\(3 \times 2 \times 2 = 12\ \text{m}^3\)
Each cubic metre holds \(1000\) litres.
\(12 \times 1000 = 12000\) litres
Every minute puts \(50\) litres in, so divide the capacity by the rate to find the number of minutes.
\(12000 \div 50 = 240\) minutes
There are \(60\) minutes in an hour.
\(240 \div 60 = 4\) hours
- (a) M1volume \(3 \times 2 \times 2 = 12\) m³
- (a) A1capacity \(12000\) litres
- (b) M1capacity divided by rate, \(12000 \div 50\)
- (b) M1minutes \(240\) changed to hours by dividing by \(60\)
- (b) A1time \(4\) hours
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Either paper 5 marks(a)
Water comes out of a outlet at a rate of \(350\) cm\(^3\) per second.
\(1\) litre is \(1000\) cm\(^3\).
Work out the rate in litres per minute.
(b)The outlet runs for \(16\) minutes.
Work out the volume of water it delivers.
Give your answer in litres.
Show answer
(a) \(21\ \text{litres per minute}\)
(b) \(336\ \text{litres}\)Change one unit at a time. First seconds to minutes, by multiplying by \(60\).
\(350 \times 60 = 21000\ \text{cm}^3\) per minute
Now cubic centimetres to litres, by dividing by \(1000\).
\(21000 \div 1000 = 21\) litres per minute
The rate is now in litres per minute, so multiply it by the number of minutes.
\(21 \times 16 = 336\) litres
- (a) M1seconds to minutes: \(350 \times 60 = 21000\) cm³ per minute
- (a) M1cm³ to litres: dividing by \(1000\)
- (a) A1rate \(21\) litres per minute
- (b) M1rate multiplied by time, \(21 \times 16\)
- (b) A1volume \(336\) litres
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Either paper 6 marks
A tank has a capacity of \(480\) litres. Tap A fills it at a rate of \(11\) litres per minute, and tap B fills it at a rate of \(13\) litres per minute.
(a)Tap A and tap B are turned on together.
Calculate how long they take to fill the tank between them.
Give your answer in minutes.
(b)Tap B is now replaced with a different tap, tap C. With tap A and tap C running together, the tank needs to fill in \(30\) minutes instead.
Find the rate tap C must supply, in litres per minute.
Show answer
(a) \(20\ \text{minutes}\)
(b) \(5\ \text{litres per minute}\)When both taps run together, their rates add.
\(11 + 13 = 24\) litres per minute
Divide the capacity by this combined rate.
\(480 \div 24 = 20\) minutes
In \(30\) minutes the two taps together must deliver the whole \(480\) litres, so their combined rate is
\(480 \div 30 = 16\) litres per minute
Tap A already supplies \(11\) litres per minute, so tap C must supply the rest.
\(16 - 11 = 5\) litres per minute
- (a) M1rates added: \(11\) + \(13\) = \(24\) litres per minute
- (a) M1capacity ÷ combined rate: \(480\) ÷ \(24\)
- (a) A1\(20\) minutes
- (b) M1combined rate needed: \(480\) ÷ \(30\) = \(16\) litres per minute
- (b) M1takes off tap A's rate: \(16\) − \(11\)
- (b) A1\(5\) litres per minute
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Either paper 5 marks
A tank is being filled by a pump at a rate of \(28\) litres per minute.
At the same time, an outlet valve at the base is left open, and water drains out through it at a rate of \(7\) litres per minute.
The tank has a capacity of \(420\) litres.
(a)Work out how long the tank takes to fill.
Give your answer in minutes.
(b)Work out how many litres leak away while the tank is filling.
Show answer
(a) \(20\ \text{minutes}\)
(b) \(140\ \text{litres}\)Subtract the drain rate from the fill rate to find the net rate at which the tank actually fills.
\(28 - 7 = 21\) litres per minute
Divide the capacity by this net rate.
\(420 \div 21 = 20\) minutes
Multiply the drain rate by the time taken to fill, found in part (a).
\(7 \times 20 = 140\) litres
- (a) M1net rate \(28 - 7 = 21\) litres per minute
- (a) M1capacity divided by the net rate, \(420 \div 21\)
- (a) A1time \(20\) minutes
- (b) M1drain rate multiplied by time taken, \(7 \times 20\)
- (b) A1volume \(140\) litres