-
Either paper 1 mark
\(a\) is inversely proportional to \(b\).
Circle the statement that is equivalent to this.
A) \(a\) is directly proportional to \(b\)
B) \(a\) is directly proportional to \(\dfrac{1}{ b }\)
C) \(a\) is directly proportional to \(b^2\)
D) \(a\) is directly proportional to \(\dfrac{1}{ b^2 }\)
Show answer
B) \(a\) is directly proportional to \(\dfrac{1}{ b }\)\(a\) is inversely proportional to \(b\) means \(a = \dfrac{ k }{ b }\) for some non-zero constant \(k\).
Writing this as \(a = k \times \dfrac{1}{ b }\) shows \(a\) is directly proportional to \(\dfrac{1}{ b }\) — not to \(b\) itself, and not to \(b^2\).
The correct statement is B) \(a\) is directly proportional to \(\dfrac{1}{ b }\)
-
Either paper 4 marks
The diagram shows the position of a tap's dial. The tap is off at \(0^\circ\) and fully on at \(180^\circ\).
When fully on, water flows out of the tap at 24 litres per minute. The rate at which water flows out is in direct proportion to the angle of turn shown on the dial.
The water flows into a tank with a capacity of 115 litres.
Will it take less than 10 minutes to fill the tank? You must show your working.
Show answer
No — the rate at \(75^\circ\) is \(10\) litres per minute, so \(10\) minutes delivers \(100\) litres, compared with the \(115\) litre tank.The flow rate is directly proportional to the angle of turn, and \(180^\circ\) gives 24 litres per minute, so the rate at \(75^\circ\) is:
\(\dfrac{ 75 }{180} \times 24 = 10\) litres per minute
At this rate, the volume that flows out in 10 minutes is:
\(10 \times 10 = 100\) litres
The tank holds 115 litres, so comparing this with 100 litres: No.
- M1flow rate at \(75^\circ\): \(24 \times \dfrac{ 75 }{180}\)
- A1\(10\) litres per minute
- M1volume in 10 minutes at this rate: \(10 \times 10 = 100\) litres, or time to fill: \(115 \div 10 = 11.50\) minutes
- A1No — \(100\) litres compared with the \(115\) litre tank, or \(11.50\) minutes compared with 10 minutes
-
Either paper 3 marks
A chef has a tub of blueberries.
The chef uses all the blueberries and puts the same number on each dessert.
\(d\) is the number of desserts.
\(b\) is the number of blueberries on each dessert.
\(b\) is inversely proportional to \(d\), so
\(b = \dfrac{k}{ d }\)
where \(k\) is a constant.
(a)What does the constant \(k\) represent?
Tick the correct box.
☐ the number of blueberries in the tub
☐ the number of desserts
☐ the number of blueberries on each dessert
☐ none of the above
(b) Complete the table.b 3 d 80 48 8 Show answer
(a) the number of blueberries in the tub
(b) \(b = 5\) and \(b = 30\)Multiply both sides of \(b = \dfrac{k}{ d }\) by \(d\):
\(k = b \times d\)
This is the number of blueberries on each dessert multiplied by the number of desserts, which counts every one of the blueberries in the tub, however they are shared out.
So \(k\) is the number of blueberries in the tub.
The first column is complete, so it gives the constant:
\(k = b \times d = 3 \times 80 = 240\)
Every column has \(b \times d = 240\), so each blank is 240 divided by the entry beside it.
Second column: \(240 \div 48 = 5\)
Third column: \(240 \div 8 = 30\)
- (b) B1second column \(b = 5\)
- (b) B1third column \(b = 30\)
-
Either paper 3 marksA train travels 42 miles. The table shows some values of the average speed of the train, in miles per hour, for different times taken.
Time (hours) Speed (mph) 1 42 2 21 3 4 5 6 7 (a)Three speeds are missing from the table. Work them out, and write the three in order: for 3 hours, for 4 hours and for 5 hours.
Give any decimal values correct to 1 decimal place.
(b)Draw a graph to show the average speed of the train for times taken between 1 hour and 6 hours. Plot the six points from your table on the grid, then join them with a smooth curve.
You cannot draw on this page. Draw (b) on paper, then show the answer and check your drawing against the model.
Show answer
(a) \(14\), \(10.5\) and \(8.4\) mph at 3, 4 and 5 hours
(b) The points \((1, 42)\), \((2, 21)\), \((3, 14)\), \((4, 10.5)\), \((5, 8.4)\) and \((6, 7)\), joined with one smooth curve that falls steeply and then levels off(b)Every entry in the table is the distance divided by the time taken:
\(42 \div 3 = 14\) mph, \(42 \div 4 \approx 10.5\) mph, \(42 \div 5 \approx 8.4\) mph
The table is not built by halving each entry along the row. Halving the 2-hour entry would give \(10.5\) at 3 hours instead of \(14\).
Plot each time against its speed from the completed table. A speed that is not on a grid line goes between the two lines either side of it.
Joining the six points with a single smooth curve gives a graph that falls steeply between 1 and 2 hours and then levels off, never touching either axis.
- (b) B1at least five of the six points plotted, each within half a square of its speed
- (b) B1all six points plotted within half a square, and joined with one smooth curve
-
Either paper 6 marks
\(T\) is directly proportional to \(M\).
When \(M = 5\), \(T = 25\).
(a)Find a formula for \(T\) in terms of \(M\).
(b)Work out the value of \(T\) when \(M = 2\).
(c)Work out the value of \(M\) when \(T = 85\).
Show answer
(a) \(T = 5M\)
(b) \(10\)
(c) \(17\)"Directly proportional" means \(T = kM\) for some fixed number \(k\), so put the pair you are given into that:
\(25 = k \times 5\)
\(k = \dfrac{ 25 }{ 5 } = 5\)
\(T = 5M\)
Put \(M = 2\) into the formula:
\(T = 5 \times 2 = 10\)
This time \(T\) is known and \(M\) is not, so put 85 into the formula and solve:
\(85 = 5M\)
\(M = \dfrac{ 85 }{ 5 } = 17\)
The same formula works in both directions — that is why it is worth writing down in part (a) before doing anything else.
- (a) M1finds the constant \(k = 25 \div 5 = 5\)
- (a) A1formula \(T = 5M\)
- (b) M1substitutes into the formula, \(5 \times 2\)
- (b) A1\(T = 10\)
- (c) M1divides \(85\) by \(5\)
- (c) A1\(M = 17\)
-
Either paper 4 marks
Here are two pairs of values of \(x\) and \(y\).
\(x = 8\), \(y = 32\)
\(x = 4\), \(y = 16\)
(a)Write down the value of \(y \div x\) for the first pair.
(b)Write down the value of \(y \div x\) for the second pair.
(c)Is \(y\) directly proportional to \(x\)? Show how you decide.
Show answer
(a) \(4\)
(b) \(4\)
(c) Yes — \(\dfrac{ 32 }{ 8 } = 4\) and \(\dfrac{ 16 }{ 4 } = 4\), which are equal, so \(y\) is directly proportional to \(x\).\(y \div x = 32 \div 8 = 4\)
\(y \div x = 16 \div 4 = 4\)
Compare the two ratios found in parts (a) and (b): \(4\) and \(4\).
\(y\) is directly proportional to \(x\) only when \(y \div x\) gives the *same* number for every pair — a table where \(y\) simply grows as \(x\) grows is not enough on its own; the ratio itself has to stay fixed.
Here the two ratios are equal, so \(y\) is directly proportional to \(x\).
- (c) M1compare the ratios, \(4\) and \(4\)
- (c) A1Yes — the ratios are equal, so \(y\) is directly proportional to \(x\)
-
Either paper 1 mark
\(a\) is directly proportional to \(\sqrt{b}\).
The value of \(b\) is multiplied by 64.
Marcus says: "\(a\) will be 32 times bigger."
Is he correct? Tick Yes or No. Give a reason for your answer.
Show answer
No — the correct change is 8 times bigger, which does not match the claim of 32 times bigger.\(a\) is directly proportional to \(\sqrt{b}\).
So the correct multiplier for \(a\) is the square root of the multiplier for \(b\): \(\sqrt{64} = 8\).
Marcus claims \(a\) will be 32 times bigger. 8 times bigger does not match that claim, so the answer is No.
-
Either paper 3 marks
In this question, assume that each person works at the same rate.
14 people can unload a delivery lorry in 12 hours.
(a)If 8 people work on the same job, how many hours will it take?
(b)In fact, of the 8 people, 4 of them work at a slower rate. Everyone else works at the rate assumed in part (a).
What does this mean about the number of hours it will take? Tick one box.
☐ It is greater than the answer to (a)
☐ It is the same as the answer to (a)
☐ It is less than the answer to (a)
☐ It is not possible to say
Show answer
(a) \(21\ \text{hours}\)
(b) It is greater than the answer to (a)The job needs the same total, so \(\text{people} \times \text{hours}\) is constant.
\(14 \times 12 = 168\) person-hours
\(168 \div 8 = 21\) hours
Part (a) assumed every person works at the same rate, so the answer to (a) is the time at that rate.
The slower workers can only lengthen the job.
- (a) M1\(14 \times 12 = 168\) person-hours
- (a) A1\(168 \div 8 = 21\) hours
-
Either paper 7 marks
\(D\) is directly proportional to \(s^2\).
When \(s = 6\), \(D = 288\).
(a)Find a formula for \(D\) in terms of \(s\).
(b)Work out the value of \(D\) when \(s = 9\).
(c)Work out the positive value of \(s\) when \(D = 512\).
Show answer
(a) \(D = 8s^2\)
(b) \(648\)
(c) \(8\)\(D = ks^2\), so square the \(s\) first and then divide:
\(288 = k \times 6^2 = k \times 36\)
\(k = \dfrac{ 288 }{ 36 } = 8\)
\(D = 8s^2\)
Square first, multiply second — the square only applies to \(s\), not to the whole of \(8s\):
\(D = 8 \times 9^2 = 8 \times 81 = 648\)
Put 512 into the formula and undo the two steps in reverse order — divide, then square root:
\(512 = 8s^2\)
\(s^2 = \dfrac{ 512 }{ 8 } = 64\)
\(s = \sqrt{ 64 } = 8\)
\(-8\) also squares to 64, which is why the question asks for the positive value.
- (a) M1finds the constant \(k = 288 \div 36 = 8\)
- (a) A1formula \(D = 8s^2\)
- (b) M1squares first, \(8 \times 81\)
- (b) A1\(D = 648\)
- (c) M1divides 512 by 8 to get \(s^2 = 64\)
- (c) M1square root of \(64\)
- (c) A1\(s = 8\)
-
Either paper 6 marks
\(y\) is inversely proportional to \(x\).
When \(x = 12\), \(y = 5\).
(a)Find a formula for \(y\) in terms of \(x\).
(b)Calculate the value of \(y\) when \(x = 5\).
(c)Calculate the value of \(x\) when \(y = 6\).
Show answer
(a) \(y = \dfrac{ 60 }{ x }\)
(b) \(12\)
(c) \(10\)"Inversely proportional" means \(y = \dfrac{ k }{ x }\), so \(k = xy\) — multiply the pair together rather than dividing them.
\(k = 12 \times 5 = 60\)
\(y = \dfrac{ 60 }{ x }\)
The test for inverse proportion is that \(x \times y\) is the same for every pair, where the test for direct proportion is that \(y \div x\) is.
\(y = \dfrac{ 60 }{ 5 } = 12\)
Check it makes sense: \(5\) is smaller than \(12\), so \(y\) should be bigger, and 12 is.
\(6 = \dfrac{ 60 }{ x }\)
Multiply both sides by \(x\):
\(6x = 60\)
\(x = \dfrac{ 60 }{ 6 } = 10\)
Because \(xy\) is always 60, this step is the same one as part (b) with the two letters swapped over.
- (a) M1finds the constant \(k = xy = 12 \times 5\)
- (a) A1formula \(y = \dfrac{ 60 }{ x }\)
- (b) M1divides \(60\) by \(5\)
- (b) A1\(y = 12\)
- (c) M1rearranges to \(x = \dfrac{ 60 }{ 6 }\)
- (c) A1\(x = 10\)