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GCSE · Algebra

5.6D Changing the Subject of a Formula: Fluency and Exam Drill

EDEXCEL 1MA1 1MA1 · Calculators not allowed · about 80 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

Make x the subject of the formula: y = x + 12

(Total for Question 1 is 1 mark)

Make x the subject of the formula: y = x - 15

(Total for Question 2 is 1 mark)

Make x the subject of the formula: y = 7x

(Total for Question 3 is 1 mark)

Make x the subject of the formula: y = x/9

(Total for Question 4 is 1 mark)

Make x the subject of the formula: y = 5x + 4

(Total for Question 5 is 2 marks)

Make x the subject of the formula: y = 6x - 13

(Total for Question 6 is 2 marks)

Make x the subject of the formula: y = (x - 8)/4

(Total for Question 7 is 2 marks)

Make x the subject of the formula: y = 2(x + 9)

(Total for Question 8 is 2 marks)

Make x the subject of the formula: y = x/5 - 6

(Total for Question 9 is 2 marks)

Make b the subject of the formula: a = (b + 11)/6

(Total for Question 10 is 2 marks)

Make x the subject of the formula: y = 4x - 3z

(Total for Question 11 is 2 marks)

Make x the subject of the formula: w = 5(2x - 1)

(Total for Question 12 is 2 marks)

Make m the subject of the formula: k = 3m + 2n

(Total for Question 13 is 2 marks)

Make x the subject of the formula: ax - b = cx + d

(Total for Question 14 is 3 marks)

Make x the subject of the formula: y = x² - 7

(Total for Question 15 is 2 marks)

A car park charges according to the formula C = 2.50 + 1.20h, where C is the total cost in pounds sterling and h is the number of hours parked.

(a)Make h the subject of the formula.(2)

(b)Sasha pays 8.50 pounds sterling to park her car. Work out how many hours she parked for.(3)

(Total for Question 16 is 5 marks)

To make x the subject of the formula y = 5x - 6, Priya writes the following working:
y - 6 = 5x
x = (y - 6)/5

(a)Identify the error in Priya's working.(1)

(b)Write down the correct expression for x in terms of y.(2)

(Total for Question 17 is 3 marks)

The formula connecting a and b is a = (3b + 7)/(b - 2). Show that b = (7 + 2a)/(a - 3)

(Total for Question 18 is 4 marks)

A wedding venue charges a fixed booking fee of 150 pounds sterling plus 40 pounds sterling per guest. The total cost, C pounds sterling, for n guests is given by the formula C = 150 + 40n

(a)Make n the subject of the formula.(2)

(b)A couple have a budget of 3000 pounds sterling for the venue. Work out the maximum number of guests they can invite.(3)

(Total for Question 19 is 5 marks)

The formula connecting x, y and the non-zero constants p, q and r is px + qy = r

(a)Show that y = (r - px)/q(2)

(b)State the one value that q cannot take in the rearranged formula, and explain why.(2)

(Total for Question 20 is 4 marks)

The time, T seconds, taken for a ball dropped from height h metres to hit the ground is given by the formula T = √2h/g, where g is the gravitational field strength.

(a)Make h the subject of the formula.(3)

(b)A ball is dropped from a window and takes exactly 2 seconds to hit the ground. Use g = 9.8 m/s² and your formula from part (a) to work out the height of the window above the ground.(2)

(Total for Question 21 is 5 marks)

A shop increases the price of an item by a fixed percentage, p%. The new price, N pounds sterling, is related to the original price, P pounds sterling, by the formula N = P(1 + p/100)

(a)Make P the subject of the formula.(2)

(b)Make p the subject of the formula.(3)

(Total for Question 22 is 5 marks)

Make x the subject of the formula: q = √(2x - 1)/(x + 4)

(Total for Question 23 is 4 marks)

The formula connecting m and n is m = (4n + 5)/(3 - n). Show that n = (3m - 5)/(m + 4)

(Total for Question 24 is 4 marks)

Mark scheme · 5.6D Changing the Subject of a Formula: Fluency and Exam Drill

Question 1

B1 x = y - 12 oe
Answer: x = y - 12

Question 2

B1 x = y + 15 oe
Answer: x = y + 15

Question 3

B1 x = y/7 oe
Answer: x = y/7

Question 4

B1 x = 9y oe
Answer: x = 9y

Question 5

M1 subtracts 4 from both sides, e.g. y - 4 = 5x
A1 x = (y - 4)/5 oe, cao
Answer: x = (y - 4)/5

Question 6

M1 adds 13 to both sides, e.g. y + 13 = 6x
A1 x = (y + 13)/6 oe, cao
Answer: x = (y + 13)/6

Question 7

M1 multiplies both sides by 4, e.g. 4y = x - 8
A1 x = 4y + 8 oe, cao
Answer: x = 4y + 8

Question 8

M1 divides both sides by 2 (or expands the bracket), e.g. y/2 = x + 9
A1 x = (y - 18)/2 oe, e.g. x = y/2 - 9, cao
Answer: x = (y - 18)/2

Question 9

M1 adds 6 to both sides, e.g. y + 6 = x/5
A1 x = 5y + 30 oe, cao
Answer: x = 5y + 30

Question 10

M1 multiplies both sides by 6, e.g. 6a = b + 11
A1 b = 6a - 11 oe, cao
Answer: b = 6a - 11

Question 11

M1 adds 3z to both sides, e.g. y + 3z = 4x
A1 x = (y + 3z)/4 oe, cao
Answer: x = (y + 3z)/4

Question 12

M1 divides both sides by 5 (or expands the bracket), e.g. w/5 = 2x - 1
A1 x = (w + 5)/10 oe, cao
Answer: x = (w + 5)/10

Question 13

M1 subtracts 2n from both sides, e.g. k - 2n = 3m
A1 m = (k - 2n)/3 oe, cao
Answer: m = (k - 2n)/3

Question 14

M1 collects the terms in x on one side, e.g. ax - cx = d + b
dM1 factorises out x, e.g. x(a - c) = d + b, dependent on the previous method mark
A1 x = (d + b)/(a - c) oe, cao
Answer: x = (d + b)/(a - c)

Question 15

M1 adds 7 to both sides, e.g. y + 7 = x²
A1 x = √y + 7 oe, cao (allow ±)
Answer: x = √y + 7

Question 16

(a) M1 subtracts 2.50 from both sides, e.g. C - 2.50 = 1.20h
(a) A1 h = (C - 2.50)/1.20 oe, cao
(a) Answer: h = (C - 2.50)/1.20
(b) M1 substitutes C = 8.50 into the formula from part (a), ft, e.g. h = (8.50 - 2.50)/1.20
(b) M1 simplifies to h = 6/1.20 (oe)
(b) A1 5 (hours), cao
(b) Answer: 5 hours

Question 17

(a) C1 identifies that Priya subtracted 6 from both sides instead of adding 6 (oe, e.g. states the sign on the 6 should be reversed)
(a) Answer: Priya should have added 6 to both sides (since the formula has -6, not +6), not subtracted 6.
(b) M1 adds 6 to both sides, e.g. y + 6 = 5x
(b) A1 x = (y + 6)/5 oe, cao
(b) Answer: x = (y + 6)/5

Question 18

M1 multiplies both sides by (b - 2), e.g. a(b - 2) = 3b + 7
M1 expands and collects all terms in b on one side, e.g. ab - 3b = 7 + 2a
dM1 factorises out b, e.g. b(a - 3) = 7 + 2a, dependent on the previous method mark
A1 correct rearrangement with all steps shown leading to b = (7 + 2a)/(a - 3), cso
Answer: b = (7 + 2a)/(a - 3) (printed answer, shown)

Question 19

(a) M1 subtracts 150 from both sides, e.g. C - 150 = 40n
(a) A1 n = (C - 150)/40 oe, cao
(a) Answer: n = (C - 150)/40
(b) M1 substitutes C = 3000 into the formula from part (a), ft, e.g. n = (3000 - 150)/40
(b) M1 evaluates n = 71.25 (or 2850/40) and recognises the number of guests must be a whole number that does not exceed the budget
(b) A1 71 (guests), cao
(b) Answer: 71 guests

Question 20

(a) M1 subtracts px from both sides, e.g. qy = r - px
(a) A1 divides both sides by q to reach y = (r - px)/q, cso
(a) Answer: y = (r - px)/q (shown)
(b) B1 states q = 0
(b) B1 gives a correct reason, e.g. dividing by zero is not defined, so the formula would be meaningless if q = 0
(b) Answer: q cannot equal 0, because dividing by zero is not defined.

Question 21

(a) M1 squares both sides, e.g. T² = 2h/g
(a) M1 multiplies both sides by g, e.g. gT² = 2h
(a) A1 h = gT²/2 oe, cao
(a) Answer: h = gT²/2
(b) M1 substitutes g = 9.8 and T = 2 into the formula from part (a), ft, e.g. h = 9.8 x 2²/2
(b) A1 h = 19.6 (m), cao
(b) Answer: h = 19.6 m

Question 22

(a) M1 divides both sides by (1 + p/100), e.g. P = N/(1 + p/100)
(a) A1 P = 100N/(100 + p) oe, cao
(a) Answer: P = 100N/(100 + p)
(b) M1 expands the bracket, e.g. N = P + Pp/100
(b) M1 rearranges to isolate the term in p, e.g. N - P = Pp/100
(b) A1 p = 100(N - P)/P oe, cao
(b) Answer: p = 100(N - P)/P

Question 23

M1 squares both sides, e.g. q² = (2x - 1)/(x + 4)
M1 multiplies both sides by (x + 4), e.g. q²(x + 4) = 2x - 1
dM1 expands and collects all terms in x on one side, e.g. q² x - 2x = -1 - 4q², dependent on the previous method mark
A1 x = (1 + 4q²)/(2 - q²) oe, cao
Answer: x = (1 + 4q²)/(2 - q²)

Question 24

M1 multiplies both sides by (3 - n), e.g. m(3 - n) = 4n + 5
M1 expands the bracket, e.g. 3m - mn = 4n + 5
dM1 collects all terms in n on one side and factorises, e.g. 3m - 5 = n(4 + m), dependent on the previous method mark
A1 correct rearrangement with all steps shown leading to n = (3m - 5)/(m + 4), cso
Answer: n = (3m - 5)/(m + 4) (printed answer, shown)

Changing the Subject of a Formula: Fluency and Exam Drill - Worksheets, Questions and Revision