Printable · GCSE Foundation · ages 14-16
Rearranging formulae worksheet — GCSE Foundation
Fifteen questions on "rearranging formulae" — DfE statement A5. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
Answer key: Rearranging formulae worksheet — GCSE Foundation
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- (a) x = (y + 9) / 2 — Method: undo the subtraction of 9 first, then undo the multiplication by 2. Working: y = 2x − 9, so adding 9 to both sides gives y + 9 = 2x, then dividing both sides by 2 gives x = (y + 9) / 2. The value x = (y − 9) / 2 comes from a sign error, keeping the −9 instead of moving it to +9. The value x = 2(y + 9) comes from multiplying by 2 instead of dividing. The value x = y / 2 + 9 comes from dividing by 2 before adding 9, the wrong order of operations, instead of dividing the whole bracket.
- (b) 158 minutes — Rearranging C = 15 + 0.2m for m: subtract 15 from both sides to get C − 15 = 0.2m, then divide by 0.2: m = (C − 15)/0.2. Substituting C = 46.60: m = (46.60 − 15)/0.2 = 31.60/0.2 = 158. Answering 233 minutes comes from dividing the whole £46.60 by 0.2 without first taking off the £15 fixed charge. Answering 308 minutes comes from adding the £15 instead of subtracting it: (46.60 + 15)/0.2. Answering 218 minutes divides first and subtracts 15 afterwards, in the wrong order: 46.60/0.2 − 15 = 233 − 15 = 218. Ali made 158 minutes of calls.
- (b) p = 3n — n = p/3 means p has been divided by 3, so to make p the subject, multiply both sides by 3: p = 3n. Writing p = n/3 leaves the formula exactly as it was, without undoing the division. Writing p = n − 3 mistakes division for subtraction and takes 3 away from n instead of multiplying. Writing p = 3/n incorrectly flips the fraction upside down rather than multiplying n by 3. The correct rearrangement is p = 3n.
- (c) x = 2y + 10 — Method: undo the operations done to x in reverse order — add 5, then multiply by 2. Working: y = x/2 − 5, so y + 5 = x/2, so x = 2(y + 5) = 2y + 10. Answer: x = 2y + 10. x = 2y + 5 comes from multiplying only the x/2 term by 2 and forgetting to multiply the 5 as well. x = 2y − 10 comes from a sign error, subtracting 5 instead of adding it before multiplying by 2. x = (y + 5)/2 comes from dividing by 2 instead of multiplying, the wrong operation to undo a division.
- (c) s = P / 4 — Method: undo the multiplication by 4 by dividing both sides by 4. Working: P = 4s, so dividing both sides by 4 gives s = P / 4. The value s = 4P comes from multiplying by 4 instead of dividing — using the wrong inverse operation. The value s = P + 4 comes from adding 4 instead of dividing. The value s = P − 4 comes from subtracting 4 instead of dividing.
- (c) x = (y + 24)/4 — Method: expand the bracket first, then undo the operations done to x in reverse order. Working: y = 4(x − 6) = 4x − 24, so y + 24 = 4x, so x = (y + 24)/4. Answer: x = (y + 24)/4. x = (y + 6)/4 comes from expanding the bracket incorrectly, treating 4(x − 6) as 4x − 6 instead of 4x − 24. x = 4y + 96 comes from multiplying by 4 instead of dividing by 4 to undo the multiplication, giving 4(y + 24) = 4y + 96. x = y/4 − 6 comes from dividing by 4 first without expanding the bracket, then subtracting 6 as if the bracket's operation still applied afterwards.
- (a) P = 100I / (RT) — Method: undo the operations done to P in reverse order — multiply by 100, then divide by R and by T. Working: I = PRT/100, so 100I = PRT, so P = 100I / (RT). Answer: P = 100I / (RT). P = IRT/100 comes from leaving R and T in the numerator instead of moving them to the denominator. P = 100RT/I comes from swapping P and I when rearranging. P = 100I/R comes from dividing by R only and forgetting to also divide by T.
- (d) x = (y + 8) / 3 — Method: add 8 to both sides, then divide the whole side by 3. Working: y = 3x − 8, so adding 8 to both sides gives y + 8 = 3x, then dividing both sides by 3 gives x = (y + 8) / 3. The value x = y + 8 / 3 is Priya's version, which comes from dividing only the 8 by 3 instead of dividing the whole expression y + 8 by 3. The value x = (y − 8) / 3 comes from a sign error, keeping −8 instead of moving it to +8. The value x = 3(y + 8) comes from multiplying by 3 instead of dividing.
- (c) 22 m — Rearranging P = 2l + 2w for l gives l = (P − 2w)/2. Substituting P = 60 and w = 8: l = (60 − 16)/2 = 44/2 = 22 m. Answering 30 m comes from ignoring the width altogether and working out l = P/2, which only holds when w = 0. Answering 38 m comes from adding 2w instead of subtracting it: (60 + 16)/2 = 38. Answering 11 m comes from dividing by 4 instead of 2, as if the formula were P = 4l + 4w. The pen's length is 22 m.
- (c) r = √(A/π) — A = πr² means r has been squared and then multiplied by π. To make r the subject, first divide both sides by π to get A/π = r², then take the square root of both sides: r = √(A/π). Writing r = A/π stops after dividing by π and forgets that r is still squared — it never undoes the square. Writing r = √A/π takes the square root before dividing by π, which square-roots only the A and not the whole of A/π. Writing r = (A/π)² squares A/π instead of taking its square root — the opposite of what is needed to undo r². The correct rearrangement is r = √(A/π).
- (d) 6 — Method: rearrange the formula to make d the subject, then substitute C = 133. Working: C = 25 + 18d, so subtracting 25 from both sides gives C − 25 = 18d, then dividing by 18 gives d = (C − 25) / 18. Substituting C = 133: d = (133 − 25) / 18 = 108 / 18 = 6. The value 7.4 comes from dividing 133 by 18 without subtracting the fixed £25 first (133 / 18 ≈ 7.4). The value 4.6 comes from pairing the numbers the wrong way round, working out (133 − 18) / 25 = 4.6. The value 90 comes from subtracting both 25 and 18 from 133 instead of dividing by 18.
- (b) x = (y − 3)/5 — To make x the subject of y = 5x + 3, first subtract 3 from both sides to get y − 3 = 5x, then divide both sides by 5: x = (y − 3)/5. Writing x = (y + 3)/5 keeps the division correct but does not change the sign of the 3 when moving it across. Writing x = y/5 − 3 divides only the y term by 5 and leaves the 3 as a separate subtraction, instead of subtracting first and dividing the whole expression. Writing x = 5(y − 3) applies the correct order of subtracting 3 first, but then multiplies by 5 instead of dividing — the inverse of 5x is division, not multiplication. The correct rearrangement is x = (y − 3)/5.
- (d) x = 5y + 4 — To make x the subject of y = (x − 4)/5, first multiply both sides by 5 to clear the fraction: 5y = x − 4, then add 4 to both sides: x = 5y + 4. Writing x = 5y − 4 multiplies correctly but keeps the minus sign on the 4 instead of changing it to a plus when moving it across. Writing x = y/5 + 4 divides by 5 instead of multiplying, the wrong inverse of the fraction. Writing x = 5(y + 4) adds 4 before multiplying by 5, reversing the correct order of the two steps. The correct rearrangement is x = 5y + 4.
- (d) h = 2A / (a + b) — Method: undo the multiplication by 1/2 by multiplying both sides by 2, then undo the multiplication by (a + b) by dividing both sides by it. Working: A = (a + b)h / 2, so multiplying both sides by 2 gives 2A = (a + b)h, then dividing both sides by (a + b) gives h = 2A / (a + b). The value h = A / (a + b) comes from forgetting to multiply by 2 to clear the 1/2 first. The value h = 2A / a + b comes from dividing only by a and leaving b outside the fraction, instead of dividing by the whole bracket (a + b). The value h = (a + b) / (2A) comes from inverting the fraction.
- (b) c = k/6 — k = 6c means c has been multiplied by 6, so to make c the subject, divide both sides by 6: c = k/6. Writing c = 6k multiplies by 6 again instead of undoing the multiplication. Writing c = k − 6 mistakes multiplying for adding, and subtracts 6 rather than dividing. Writing c = 6 − k reverses the order as well as the operation. The correct rearrangement is c = k/6.
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