Rearranging Exam Answers: Grade A

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1.

4(s – 8u) = a

2

s – 8u = a/4

a M1 for s – 8u = 4 OR 4s = a + 32u A1 [2]

2.

a=

7c + 1 6

3

6a – 2c = 5c + 1 6a = 5c + 2c + 1

5c + 1 M1 for either 6a – 2c = 5c + 1 OR 3a – c = 2 oe RHS M1 ft (indep) for correct process to isolate term in a A1 cao [3]

2

3.

2

(a) x – xy – xy + y 2 2 x – 2xy + y 2 M1 for 3 terms correct with sign, or 4 terms correct ignoring 2 2 signs, or x – 2xy – y A1 cao (b)

aq – ac = d aq = ac + d ac + d a 3 B1 aq – ac M1 for +ac or ÷ a both sides A1 oe OR d q–c= a B2 d B1 q = a + c, q = d + a + c oe [5]

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4.

D – kt 2 T

2

D – kt² = ut D – kt 2 T u= D – kt 2 T B2 for oe

D ut + kt 2 = t (B1 for t or D – kt² = ut or one of two steps correct) [2]

5.

(a)

24.8 3

84 = 6.7π + 2 × 6.7 + 2a 2a + 13.4 = 62.95…. or 2a + 34.44 = 84 M1 for substituting correctly, π may be left M1 for correct rearrangement as far as ± 2a A1 for 24.7-24.8

(b)

P – 2a π +2

3

P = πr + 2r + 2a P – 2a = πr + 2r P – 2a = (π + 2)r M1 subtracting 2a from each side M1 for factorising to get (π + 2)r P – 2a A1 for π + 2 oe p − 2a S.C 5.14 oe is M1 M1 A0 [6]

6.

(a)

–5

7.75 × 10 –5 B2 for 7.75 × 10 5 9 (B1 for either 7.75 or 3.875 × 10 or 5 × 10 )

2


(b)

p xp + 1

4

xpq = p – q xpq + q = p q(xp + 1) = p M1 for correct elimination of fractions M1 for a correct isolation M1 for q(xp + 1) = p p A1 for xp + 1 [6]

7.

(a)

60

3w + 20 200

3

=1

3w + 20 = 200 M1 p = 1 stated or used M1dep 3w + 20 = 200 oe A1 cao 200 p − 20 3

(b)

3

200p = 3w + 20 3w = 200p – 20 3w 20 + 200 200 M1 for 200p = 3w + 20 or p =

M1 3w = 200p – 20 or correct ft to isolate the w term 200 p − 20

A1 for

3

oe

4( 3w + 20)

(c)

60 − w

4

(3w + 20)(A + 12) = 200A 3wA + 36w + 20A +240 36w + 240 = 3A (60 − w) or eg. 36w + 240 = A (180 − 3w)

(= 200A)

Alternative for (c) A + 12

1+

A 12

= =

200 3w + 20 200

3w + 20 200 = −1 A 3w + 20 A

12

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A 12

1

=

A=

200

−1 3w + 20 12( 3w + 20) 180 − 3w

M1 for (3w + 20)(A + 12) = 200A B1 for correct expansion of brackets M1for isolating A terms and factorising; condone one arithmetic/sign slip in total during these two processes 4( 3w + 20 ) 60 − w

A1 for M1 for

12

1+ 12

A1 for A

A =

A 12

M1 for A1

A=

36w + 240

=

=

oe e.g. 180 − 3w 200 3w + 20

200 3w + 20

−1

1 200 3w + 20

−1

12( 3w + 20) 180 − 3w

or 12(3w + 20) = A [200 − (3w + 20)]

oe [10]

8.

5x – 15 = 4y – 3xy 5x + 3xy = 4y + 15 x(5 + 3y) = 4y + 15 4 y + 15 x= 5 + 3y

4 M1 for expanding into four terms three of which are correct M1(indep) for rearranging correctly to isolate x terms M1(indep) for factorising x from 2 terms with one factor involving y 4 y + 15 x= 5 + 3 y oe A1 cao for final answer [4]

9.

2

(n + a)P = n + a 2 nP + aP = n + a 2 a(P – 1) = n – nP n 2 – nP a= P –1

4 2

M1 ( n + a)P = n + a 2 M1 nP + aP = n + a 2 2 M1a(P–1) = n – nP or a(1–P) = nP – n A1 for

a=

n 2 – nP P – 1 oe [4]


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