Mathematics · Oct / Nov 2023 · Paper 11 · Question 3
YMA01
As printed in the paper
The question as text
3. In this question you must show all stages of your working. Solutions relying on calculator technology are not acceptable.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(a)[3]
8
15
2 3
5
16 3
8 5
2 45
75
−
−
−
−
+
×
=
2 3
5
2 3
5
12
−
5
+
−
e.g. 21 3 14 5
−
7
3 3
−
2 5
(
)
3(b)[3]
+
5 3
5
=
40
−
2
3
5
+
2
3
=
40 5 15
−
x
x
x
x
) 40
5 15
−
(
=
x
2 3
5
+
(
) 15 3 10 5
x =
−
(a) Note that 3 3
2 5
−
with no working scores M0dM0A0
Attempts to rationalise the denominator by multiplying both numerator and denominator by (
M1:
k is an integer usually 1, and proceeds to a fraction such as ... 3
±
... 5
±
... 45
±
... 75
or
...
...
...
but not for e.g.
or
as these still have surds on the denominator.
(
)
2
(
)
2
12
+
2 15
−
2 15
−
5
2 3
−
5
Allow 45 and 75 to be written in terms of 3 and 5 as well.
Condone slips in multiplying out as well as miscopying errors.
Note 21 3 14 5
−
with no intermediate working can still score M1 for the denominator.
7
Attempting to multiply by a multiple of 2 3
+
5
is M0dM0A0
2 3
+
5
dM1:
Attempts to simplify surds and may collect terms to achieve a fraction where
3 and 5 only.
the numerator is in terms of
the denominator is a multiple of 7 (which may be unsimplified)
It is dependent on the previous method mark.
If they have not fully multiplied out the numerator (or implied) then this mark cannot be scored. e.g.
(8
−
15)(2 3
−
5)
16 3
+
8 5
=
scores a maximum of M1dM0A0.
...
...
Note 21 3 14 5
−
with no previous working seen scores M1dM0A0 because they have not shown any
7
multiplying out of the brackets on the numerator.
Note 8
−
15
16 3
−
8 5
−
2 45
+
75
=
=
3 3
−
2 5
is M1dM0A0 because they have not collected terms
2 3
+
5
7
on the numerator (or changed them all to be in terms of 3 and 5 before simplifying to the final answer).
3 3
−
2 5
A1:
. The answer does not imply the method marks. Do not withhold this mark for slips in working such as
invisible brackets provided they are recovered/implied by further work.
Note that a number of candidates are misreading 8
−
15
2 3
−
5
16 3
−
40
−
2 45
+
5 15
×
=
2 3
+
5
2 3
−
5
12
−
25
condoning one sign slip in their multiplying out of the brackets in the numerator.
15 3 10 5
(b) Note that
x =
−
with no working scores M0A0A0
M1:
Multiplies out the brackets and isolates the x terms on one side. Condone for this mark if the surds are converted to
rounded decimals.
e.g. ( 5
+
2 3)
x
=
5(8
−
15)
scores M1
Alternatively divides both sides by 5 and isolates the x terms on one side.
3
40
e.g.
+
2
=
−
5 3
x
x
5
5
Condone slips in their working and invisible brackets, but there must be two terms on each side of the equation.
40
−
5 15
or exact equivalent (which cannot be 15 3 10 5
−
A1:
x
=
).
2 3
+
5
40
−
5 3
5
In the alternative method it may be seen as
x
=
which can score this mark.
3
1
+
2
5
Do not accept proceeding from ( 5
+
2 3)
=
5(8
−
15)
=
15 3 10 5
−
x
x
without any intermediate
expression for x. This scores A0 as the use of calculator technology is not acceptable.
An expression in decimals only is A0.
(
)
15 3 10 5
x =
−
x =
5 3 3
−
2 5
A1ft:
only (or exact simplified equivalent) e.g.
This mark cannot be awarded
without the previous A mark being scored.
3
+
5
so award for 5
3
+
5
5
Follow through on their part (a) answer of the form
a
b
a
b
(
)
5
3
+
5
a
b
where a and b may be fractions. Isw once they have achieved the correct answer.
Note: If a candidate has an unsimplified answer in part (a) e.g. 21 3 14 5
−
then do not withhold this mark in
7
part (b) if they proceed to e.g. 105 3
−
70 5
as they have already been penalised once for not giving an answer
7
in simplest form.
Alternative method – squaring both sides (send to review if unsure)
M1:
Attempts to square both sides and proceeds to a three term quadratic in x with terms collected on one side of the
equation (condone slips in multiplying out and collecting terms.
A1:
As above in main scheme
A1ft:
As above in main scheme
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