Mathematics · May / June 2022 · 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. 180 − 80
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
3(i)[2]
6
5
−
4 5
2
5
Concludes working
=
= 2
5
5
4 5
−
5
4 5
−
5
7
+
3 5
3(ii)[3]
=
×
=
...
7
−
3 5
7
−
3 5
7
+
3 5
(
)
2
28
5
−
35 12
+
5
−
15
5
=
49
−
9
×
5
25
13
=
+
5
4
4
Notes: This is a non calculator question and all stages of working must be shown
(i) Main method
M1: Writes 180 as 6 5 or 80 as 4 5 .
Additional lines of working may be seen but they are not required for this mark
A1: Correct work leading to the answer 2. The M mark must have been awarded.
The solution shown in the main mark scheme is the minimum evidence required.
180
−
80
6
5
−
4 5
6 5
4 5
A valid alternative is
=
=
−
=
6
−
4
=
2
5
5
5
5
(i) Alt method I 180
−
80
180
80
36
16
6
4
2
=
−
=
−
=
−
=
5
5
5
M1: Writes 180 as
80 as 16
36
or
5
5
Additional lines of working may be seen but they are not required
A1: Correct work leading to the answer 2. The solution shown below is the minimum evidence required
E.g. 180
80
−
=
36
−
16
=
6
−
4
=
2
5
5
(i) Alt method II 180
−
80
5
900
−
400
30
−
20
2
×
=
=
=
5
5
5
5
M1: Multiplies numerator and denominator by 5 and writes 180
×
5
as 900 or 80
×
5
A1: Correct work leading to the answer 2
Do NOT allow solutions that rely on huge jumps. The M marks must be awarded as above.
For example
(
)
180
−
80
×
5
180
−
8
0
30
−
20
=
=
=
2
scores M0 A0
5
5
5
×
5
without sight of 180
×
5
as 900 or 80
×
5
as
400 o.e.
Or
180
−
80
2
5
=
=
2
scores M0 A0 without sight of 180 as 6 5 or 80 as 4 5
5
5
(ii)
Main method
M1: Correct attempt to rationalise. Look for attempt at multiplying numerator and denominator by 7
The multiplications need not be carried out for this mark.
dM1: It is dependent upon the previous mark. Scored for
Correct expression for denominator (without surds) Accept as shown, or as
7
2
−
3
2
× or just 4
5
and attempts to multiply out the numerator with the sight of 4 terms . Condone slips here. See ***
6.25 3.25
+
5
A1: Correct answer any way around. Allow
25
13
Do NOT allow in the form 25 13
+
5
a
=
,
b
=
but accept if followed by ''hence
''
4
4
4
(ii) Alt method
4 5
−
5
(
)(
)
M1: Sets
=
a
+
b
5
4 5
−
5
=
a
+
b
5
7
−
3 5
and attempts to set up simultaneous
7
3 5
−
equations in a and b. FYI the correct equations are 7
15
5, 7
3
4
a
−
b
= −
b
−
a
=
dM1: Attempts to solve their equations. They must have at least one correct equation
A1: Correct answer. Must be 25
13
+
5
and not just values of a and b
4
4
The following solution is a minimally acceptable solution for 3 marks, the first M1 being implied.
(
) (
)
4 5
−
5
×
7
+
3 5
4 5
5
28 5
35
60 15 5
13
25
−
−
+
−
=
=
=
5
+
***
4
4
4
4
7
−
3 5
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