Download QAHS Prelim 2015 Solutions

National 5 All Units Paper 1
Qu
1
Mar 2015
Marking Scheme
Give one mark for each ●
ans: 3x² – 4x - 15
2 marks
Illustrations for awarding mark
1 starts to multiply brackets

2
2a
ans:
1
2
b
ans:
completes multiplication and simplifies
(x + 3)² – 13
2 marks
1
starts to complete square
completes and answer
(-3, 13) ; minimum
3
1
first operation
2
second operation
4
ans:
5
●1 realises 120%
●2 finds 20%
3 finds 100%
 4 


ans:   12 
 6 


●1
2
( x  3) 2 .....
…..  9  4  ( x  3) 2  13
1
2
(-3,13)
minimum
2 marks
1
2
correct coordinates
states nature
6
ans:
7
1 3 x 2  5 x.....
2 ....  9 x  15 x  3x 2  4 x  15
2 marks
£35 000
1
2
3 marks
●1 120% = £42 000
●2 £42 000 ÷ 6 = £7 000
3 £7 000 × 5 = £35 000
2 marks
knows how to find components

1
●2
states components
QAHS March 2015
4 4 48
 
7 5 35
5 48 240 6



8 35 280 7
●2
National 5
 3   5 
   
3  5     3 
 2   0 
   
 4 


  12 
 6 


Prelim – all units
Qu
6
7
8
ans:
10
Illustrations for awarding mark
1 uses b 2  4ac
2 finds an answer of 13
3 writes 2 roots since b 2  4ac > 0
ans: 4x – 3y = 17
3 marks
1 uses b 2  4ac
2 finds an answer of 13
3 writes 2 roots since b 2  4ac > 0
1
uses gradient of 4/3
1
m
2
subs into straight line equation
2
3
rearranges to required form
3
4
4
( x  2) y  3  ( x  2)
3
3
4 x  3 y  17
ans:
1
2
3
9
Give one mark for each ●
see below for 3 marks
ans:
3
1
2
3
1
2
2
3
correct x term
correct y term
3
0·8
uses sine rule
1
2
rearranges formula
2
3
simplifies
3
●1
●2
●3
4 3 3 3  7 3
R
3Q
P
cross multiplies
divides both sides by P
takes square root
QAHS March 2015
2
x7
.....
y
3 marks
1
ans:
48  16  3  4 3
3 marks
correct number
11
6  8  48
7
1
ans:
y3
3 marks
first operation
simplifies surd
simplifies expressions
2x
y
4
3
16
8

sin R 0  4
16  0  4
sin R 
8
0·8
3 marks
●1 PR 2  3Q
3Q
●2 R 2 
P
3Q
●3 R 
P
National 5
Prelim – all units
Qu
12
Give one mark for each ●
ans:
1
2
3
13
45o
Illustrations for awarding mark
4 marks
1
2
uses correct diameter
finds circumference of circle
knows how to find angle
4 answer
ans: sinxo
4
d = 16 [may be formula]
C   16  16
2
x 
 360
16
45o
cos x 
3
2 marks
1
substitutes for tanxo
1
2
simplifies
2 sin x
Total
QAHS March 2015
National 5
sin x
cos x
37 marks
Prelim – all units
National 5 All units Paper 2 March 2015
Qu
1
Give one mark for each ●
yes with reason
3 marks
ans:
1
2
3
2
correct multiplier
correct method
conclusion
4x 2
ans :
7
1
Marking Scheme
Illustrations for awarding mark
1
2
3
108
40 000 × 1·083 = £50 388.48
yes since £50 388.48 > £50 000
1
........
3 marks
inverts divisor and multiplies

2
2
3
4
3 simplifies
ans : 16·3cm
6
3
changes to cm³ and subs into formula
1
2
starts to solve for r
2
3
finds r
3
ans: 78∙5o, 281∙5o
2 equations, £5
1
5000     r 2  18
3
5000
r2 
6
5000
r
 16  3
6
3 marks
solves for cosxo
finds one solution
finds second solution
ans :
4bx 3
7 xb
4x 2
7
3 marks
1
1
2
3
5
multiplies
x3
b
1
2
3
cosxo = 0·2
78∙5o
281∙5o
5 marks
1
2
3
4
3N + 2B = 145
5N + 3B = 240
evidence of scaling equations
Solves for B
 States “Breakfast cost £5”
ans: see below
5 marks
1
2
3 evidence of scaling equations
4 B = 5
 States “Breakfast cost £5”
1 Mean = 76.5
2 Correct numbers entered in SD formula
 Correct SD calculated SD = 6.75
 Children’s pulse rates tend to be higher
 And to vary less than adults pulse rates
1
QAHS March 2015
National 5
Mean = 459/6 = 76.5
 Correct numbers entered in SD formula
227.5
SD 
5
 SD = 6.75
 Children’s pulse rates tend to be higher
 And to vary less than adults pulse rates
2
Prelim – all units
Give one mark for each ●
1687.5 ml
3 marks
Qu
7
ans:
1
finds linear scale factor
1
Illustrations for awarding mark
9 3
L.S.F. = 
6 2

3
V.S.F. =   [or equivalent]
2
3
2
2
finds volume scale factor
3
3
8
finds volume of larger bottle
ans :
10
a
0.7, -1.4
finds a, b & c & subs into formula


calculates b2 – 4ac
finds roots
correct rounding
ans:
3
   500  1687  5 ml
2
4 marks


9

3
angle not right with reason
 a=3, b =2, c = -3
 2  40
23
 40
 0.72, -1.38
 0·7, –1·4
4 marks
18cm
18cm
1 assembles facts in triangle
2 knows condition for right angle
3 tests both sides
4 valid conclusion
ans : 120cm
3 marks
1
27cm
2 statement e.g. if 18² + 18² = 27² angle is right
3 18² + 18² = 648; 27² = 729
4 since 18² + 18² ≠ 27² angle is not right
1
2
3
1
2
3
5  4x  x 2  0
(5  x)(1  x)  0; x  5 or –1
6 × 20cm = 120cm [or 1·2 m]
1
2
3
4
evidence of finding middle of 5 and –1
y  5  4(2)  (2) 2  9 units
9 × 20 = 180cm
Yes since 180cm > 1·75m
sets equation = 0
solves quadratic equation
finds distance
ans: Yes since 180cm > 1·75m
4 mark
b
1
2
3
4
11
ans:
1
2
3
4
knows to find maximum TP
knowing to sub into equation/evaluates
uses scale
valid conclusion
proof
4 marks
correct strategy
uses Cosine Rule
evaluates side
simplifies answer
1
2
3
4
angle ACB = 80o
AB 2  6 2  6 2  2  6  6  cos 80
AB 2  72  72 cos 80 0
AB  72(1  cos 80)
Total 44 marks
QAHS March 2015
National 5
Prelim – all units