Question 6:
6 | (a) | ( )
Area= 12q+1 pq
2
x  6  12+1 p
( 12q+1 2 pq )  y   =12q  1 2 q   +1 2 pq   3 2q 3   
72+6p+1 p2  432+36p+ p2 
x = 6 = 
 
12+ 1 p  72+3p 
2
6q+1 pq  36q+2pq
 y = 3 = 
12+1 p  72+3p 
2 | B1
M1
A1
A1
[4] | 1.1
3.3
1.1
1.1 | soi Award if correct ratios 24 : p : (24+p) oe used
Equation for one component Allow one error
Count 𝑥̅ and 𝑦̅ interchanged as one error
Equation for one component correct FT their area
Condone missing brackets if the intention is clear
This M1A1 can be awarded for a frame (see below)
cao Both components (any correct form)
SC Considered as a uniform frame (Max 2/4)
1
𝑥̅ 6 12+ 𝑝 6+ 1 𝑝 0
(24+𝑝+𝑞+√𝑝2+𝑞2 )( ) = (12)( )+√𝑝2+𝑞2( 2 )+(12+𝑝)( 2 )+𝑞(1 )
𝑦̅ 0 1 𝑞 𝑞 𝑞
2
2 | 1
𝑥̅ 6 12+ 𝑝 6+ 1 𝑝 0
(24+𝑝+𝑞+√𝑝2+𝑞2 )( ) = (12)( )+√𝑝2+𝑞2( 2 )+(12+𝑝)( 2 )+𝑞(1 )
𝑦̅ 0 1 𝑞 𝑞 𝑞
2
2 | M1 Equation for one component Allow one error
A1 Equation for one component correct | M1 Equation for one component Allow one error
A1 Equation for one component correct
(b) | x =7.6
432+36p+ p2 =7.6(72+3p)
 p2 +13.2p−115.2=0
 p=6 | B1
M1
A1
[3] | 3.1b
1.1
1.1 | Stated or implied, e.g. by equating their expression
for 𝑥̅ in part (a) to 7.6. Ignore incorrect value of y ,
e.g. ‘(7.6, q) is the CM’ earns B1
Equating their expression for 𝑥̅ in part (a) to 7.6, and
obtaining a three term quadratic equation for p
(Not necessarily …=0) Implied by 𝑝 = 6 www
BC
(c) | 6q+16q
y = 3 ( = 81 q )
12+16 5
2
𝑞−𝑦̅ = 7.6tan35° (= 5.32)
𝑞 = 11.4 | M1
M1
A1
[3] | 3.4
3.1b
1.1 | Substituting their p into their expression for y
M0 if their y does not contain p
soi Allow 7.6/tan35° or equivalent
If their 𝑦̅ = ½𝑞, then ½𝑞 = 5.32, by itself, is M0
Allow 30° as a misread for 35°
Awrt 11.4 (11.403379…)
PMT
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