Moderate -0.8 This is a straightforward application of the trapezium rule to find area, then multiply by length for volume. The calculation is routine with values provided in a table, requiring only careful arithmetic and unit conversion (cm² to cm³, accounting for the 1.5m length). No problem-solving insight needed, just methodical application of a standard formula.
Figure 2 below shows a 1.5 metre length of pipe.
\includegraphics{figure_16}
The symmetrical cross-section of the pipe is shown below, in Figure 3, where \(x\) and \(y\) are measured in centimetres.
\includegraphics{figure_16_cross_section}
Use the trapezium rule, with the values shown in the table below, to find the best estimate for the volume of the pipe.
\begin{array}{|c|c|c|c|c|c|c|}
\hline
x & 0 & 0.4 & 0.8 & 1.2 & 1.6 & 2
\hline
y & -3 & -2.943 & -2.752 & -2.353 & -1.572 & 0
\hline
\end{array}
[5 marks]
Question 16:
16 | Uses the symmetry of the curve.
Evidenced by doubling area
from x = 0 to x = 2
Or considering the whole region
from x = – 2 to x = 2 | 3.1a | M1 | Area 0.4( 3+0+22.943+2.752+2.353+1.572)
2
=4.448
T o t a l a r e a = 8 .8 9 6
V o lu m e = 8 .8 9 6 1 5 0
= 1 3 3 4 .4 c m 3
1 3 0 0 c m 3
States or uses h = 0.4 OE
Accept 0.2 as the multiplier.
PI by 4.448 or 8.896
Accept use of h=0.8 or multiplier
of 0.4 provided their answer is
not then doubled. | 2.2a | B1
Substitutes given y values or
absolute y values to achieve
3+0+2 2.943+2.752+2.353+1.572
or
− 3 + 0 + 2 − 2 .9 4 3 − 2 .7 5 2 − 2 .3 5 3 − 1 .5 7 2
or
2.943+2.752+2.353+1.572
0+0+2
+3+2.943+2.752+2.353+1.572
or
−2.943−2.752−2.353−1.572
0+0+2
−3−2.943−2.752−2.353−1.572
Condone missing or misplaced
zeros
PI by 22.24 or 44.48 | 1.1a | M1
Obtains 8.896 or 4.448
Do not award this mark if they
go on to obtain 17.792 | 1.1b | A1
Obtains AWRT 1300cm3
Or AWRT 0.0013m3
Must include units | 3.2a | A1
Question 16 Total | 5
Q | Marking instructions | AO | Marks | Typical solution
Figure 2 below shows a 1.5 metre length of pipe.
\includegraphics{figure_16}
The symmetrical cross-section of the pipe is shown below, in Figure 3, where $x$ and $y$ are measured in centimetres.
\includegraphics{figure_16_cross_section}
Use the trapezium rule, with the values shown in the table below, to find the best estimate for the volume of the pipe.
\begin{array}{|c|c|c|c|c|c|c|}
\hline
x & 0 & 0.4 & 0.8 & 1.2 & 1.6 & 2 \\
\hline
y & -3 & -2.943 & -2.752 & -2.353 & -1.572 & 0 \\
\hline
\end{array}
[5 marks]
\hfill \mbox{\textit{AQA Paper 1 2024 Q16 [5]}}