Easy -1.8 This is a straightforward substitution and comparison question requiring only plugging values into a given formula (r=2.7, h=12) to get V≈92cm³, then comparing to 75cm³. No derivation, integration, or problem-solving needed—purely arithmetic verification with basic commentary.
15 A typical tube of toothpaste measures 5.4 cm across the straight edge at the top and is 12 cm high. It contains 75 ml of toothpaste so it needs to have an internal volume of \(75 \mathrm {~cm} ^ { 3 }\).
Comment on the accuracy of the formula \(V = \frac { 2 } { 3 } \pi r ^ { 2 } h\), as given in line 41 , for the volume in this case.
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\(V = 74.2\) to \(74.3\), or \(r = 1.727\) or \(1.73\)
B1
1.1
Suitable comment consistent with working, e.g.: not exactly right but the measurements may not be exact; the volume/radius is too small but it's close; allowing for approximation the formula seems accurate
E1
3.5b
[3 marks]
## Question 15:
$r = \dfrac{5.4}{\pi}$ | B1 | 3.1a | soi by 1.71[88], 1.72 |
$V = 74.2$ to $74.3$, or $r = 1.727$ or $1.73$ | B1 | 1.1 | If $r$ from $V = 75$ and $h = 12$ |
Suitable comment consistent with working, e.g.: not exactly right but the measurements may not be exact; the volume/radius is too small but it's close; allowing for approximation the formula seems accurate | E1 | 3.5b | Dependent on both B marks |
**[3 marks]**
15 A typical tube of toothpaste measures 5.4 cm across the straight edge at the top and is 12 cm high. It contains 75 ml of toothpaste so it needs to have an internal volume of $75 \mathrm {~cm} ^ { 3 }$.
Comment on the accuracy of the formula $V = \frac { 2 } { 3 } \pi r ^ { 2 } h$, as given in line 41 , for the volume in this case.
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\hfill \mbox{\textit{OCR MEI Paper 3 2019 Q15 [3]}}