IB Maths AI SL · Paper 1

1. Change the land figure to km${}^2$: $148.9$ million is $148900000$.
2. Subtract from the total to get the water area, then divide by the total and multiply by 100.

Another way. Find the land percentage (about $29.2\%$) and take it away from $100\%$.

1. The known angle at C sits opposite side AB, and the angle you want at A sits opposite BC. That is a matching pair, so use the sine rule.
2. Rearrange for $\sin \hat{A}$ and take the inverse sine.

1. Find the third angle at B from the angle sum, using the unrounded value from (a).
2. Use $\frac{1}{2}ab\sin C$ with the two sides that meet at B.

1. Use the midpoint of each class ($70,90,110,130,150$) as the data values and the frequencies as weights.
2. Enter both lists in the GDC's one-variable statistics and read off the mean and the population standard deviation $\sigma$.

1. Put $T$ in list 1 and $n$ in list 2, then run a linear regression on the GDC.
2. (i) Read off $r$. (ii) Write the line with $n$ as the subject, using the gradient and intercept to 3 s.f.

1. The start of 2034 is ten years after the start of 2024, so $t=10$.
2. Round to the nearest whole number as asked.

Another way. Find FB in the horizontal plane ($\sqrt{1300}$), then combine it with the vertical drop SB $=80$ using Pythagoras.

1. The condition cuts the sample down to the over-50 column only: $19+34=53$ people.
2. Of those, 19 chose hot.

1. Enter the observed counts as a $2\times 3$ matrix and run the ${\chi }^2$ test for independence on the GDC.
2. Read off the statistic and the $p$-value.

1. Compare the statistic with the critical value (or the $p$-value with $0.01$).
2. The statistic is smaller, so there is not enough evidence to reject $H_0$.

1. A $14\%$ daily increase means multiplying by $1.14$ each day, so the heights form a geometric sequence with $u_1=3$ and $r=1.14$.
2. Use $u_n=u_1{r}^{n-1}$ with $n=10$.

1. Change $1.9$ m to $190$ cm so the units match.
2. Solve $3\times 1.14^{n-1}>190$ on the GDC (or with logs), then round up to a whole day.

Another way. Check the terms either side: $u_{32}\approx 174$ cm and $u_{33}\approx 199$ cm, so day 33 is the first above 190 cm.

1. Monthly compounding over 5 years is $60$ periods at $\frac{1.8}{12}\%$ each.
2. Use the TVM solver or the compound interest formula.

1. Set $FV=5000$, keep the other settings, and solve for $N$ (in months).
2. Convert months to years and round to the nearest year.

1. The amount borrowed is $4000-1400=2600$.
2. In the TVM solver use $N=24$, $I\%=3.8$, $PV=2600$, $FV=0$, $P/Y=C/Y=12$, and solve for $PMT$.

1. Set the sum formula $S_n=\frac{n}{2}\left(2u_1+(n-1)d\right)$ equal to $9475$.
2. Expand and move everything to one side.

Another way. Leaving the halves in gives $3.5n^2+14.5n-9475=0$, which is equally valid.

1. Solve the quadratic with the GDC polynomial solver.
2. Reject the negative root, since $n$ counts terms.

1. Integrate each term of $f^{\prime }(x)$ and add a constant $c$.
2. Substitute the point $(2,12)$ to find $c$.
3. Write out $f(x)$ in full with the value of $c$.

1. Find the gradient and equation of the road through A and B.
2. The closest point D is where the perpendicular from C meets the road, so the line CD has the negative reciprocal gradient.
3. Write the equation of CD through $C(4,5)$ and solve the two equations together.

Another way. Write the squared distance from C to a general point $(x,-2x+12)$ on the road and use the GDC to find the $x$ that minimises it.