WAEC 2023 · Paper 2 · Q6

1. List the heights from shortest to tallest.
2. They form an A.P. with first term $a$ (the shortest, unknown), $d = 0.05$, last term $l = 1.90$ and $n$ terms (the number of students).
3. From the last term: $a + 0.05(n - 1) = 1.90$, so $a = 1.95 - 0.05n$.
4. From the sum, $S_n = \frac{n}{2}(a + l)$: $\frac{n}{2}(a + 1.90) = 23.25$.
5. Put in $a$: $n(1.95 - 0.05n + 1.90) = 46.5$, that is $3.85n - 0.05n^2 = 46.5$.

(a)

1. Multiply by 20 and rearrange: $n^2 - 77n + 930 = 0$.
2. Factorise: $(n - 15)(n - 62) = 0$, so $n = 15$ or $n = 62$.
3. $n = 62$ gives $a = 1.95 - 3.10 = -1.15$ m, a negative height, so reject it.
4. There are 15 students.

(b)

1. $a = 1.95 - 0.05 \times 15$
   $= 1.95 - 0.75$
   $= 1.20$.
2. The shortest student is $1.20$ m tall.
3. (Check: $\frac{15}{2}(1.20 + 1.90) = 23.25$ ✓.)