WAEC 2012 · Paper 2 · Q13

(a)
The internal diameter of a spherical bowl, half full of water, is $20\text{ cm}$ . The content is poured into an empty cylindrical vessel with internal diameter $10\text{ cm}$ . Calculate, correct to one decimal place, the depth of water in the vessel. $\left[\text{Take }\pi = \frac{22}{7}\right]$
(b)
The fourth term of an Arithmetic Progression (A.P.) is 1 less than twice the second term. If the sixth term is 7, find the first term.

Worked solution (try it first)

(a)

The bowl is a sphere of radius 10 cm, half full, so the water is half a sphere: $\frac12 \times \frac43\pi \times 10^3 = \frac{2000}{3}\pi\text{ cm}^3$ .
In the cylinder (radius 5 cm) the same volume stands to a depth $h$ : $\pi \times 5^2 \times h = \frac{2000}{3}\pi$ .
The $\pi$ cancels: $25h = \frac{2000}{3}$ , so $h = \frac{80}{3} \approx 26.7$ cm.

(b)

$T_4 = a + 3d$ and $T_2 = a + d$ , so $a + 3d = 2(a + d) - 1$ , which gives $a - d = 1$ .
$T_6 = a + 5d = 7$ .
Subtract: $6d = 6$ , so $d = 1$ and $a = 2$ .
The first term is 2.

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