NECO 2023 · Paper 2 · Q7

(a)
The sum of the ages of a man and his daughter is 60 years. Six years ago, the man's age was three times that of his daughter. Find their present ages (man, daughter).
Separate values with commas, e.g. 3, −2
(b)
Find the equation whose roots are $-\frac34$ and $\frac56$ .
Show the answer
$24x^2 - 2x - 15 = 0$
(c)
Evaluate $4\left(1 - \dfrac{144}{169}\right)^{\frac12} \times \left(\dfrac{2}{13}\right)^{-1}$ .

Worked solution (try it first)

Let the man's age now be $m$ years and his daughter's age be $d$ years.
Their ages add up to 60, so $m + d = 60$ .
Six years ago they were $m - 6$ and $d - 6$ , and the man was three times as old: $m - 6 = 3(d - 6)$ .
Expanding, $m - 6 = 3d - 18$ , so $m = 3d - 12$ .
Substitute into the first equation: $3d - 12 + d = 60$ , so $4d = 72$ and $d = 18$ .
Then $m = 60 - 18 = 42$ .
The man is 42 and his daughter is 18.
Check: six years ago they were 36 and 12, and $36 = 3 \times 12$ ✓.

Inside the bracket: $1 - \frac{144}{169} = \frac{25}{169}$ .
The power $\frac12$ is a square root: $\left(\frac{25}{169}\right)^{\frac12} = \frac{5}{13}$ .
The power $-1$ turns a fraction upside down: $\left(\frac{2}{13}\right)^{-1} = \frac{13}{2}$ .
So the value is $4 \times \frac{5}{13} \times \frac{13}{2} = 10$ .