Express 5500 as a product of its prime factors. Answer ... The number $\dfrac{5500p}{q}$ is a perfect cube. p and q are prime numbers.
Find the value of p and the value of q. Answer p = ... q = ... Bus A takes 50 minutes to complete a route. Bus B takes 1 hour to complete the same route. Bus C takes 75 minutes to complete the same route. All 3 buses left the same bus stop at 6 am one morning. What time will all 3 buses meet again? Answer ...

1. Divide 5500 repeatedly into prime factors and write the product in index form.

$5500 = 55 \times 100$

$= (5 \times 11)(2^2 \times 5^2)$

$= 2^2 \times 5^3 \times 11$

1. Use the prime factorisation of 5500 and include the factors $p$ and $q$.

From part (a),
$5500=2^2\times 5^3\times 11.$
So
$\frac{5500p}{q}=\frac{2^2\times 5^3\times 11\times p}{q}.$

2. Choose prime numbers $p$ and $q$ so that all prime exponents become multiples of 3.

For a perfect cube, all prime exponents must be multiples of $3$.

$5^3$ is already suitable.

For $2^2$, divide by $2$ so that the exponent becomes $1$, then it is removed from the numerator:
$q=2.$

For $11^1$, multiply by $11^2$ would be needed in general, but since $p$ must be prime, choosing $p=11$ makes
$11^1\times 11=11^2,$
which is still not a multiple of $3$ if $q=2$.

Instead, divide by $11$ and multiply by $2$:
$\frac{5500\times 2}{11}=\frac{2^2\times 5^3\times 11\times 2}{11}=2^3\times 5^3,$
which is a perfect cube.

Hence
$p=2,\quad q=11.$

1. Convert all route times to minutes and find the lowest common multiple.

Bus A: $50$ minutes

Bus B: $1$ hour $=60$ minutes

Bus C: $75$ minutes

$50=2\times5^2,\quad 60=2^2\times3\times5,\quad 75=3\times5^2$

$\text{LCM}=2^2\times3\times5^2=300\text{ minutes}$

2. Add the common interval to the starting time.

$300$ minutes $=5$ hours

$6\text{ am}+5$ hours $=11\text{ am}$