Please help, P6 Maths Problem Sums

kkhing SG

1. Mrs. Tan has drawn up a schedule to have her home cleaned by 3 part-time workers. The cleaner goes to her home once every 3 days, the sweeper once every 4 days and the gardener once every 6 days. If the 3 workers first met on 28 July, when was the earliest date the cleaner has to start work?

2. Mary is meeting a friend at a certain time. If she drives at 80 km/h, she will be 1/3 hour late. If she drives at 60 km/h, she wil be 3/4 hour late. How long will the journey take if she drives at 90 km/h?

rurumoo SG

Let's understand the sum.

All three workers share a COMMON working day (their first common working day) and that is 28th July.

Let's assume that 28th July was their SECOND meeting and they met before once, too.

Now 3, 4, and 6 have LCM to be 12.

28-12=16 July (the day they 'first' met, if you take 28th July to be their second meeting)

But they did NOT meet at 16 July. that was just an assumption, remember?

So, if 28th July was their second meeting, they would first meet on 16th July.

But 28th July is their first meeting, so 16th july cannot be a day the cleaner works.

Instead, the cleaner starts work AFTER that. How early can we go to make it the earliest?

16+3 = 19 July 9 (the answer)

______________

Take x to be the distance

Take y to be the certain time she needs to meet.

x/80 = y+ 1/3h

x/60 =y+ 3/4h

Now, let's make x equal/

x = 80y + 80/3h

x = 60y + 45h

80y + 80/3 h = 60 y + 45h

20y = 18 and 1/3

y= 11/12 h = 55 minutes.

Time needed to meet friend = 55 minutes

55minutes+20minutes= 75 minutes = 1 and 1/4 h = 5/4 h (time taken travelling at 8okm/h)

80x5/4= 100km (distance)

100km / 90 km = 10/9 = 1 and 1/9 h

:D sorry if it's too long, but that's what i do to solve :)