Math Is Fun Forum

I had at a look at the long division page. While I can understand the use of the take away rows, they make it look awfully complicated. Why not just learn to transfer the remainder to make a new ten?

EG

25 ) 425

25 into 42 goes 1, remainder 17, (write a small 17 to the left of the 5) which then becomes 175, etc...

25 into 175 goes 7, remainder 0

Much quicker...

Mind you, they need to test answers in the old way: 25 x 6 is too small, etc.

I'm in Sydney, teaching selective school trial test Maths.

They do practice papers for one year before the actual test, in Maths, English and General Ability.

They are mostly Chinese and Korean students. Gosh, no surprise there.

Yes, I did draw a graph to help me visualise the problem.

Can I assume that the 35t assumes a time of one hour?

So, '35/60 x t' becomes 35t, right?

This is not an easy question for this age group. I want to make it as simple as possible.

Car A travels at 35 kph plus a head start of 35kph for 3/4h

I need to explain how I get from s = d/t to d = 35t, so 35km/1hour helps to simplify the term to 35, not 35/60.

35kph = d/t

d = 35kph x t (simplify by ignoring 'kph')

35t plus the extra 3/4hr at 35kph = 50t

(We don't write 35 x 3/4t, because 't' refers to an unknown time, where the 3/4hr is a known time - is that right?)

15t = 35 x 3/4

t = 35/15 x 3/4

t = 7/3 x 3/4

t = 7/4

7/4 hours from zero hours, or 10/4 hours from zero hours?

Is it 1 3/4 hours from the start of the race, or  2 1/2 hours?

I see you asked the question, "how long will it take Car B to catch up with Car A" (you got the A and B mixed up, I think), so that would imply we have to add the 3/4 hour to find the time from the start.

Can you explain in words what you have written in math terms?

What is t in 35t? Time?

I know S = D/T, so D = SxT, right?