Hard maths question (1 Viewer)

[RealiseNothing]

Well then yer, it's 12:40pm lol.

 

[Demento1]

1) A bus and a truck left town A at 10.00am and went along the road between towns A and B at constant speeds. At the same time, a car set off from B along the same road at a constant speed. The car first met the truck 30*** minutes later and then after 10 more minutes it met the bus. The truck reached B, immediately turned back and, on its way to A, met the bus 2 hours since it left A. When will the truck get back to A?

2) A student rows at a constant speed (relative to the water) downstream in the Nambucca River from Bowraville to Macksville in 3 hours and back upstream in 4 hours. If the river flows at a constant rate, find the number of hours it would take a piece of driftwood to float downstream from Bowraville to Macksville.

3) (Is it possible to work this question out with simultaneous equations and how?) A certain integer is between 10 and 100. Its value is 8 times the sum of its digits and if it is reduced by 45, its digits are reversed. Find the integer.

Has anybody solved this question? The question is quite odd...

 

[bleakarcher]

naa, I haven't tried that yet. I look at it later. By then, someone else would have done it though.

 

[barbernator]

t=24 i think. ill put up my answer and see what u guys think

 

[bleakarcher]

yeh just tried it and got 24 hours.

 

[barbernator]

<a href="http://www.codecogs.com/eqnedit.php?latex=\inline s=\frac{d}{t}\\ t=\frac{d}{s}\\ \\ downstream\\ \\ 3=\frac{d}{a@plus;b}~where~a~is~rowspeed~and~b~is~riverspeed\\ \\ upstream\\ 4=\frac{d}{a-b}\\ \\ simultaneously\\ \frac{3}{4}=\frac{d(a-b)}{d(a@plus;b)}\\ simplifying,~and~skipping~steps\\ a=7b\\ \frac{a}{7}=b\\ \\ we~know~now~that~the~time~taken~is\\ t=\frac{d}{b}\\ and~that~\\ 3=\frac{d}{8b}\\ simultaneously\\ t=24b/b=24hours" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\inline s=\frac{d}{t}\\ t=\frac{d}{s}\\ \\ downstream\\ \\ 3=\frac{d}{a+b}~where~a~is~rowspeed~and~b~is~riverspeed\\ \\ upstream\\ 4=\frac{d}{a-b}\\ \\ simultaneously\\ \frac{3}{4}=\frac{d(a-b)}{d(a+b)}\\ simplifying,~and~skipping~steps\\ a=7b\\ \frac{a}{7}=b\\ \\ we~know~now~that~the~time~taken~is\\ t=\frac{d}{b}\\ and~that~\\ 3=\frac{d}{8b}\\ simultaneously\\ t=24b/b=24hours" title="\inline s=\frac{d}{t}\\ t=\frac{d}{s}\\ \\ downstream\\ \\ 3=\frac{d}{a+b}~where~a~is~rowspeed~and~b~is~riverspeed\\ \\ upstream\\ 4=\frac{d}{a-b}\\ \\ simultaneously\\ \frac{3}{4}=\frac{d(a-b)}{d(a+b)}\\ simplifying,~and~skipping~steps\\ a=7b\\ \frac{a}{7}=b\\ \\ we~know~now~that~the~time~taken~is\\ t=\frac{d}{b}\\ and~that~\\ 3=\frac{d}{8b}\\ simultaneously\\ t=24b/b=24hours" /></a>

 

[wantingtoknow]

thanks guyssss, homework completed

 

[Demento1]

thanks guyssss, homework completed

Surely you can't be year 10...

 

[wantingtoknow]

Surely you can't be year 10...

? why year 10? and why can't I be lol

 

[Demento1]

? why year 10? and why can't I be lol

Just how you posted the questions in the year 9 and 10 section.

 

[wantingtoknow]

Just how you posted the questions in the year 9 and 10 section.

found it in a year 9 person's homework and I wasn't able to help. so I thought it was appropriate to post in this section.

 

[nightweaver066]

found it in a year 9 person's homework and I wasn't able to help. so I thought it was appropriate to post in this section.

A bit tough for a year 9 student if you ask me lol.

I think i just learnt simultaneous equations in year 9 let alone use it extensively like how i solved the 3rd question.

 

[Demento1]

A bit tough for a year 9 student if you ask me lol.

I think i just learnt simultaneous equations in year 9 let alone use it extensively like how i solved the 3rd question.

Requires some extended application from a year 9, although a good question for perhaps a year 10.

 

[abecina]

3 was fun.

I used simultaneous equations (had to force myself to remember little rules i set so that i don't get them mixed up) and i ended up with 72.







































Although you can just use the first generated equation,



and notice how X must be integral, so Y = 2, 4, 6, 8, ...

Test the first value for Y and you get the answer as well haha.

Here is another way. Rather than have your number as XY, write it as

.

Then using the second piece of info







Since is divisible by 8, it is an even number (), and must be a multiple of 8.
Since