GCSE Maths question - Help...?

Dunroamin

foofi22 wrote: »

Yep, this is more A Level

These days, you could be right!:rotfl:

whitewing

I found it a very useful example of how maths can be applied to real life.

pooch

stamina9008 wrote: »

Hi, my dad uses this site but i thought it might be a good place to ask this question..
Please help me answer it, it's part of my homework and i cant understand it..
I dont think the answer is 3/5, so..please help..

4. Challenge!
Simon ate 1/5 of a cake and his sister ate 1/4 of what was left.
What fraction of the cake is now left? (hint: a diagram may help!)

I understand the four rules of fractions, but this question is really annoying me..

Have a nice day

I understand how annoying some mathematical problems can be to solve, but there is a hint in the question to help you

So lets use that hint and start by dividing a cake (or a pie as mathematicans always prefer!) into 5 equal pieces.



Now if Simon eats a piece, the remainder will look like this:



If Simon's sister now eats a quarter of this, the remaining cake will look like this:



Which obviously leaves 3/5 of the original cake remaining

pooch

mrcow wrote: »

PS. I'm a maths boffin

Oh goody, perhaps you could supply the answer to this little question I have for my assignment

AnnBar

Dunroamin wrote: »

Isn't anyone else shocked that a question like this is being set at GCSE level?

Yes. To me it should be primary 7 work!

LTL

pooch wrote: »

Oh goody, perhaps you could supply the answer to this little question I have for my assignment

The trench is a "truncated pyramid"

Area = Area of a trapezium * length = 1/2(a+b)hl

a = 11 1/2
b = 3 1/2
h = 4
l = 140

Area = 1/2(11 1/2 + 3 1/2)* 4 * 140
Area = 8,400 cu ft.

Area of Cylinder = Pie * r squared * h (or in this case l)

Area = 22/7 * 3/4 squared * 140
Area = 247.50
Rubble required = 8,400 - 247.50
Rubble required = 8,152.50 cu ft.

Weight = 28 tonnes = 551.155655462194 cwt
1 cu ft. = 0.037037037 cu yds
8,152.50 cu ft. = 301.9444 cu yds
weight of 1 cu yd = 301.9444 / 551.155655462194
Weight = 1.82535451853547
Weight = 1.8 cwt to the nearest tenth of a cwt

LittleVoice

pooch wrote: »

Oh goody, perhaps you could supply the answer to this little question I have for my assignment

Could I hazard a guess that that is part of a question from a GCE 'O'level paper from the 1960s?

agrinnall

LittleVoice wrote: »

Could I hazard a guess that that is part of a question from a GCE 'O'level paper from the 1960s?

11 plus, 1954 :rotfl:

Rhenser

LTL wrote: »

The trench is a "truncated pyramid"

Area = Area of a trapezium * length = 1/2(a+b)hl

a = 11 1/2
b = 3 1/2
h = 4
l = 140

Area = 1/2(11 1/2 + 3 1/2)* 4 * 140
Area = 8,400 cu ft.

Area of Cylinder = Pie * r squared * h (or in this case l)

Area = 22/7 * 3/4 squared * 140
Area = 247.50
Rubble required = 8,400 - 247.50
Rubble required = 8,152.50 cu ft.

Weight = 28 tonnes = 551.155655462194 cwt
1 cu ft. = 0.037037037 cu yds
8,152.50 cu ft. = 301.9444 cu yds
weight of 1 cu yd = 301.9444 / 551.155655462194
Weight = 1.82535451853547
Weight = 1.8 cwt to the nearest tenth of a cwt

Wow! That much? :eek:
I'm glad you don't work in our building company - we'll be bust by the end of the month if we went ordering that amount of rubble.

Put the calculator down and think about the answer a little.

The trench is 4 ft deep and, on average, less than 3ft wide.
So lets assume, roughly, 10 cubic foot per foot of length to keep the numbers simple.
Now the length is 140ft.

So I would expect the total excavated volume to only be about 1400 cu.ft.
Then there's the volume of the pipe to deduct, so an answer about 1150 cu.ft is what I would be expecting.

But what do I know? I'm just a poor navvy.

Btw, there were only 20 cwt to a ton when I went to school
Perhaps you've allowed for inflation?

LTL

Rhenser wrote: »

Wow! That much? :eek:
I'm glad you don't work in our building company - we'll be bust by the end of the month if we went ordering that amount of rubble.

Put the calculator down and think about the answer a little.

The trench is 4 ft deep and, on average, less than 3ft wide.
So lets assume, roughly, 10 cubic foot per foot of length to keep the numbers simple.
Now the length is 140ft.

So I would expect the total excavated volume to only be about 1400 cu.ft.
Then there's the volume of the pipe to deduct, so an answer about 1150 cu.ft is what I would be expecting.

But what do I know? I'm just a poor navvy.

Btw, there were only 20 cwt to a ton when I went to school
Perhaps you've allowed for inflation?

OMG! I thought it said 11 1/2 ft at the bottom not 1 1/2 foot

TO BE FAIR! You would need that much rubble for the ridiculously big hole I just made someone dig unnecessarily

(how many cwt to tons could be different I guess as there could be a miscommunication between metric and imperial - get into the 21st century guys! )