Edexcel: 11th and 14th June 2013 (Linear) 1MA0/1H + 2H (Official Thread)

How do you do this question..

21. A field is in the shape of a rectangle.
The width of the field is 28 metres, measured to the nearest metre.
(a) Work out the upper bound of the width of the field.
......................... metres

The length of the field is 145 metres, measured to the nearest 5 metres.
(b) Work out the upper bound for the perimeter of the field.
......................... metres

I know how to do it.. but the answer's weird. :/

a) 28.5 b)145.5?

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Reply 501

lolface32

Original post by lolface32

a) 28.5 b)145.5?

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Oh lol, did not read it say the perimeter -_-

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Reply 502

Tinkletea

8

145 to the nearest 5m isn't 145.5

Is it 147.5

The ub is yes

Original post by PriyaPaul

How do you do this question..

21. A field is in the shape of a rectangle.
The width of the field is 28 metres, measured to the nearest metre.
(a) Work out the upper bound of the width of the field.
......................... metres

The length of the field is 145 metres, measured to the nearest 5 metres.
(b) Work out the upper bound for the perimeter of the field.
......................... metres

I know how to do it.. but the answer's weird. :/

a) is simply 28.5.
For b), the upper bound for the length is 147.5, so to find the upper bound for the perimeter, work out the perimeter using the upper bounds of the length and width. That is, 28.5+28.5+147.5+147.5=352

Reply 506

Yellow Messiah

Does anyone know if we need to remember metric to imperial conversions e.g 30cm = 1ft or will they be given in the question.

Reply 507

PriyaPaul

Original post by qwepoizxcmnb

a) is simply 28.5.
For b), the upper bound for the length is 147.5, so to find the upper bound for the perimeter, work out the perimeter using the upper bounds of the length and width. That is, 28.5+28.5+147.5+147.5=352

Thanks

Reply 508

Radhika0303

Does anyone know the topics to revise for the Calculator? And, are the topics that werent on the Non-calculator paper going to be on the calculator paper?

Reply 509

Student08

1

I think i did really bad on the non calc paper.. i really want to do good on this paper,but do you think its going to be easier then the first paper??

Reply 510

Student08

1

I really think getting 120 marks should be able to get you an A, right?

Reply 511

cmorga1

I need around 80 for an a* tommorow, very confident with trigonometry but nervous about graph functions etc

Reply 512

Captain Anonymous

14

Original post by cmorga1

I need around 80 for an a* tommorow, very confident with trigonometry but nervous about graph functions etc

how did you do on the first paper?

Reply 513

jack97

10

what will the bounderies be for an A*?

Got around 78 marks on the first paper, so what do you reckon i need?

Reply 514

Captain Anonymous

14

A* around 164
A around 128

Reply 515

Ms_laviscount

1

Who thinks the next exam will be easy? I need some reassurance lol 😖
I got 73 in paper 1 what do I need in paper 2 to secure A*? I just don't want an A again.

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Reply 516

Tinkletea

8

Original post by cmorga1

I need around 80 for an a* tommorow, very confident with trigonometry but nervous about graph functions etc

Graph functions are easy you just need to memorise the few rules -f(X) is a reflection in the X axis. f(-x) is a reflection in the Y axis.
f(x)+3 is moving 3 up the y axis, or f(x)-1 is moving 1 down the y axis.
f(x+1) is moving one to the LEFT of the x axis, not right. f(x-2) is moving 2 to the right.
2f(x) is simply multiplying the y axis points by 2, the x axis points do not change. I.e if you have a point as (1,2). The tranformed point will be (1,4). If you have 1/2f(x) and your original co-ordinates are (1,2), your transformed co-ordinates will be (1,1).
f(2x) is compressing in the direction of the x axis and you need to remember that the applied function's recipical is applied to X. So if you have the function f(2x) and you have co-ordinates (2,2) your new co-ordinates would be (1,2) (since the y axis does not change, and 1 is 1/2 of 2 and the recipical of 2 is 1/2) so if you have f(4x) then your transformed x co-ordinates will be 1/4 of what they were originally