[librastar28]

can anyone tell me whether this answer is wrong question 7a

question paper
http://downloads.bbc.co.uk/schools/g...kh1_nocalc.pdf

mark scheme
http://downloads.bbc.co.uk/schools/g...1ms_nocalc.pdf

plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz zz help

[librastar28]

(Original post by klaidia)
wait is the first one in the afternoon? and the second one in the morning?

yeah answer my question plzzzzzzzzzzzzzzzz

[Jigglypuff786]

(Original post by mia_hilton)
Hi guys am stuck yet again can someone explain the answer to me please?
Attachment 155253 I cant access the answers to the 1MAO practice papers Set, A,B,C...help please

Basically you have to split the trapezium into a rectangle and a triangle by drawing a line through it.
Then work out the area of the rectangle and the triangle separately.
Area of rectangle = (x+2)(x-5)
= x² -3x -10
To work out the area of the triangle you must first work out the base length which is obviously 4, and the height is given which is (x-5)
So the area of the triangle = 2(x-5)
= 2x-10
Now add both of them together to get: x² - 3x - 10 + 2x- 10 = 36
Tidy it up and you get: x² - x - 20 = 36
This can also be written as x² - x - 56 = 0.

[usycool1]

(Original post by librastar28)
can anyone tell me whether this answer is wrong question 7a

question paper
http://downloads.bbc.co.uk/schools/g...kh1_nocalc.pdf

mark scheme
http://downloads.bbc.co.uk/schools/g...1ms_nocalc.pdf

plzzzzzzzzzzzzzzzzzzzzzzzzzzzzzz zz help

The links give me a 404 error...

And also, in the nicest possible way, don't post the same thing about 3 times. It just annoys people and leads them to not helping you (and it can be classified as spamming)

[mia_hilton]

(Original post by Jigglypuff786)
Basically you have to split the trapezium into a rectangle and a triangle by drawing a line through it.
Then work out the area of the rectangle and the triangle separately.
Area of rectangle = (x+2)(x-5)
= x² -3x -10
To work out the area of the triangle you must first work out the base length which is obviously 4, and the height is given which is (x-5)
So the area of the triangle = 2(x-5)
= 2x-10
Now add both of them together to get: x² - 3x - 10 + 2x- 10 = 36
Tidy it up and you get: x² - x - 20 = 36
This can also be written as x² - x - 56 = 0.

Thank you! This is great - i understand! That would mean that x=8 and that the shortest side is x-5, thanks again I'd give you 10 reps if i could

[Sam_1996]

So just to confirm that the mock papers on page 1 and page 6,are going to be the same style as the June exam?

[mia_hilton]

(Original post by gdunne42)
Sorry about that. They had been working a while ago.

They are also here as MathsMind had said
http://monmouth.monmoodle.co.uk/mod/...ew.php?id=3424

Hi i cant find the mark schemes either and they don't appear on this link, sorry can you please let me know where to find them , thank you

[rmpr97]

(Original post by lil-mazie)
Can one of you nice and smart students care to help me i have a question i'm stuck on it's reverse percentage intrest
In a sale, a TV is reduced to £322. This is 8% reduction from the orginal price. What was the orginal price
Ok basically i did this:-
100+8=108 then i divided it with 100 so it equals 1.08
I then multiplied it to 322 which gave me £347.76 I'm not sure it's correct and am I suppose to round it up???
Please help me out Thanks in advance

I know you already have an explanation but I couldn't understand someone and for me a good revision technique is explaining something.

So basically its a reverse percentage.

£322 = 92% of the price, as its 8% off.
Then you want to find 1% of the price which would be £322/92 = £3.50
Then multiply that by 100 to get 100% = £350

[swiftz]

just want to give a quick question i have just opened up this paper which is attached and i just want to know if this is the specimen paper for the new unit if it is then do you think this would be similar to the exam, but if it is not a specimen can someone do me a huge favor and send me the papers thankksss!!!!!!!

[l-g]

Could someone please upload the mark scheme for set b 4H, i would really appreciate it

[Freddy-Francis]

(Original post by mia_hilton)
Hi i cant find the mark schemes either and they don't appear on this link, sorry can you please let me know where to find them , thank you

http://www.whitefield.barnet.sch.uk/...ark_scheme.pdf

[Freddy-Francis]

(Original post by JadeAng)
Do you have the mark schemes by any chance?

Yeah i do now. PM me ur email

[Freddy-Francis]

(Original post by Jigglypuff786)
Basically you have to split the trapezium into a rectangle and a triangle by drawing a line through it.
Then work out the area of the rectangle and the triangle separately.
Area of rectangle = (x+2)(x-5)
= x² -3x -10
To work out the area of the triangle you must first work out the base length which is obviously 4, and the height is given which is (x-5)
So the area of the triangle = 2(x-5)
= 2x-10
Now add both of them together to get: x² - 3x - 10 + 2x- 10 = 36
Tidy it up and you get: x² - x - 20 = 36
This can also be written as x² - x - 56 = 0.

I am confused. Where did the 4 go. It became 2(x-5)
Sorry i dont get it. Please explain

[Jigglypuff786]

(Original post by Freddy-Francis)
I am confused. Where did the 4 go. It became 2(x-5)
Sorry i dont get it. Please explain

Because to find the area of a triangle you need to do "half-base x height" the base was 4cm and the height was (x-5). So you do 2(x-5)

[>Username<]

(Original post by Freddy-Francis)
http://www.worsleyschool.net/science...daccuracy.html
Check the website out. may help you

I had a look at this website, but I still don't get it. I get what was the website was saying, but it didn't have the difference between the nearest tenth of a cm and to the nearest centimetre...

Thanks, anyway.

[>Username<]

(Original post by Alexx53)
You need to create a set of simultaneous equations.

b=bus fares
f=food

b+f=£10
b=f+£1.50

then solve them

Spoiler:

Show

f+£1.50+f=10
2f=£8.50
f=£4.25

b+£4.25=£10
b=£5.75

So you need to do all that for a 2 mark question right at the start of the paper? I'm also confused by your working out, but I get when you say work it out by using simultaneous equations...

Is there not another method?

[>Username<]

(Original post by Jigglypuff786)
Uhhh can you link me that paper please?

This is the link, and it's question 1c.

Thanks.

[>Username<]

Please, could someone help me out with this question? I've done both parts... But I'm struggling with part b, I've done everything but I'm finding it hard to simplify the vector...

[>Username<]

I have done this question several times now. And every time I get the same answer, which according to the mark scheme is wrong. However, I think the mark scheme is wrong, because in a quadratic where something = 0 eg, 4x + 6x +9 =0. When you've worked it out the brackets must = to 0, let's say the answer was (x+9) (7+x), x must equal -9 and -7 because both brackets must equal 0... However in the mark scheme it says one of the answers is 5 when the bracket is (x+5). I'm utterly confused.

Edexcel have made mistakes in the past, but this a proper mark scheme which makes me think I've got the wrong answer.

Please, can someone help me out, I've been doing this for ages now...

[Alexx53]

(Original post by >Username<)
So you need to do all that for a 2 mark question right at the start of the paper? I'm also confused by your working out, but I get when you say work it out by using simultaneous equations...

Is there not another method?

What are you confused about? I just subbed in the B value into the other equation. You might be able to use trial and error, but I don't think there's another method.

(Original post by >Username<)


Please, could someone help me out with this question? I've done both parts... But I'm struggling with part b, I've done everything but I'm finding it hard to simplify the vector...

For part b

so
so
so
However so

So overall, I think

(Original post by >Username<)


I have done this question several times now. And every time I get the same answer, which according to the mark scheme is wrong. However, I think the mark scheme is wrong, because in a quadratic where something = 0 eg, 4x + 6x +9 =0. When you've worked it out the brackets must = to 0, let's say the answer was (x+9) (7+x), x must equal -9 and -7 because both brackets must equal 0... However in the mark scheme it says one of the answers is 5 when the bracket is (x+5). I'm utterly confused.

Edexcel have made mistakes in the past, but this a proper mark scheme which makes me think I've got the wrong answer.

Please, can someone help me out, I've been doing this for ages now...

The mark scheme is right.



Multiply through by the denominators



Then cancel



Expand the brackets



Put one side



Factorise



So