Mathematics: Post your doubts here!

[loaie Amgad]

S

stretch and shear were out of 2015 syllabus

Thats right , no more stretch and shear but do u know any additional topics to the syllabus ???????
Because I ve heard that log function is now a part of our syllabus
Provide me with any news pleaseeeee
Thanks in advance........

 

[Zohaa Faiz]

Hey , does anyone know how to solve question no 5 in maths may june 2006 paer 2 ,with a logical explaination?http://www.examtestprep.com/CIE/IGC...SE-Mathematics-Mark-Scheme-2-Winter-2006.html

 

[NIM]

Guyzz need help for paper 62 .. M/J 2013
Question no 1

 

[loaie Amgad]

Hey , does anyone know how to solve question no 5 in maths may june 2006 paer 2 ,with a logical explaination?http://www.examtestprep.com/CIE/IGC...SE-Mathematics-Mark-Scheme-2-Winter-2006.html

oh man , thats an easy question
you just need to use area of triangle formula
Area=0.5 x side A x side B x sin angle C......
so answer is 21.3 cm2

 

[masterex567]

Thank you so much. Your explanation was amazing!
Although I do have one doubt, what if he gets 2 in his first throw and lands on 96 and gets a 3 again? Why don't we take that into account? Or is it because of the fact that we have to consider ONLY the possibilities which will make sure that he wins in his NEXT 2 throws?

Thanks once again!

Yes, exactly we only we take the probabilities into account that will make sure he reaches 100 in next 2 throws.

 

[AnonymousX9]

One more, O/N 2013, Paper 43, Question 1 (b), please.

 

[masterex567]

QUESTION 11 Part b(i )and question 6 probability part c please help this MAY JUNE 2014 V 43

IN MAY JUNE 2014 V 42 I do have doubts too in

Question 11 I can't understand what does the question even want me to do please I need full explanation roughly written on paper
Question 9 probability part (E) My answer was (3/4)^4 *1/4 as the question said at least one day the weather won't be fine I simply got the probability of being fine for four days and multiplied it by the probability of one day of not being fine and sadly that's was wrong.

I hope someone could help me I'll really appreciate it.

11 part b)i)
The numbers of smallest triangles as power of 2 are increasing powers by 2.
Hence after 2^4, it'll be 2^6, 2^8 etc.
Work out 2^6, and 2^8 and put its values in number of smallest triangles row.

6c)
Probability of choosing a N is 2/6. So probabibility of NOT getting N is (1-2/6) which is 4/6.
Hence till 4th card probability will be:
4/6 * 3/5 * 2/4 * 2/3, which is 2/15. The denominators are decreasing, because cards are being chosen without replacement.

_______________________________________________________________________
Question 11:
Okay, if the total area of the shapes is X. And the area of the shaded part is k. It means k is a number that is multiplied with the area to give shaded area.
For example in the first one, if the total area of a triangle is X, and the shaded area is 1/3 of the total area (of X). Then to calculate this shaded area it'll be 1/3 * X. In this case k is 1/3. It's a number multiplied by X (total area) to give the shaded area.
Now the second one, We can calculate the area of the sector by 72/360 * pi * radius squared. If the total area (X) of the circle is pi * radius squared, what are you multiplying with it to get shaded area? That's 72/360. So 72/360 is k.
For the third one, EF = FG, hence if we assume EF is 1, then FG will also be 1. Therefore EG will be 2. Now since they are similar triangles, (EF/EG)squared will give area. Therefore 1squared/2squared is answer. which is equal to 1/4.
For the fourth one, a hexagon is made of 6 equilateral triangles. Hence each angle will be 60 degrees. So to find total area (X) it'll be 6 * length * length * sin60. now the area that is shaded has the same area as any triangle from the center to the two sides. So they want to find area of only 1 triangle out of the 6 triangles. Hence k will be 1/6.
Last one, shaded area is area of sector - area of triangle. Which is 90/360*pi*radius squared - 1/2 * radius squared. Work this out and factorise, it'll give 1/4*radius squared (pi - 2). Now this is the shaded area. They want k which is the value you multiply with the whole sector to give this shaded area. so divide the shaded area by area of whole sector. the answer will be pi - 2/pi.

Question 9)e):
Now probability that the weather is fine is 3/4. Now if the weather is fine for 5 consecutive days the probability will be (3/4)^5. So probability that it is NOT fine for 5 consecutive days will be 1 - (3/4)^5. Your working was wrong because it depicts that weather was fine for the first four days, then fifth day it was not fine. You didn't take to account that the weather could not be fine on the 1st, 2nd, 3rd or 4th day even. Hence this is the best method.

Cheers

 

[masterex567]

One more, O/N 2013, Paper 43, Question 1 (b), please.

Alright, in part a)iii) we calculated the area of this trapezium as 65.8.

Now in b) a similiar trapezium has height 9.4 cm.
Which means (4.7 which is height of our trapezium/9.4)^2 = 65.8/x.
Now we cross multiply and find x.
65.8 * 9.4^2 = 4.7^2*x.
therefore x is 263.2!

we square the heights because we want to find area. If we wanted to find volume for example, you cube the heights

 

[NOneed2speedd]

okay :
yours are same time as mine too? (am/pm)

Yea both am

 

[AnonymousX9]

Alright, in part a)iii) we calculated the area of this trapezium as 65.8.

Now in b) a similiar trapezium has height 9.4 cm.
Which means (4.7 which is height of our trapezium/9.4)^2 = 65.8/x.
Now we cross multiply and find x.
65.8 * 9.4^2 = 4.7^2*x.
therefore x is 263.2!

we square the heights because we want to find area. If we wanted to find volume for example, you cube the heights

Umm I guess you checked the wrong question, it's not this one.

 

[NOneed2speedd]

Hey , does anyone know how to solve question no 5 in maths may june 2006 paer 2 ,with a logical explaination?http://www.examtestprep.com/CIE/IGC...SE-Mathematics-Mark-Scheme-2-Winter-2006.html

u mean this one?

 

[masterex567]

Umm I guess you checked the wrong question, it's not this one.

Ohh sorry about that.
Okay, the way i did it was like this:
1. 25% is used for second hand cars.
2. 62.5% of remaining pages used for features,
The remaining pages here is 100-25 that we used for second hand cars. Leaving 75% as remaining pages.
So 62.5% of the remaining pages is 62.5% * 75 % which is 46.875% for features.
3. now other 36 pages are used for reviews,
Now out of 100%, 25% and 46.875% are used. We're left with 100-25-46.875, which is 28.125% for reviews.

If x is the total number of pages:
28.125/100 of x = 36 pages.
therefore x is 128.

 

[AnonymousX9]

Ohh sorry about that.
Okay, the way i did it was like this:
1. 25% is used for second hand cars.
2. 62.5% of remaining pages used for features,
The remaining pages here is 100-25 that we used for second hand cars. Leaving 75% as remaining pages.
So 62.5% of the remaining pages is 62.5% * 75 % which is 46.875% for features.
3. now other 36 pages are used for reviews,
Now out of 100%, 25% and 46.875% are used. We're left with 100-25-46.875, which is 28.125% for reviews.

If x is the total number of pages:
28.125/100 of x = 36 pages.
therefore x is 128.

omg, thank you so much! :') I finally understood it. How is your preparation going on for the Math exam?

 

[TRAMANZ]

Thats right , no more stretch and shear but do u know any additional topics to the syllabus ???????
Because I ve heard that log function is now a part of our syllabus
Provide me with any news pleaseeeee
Thanks in advance........

This should help you.

 

[masterex567]

omg, thank you so much! :') I finally understood it. How is your preparation going on for the Math exam?

Hahaha no prob! Not bad, although i keep making these petty silly mistakes . It's only 6 days away!

 

[misscute406]

(c) (i) Find the area of the shaded segment CAD.

Hello...I need help!! 0580/21/o/n/14...Question 19!! Can anyone help me out!!

 

[Saad the Paki]

Hahaha no prob! Not bad, although i keep making these petty silly mistakes . It's only 6 days away!

so will u be giving variant 22 and 42?

 

[NOneed2speedd]

Can any one help me in part ii
i wrote answer frm 2 to 10 but answer is frm -4 to 2
can someone explain how....... i cannot understand

 

[Dsilentkila]

salam folks.
got a question ,
Paper 4 (41) , MJ 2010.
Q8.d.iii .
thanks alot in advance

 

[Saad the Paki]

Can any one help me in part ii
i wrote answer frm 2 to 10 but answer is frm -4 to 2
can someone explain how....... i cannot understand

I think u didmt understand the question properly.. they asked for the MINIMUM values.. other wise the graph can go on forever and the range will be huge.. so from the graph approx 2 and -4 r the minimum values from wich u can get only one solution