Maths, Addmaths and Statistics: Post your doubts here!

[robinhoodmustafa]

Yes. Always look for the from. You know how bad-*** these CIE peepalz can be sometimes.
Yeah, revise circle theorems, probability, rates & ratios, loci and transformations. These are, what I believe, to be the most important P1 topics.

Ok.. Please help me in probability

See they ask us that 2 balls are chosen at random without replacement and give us a chart to fill so how do we find the value of the incomplete blanks?

and yes.. i am on to Transformation.

 

[Saad Mughal]

2005 June is different. It is hard

It was a little tougher but not that hard. In fact, all the latest papers (of this century ) are easy. Unlike the 90's.
I did the 2005 paper just earlier, found it pretty easy except a certain AE vector part that I got wrong.

 

[Saad Mughal]

Ok.. Please help me in probability

See they ask us that 2 balls are chosen at random without replacement and give us a chart to fill so how do we find the value of the incomplete blanks?

and yes.. i am on to Transformation.

If the two balls are of different colors then the probability of choosing one of the colors is 1/2; if you're asked to choose again from the same bag (with the two balls), the choice is again 1/2. It's not of 2 balls waise, it's usually of dices with 6 (or four) faces.

 

[Tulaib]

Oh Right.. Got It Now Thanks!!

ma pleasure dude

 

[robinhoodmustafa]

It was a little tougher but not that hard. In fact, all the latest papers (of this century ) are easy. Unlike the 90's.
I did the 2005 paper just earlier, found it pretty easy except a certain AE vector part that I got wrong.

I suck at congruency and similarities.
Wish Circular theorem comes

 

[Saad Mughal]

I suck at congruency and similarities.
Wish Circular theorem comes

Everything comes. The 25 questions (mostly) are almost always from different topics so there is little chance that anything is left out (except pure trigonometry and coordinate geometry of course). So, be ready for everything. THE BEST WAY to learn the congruency and similarities at this time is to read mark schemes and look at the REASONS provided for the answer, just learn them and use them to prove your triangles.

 

[robinhoodmustafa]

Everything comes. The 25 questions (mostly) are almost always from different topics so there is little chance that anything is left out (except pure trigonometry and coordinate geometry of course). So, be ready for everything. THE BEST WAY to learn the congruency and similarities at this time is to read mark schemes and look at the REASONS provided for the answer, just learn them and use them to prove your triangles.

Help me in this question... It is loci
They say that 1 cm represent 10m

so part a ask.. Express the scale in the form 1:n

 

[robinhoodmustafa]

Saad Mughal

Hence evaluate "Root on 7056"

 

[Saad Mughal]

Help me in this question... It is loci
They say that 1 cm represent 10m

so part a ask.. Express the scale in the form 1:n

You always express both the quantities in the same unit for 1:n ratio, so,
10 m = 1000 cm
Therefore,
1 cm : 10 m = 1 : 1000.

 

[Saad Mughal]

Saad Mughal

Hence evaluate "Root on 7056"

What was the first part?

 

[robinhoodmustafa]

What was the first part?

Find 7056 as a product of it prime factor

 

[robinhoodmustafa]

You always express both the quantities in the same unit for 1:n ratio, so,
10 m = 1000 cm
Therefore,
1 cm : 10 m = 1 : 1000.

Thanks..

Now if they say " Nearer to A than C" and they said draw this.. So Hum iss mai Perpendicular bisector banataay hai or what?

 

[Saad Mughal]

Find 7056 as a product of it prime factor

The prime factors must have been 3^2 * 4^2 * 7^2.
So, from this, you can easily evaluate the root,
root 7056 = 3 * 4 * 7 = 84.

 

[usama321]

1/4

oops it would be 3/4 into 1/3 = 3/12 = 1/4

 

[Saad Mughal]

Thanks..

Now if they say " Nearer to A than C" and they said draw this.. So Hum iss mai Perpendicular bisector banataay hai or what?

Loci;
Equidistant from ONE line = Parallel line to that line.
Equidistant from ONE point = Circle.
Equidistant from TWO points = Perpendicular Bisector.
Equidistant from TWO lines (with common vertice) = Angle Bisector.
These are the four most common and widely asked loci.

 

[robinhoodmustafa]

Loci;
Equidistant from ONE line = Parallel line to that line.
Equidistant from ONE point = Circle.
Equidistant from TWO points = Perpendicular Bisector.
Equidistant from TWO lines (with common vertice) = Angle Bisector.
These are the four most common and widely asked loci.

ok.. so this will be a perpendicular bisector but what should I take in length

and they say equidistant from AB to AC so this means an Angel bisector and A is the common point

 

[robinhoodmustafa]

The prime factors must have been 3^2 * 4^2 * 7^2.
So, from this, you can easily evaluate the root,
root 7056 = 3 * 4 * 7 = 84.

What if it is like this??

2^2 * 3^2 *7^3

Than we will take all common pairs outside the root

 

[Saad Mughal]

ok.. so this will be a perpendicular bisector but what should I take in length

and they say equidistant from AB to AC so this means an Angel bisector and A is the common point

Yes. What you mean by length?

 

[robinhoodmustafa]

Yes. What you mean by length?

Measurements on the compass?

 

[Saad Mughal]

What if it is like this??

2^2 * 3^2 *7^3

Than we will take all common pairs outside the root

Questions like these don't come very often in paper one. If it does come then yes, take, 2 * 3 * (root 7^3) which means 6 (root 7^3). Either express it this way or use estimation to calculate a rough value.