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Mathematics: Post your doubts here!

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Well I tlhought of dividing 85 first by 3 then I got the limits I am not convinced that I should directly divide it by 3 after getting the limits of perimeter

You already got the upper and lower bound for the entire perimeter, you needn't meddle with each side now, and just simple divide it by three.

Using your method : 85/3 = 28.3333. The upper and lower limits still don't account to 28.7 and 28.8 if you work out with them.
 
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how to solve question 18 m/j 2014 p22 plzz help!!View attachment 48349
OMG this question had come in our exams and i loved it :p

See, throughout, we apply the principle that Volume = Area * length ( whatever length of cross-section... ), and step-by-step, we find each of the variables for both the containers.

For the large container, we have the volume and area, so let's find its length-
Volume = area * length
3456 = 1024 * length
length = 3.375 cm

For the smaller container, we will first find its length in order to get its area, by comparing its values with the bigger shape as they both are similar.
Using the formula V1/v2 = (length1/length2)^3, we substitute (the formula is somewhat that, i don't know how you write it)
3456/1458 = (3.375/length2)^3
length2 = 2.53125 (don't cut off decimal numbers even after three decimal places. Use the calculator to work out the answer)

Now you have the length as well as the volume for the smaller container,
Volume = area * length
1458 = area * 2.53125
area = 576

Erm... i hope you got it :/
 
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A box contains 3 red pencils, 2 blue pencils and 4 green pencils.
Raj chooses 2 pencils at random, without replacement.

Calculate the probability that: Exactly one for the two chosen pencils is green.
Thought blocker
There are 9 pencils, so there are 9C2 = 9x8/2 = 36 ways to pick 2 pencils.
To get exactly 1 green pencil, there is 1 green and 1 other.
Raj picks from 4 green and 5 others.
There are 4 ways to pick one green and 5 ways to pick the other.
4x5 = 20 ways to pick exactly 1 green
20/36 = 5/9
 
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How much do the examiners care about working out?
There's hella funky stuff in paper 2 in terms of working.
Would you get away with just showing what you did in your calculator and getting the answer correct?
 
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OMG this question had come in our exams and i loved it :p

See, throughout, we apply the principle that Volume = Area * length ( whatever length of cross-section... ), and step-by-step, we find each of the variables for both the containers.

For the large container, we have the volume and area, so let's find its length-
Volume = area * length
3456 = 1024 * length
length = 3.375 cm

For the smaller container, we will first find its length in order to get its area, by comparing its values with the bigger shape as they both are similar.
Using the formula V1/v2 = (length1/length2)^3, we substitute (the formula is somewhat that, i don't know how you write it)
3456/1458 = (3.375/length2)^3
length2 = 2.53125 (don't cut off decimal numbers even after three decimal places. Use the calculator to work out the answer)

Now you have the length as well as the volume for the smaller container,
Volume = area * length
1458 = area * 2.53125
area = 576

Erm... i hope you got it :/

That would probably get you a right answer wrong working mark. The correct method is to calculate the volume scale factor (k^3) and use that to find the area scale factor (k^2)

k^3 = 1458/3456
k^3 = 27/64
k = 3/4
k^2 = 9/16

Hence the surface area of the smaller shape is
A = 1024*9/16 = 576cm^2
 
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