they are about differentiating academic ability, is that right?
if that is the case, if academic ability is about brute force memorisation, then it can encompass brute force trial and error methods too, right?
i fail to see how setting a goal for primary school students where they could get full marks in mathematics examinations by studying enough will aid them in the long run. it's high time that kids get a taste of real life earlier on, so that the singaporean education system doesn't just produce a bunch of kids waltzing into secondary school, falling flat on their face when they can't get 100 marks for maths, and then start crying in class.
Last edited by night86mare; 15th July 2010 at 09:43 PM.
Wahlao...I totally stone. Is this math or is this a letter to aunt angie?
I love A, but A love C while C hate K but C loves K. LOL
this one was familiar cause i encountered when i was p4-6 that time
u can imagine. i used to be a good boy and do 10 pages per assessment book per topic per day one
encounter this always use logic. what are the stuff i can link and limit first?
then brute force
Last edited by allenleonhart; 15th July 2010 at 09:46 PM.
essentially, the skills needed here are:
2) determination to get the answer
3) simple addition skills
i believe 1 and 2 are part and parcel of examinations, 3 is taught by primary 4. i hope.
Here is another possible answer:
So as you can see, the answer is not exact, and there are many permutations. The question looks like something that was conjured up by a teacher too creative. And it looks more like IQ test than math.
Last edited by ziploc; 15th July 2010 at 09:51 PM.
Sure, some logic is involved, but nothing much more than that. I thought they were already trying to discourage the "trial and error" method since the start of our education, why else would they teach us ways to work out "number pattern" style questions?
i think as long as the entire examination doesn't centre around 50 questions of that sort, it should be fine.
such questions can be a way to differentiate better students from the poorer ones - or perhaps how exam-smart you are. obviously, if i were a pri 4 student and i got this for my final examinations, i would realise quickly that this would be a waste of my time relative to the other "safer" questions, and leave it for last. then if i check my answers and i still have time, then can play with numbers for fun.
Last edited by allenleonhart; 15th July 2010 at 10:00 PM.
Right, go for it, math geek!
G = remainder((H + D) / 10), X = quotient((H + D) / 10)
K = remainder((A + E + X) / 10), Y = quotient((A + E + X) / 10)
F = remainder((K + J + Y) / 10), Z = quotient((K + J + Y) / 10)
H = K + K + Z
Last edited by ziploc; 15th July 2010 at 10:11 PM.
LOL WHERE GOT PRIMARY 4!! If really then i really fail......
Pass PSLE last year, also no such question....
My answer K = 3
using logic K + K = H, H <=9 thus K can only be 1-4
so if K = 1, then H = 2 11A2 + 2JED = 2F1G
since 1 and 2 is taken, the rest of the letters = 3,4,5,6,7,8,9,0
using the additional of the second digit 'A' and 'E' A+E = 1 or 11 there for A,E = 5,6/4,7,3/8 (combination to form 11)
if we choose A, E as 5, 6 then the 1st digit 2 + D must be less than 10 else the 2nd digit is not true, thus if we sub D as 7 then 2 + 7 = G = 9
now left with 4,3,8,0
3rd digit K + J = F, since 2nd digit A+E we set as 11 therefore the 3rd digit will have a carry over, become 1+1+J = F === 2+J = F (F = 3/4/8/0) F cannot = 3 cause J is not 1, F cannot = 4 cause J is not 2 (K = 1 and H = 2) leaving 8 and 0 but 2 + J = F cannot have a carry over thus F cannot be 8 as A, E is 5, 6 and 0 is out
there for K = 1 is not possible, continue with K = 2 and H = 4, bah bah bah
when u get to K = 3 33A6 + 6JED = 6F3G
Numbers left = 1,2,4,5,6,7,8,9,0
same method start with A, E A + E = 13 (combination for A, E = 4,9/5,8/6,7)
set A, E as 4, 9 then your 3rd digit must be 1+3+J = F (F must be <9 so that there is no carry over) let J = 1 your 3d digit 1+3+1 = F (F=5)
so far numbers left are ( 2,7,8,0 )
last set of digits 6 + D = G (G < 10) thus 6 + D(D=2) = G(G=8)
numbers left ( 7,0 )
D80/D700 18-135/17-50/14-24/24-70/70-200/70-300/50/60 SB600/900