Questions are from Xinmin sec prelim 2008 Paper 1
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Question 1:
A construction project consisting of 12 semi-detached houses can
be completed in 18 months by 54men. Find the number of men required
to complete another similar project consisting of 35 semi detached
houses which must be built in 12 months
My Method:
12 hosues-----18months-------54men
1 house------1.5month--------54men (i use 18
divide by 12 to get 1.5)
36 houses------54months-------54 men (1.5 x 36 will get 54,this is
working with 54men)
So if i want to complete the hosue in 12 months,
36house----54months-----54men
36hosue----12months----X men (54/12=4.5, need to work in 4.5 times
more)
36house----12months----54 x 4.5=243men
54 to 12 months is 4.5 times faster, in order to complete, u
need 4.5 times more men, thus 54 x 4.5, getiing 243 men.
This is my way of doing, but the question is that how am i going
to make use of y=k/x formula?
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Question 2:(originnaly qn 6 on
paper)
Light travels in 3.3 nano seconds
and soud travels 1m in 3.03 mili seconds. Estimate how far away is
the storm site if it takes 12 seconds btw the flash and
thunder.
I dunno how to do. Answer is
3960m
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Question 3:(originally qn 8 on paper)
John invest $4750 in a bank that pays a compound interest at the
rate of n% per year, Find the value of n if john has $6250 after 7
years.
P(1+r/100)^n
Principle= $4750
Rate=n
Years=7
thus,
4750(1+n/100)^7 = 6250
(1+n/100)^7=6250/4750
1+n/100=7√(6250/4750)
n/100=1.039988 - 1
n= 0.039988 x 100
n=3.9988, approx 4%
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Question 4:(Originnaly qn 9 on
paper)
The
size if singpaore on a map with a scale of 1:50000 is reflected as
2730.8cm2 and the coastline is relfected as 386cm. The size of
singpaore when drawn on another map of a diff scale is 10923.2cm2.
Find the length of coastline as reflected on this second
map
Answer is 772 cm, i duno how to
do
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Question 5:(Originally qn
11 on paper)
(i) The nth term of a
sequence S is n^3+2. Write down the first 4 terms of the
sequence
(ii) The first four terms
of another sequence T are 4,12,32 and 70
By comparing S&T, find
the 5th term of T
(iii) Find, in terms of n,
an expression for the nth term of sequence T
Solution:
(i)
1st term --------- 1^3+2 =
3
2nd term----------2^3+2= 10
3rd term-----------3^3+2= 29
4th term------------4^3+2= 66
Thus, the number sequence
is 3,10,29,66
(ii)
Compare:
1. 3,10,29,66 ---- S
2. 4,12,32,70 ----- T
You realise that the relationship of the 1st term of S and T is
answer for 1st term of S+1, the relationship of the 2nd term of S
and T is answer for 2nd term of S +2. Third term is answer for
3rd term of S+3
1st term is S +1
2nd term is S+2
3rd term is S+3
n term
is S + n
5th term= 5^3 + 2 +5 =
132
(iii) now u see that n
term is S + n, and the equastion for S is n^3+n, u can say that the
expression for Nth term of T is n^3+2+n
AFter shifting the like
terms tgt, u will get n^3+n+2
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i will post more later, i
need to scan in the question