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
Thanks for helping to contribute to ExamWorld :D
I will help in the questions in red first, then look at the rest later.
Question 2:(originnaly qn 6 on paper)
Light travels in 1 m in 3.3 nano seconds and sound 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
I changed to light travels 1m in 3.3 ns (added extra info... cuz I think you missed it out)
Let the distance be d metres
time light took to travel = d *3.3 ns
time sound took to travel = d * 3.03 ms
So d * 3.03 ms - d * 3.3 ns = 12
d (3.03 * 10^-3 - 3.3 * 10^-9) = 12
d (0.0030299967) = 12
so d = 3960 m
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 reflected 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
the difference in ratio between the two maps is
sqrt( 10923.2 / 2730.8 ) = 2
so, coastline of second map / 386 = 2 (not area, so dun need to square it)
coastline of second map = 2 * 386 = 772
Regards,
Eagle
ExamWorld
Originally posted by MyPillowTalks: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 36 semi detached houses which must be built in 12 months
This is my way of doing, but the question is that how am i going to make use of y=k/x formula?
Concept
1. Determine the 2 key variables.
2. Is it direct or inverse relationship between the 2 variables?
3. Solve unknowns.
Concept in action
1. Nos of Houses (H) and Man-Months taken (M)
2. More house more man-months. Thus direct variation.
H varies directly as M
H = kM
When H= 12, M = 972
12 = k(972)
k = 1/81
Thus H= (1/81)M
When H= 36 and M = 12x (where x is the number of men)
36 = (1/81)(12x)
x= 243