Question

Suppose the number of babies born in Hong Kong in 1994 is 70000 and in subsequent years, the number of babies born each year increases by 2% of that of the previous year.
(a) Find the number of babies born in HK
(i) in the first year after 1994. [71400]
(ii) in the nth year after 1994. [70000*1.02^(n)]
(b) In which year will the number of babies born in HK first exceed 90000? [n=12]
>>>>(c) Find the total number of babies born in HK from 1997 to 2046 inclusive.<<<<
>>>>(d) It is known that from 1901 to 2099, a year is a leap year if its number is divisible by 4.
(i) Find the number of leap years between 1997 and 2046.
(ii) Find the total number of babies born in Hong Kong in the leap years between 1997 and 2046.<<<<

Answer 1

@Directrix @Hero

Answer 2

\(100\%+2\%=1.02\)

Answer 3

i know

Answer 4

\(1.2^n\)?

Answer 5

yes, it's a geometric sequence

Answer 6

Where is \(1.2\) from? I only see \(1.02\)

Answer 7

ohh, i type it wrongly.

Answer 8

my problem is in part c

Answer 9

Okay, first find \(n\)

Answer 10

\[
\sum_{k=0}^{n}ar^k=a\frac{1-r^{n+1}}{1-r}
\]

Answer 11

ohh, wait

Answer 12

n=12

Answer 13

\[
n=2046-1997
\]

Answer 14

The only problem is that \(k\) starts at \(4\)

Answer 15

ohh, you are talking about part c @@
n=49

Answer 16

what's k?

Answer 17

I mean \(3\)

Answer 18

i am thinking there are 49years or 50 years..

Answer 19

Okay let's actually do \(n_1= 1997-1994\) and \(n_2=2046-1994\)

Answer 20

\[
\sum_{k=0}^{n_2}ar^k -\sum_{k=0}^{n_1}ar^k
\]

Answer 21

Basically it is telling you how to add up geometric series.

Answer 22

okk...

Answer 23

So basically add up all children from 1994 to 2046 and then subtract all children from 1994 to 1996

Answer 24

oh, i see

Answer 25

\(\color{blue}{\text{Originally Posted by}}\) @wio
\[
\sum_{k=0}^{n}ar^k=a\frac{1-r^{n+1}}{1-r}
\]
\(\color{blue}{\text{End of Quote}}\)
This is the formula for adding up form year 0 to year n

Answer 26

year 0 is 1994

Answer 27

70000*(1+2%)^2=72828
70000*(1+2%)^52=196022.973

196022.973-72828=123194.973 ?!

Answer 28

No, look at the formula. You only calculating for a particular year.

Answer 29

what do you mean..

Answer 30

@hartnn

Answer 31

You need to calculate the sum of each year.

Answer 32

ohh my god, so complicated

Answer 33

You could at least try the formula I gave.

Answer 34

I mean... we know \(a=70000\), \(r=1.02\)

Answer 35

S(52)=6301148.649

Answer 36

Going from 1994 to 1996 we have \(n=2\)
Going from 1994 to 2046 we have \(n=52\)

Answer 37

find the sum from 1994 to 2046 and subtract that with the sum from 1994 to 1997
you will get sum for 1997 to 2046

Answer 38

\[
70000\frac{1-(1.02)^{52+1}}{1-1.02}- 70000\frac{1-(1.02)^{2+1}}{1-1.02}
\]

Answer 39

i see thats already been proposed

Answer 40

why +1?

Answer 41

Because the formula has \(n+1\) in it.

Answer 42

not n-1?

Answer 43

\(\color{blue}{\text{Originally Posted by}}\) @wio
\[
\sum_{k=0}^{n}ar^k=a\frac{1-r^{n+1}}{1-r}
\]
\(\color{blue}{\text{End of Quote}}\)

Answer 44

ok...

Answer 45

how about part d?

Answer 46

The first problem is division

Answer 47

eg 1904, 2096 are leap years?

Answer 48

The second problem requires that you create a new formula which is just number of children in a leap year.

Answer 49

So the new \(r\) is going to be \(1.02^4\)

Answer 50

i got 27 lol

Answer 51

@ganeshie8

Answer 52

(i) Find the number of leap years between 1997 and 2046.

if we start and end on a leap year, for example
2000 to 2008 (where we know we have 3 leap years)
we do the following : 2008 - 2000 = 8 years
divide by 4, to get 2
add 1 to get 3 leap years between 2000 and 2008 inclusive.
you can test other years, and see the formula is

(last_year - first_year) /4 + 1
as long as both the last_year and first_year are leap years.

in your case
1997 and 2046.

the first leap year in this interval is 2000 ( you can use the rule: if the last 2 digits are evenly divisible by 4, the whole number is divisible by 4)

the last leap year in this range is 2044 (44 is evenly divisible by 4)

now use the formula: 2044- 2000 = 44
divide by 4 to get 11
add 1
we find that there are 12 leap years between 1997 and 2046.

Answer 53

(ii) Find the total number of babies born in Hong Kong in the leap years between 1997 and 2046.

from part (i), we know the range will be leap years from 2000 to 2044 inclusive

We can use the result of (a, ii) to find the number born in the year 2000
call this number P2000
the number born in "0th leap yr" is P2000
the number born in 2004 "1st leap yr" P2000 * (1.02)^4
the number born in 2008 "2nd leap yr" P2004* (1.02)^4 or P2000* (1.02^4)^2
if you carry out this pattern you see the number born in the nth leap year is
P2000* (1.02^4)^n

if we let 1.02^4 = r
then the sum of the those born in leap years from number 0 (yr 2000) to number 11 (yr 2044) is
\[\sum_{n=0}^{11} r^n= \frac{1-r^{12}}{1-r}\]

Answer 54

how come i get 6255546.587?? @phi @hartnn

Answer 55

@Hero

Answer 56

what did you get for (1- r^12)/(1-r) ?

Answer 57

I get r= 1.02^4 = 1.0824

(1-r^12)/(1-r) = 19.2530

the population in the year 2000 is 70000*1.02^6 = 78831.37

times 19.253 gives 1,517,744.188

Answer 58

i got it now. Thanks @phi