maths

2007-03-14 5:04 am
1.last week,the exchang rate of Australian dollars to Hong Kong dollars was AUS $1=HK$5.tHIS WEEK is AUS$1=HK$4.5.
1a.Find the % decrease of the exchange rate of Australian dollars to Hong Kong dollars.
b.Daniel has AUS $1000.If he wants to get more Honk Kong dollars ,should he exchange the money last week or this week?What is the difference between the two?
2.if each side of a cube decreases by 10%,find the % des in its volume.

回答 (2)

2007-03-14 7:58 pm
✔ 最佳答案
Reemember the formula when you are calculating % change, and you will get the right answer:
("New" - "Old") / "Old" * 100%

1a. The "new" exchange rate of Australian dollors to HK dollar is 4.5 this week, and the "old" one is 5
so the % change of the exchange rate of Australian dollars to Hong Kong dollars
= (4.5 - 5) / 5 * 100%
= -10%

The % is decreased by 10%

1b. Last week, Daniel can exchange at rate A$1=HK$5
so he can get HK$
= 1000 * 5
=$5,000

This week, he can only get HK$
= 1000 * 4.5
=$4,500

So Daniel should exchange the money last week to get more HK$, and the difference is
=$5,000 - $4,500
=$500

2. Let x be the side of a cube ("old" figure)
and the "new" side of the cube
= x * (1-10%)
=0.9x

Volume of "old" cube
= x*x*x
=x^3

Volume of "new' cube
= 0.9x * 0.9x *0.9x
= 0.729x^3

% change in its volume (remember to use the same formula):
= (0.729x^3 - x^3) / x^3 *100%
= -27.1%
and so the % decreased in its volume is 27.1%
2007-03-14 6:26 am
1(a) percentage decrease=(5-4.5)/5X100%=10%
(b)Last week. Last week, he can get 1000X5=$5000
THis week, he get 1000X4.5=$4500
So the difference is $500
2.let the distance of original side be x , so the new one would become 0.9x
the % decrease in volume= x^3-(0.9x)^3/x^3X100%=72.9%
Remarks:(1) % decrease= (Big -Small)/ orginal X100%
(2) For question (2) , x^3 is the volume of a cube,and the x^3 約走左


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