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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE General & Further Mathematics => Topic started by: Sanguinne on May 18, 2014, 07:42:28 pm

Title: VCAA Number Patterns Question
Post by: Sanguinne on May 18, 2014, 07:42:28 pm

The number of water fowl living in a wetlands area has decreased by 4% each year since 2003.
At the start of 2003 the number of waterfowl was 680.

If this percentage decrease continues at the same rate, the number of waterfowl in the wetlands at the start of 2008 will be closest to

a) 532
b) 544
c) 554
d) 571
e) 578

Why is the answer c and not e. The way I solved it was t₅ = 680 x 0.96^5-1 which equals 578

Title: Re: VCAA Number Patterns Question
Post by: Zealous on May 18, 2014, 07:55:37 pm

In this question, the "n-1" above the ratio is not necessary. From 2003 to 2008, we need the population to decrease 5 times, and hence 0.96 is raised to the power of 5.

Title: Re: VCAA Number Patterns Question
Post by: plato on May 19, 2014, 10:08:19 pm

While you can use the rule as you did, you need to be careful.
In this question.
t₂₀₀₃ = t₁
t₂₀₀₄ = t₂
t₂₀₀₅ = t₃
t₂₀₀₆ = t₄
t₂₀₀₇ = t₅
t₂₀₀₈ = t₆

If using the rule, you need to find t₆ rather than t₅

t₆ = 680 x 0.96^6-1
=680 x 0.96⁵ which is the same as used by Zealous