APPRECIATION AND INFLATION WORD PROBLEMS

When the value of an item such as a house or investment increases, we say it appreciates in value. You may have noticed that the prices of everyday goods and services also increase over time due to inflation.

Problem 1 :

The inflation rate over the next three years is predicted to be 2% then 3% then 3.5%. What will be the cost in three years time of an item which currently costs $140 if the cost rises in line with inflation?

Solution:

A = P1+r100TA=1401+21001+31001+3.5100A = 140102100103100103.5100A = 140×1.02×1.03×1.035A=$152.23

Cost in 3 years = $152.23

Problem 2 :

What would you expect a bicycle costing $500 today to cost in four years time, if the inflation rate is predicted to be 3%, 3.5%, 4% and then 5% over the four years?

Solution:

A = P1+r100TA=5001+31001+3.51001+41001+5100A = 500103100103.5100104100105100A = 500×1.03×1.035×1.04×1.05A=$582

Cost in 4 years = $582

Problem 3 :

If the rate of inflation is expected to remain constant at 4% per year for the next 5 years, what would you expect a $20000 car to cost in 5 years time?

Solution:

A = P1+r100TA=200001+41005A = 200001041005A = 20000×(1.04)5A=$24333

Cost in 5 years = $24333

Problem 4 :

An investment of $18250 is left to accumulate interest over a 4 year period. During the first year the interest paid was 9.2%, and in successive years the rates paid were 8.6%, 7.5% and 5.6%. Find the value of the investment after 4 years.

Solution:

Given, P = $18250, T = 4 years

r = 9.2% for first year and 

r = 8.6% for second years and

r = 7.5% for third years and

r = 5.6% for fourth years.

A = P1+r100TA=182501+9.21001+8.61001+7.51001+5.6100A = 18250109.2100108.6100107.5100105.6100A = 18250×1.092×1.086×1.075×1.056A=$24569

Problem 5 :

An amount of money is invested at 8% for the first year, 11% for the second year and 7% for the third year. What is the percentage increase in the value of the investment over this period?

Solution:

A = P1+r100TA=x1+81001+111001+7100A = x108100111100107100A = x×1.08×1.11×1.07A=1.282716 x % = (1.282716 x %)×100= 128.2716 x %Percentage increase=128.2716 %- 100%= 28.27 %

= 28.3 % increase

Problem 6 :

A politician's wage increases by 3%, 8%, 5% and then 10% over a four year period. What is the percentage increase in the wage over this period of time?

Solution:

A = P1+r100TA=x1+31001+81001+51001+10100A = x103100108100105100110100A = x×1.03×1.08×1.05×1.1A=1.284822 x % = (1.284822 x %)×100= 128.4822 x %Percentage increase=128.4822 %- 100%= 28.48 %= 28.5 %

= 28.5 % increase

Problem 7 :

A share fund reported a 9% increase in value for year 1, a 13% decrease in value for year 2 and a 4% increase in value for year 3. What was the overall percentage increase or decrease of the share fund over the 3 years?

Solution:

A = P1+r100TA=x1+91001-131001+4100A = x10910087100104100A = x×1.09×0.87×1.04A=0.986232 x % = (0.986232 x %)×100= 98.6232 x %Percentage decrease=98.6232 %- 100%= -1.3768%= 1.38 % decrease

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