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A new machine is bought for €30000 - Leaving Cert Mathematics - Question 1 - 2017

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A new machine is bought for €30000. Its value depreciates by 15% each year for five years. Find the value of the machine at the end of the five years. A sum of mone... show full transcript

Worked Solution & Example Answer:A new machine is bought for €30000 - Leaving Cert Mathematics - Question 1 - 2017

Step 1

Find the value of the machine at the end of the five years.

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Answer

To find the depreciated value of the machine after five years, we can use the formula:

A=P(1r)tA = P(1 - r)^t
where:

  • AA is the amount after time tt,
  • PP is the original price (€30000),
  • rr is the annual depreciation rate (15% or 0.15), and
  • tt is the number of years (5).

Substituting the values, we get:

A=30000(10.15)5A = 30000(1 - 0.15)^5

Calculating step-by-step:

  • 10.15=0.851 - 0.15 = 0.85
  • 0.8550.44370.85^5 \approx 0.4437
  • Therefore,
    A30000×0.443713311.16A \approx 30000 \times 0.4437 \approx 13311.16

Thus, the value of the machine at the end of five years is approximately €13,311.16.

Step 2

Find the sum invested.

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Answer

To find the initial sum invested, we use the formula for compound interest:

A=P(1+r)tA = P(1 + r)^t
where:

  • A=30000A = 30000
  • r=0.03r = 0.03 (3% interest)
  • t=2t = 2 (duration in years).

Rearranging gives us:

P=A(1+r)tP = \frac{A}{(1 + r)^t}
Substituting the values:
P=30000(1+0.03)2P = \frac{30000}{(1 + 0.03)^2}
Calculating this value:

  • (1+0.03)2=1.0609(1 + 0.03)^2 = 1.0609
  • Therefore,
    P300001.060928277.88P \approx \frac{30000}{1.0609} \approx 28277.88

Thus, the sum invested was approximately €28,277.88.

Step 3

Find the rate of interest.

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Answer

To find the rate of interest, we also use the compound interest formula, rearranging it. The formula is given by:

A=P(1+i)tA = P(1 + i)^t
To solve for ii, we need to rearrange:

1+i=(AP)1t1 + i = \left( \frac{A}{P} \right)^{\frac{1}{t}}
So,
i=(AP)1t1i = \left( \frac{A}{P} \right)^{\frac{1}{t}} - 1
Substituting in our values:

  • A=26530.20A = 26530.20
  • P=25000P = 25000
  • t=3t = 3
    We find:
    i=(26530.2025000)131i = \left( \frac{26530.20}{25000} \right)^{\frac{1}{3}} - 1
    Calculating gives us:
  • 26530.20250001.061208\frac{26530.20}{25000} \approx 1.061208
  • Taking the cube root:
    (1.061208)131.02(1.061208)^{\frac{1}{3}} \approx 1.02
    Thus, i0.02i \approx 0.02, which translates to a rate of interest of approximately 2%.

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