The power (P) of an engine is measured in horsepower using the formula: $$ P = \frac{R \times T}{5252} $$ where R is the engine speed measured in revolutions per minute (RPM) and T is the torque measured in appropriate units. (i) Find the power of an engine that generates 480 units of torque at 2500 RPM. Give your answer correct to the nearest whole number. (ii) Rearrange the formula to write R in terms of P and T. A company was set up in January 2016 to repair engines. In the first month of its existence the company made a loss of €4000. This loss reduced by €250 a month for each month that the company traded. (i) Complete the table below to show the company's loss/profit for each of the first six months of trading. | Month | 1 | 2 | 3 | 4 | 5 | 6 | |-------|---|---|---|---|---|---| | Profit (€) | -4000 | -3750 | -3250 | | | | (ii) Show that the profit the company makes in month n is given by the formula $$ T_n = 250n - 4250 $$ (iii) What profit does the company make in January 2018 (i.e. month 25)? (iv) Find the month in which the company breaks even (i.e. €0 profit). (v) Find $S_n$, the general term for the total profit of the company after n months. (vi) Hence, or otherwise, find the total profit of the company at the end of January 2019 (i.e. month 37).

Leaving Cert Mathematics Sequences & Series: The power (P) of an engine is me

The power (P) of an engine is measured in horsepower using the formula: $$ P = \frac{R \times T}{5252} $$ where R is the engine speed measured in revolutions per minute (RPM) and T is the torque measured in appropriate units - Leaving Cert Mathematics - Question 8 - 2019

Question 8

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The power (P) of an engine is measured in horsepower using the formula: $$ P = \frac{R \times T}{5252} $$ where R is the engine speed measured in revolutions per m... show full transcript

Worked Solution & Example Answer:The power (P) of an engine is measured in horsepower using the formula: $$ P = \frac{R \times T}{5252} $$ where R is the engine speed measured in revolutions per minute (RPM) and T is the torque measured in appropriate units - Leaving Cert Mathematics - Question 8 - 2019

Step 1

Find the power of an engine that generates 480 units of torque at 2500 RPM. Give your answer correct to the nearest whole number.

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Answer

To find the power, we substitute the values into the formula:

P = \frac{480 \times 2500}{5252}

Calculating this gives:

P = \frac{1200000}{5252} \approx 228.48

Rounding to the nearest whole number, the power is 228 horsepower.

Step 2

Rearrange the formula to write R in terms of P and T.

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Answer

Starting with the original formula:

P = \frac{R \times T}{5252}

We rearrange it to solve for R:

R = \frac{P \times 5252}{T}

Step 3

Complete the table below to show the company's loss/profit for each of the first six months of trading.

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Answer

Month	1	2	3	4	5	6
Profit (€)	-4000	-3750	-3500	-3250	-3000	-2750

Step 4

Show that the profit the company makes in month n is given by the formula Tn = 250n - 4250.

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Answer

The profit for month n can be calculated as follows:

Initial loss of €4000 reduces by €250 each month, so:

T_n = -4000 + (n - 1)250

Simplifying gives:

T_n = 250n - 4250

Step 5

What profit does the company make in January 2018 (i.e. month 25)?

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Answer

To find the profit in month 25, substitute n = 25 into the formula:

T_{25} = 250(25) - 4250

Calculating this gives:

T_{25} = 6250 - 4250 = 2000

Thus, the company makes a profit of €2000 in January 2018.

Step 6

Find the month in which the company breaks even (i.e. €0 profit).

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Answer

Setting the profit formula to zero:

0 = 250n - 4250

Solving for n gives:

250n = 4250 \Rightarrow n = 17

Thus, the company breaks even in month 17.

Step 7

Find Sn, the general term for the total profit of the company after n months.

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Answer

The total profit after n months can be found by summing the profit for each month:

S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (250k - 4250)

This simplifies to:

S_n = 250 \left( \frac{n(n + 1)}{2} \right) - 4250n = \frac{250n(n + 1)}{2} - 4250n

Step 8

Hence, find the total profit of the company at the end of January 2019 (i.e. month 37).

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Answer

Using the general term derived:

S_{37} = \frac{37}{2} [250n - 4250] = \frac{37}{2}[250(37) - 4250] = \frac{37}{2}[9250 - 4250] = \frac{37}{2}[5000] = 18500

Thus, the total profit at the end of January 2019 is €18500.

The power (P) of an engine is measured in horsepower using the formula: $$ P = \frac{R \times T}{5252} $$ where R is the engine speed measured in revolutions per minute (RPM) and T is the torque measured in appropriate units - Leaving Cert Mathematics - Question 8 - 2019

Find the power of an engine that generates 480 units of torque at 2500 RPM. Give your answer correct to the nearest whole number.

Rearrange the formula to write R in terms of P and T.

Complete the table below to show the company's loss/profit for each of the first six months of trading.

Show that the profit the company makes in month n is given by the formula Tn = 250n - 4250.

What profit does the company make in January 2018 (i.e. month 25)?

Find the month in which the company breaks even (i.e. €0 profit).

Find Sn, the general term for the total profit of the company after n months.

Hence, find the total profit of the company at the end of January 2019 (i.e. month 37).

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