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3.1 Calculate, in rand per minute, the rate for solo flying - NSC Mathematical Literacy - Question 3 - 2019 - Paper 2

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3.1 Calculate, in rand per minute, the rate for solo flying. 3.1.2 Calculate the total approximate cost to get a private pilot's licence. 3.2 Two years ago Franco ... show full transcript

Worked Solution & Example Answer:3.1 Calculate, in rand per minute, the rate for solo flying - NSC Mathematical Literacy - Question 3 - 2019 - Paper 2

Step 1

Calculate, in rand per minute, the rate for solo flying.

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Answer

To find the rate per minute for solo flying, divide the total cost of solo flying by the total number of minutes:

  1. Total cost for solo flying: R31 050
  2. Total hours of solo flying: 18 hours = 18 * 60 minutes = 1080 minutes

Rate per minute: extRate=R310501080=R28,75/min ext{Rate} = \frac{R31 050}{1080} = R28,75/\text{min}

Step 2

Calculate the total approximate cost to get a private pilot's licence.

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Answer

The total approximate cost for a private pilot's licence is calculated by adding all components:

  1. Flight with instructor: 28 hours at R2 050 per hour = R57 400
  2. Solo flying: R31 050
  3. Ground lessons: 15 hours at R1 242,42 per hour = R18 636,30
  4. Ground school fees: R700
  5. Flying kit: R6 544
  6. Examinations: 7 at R190 each = R1 330

Total Cost: R57,400+R31,050+R18,636.30+R700+R6,544+R1,330=R115,660.30R57,400 + R31,050 + R18,636.30 + R700 + R6,544 + R1,330 = R115,660.30

Step 3

Verify, with calculations, whether this total amount with interest is enough to pay for a private pilot's licence.

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Answer

We need to calculate the total amount after two years with compound interest:

Given:

  • Principal (P): R90 000
  • Rate (r): 8.5% per annum
  • Time (t): 2 years

The formula for compound interest is: A=P(1+r/n)ntA = P(1 + r/n)^{nt} Assuming interest is compounded yearly (n=1): A=R90000(1+0.085)2A = R90 000(1 + 0.085)^2 Calculating gives: A=R90000(1.085)2=R90000(1.177225)=R105950,25A = R90 000(1.085)^2 = R90 000(1.177225) = R105 950,25

Since R105 950,25 is less than the total cost of R115,660.30, the total amount is not enough.

Step 4

Give a possible reason why the probability of passing increased after the second attempt.

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Answer

Students might have gained more experience from their first attempt and were better prepared for the second. They may have learned from their mistakes and had more practice before retaking the examination. This additional preparation likely improved their chances of passing.

Step 5

Determine, showing ALL calculations, the missing values A, B, C and D to calculate the total number of students that passed after both attempts.

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Answer

From TABLE 2:

  • A's pass rate is 20%, so: A=24/0.2=120A = 24/0.2 = 120
  • B’s pass rate from the second attempt is 30%, so: B=67/0.3=29B = 67/0.3 = 29
  • C = total number of students = 96
  • D = number who passed = 96 - 29 = 67.

Thus, the total number of students that passed after both attempts: 24+29=5224 + 29 = 52.

Step 6

Verify, showing ALL calculations, whether this statement is valid.

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Answer

To verify the conversion of 26 000 flying hours:

  1. Convert total hours to days: 26,000 hours =26,000241083.33 days26,000 \text{ hours } = \frac{26,000}{24} \approx 1083.33 \text{ days}
  2. Days in weeks: 1083.337154.76 weeks\frac{1083.33}{7} \approx 154.76 \text{ weeks}
  3. It confirms that the statement is valid since both calculations align with the given data.

Step 7

Write down the total number of loose parts in a box.

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Answer

The total number of loose parts in a box is derived from ANNEXURE E. Counting all the visible parts in the illustration yields a total of 33 loose parts.

Step 8

Write down TWO instructions that will match the illustration in step 2.

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Answer

  1. Place the bench face down.
  2. Attach the legs to the bench seat correctly, ensuring they align with the pre-drilled holes.

Step 9

Explain the purpose of the long panel.

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Answer

The long panel serves as the main support for the bench structure. It connects the legs and provides stability, ensuring that the assembled bench is sturdy and capable of bearing weight.

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