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Define the following terms: (i) measurement (ii) quantity (iii) unit (iv) accuracy (v) precision

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GSE 212: BASIC GENERAL MATHEMATICS III (1C)

Question 1:

a) Define the following terms:

  • (i) Measurement: The process of assigning a numerical value to a physical quantity using a standard unit.
  • (ii) Quantity: A property of a phenomenon, body, or substance that can be quantified by measurement.
  • (iii) Unit: A standard amount of a physical quantity, such as a meter for length or a kilogram for mass, used for comparison.
  • (iv) Accuracy: How close a measured value is to the true or actual value of the quantity being measured.
  • (v) Precision: The degree to which repeated measurements under unchanged conditions show the same results; it indicates the reproducibility of a measurement.

b) Given 3x+4y=263x + 4y = 26, express xx in terms of yy and find xx when y=5y = 5 and y=10y = 10.

  1. Express xx in terms of yy.

    Subtract 4y4y from both sides, then divide by 3.

    3x+4y=263x + 4y = 26 3x=26−4y3x = 26 - 4y x=26−4y3x = \frac{26 - 4y}{3}

    The expression for xx in terms of yy is x=26−4y3\boxed{x = \frac{26 - 4y}{3}}.

  2. Find xx when y=5y = 5.

    Substitute y=5y = 5 into the expression for xx.

    x=26−4(5)3x = \frac{26 - 4(5)}{3} x=26−203x = \frac{26 - 20}{3} x=63x = \frac{6}{3} x=2x = 2

    When y=5y = 5, x=2x = \boxed{2}.

  3. Find xx when y=10y = 10.

    Substitute y=10y = 10 into the expression for xx.

    x=26−4(10)3x = \frac{26 - 4(10)}{3} x=26−403x = \frac{26 - 40}{3} x=−143x = \frac{-14}{3}

    When y=10y = 10, x=−143x = \boxed{-\frac{14}{3}}.

    c) In the diagram, PQST is a parallelogram, PR is a straight line. ∣TS∣=8cm|TS| = 8 \text{cm}, ∣SM∣=6cm|SM| = 6 \text{cm} and area of triangle PSR =36cm2= 36 \text{cm}^2. Find the value of ∣QR∣|QR|.

  4. Find the length of the base PR using the area of triangle PSR.

    The area of a triangle is given by 12×base×height\frac{1}{2} \times \text{base} \times \text{height}. Here, the base is PR and the height is SM.

    Area of △PSR=12×∣PR∣×∣SM∣\text{Area of } \triangle PSR = \frac{1}{2} \times |PR| \times |SM| 36 cm2=12×∣PR∣×6 cm36 \text{ cm}^2 = \frac{1}{2} \times |PR| \times 6 \text{ cm} 36=3×∣PR∣36 = 3 \times |PR| ∣PR∣=363|PR| = \frac{36}{3} ∣PR∣=12 cm|PR| = 12 \text{ cm}
  5. Determine the length of PQ.

    Since PQST is a parallelogram, opposite sides are equal in length. Therefore, ∣PQ∣=∣TS∣|PQ| = |TS|.

    ∣PQ∣=8 cm|PQ| = 8 \text{ cm}
  6. Calculate the value of $|QR|$.

    From the diagram, PR is a straight line, so ∣PR∣=∣PQ∣+∣QR∣|PR| = |PQ| + |QR|.

    12 cm=8 cm+∣QR∣12 \text{ cm} = 8 \text{ cm} + |QR| ∣QR∣=12−8|QR| = 12 - 8 ∣QR∣=4 cm|QR| = 4 \text{ cm}

    The value of ∣QR∣|QR| is 4 cm\boxed{4 \text{ cm}}.

    Question 2:

    a) The formula T=4b221(d−35b)T = \frac{4b^2}{21}(d - \frac{3}{5}b) gives the approximate mass, TT tonnes, of a ship dd metres long and bb metres wide.

    i) Make dd the subject of the formula.

  7. Multiply both sides by 21.

    21T=4b2(d−35b)21T = 4b^2 \left(d - \frac{3}{5}b\right)
  8. Divide both sides by 4b24b^2.

    21T4b2=d−35b\frac{21T}{4b^2} = d - \frac{3}{5}b
  9. Add 35b\frac{3}{5}b to both sides.

    d=21T4b2+35bd = \frac{21T}{4b^2} + \frac{3}{5}b

    The formula with dd as the subject is d=21T4b2+35b\boxed{d = \frac{21T}{4b^2} + \frac{3}{5}b}.

    ii) Calculate the length of a 4500 tonnes ship if it is 20m wide, giving your answer to the nearest 5m.

  10. Substitute the given values into the formula for dd.

    Given T=4500T = 4500 tonnes and b=20b = 20 m.

    d=21(4500)4(20)2+35(20)d = \frac{21(4500)}{4(20)^2} + \frac{3}{5}(20) d=945004(400)+605d = \frac{94500}{4(400)} + \frac{60}{5} d=945001600+12d = \frac{94500}{1600} + 12
  11. Perform the calculations.

    d=59.0625+12d = 59.0625 + 12 d=71.0625 md = 71.0625 \text{ m}
  12. Round the answer to the nearest 5m.

    71.0625 m is closer to 70 m than 75 m.
    The length of the ship is approximately 70 m\boxed{70 \text{ m}}.

    b) A wooden structure in form of a cone mounted on a hemispherical base. The vertical height of the cone is 24cm and the base radius 7cm. Calculate, correct to 3 Significant figures, the surface area of the structure. (π=227\pi = \frac{22}{7}).

  13. Calculate the slant height (ll) of the cone.

    The slant height can be found using the Pythagorean theorem: l=r2+h2l = \sqrt{r^2 + h^2}.

    l=72+242l = \sqrt{7^2 + 24^2} l=49+576l = \sqrt{49 + 576} l=625l = \sqrt{625} l=25 cml = 25 \text{ cm}
  14. Calculate the curved surface area of the cone (CSAconeCSA_{cone}).

    The formula for the curved surface area of a cone is πrl\pi r l.

    CSAcone=227×7 cm×25 cmCSA_{cone} = \frac{22}{7} \times 7 \text{ cm} \times 25 \text{ cm} CSAcone=22×25 cm2CSA_{cone} = 22 \times 25 \text{ cm}^2 CSAcone=550 cm2CSA_{cone} = 550 \text{ cm}^2
  15. Calculate the curved surface area of the hemisphere (CSAhemiCSA_{hemi}).

    The formula for the curved surface area of a hemisphere is 2πr22 \pi r^2.

    CSAhemi=2×227×(7 cm)2CSA_{hemi} = 2 \times \frac{22}{7} \times (7 \text{ cm})^2 CSAhemi=2×227×49 cm2CSA_{hemi} = 2 \times \frac{22}{7} \times 49 \text{ cm}^2 CSAhemi=2×22×7 cm2CSA_{hemi} = 2 \times 22 \times 7 \text{ cm}^2 CSAhemi=308 cm2CSA_{hemi} = 308 \text{ cm}^2
  16. Calculate the total surface area of the structure.

    The total surface area is the sum of the curved surface area of the cone and the curved surface area of the hemisphere (the base of the cone is covered by the hemisphere).

    Total Surface Area=CSAcone+CSAhemi\text{Total Surface Area} = CSA_{cone} + CSA_{hemi} Total Surface Area=550 cm2+308 cm2\text{Total Surface Area} = 550 \text{ cm}^2 + 308 \text{ cm}^2 Total Surface Area=858 cm2\text{Total Surface Area} = 858 \text{ cm}^2
  17. Express the answer correct to 3 significant figures.

    858 cm2858 \text{ cm}^2 is already expressed to 3 significant figures.
    The surface area of the structure is 858 cm2\boxed{858 \text{ cm}^2}.

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