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Calculate: 8 1/3 x 5/8

Mathematics

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Question 3

3.1 Calculate: 813×388\frac{1}{3} \times \frac{3}{8}

  1. Convert the mixed number to an improper fraction.

    813=(8×3)+13=24+13=2538\frac{1}{3} = \frac{(8 \times 3) + 1}{3} = \frac{24 + 1}{3} = \frac{25}{3}

  2. Multiply the fractions.

    253×38=25×33×8\frac{25}{3} \times \frac{3}{8} = \frac{25 \times 3}{3 \times 8}

  3. Simplify the expression by canceling out common factors.

    25×33×8=258\frac{25 \times \cancel{3}}{\cancel{3} \times 8} = \frac{25}{8}

  4. Convert the improper fraction to a mixed number (optional, but good practice).

    258=318\frac{25}{8} = 3\frac{1}{8} 258 or 318\boxed{\frac{25}{8} \text{ or } 3\frac{1}{8}}

    3.2 A school had 450 learners in January. By March, the number increased to 540 learners. In June, the numbers increased by 10% from March number. What is the number of learners in June?

  5. Identify the number of learners in March.

    Number of learners in March = 540

  6. Calculate the 10% increase from the March number.

    Increase=10% of 540=10100×540\text{Increase} = 10\% \text{ of } 540 = \frac{10}{100} \times 540 Increase=0.10×540=54\text{Increase} = 0.10 \times 540 = 54

  7. Add the increase to the March number to find the number of learners in June.

    Learners in June=Learners in March+Increase\text{Learners in June} = \text{Learners in March} + \text{Increase} Learners in June=540+54=594\text{Learners in June} = 540 + 54 = 594 594 learners\boxed{594 \text{ learners}}

    3.3 If 30% of a number is 45. What is 80% of that number?

  8. Find the original number.

    Let the number be xx.

    30% of x=4530\% \text{ of } x = 45 30100×x=45\frac{30}{100} \times x = 45 0.3x=450.3x = 45 x=450.3x = \frac{45}{0.3} x=150x = 150

  9. Calculate 80% of the original number.

    80% of 150=80100×15080\% \text{ of } 150 = \frac{80}{100} \times 150 80% of 150=0.8×15080\% \text{ of } 150 = 0.8 \times 150 80% of 150=12080\% \text{ of } 150 = 120 120\boxed{120}

    Question 4

    4.1 Calculate: 1,250×161,250 \times 16

  10. Perform the multiplication.

    1,250×16=20,0001,250 \times 16 = 20,000 20,000\boxed{20,000}

    4.2 Divide a ribbon of 9,864m9,864m into 4 equal pieces.

  11. Perform the division.

    9,864m4=2,466m\frac{9,864m}{4} = 2,466m 2,466m\boxed{2,466m}

    4.3 A hawker buys 3750g3750 g of apples at R12.40 per kg and 2.5kg2.5kg of banana at R15.60 per kg. How much does the hawker pay all together?

  12. Convert the mass of apples from grams to kilograms.

    3750g=37501000kg=3.75kg3750 g = \frac{3750}{1000} kg = 3.75 kg

  13. Calculate the cost of the apples.

    Cost of apples=3.75kg×R12.40/kg\text{Cost of apples} = 3.75 kg \times \text{R}12.40/\text{kg} Cost of apples=R46.50\text{Cost of apples} = \text{R}46.50

  14. Calculate the cost of the bananas.

    Cost of bananas=2.5kg×R15.60/kg\text{Cost of bananas} = 2.5 kg \times \text{R}15.60/\text{kg} Cost of bananas=R39.00\text{Cost of bananas} = \text{R}39.00

  15. Calculate the total amount the hawker pays.

    Total cost=Cost of apples+Cost of bananas\text{Total cost} = \text{Cost of apples} + \text{Cost of bananas} Total cost=R46.50+R39.00\text{Total cost} = \text{R}46.50 + \text{R}39.00 Total cost=R85.50\text{Total cost} = \text{R}85.50 R85.50\boxed{\text{R}85.50}

    Question 5

    5.1 Use the smallest base to write 81 in exponential form.

  16. Find the prime factors of 81.

    81=3×2781 = 3 \times 27 27=3×927 = 3 \times 9 9=3×39 = 3 \times 3

    So, 81=3×3×3×381 = 3 \times 3 \times 3 \times 3.

  17. Express 81 as a power of its smallest prime base.

    81=3481 = 3^4 34\boxed{3^4}

    5.2 Calculate: 273+32\sqrt[3]{27} + 3^2

  18. Calculate the cube root of 27.

    273=3(since 3×3×3=27)\sqrt[3]{27} = 3 \quad (\text{since } 3 \times 3 \times 3 = 27)

  19. Calculate the square of 3.

    32=3×3=93^2 = 3 \times 3 = 9

  20. Add the results.

    3+9=123 + 9 = 12 12\boxed{12}

    Question 6: INTEGERS

    6.1 Calculate: −4+(−6)−15+(−5+13)-4 + (-6) - 15 + (-5 + 13)

  21. Simplify the expression inside the parentheses first.

    −5+13=8-5 + 13 = 8

  22. Rewrite the expression with the simplified term.

    −4+(−6)−15+8-4 + (-6) - 15 + 8 −4−6−15+8-4 - 6 - 15 + 8

  23. Perform additions and subtractions from left to right.

    (−4−6)−15+8(-4 - 6) - 15 + 8 −10−15+8-10 - 15 + 8 (−10−15)+8(-10 - 15) + 8 −25+8-25 + 8 −17-17 −17\boxed{-17}

    6.2 Complete:

    6.2.1 −23+17=⋯+(−23)-23 + 17 = \dots + (-23)

  24. Calculate the left side of the equation.

    −23+17=−6-23 + 17 = -6

  25. Use the commutative property of addition (a+b=b+aa + b = b + a) to find the missing number.

    −6=⋯+(−23)-6 = \dots + (-23)

    The missing number is 17.

    −23+17=17+(−23)-23 + 17 = 17 + (-23) 17\boxed{17}

    6.2.2 (2+5)+⋯=⋯+(5+(−3))(2 + 5) + \dots = \dots + (5 + (-3))

  26. Simplify the known parts of the equation.

    (2+5)=7(2 + 5) = 7 5+(−3)=5−3=25 + (-3) = 5 - 3 = 2

  27. Rewrite the equation with the simplified parts.

    7+⋯=⋯+27 + \dots = \dots + 2

  28. Use the associative property of addition ((a+b)+c=a+(b+c)(a + b) + c = a + (b + c)).

    If we let the missing number on the left be cc and the missing number on the right be aa, then:

    (2+5)+(−3)=2+(5+(−3))(2 + 5) + (-3) = 2 + (5 + (-3)) 7+(−3)=2+27 + (-3) = 2 + 2 4=44 = 4

    So, the missing number on the left is −3-3, and the missing number on the right is 2.

    (2+5)+(−3)=2+(5+(−3))\boxed{(2 + 5) + \mathbf{(-3)} = \mathbf{2} + (5 + (-3))}
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