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Write in figures: One and a half million. 2) Express % as a decimal. 3) Expand 72 x 53. 4) Which of these years was a leap year: 2020, 2021, 2022?

Mathematics

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1) Numbers can either be written in figures, words or Roman numerals. Write in figures: One and a half million

  1. Understand "one million".

    One million is 1,000,000.

  2. Understand "half million".

    Half a million is 12×1,000,000=500,000\frac{1}{2} \times 1,000,000 = 500,000.

  3. Combine "one million" and "half million".

    1,000,000+500,000=1,500,0001,000,000 + 500,000 = 1,500,000.
    In figures, one and a half million is 1,500,000\boxed{1,500,000}.

    2) Fractions can be expressed as decimals. Express % as a decimal
    This question asks how to express a percentage as a decimal in general.

  4. Understand that "percent" means "per hundred" or "out of 100".

  5. To convert a percentage to a decimal, divide the percentage value by 100.

    For example, if you have X%X\%, it is equivalent to X100\frac{X}{100} as a decimal.
    So, to express a percentage as a decimal, you divide the number by 100. For example, 25%=25100=0.2525\% = \frac{25}{100} = \boxed{0.25}.

    3) Numbers written in index form can be expanded. Expand 72×537^2 \times 5^3

  6. Expand 727^2.

    72=7×7=497^2 = 7 \times 7 = 49.

  7. Expand 535^3.

    53=5×5×5=25×5=1255^3 = 5 \times 5 \times 5 = 25 \times 5 = 125.

  8. Multiply the expanded values.

    49×125=612549 \times 125 = 6125.
    The expanded form is 49×125=6125\boxed{49 \times 125 = 6125}.

    4) Some years are called leap years. Which of these years was a leap year: 2020, 2021, 2022?

  9. Recall the rules for a leap year.

    A year is a leap year if it is divisible by 4, unless it is a century year (like 1900) which must be divisible by 400.

  10. Check each given year.

    • For 2020: 2020÷4=5052020 \div 4 = 505. Since 2020 is divisible by 4, it is a leap year.
    • For 2021: 2021÷4=5052021 \div 4 = 505 with a remainder of 1. So, 2021 is not a leap year.
    • For 2022: 2022÷4=5052022 \div 4 = 505 with a remainder of 2. So, 2022 is not a leap year. The leap year among the given years is 2020\boxed{2020}.

    5) A mixed fraction can be converted to an improper fraction. Convert 141214 \frac{1}{2} to an improper fraction.

  11. Multiply the whole number by the denominator.

    14×2=2814 \times 2 = 28.

  12. Add the numerator to the result from Step 1.

    28+1=2928 + 1 = 29.

  13. Place this sum over the original denominator.

    The improper fraction is 292\boxed{\frac{29}{2}}.

    6) Write 16.53675 correct to 4 significant figures.

  14. Identify the first four significant figures.

    These are 1, 6, 5, 3.

  15. Look at the fifth significant figure, which is 6.

  16. Since the fifth significant figure (6) is 5 or greater, round up the fourth significant figure (3).

    The number becomes 16.54.
    16.53675 correct to 4 significant figures is 16.54\boxed{16.54}.

    7) Given that x=5x = 5 and y=4y = 4, calculate (x+y)2+(2x−y)2(x+y)^2 + (2x-y)^2

  17. Substitute the values of xx and yy into the expression.

    (5+4)2+(2×5−4)2(5+4)^2 + (2 \times 5 - 4)^2
  18. Calculate the values inside the parentheses.

    (9)2+(10−4)2(9)^2 + (10 - 4)^2 (9)2+(6)2(9)^2 + (6)^2
  19. Square the numbers.

    81+3681 + 36
  20. Add the results.

    81+36=11781 + 36 = 117

    The value of the expression is 117\boxed{117}.

    8) The sum of two numbers is 502. If one of the numbers is 128, what is the value of the other number?

  21. Let the two numbers be aa and bb.

    Given a+b=502a + b = 502.

  22. Let one number be a=128a = 128.

    Substitute this into the equation.
    128+b=502128 + b = 502.

  23. Solve for bb.

    b=502−128b = 502 - 128.
    b=374b = 374.
    The value of the other number is 374\boxed{374}.

    9) Modular arithmetic deals with the remainder. What is 3+7(mod4)3 + 7 \pmod{4}?

  24. Perform the addition.

    3+7=103 + 7 = 10.

  25. Find the remainder when 10 is divided by 4.

    10÷4=210 \div 4 = 2 with a remainder of 10−(4×2)=10−8=210 - (4 \times 2) = 10 - 8 = 2.
    So, 10≡2(mod4)10 \equiv 2 \pmod{4}.
    3+7(mod4)3 + 7 \pmod{4} is 2\boxed{2}.

    10) A clock loses 2 minutes in every 3 hours. How many minutes does it lose in 15 hours?

  26. Determine the rate of loss per hour.

    Loss rate = 2 minutes3 hours\frac{2 \text{ minutes}}{3 \text{ hours}}.

  27. Calculate the total loss for 15 hours.

    Total loss = Loss rate ×\times Total hours

    Total loss=2 minutes3 hours×15 hours\text{Total loss} = \frac{2 \text{ minutes}}{3 \text{ hours}} \times 15 \text{ hours} Total loss=2×153 minutes\text{Total loss} = \frac{2 \times 15}{3} \text{ minutes} Total loss=303 minutes\text{Total loss} = \frac{30}{3} \text{ minutes} Total loss=10 minutes\text{Total loss} = 10 \text{ minutes}

    The clock loses 10 minutes\boxed{10 \text{ minutes}} in 15 hours.

    11) After spending 14\frac{1}{4} of my money on bread and 13\frac{1}{3} on eggs, what fraction have I left?

  28. Find the total fraction of money spent.

    Total spent = 14+13\frac{1}{4} + \frac{1}{3}.

  29. Find a common denominator for the fractions, which is 12.

    14=1×34×3=312\frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12} 13=1×43×4=412\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12}
  30. Add the fractions.

    Total spent=312+412=3+412=712\text{Total spent} = \frac{3}{12} + \frac{4}{12} = \frac{3+4}{12} = \frac{7}{12}
  31. The total money is represented by 1 (or 1212\frac{12}{12}).

    Subtract the spent fraction from 1 to find the fraction left.

    Fraction left=1−712=1212−712=12−712=512\text{Fraction left} = 1 - \frac{7}{12} = \frac{12}{12} - \frac{7}{12} = \frac{12-7}{12} = \frac{5}{12}

    The fraction of money left is 512\boxed{\frac{5}{12}}.

    12) If ξ={1,2,3,4,5,6,...,10}\xi = \{1, 2, 3, 4, 5, 6, ..., 10\}, A={1,3,6,9,10}A = \{1, 3, 6, 9, 10\} and B={2,4,6,8,10}B = \{2, 4, 6, 8, 10\}, Find (A∩B)′(A \cap B)'

  32. Identify the universal set ξ\xi.

    ξ={1,2,3,4,5,6,7,8,9,10}\xi = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}.

  33. Find the intersection of Set A and Set B (A∩BA \cap B).

    This includes elements common to both A and B.
    A={1,3,6,9,10}A = \{1, 3, 6, 9, 10\}
    B={2,4,6,8,10}B = \{2, 4, 6, 8, 10\}
    The common elements are 6 and 10.
    A∩B={6,10}A \cap B = \{6, 10\}.

  34. Find the complement of (A∩B)(A \cap B), denoted as (A∩B)′(A \cap B)'.

    This includes all elements in the universal set ξ\xi that are not in (A∩B)(A \cap B).
    Elements in ξ\xi: {1,2,3,4,5,6,7,8,9,10}\{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}
    Elements in A∩BA \cap B: {6,10}\{6, 10\}
    Remove {6,10}\{6, 10\} from ξ\xi:
    (A∩B)′={1,2,3,4,5,7,8,9}(A \cap B)' = \{1, 2, 3, 4, 5, 7, 8, 9\}.
    (A∩B)′={1,2,3,4,5,7,8,9}(A \cap B)' = \boxed{\{1, 2, 3, 4, 5, 7, 8, 9\}}.

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