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Solve the simultaneous equations: x + 2y = 7, 2x − 2y = 2. Consider the sets P = x: x is a prime number less than 15 and Q = x: x is an even number less than 15. List elements of P and Q, find P U Q,

Mathematics

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Question 1

a) Solve the simultaneous equations:

x+2y=7(1)x + 2y = 7 \quad \text{(1)} 2x−2y=2(2)2x - 2y = 2 \quad \text{(2)}
  1. Add Equation (1) and Equation (2) to eliminate yy.

    (x+2y)+(2x−2y)=7+2(x + 2y) + (2x - 2y) = 7 + 2 3x=93x = 9
  2. Solve for xx.

    x=93x = \frac{9}{3} x=3x = 3
  3. Substitute the value of xx into Equation (1) to find yy.

    3+2y=73 + 2y = 7 2y=7−32y = 7 - 3 2y=42y = 4
  4. Solve for yy.

    y=42y = \frac{4}{2} y=2y = 2

    The solution is x=3x=3 and y=2y=2.
    x=3,y=2\boxed{x=3, y=2}

    b) Consider the sets:
    P={x:x is a prime number less than 15}P = \{x: x \text{ is a prime number less than 15}\}
    Q={x:x is an even number less than 15}Q = \{x: x \text{ is an even number less than 15}\}

    i) List all the elements of sets P and Q.
    Prime numbers less than 15 are numbers greater than 1 that are only divisible by 1 and themselves.

    P={2,3,5,7,11,13}P = \{2, 3, 5, 7, 11, 13\}

    Even numbers less than 15 are integers divisible by 2.

    Q={2,4,6,8,10,12,14}Q = \{2, 4, 6, 8, 10, 12, 14\}

    ii) Find P∪QP \cup Q.
    The union of sets PP and QQ includes all elements that are in PP, or in QQ, or in both.

    P∪Q={2,3,4,5,6,7,8,10,11,12,13,14}P \cup Q = \{2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14\}

    iii) State whether PP and QQ are disjoint sets.
    Disjoint sets have no common elements (their intersection is an empty set).
    The intersection of PP and QQ is P∩Q={2}P \cap Q = \{2\}.
    Since P∩Q≠∅P \cap Q \neq \emptyset, the sets are not disjoint.
    No, P and Q are not disjoint sets.\boxed{\text{No, P and Q are not disjoint sets.}}

    Question 2

    a) Solve 2(x+4)≥5x−12(x + 4) \ge 5x - 1 and graph the solution on a number line.

  5. Distribute the 2 on the left side of the inequality.

    2x+8≥5x−12x + 8 \ge 5x - 1
  6. Subtract $2x$ from both sides.

    8≥5x−2x−18 \ge 5x - 2x - 1 8≥3x−18 \ge 3x - 1
  7. Add 1 to both sides.

    8+1≥3x8 + 1 \ge 3x 9≥3x9 \ge 3x
  8. Divide by 3.

    93≥x\frac{9}{3} \ge x 3≥x3 \ge x

    This can also be written as x≤3x \le 3.
    x≤3\boxed{x \le 3}

    Graph on a number line:
    A closed circle at 3 with an arrow extending to the left, indicating all numbers less than or equal to 3.

    <-----|-----|-----|-----|-----|-----|-----|----->
         -2    -1     0     1     2     3     4
                            [-----------
    

    b) Change 11101211101_2 to base ten numeral.

  9. Write out the place values for each digit in base 2.

    111012=(1×24)+(1×23)+(1×22)+(0×21)+(1×20)11101_2 = (1 \times 2^4) + (1 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0)
  10. Calculate the powers of 2.

    24=162^4 = 16 23=82^3 = 8 22=42^2 = 4 21=22^1 = 2 20=12^0 = 1
  11. Substitute the powers of 2 and multiply.

    111012=(1×16)+(1×8)+(1×4)+(0×2)+(1×1)11101_2 = (1 \times 16) + (1 \times 8) + (1 \times 4) + (0 \times 2) + (1 \times 1) 111012=16+8+4+0+111101_2 = 16 + 8 + 4 + 0 + 1
  12. Sum the results.

    111012=2911101_2 = 29

    2910\boxed{29_{10}}

    Question 3

    a) Three machines produce packages at regular intervals: Machine A every 12 minutes, Machine B every 18 minutes, and Machine C every 24 minutes. Find the first time (in minutes) when all three machines will produce a package at the same time.

    To find when all three machines will produce a package at the same time, we need to find the Least Common Multiple (LCM) of their intervals: 12, 18, and 24 minutes.

  13. Find the prime factorization of each number.

    12=2×2×3=22×3112 = 2 \times 2 \times 3 = 2^2 \times 3^1 18=2×3×3=21×3218 = 2 \times 3 \times 3 = 2^1 \times 3^2 24=2×2×2×3=23×3124 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1
  14. To find the LCM, take the highest power of each prime factor present in any of the factorizations.

    The highest power of 2 is 232^3.
    The highest power of 3 is 323^2.

  15. Multiply these highest powers together.

    LCM(12,18,24)=23×32=8×9=72\text{LCM}(12, 18, 24) = 2^3 \times 3^2 = 8 \times 9 = 72

    All three machines will produce a package at the same time after 72 minutes.
    72 minutes\boxed{72 \text{ minutes}}

    b) If an item is sold at a 20% profit for GH¢60, find its original cost price.

    Let the original cost price be CC.
    The profit is 20% of the cost price, which is 0.20C0.20C.
    The selling price (SP) is the cost price plus the profit.

    SP=C+0.20CSP = C + 0.20C SP=1.20CSP = 1.20C

    We are given that the selling price is GH¢60.

    60=1.20C60 = 1.20C
  16. Solve for CC.

    C=601.20C = \frac{60}{1.20} C=60120100C = \frac{60}{\frac{120}{100}} C=60×100120C = \frac{60 \times 100}{120} C=6000120C = \frac{6000}{120} C=50C = 50

    The original cost price is GH¢50.
    GH¢50\boxed{\text{GH¢}50}

    c) Arrange the numbers −20,15,−2,0,56,4-20, 15, -2, 0, 56, 4 in ascending order.

    Ascending order means arranging numbers from the smallest to the largest.

  17. Identify the negative numbers and arrange them

    −20,−2-20, -2.

  18. Identify zero

  19. Identify the positive numbers and arrange them

    4,15,564, 15, 56.

    Combining these in order:

    −20,−2,0,4,15,56-20, -2, 0, 4, 15, 56

    −20,−2,0,4,15,56\boxed{-20, -2, 0, 4, 15, 56}

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