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A pencil and 6 erasers cost Gh¢15.00. The pencil cost Gh¢6.00. What is the cost of one eraser? 2. Solve the equation 10^x = 0.000001.

Mathematics

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Answer

C

Step-by-step solution

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Here are the solutions to the problems:

1. A pencil and 6 erasers cost Gh¢15.00. The pencil cost Gh¢6.00. What is the cost of one eraser?

  1. Let PP be the cost of a pencil and EE be the cost of an eraser.

    We are given the equations:

    P+6E=15.00P + 6E = 15.00 P=6.00P = 6.00

  2. Substitute the cost of the pencil into the first equation.

    6.00+6E=15.006.00 + 6E = 15.00

  3. Subtract Gh¢6.00 from both sides to find the cost of 6 erasers.

    6E=15.00−6.006E = 15.00 - 6.00 6E=9.006E = 9.00

  4. Divide by 6 to find the cost of one eraser.

    E=9.006E = \frac{9.00}{6} E=1.50E = 1.50

    The cost of one eraser is Gh¢1.50.

    The correct option is C.

    Gh¢1.50\boxed{\text{Gh¢1.50}}

    2. Solve the equation 10x=0.00000110^x = 0.000001.

  5. Express 0.000001 as a power of 10.

    0.000001=11,000,000=1106=10−60.000001 = \frac{1}{1,000,000} = \frac{1}{10^6} = 10^{-6}

  6. Set the powers equal.

    10x=10−610^x = 10^{-6} x=−6x = -6

    Assumption: The given options are A. -3, B. -2, C. 2, D. 3. Since -6 is not an option, it is highly probable that there is a typo in the question and 0.000001 was intended to be 0.001. If 10x=0.00110^x = 0.001, then 10x=10−310^x = 10^{-3}, which means x=−3x = -3. We will proceed with this assumption to match an option.

    Step 1 (Revised with assumption): Assume the equation is 10x=0.00110^x = 0.001.
    Express 0.001 as a power of 10.

    0.001=11000=1103=10−30.001 = \frac{1}{1000} = \frac{1}{10^3} = 10^{-3}

    Step 2 (Revised): Set the powers equal.

    10x=10−310^x = 10^{-3} x=−3x = -3

    The correct option, based on the assumption of a typo, is A.

    −3\boxed{-3}

    3. If pp oranges are sold for qq pesewas each, find the total selling price in cedis.

  7. Calculate the total selling price in pesewas.

    Total selling price in pesewas = (number of oranges) ×\times (price per orange in pesewas)

    =p×q=pq pesewas= p \times q = pq \text{ pesewas}

  8. Convert the total selling price from pesewas to cedis.

    Since 1 Cedi = 100 pesewas, to convert pesewas to cedis, divide by 100.
    Total selling price in cedis = pq100\frac{pq}{100} cedis.

    The correct option is A.

    pq100\boxed{\frac{pq}{100}}

    4. Which of the basic properties of rational numbers is used in the following operation? (27×24)+(23×24)=50×24(27 \times 24) + (23 \times 24) = 50 \times 24.

    The operation shown is:

    (27×24)+(23×24)=(27+23)×24=50×24(27 \times 24) + (23 \times 24) = (27 + 23) \times 24 = 50 \times 24

    This demonstrates the Distributive Property of multiplication over addition, which states that a×c+b×c=(a+b)×ca \times c + b \times c = (a+b) \times c.

    The correct option is C.

    Distributive\boxed{\text{Distributive}}

    5. The set P={prime factors of 24}P = \{\text{prime factors of } 24\} and the set Q={prime factors of 35}Q = \{\text{prime factors of } 35\}. List the members of P∪QP \cup Q.

  9. Find the prime factors of 24.

    24=2×12=2×2×6=2×2×2×3=23×324 = 2 \times 12 = 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3

    So, the set of prime factors of 24 is P={2,3}P = \{2, 3\}.

  10. Find the prime factors of 35.

    35=5×735 = 5 \times 7

    So, the set of prime factors of 35 is Q={5,7}Q = \{5, 7\}.

  11. Find the union of sets PP and QQ, denoted as P∪QP \cup Q.

    The union includes all unique elements from both sets.

    P∪Q={2,3}∪{5,7}={2,3,5,7}P \cup Q = \{2, 3\} \cup \{5, 7\} = \{2, 3, 5, 7\}

    The correct option is C.

    {2,3,5,7}\boxed{\{2, 3, 5, 7\}}

    6. Find the median of the numbers 25, 57, 17, 33, 73, 81, 42.

  12. Arrange the numbers in ascending order.

    17,25,33,42,57,73,8117, 25, 33, 42, 57, 73, 81

  13. Count the number of values (nn).

    There are 7 numbers, so n=7n = 7.

  14. Since nn is an odd number, the median is the (n+12)\left(\frac{n+1}{2}\right)-th term.

    Median position = 7+12=82=4\frac{7+1}{2} = \frac{8}{2} = 4-th term.

  15. Identify the 4th term in the ordered list.

    The 4th term is 42.

    The correct option is B.

    42\boxed{42}

    7. Factorize x(x−2)−3xy+6yx(x-2) - 3xy + 6y.

  16. Expand the first term.

    x(x−2)−3xy+6y=x2−2x−3xy+6yx(x-2) - 3xy + 6y = x^2 - 2x - 3xy + 6y

  17. Group terms with common factors.

    (x2−2x)+(−3xy+6y)(x^2 - 2x) + (-3xy + 6y)

  18. Factor out the common factor from each group.

    x(x−2)−3y(x−2)x(x - 2) - 3y(x - 2)

  19. Factor out the common binomial factor (x−2)(x-2).

    (x−2)(x−3y)(x - 2)(x - 3y)

    The correct option is A.

    (x−2)(x−3y)\boxed{(x-2)(x-3y)}

    8. If 15−px−x2=015 - px - x^2 = 0, find pp, when x=3x = 3.

  20. Substitute x=3x=3 into the given equation.

    15−p(3)−(3)2=015 - p(3) - (3)^2 = 0

  21. Simplify the equation.

    15−3p−9=015 - 3p - 9 = 0

  22. Combine the constant terms.

    6−3p=06 - 3p = 0

  23. Solve for pp.

    6=3p6 = 3p p=63p = \frac{6}{3} p=2p = 2

    The correct option is C.

    2\boxed{2}
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