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Solve the following problems: 1. A car's value decreased by 2/7 of its original value. If the original value was $7700, what is its new value? 2. If the LCM of two numbers is 2079 and their GCF is 27,

Mathematics

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Answer

B

Step-by-step solution

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Here are the solutions to the problems.

1.

  1. Calculate the amount by which the car's value decreased.

    The value decreased by 27\frac{2}{7} of its original value. Decrease=27×$7700=2×$77007=2×$1100=$2200\text{Decrease} = \frac{2}{7} \times \$7700 = 2 \times \frac{\$7700}{7} = 2 \times \$1100 = \$2200

  2. Calculate the new value of the car.

    The new value is the original value minus the decrease. New value=$7700−$2200=$5500\text{New value} = \$7700 - \$2200 = \$5500 The correct option is B.
    The new value of the car is \boxed{\text{\5500}}$.

    2.

  3. Recall the relationship between LCM, GCF, and two numbers.

    For two numbers N1N_1 and N2N_2, their product is equal to the product of their LCM and GCF:

    N1×N2=LCM(N1,N2)×GCF(N1,N2)N_1 \times N_2 = \text{LCM}(N_1, N_2) \times \text{GCF}(N_1, N_2)

  4. Substitute the given values into the formula.

    Given N1=189N_1 = 189, LCM=2079\text{LCM} = 2079, and GCF=27\text{GCF} = 27.

    189×N2=2079×27189 \times N_2 = 2079 \times 27

  5. Solve for the other number, N2N_2.

    N2=2079×27189N_2 = \frac{2079 \times 27}{189}

    We can simplify by noting that 189=7×27189 = 7 \times 27.

    N2=2079×277×27=20797N_2 = \frac{2079 \times \cancel{27}}{7 \times \cancel{27}} = \frac{2079}{7} N2=297N_2 = 297

    The correct option is E.
    The other number is 297\boxed{297}.

    3.

  6. Evaluate the division inside the first parenthesis.

    23÷43=23×34=2×33×4=612=12\frac{2}{3} \div \frac{4}{3} = \frac{2}{3} \times \frac{3}{4} = \frac{2 \times 3}{3 \times 4} = \frac{6}{12} = \frac{1}{2}

  7. Evaluate the multiplication inside the second parenthesis.

    45×56=4×55×6=2030=23\frac{4}{5} \times \frac{5}{6} = \frac{4 \times 5}{5 \times 6} = \frac{20}{30} = \frac{2}{3}

  8. Substitute these results back into the original expression.

    12+12−23\frac{1}{2} + \frac{1}{2} - \frac{2}{3}

  9. Perform the addition and subtraction from left to right.

    (12+12)−23=1−23\left(\frac{1}{2} + \frac{1}{2}\right) - \frac{2}{3} = 1 - \frac{2}{3} 1−23=33−23=131 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3}

    The correct option is B.
    The evaluated expression is 13\boxed{\frac{1}{3}}.

    4.

  10. Define the original fraction using a variable.

    Let the numerator of the original fraction be xx.
    The denominator exceeds the numerator by 3, so the denominator is x+3x+3.
    The original fraction is xx+3\frac{x}{x+3}.

  11. Set up an equation based on the given condition.

    If 1 is added to the numerator, the new numerator is x+1x+1. The denominator remains unchanged.
    The new fraction is x+1x+3\frac{x+1}{x+3}.
    This new fraction is equal to 34\frac{3}{4}.

    x+1x+3=34\frac{x+1}{x+3} = \frac{3}{4}

  12. Solve the equation for xx.

    Cross-multiply:

    4(x+1)=3(x+3)4(x+1) = 3(x+3) 4x+4=3x+94x + 4 = 3x + 9

    Subtract 3x3x from both sides:

    x+4=9x + 4 = 9

    Subtract 4 from both sides:

    x=5x = 5

  13. Find the original fraction.

    The numerator is x=5x=5.
    The denominator is x+3=5+3=8x+3 = 5+3 = 8.
    The original fraction is 58\frac{5}{8}.
    The correct option is A (or C, as they are identical).
    The original fraction is 58\boxed{\frac{5}{8}}.

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