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Add the numbers in base 5.

Chemistry

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b) Add: 123five+43five123_{\text{five}} + 43_{\text{five}}

  1. Add the numbers in base 5.

    When adding in base 5, if the sum of digits is 5 or more, we carry over groups of 5.

    123five+43five\begin{array}{r} 123_{\text{five}} \\ +\quad 43_{\text{five}} \\ \hline \end{array}
    • Rightmost column: 3+3=63 + 3 = 6. In base 5, 6=1×5+16 = 1 \times 5 + 1. Write down 1, carry over 1.
    • Middle column: 2+4+1(carry)=72 + 4 + 1 (\text{carry}) = 7. In base 5, 7=1×5+27 = 1 \times 5 + 2. Write down 2, carry over 1.
    • Leftmost column: 1+1(carry)=21 + 1 (\text{carry}) = 2. Write down 2.

    The sum is 221five221_{\text{five}}.

    • The final answer is 221five\boxed{221_{\text{five}}}.

    c) Given that 12y=7ten12_y = 7_{\text{ten}}. Find the value of base yy.

  2. Convert the number 12y12_y from base yy to base 10.

    12y=1×y1+2×y0=y+212_y = 1 \times y^1 + 2 \times y^0 = y + 2
  3. Set up the equation using the given information.

    y+2=7y + 2 = 7
  4. Solve for yy.

    y=7−2y = 7 - 2 y=5y = 5
    • The final answer is 5\boxed{5}.

    24. The sum of 3 consecutive counting numbers is 63. Find the numbers.

  5. Let the first counting number be nn.

    The three consecutive counting numbers are nn, n+1n+1, and n+2n+2.

  6. Write the equation for their sum.

    n+(n+1)+(n+2)=63n + (n+1) + (n+2) = 63 3n+3=633n + 3 = 63
  7. Solve for nn.

    3n=63−33n = 63 - 3 3n=603n = 60 n=603n = \frac{60}{3} n=20n = 20
  8. Find the three consecutive numbers.

    The numbers are nn, n+1n+1, n+2n+2.
    20, 20+120+1, 20+220+2.
    20, 21, 22.

    • The final answer is 20,21,22\boxed{20, 21, 22}.

    25. In a group of 50 farmers, 25\frac{2}{5} grow crops and the rest rear animals.

    a) Find the fraction of the farmers that rear animals.

  9. The total fraction of farmers is 1.

    Subtract the fraction that grow crops from 1 to find the fraction that rear animals.

    Fraction rearing animals=1−Fraction growing crops\text{Fraction rearing animals} = 1 - \text{Fraction growing crops} Fraction rearing animals=1−25\text{Fraction rearing animals} = 1 - \frac{2}{5} Fraction rearing animals=55−25\text{Fraction rearing animals} = \frac{5}{5} - \frac{2}{5} Fraction rearing animals=35\text{Fraction rearing animals} = \frac{3}{5}
    • The final answer is 35\boxed{\frac{3}{5}}.

    b) How many farmers grow crops?

  10. Multiply the total number of farmers by the fraction that grow crops.

    Number of farmers growing crops=25×50\text{Number of farmers growing crops} = \frac{2}{5} \times 50 Number of farmers growing crops=2×505\text{Number of farmers growing crops} = 2 \times \frac{50}{5} Number of farmers growing crops=2×10\text{Number of farmers growing crops} = 2 \times 10 Number of farmers growing crops=20\text{Number of farmers growing crops} = 20
    • The final answer is 20 farmers\boxed{20 \text{ farmers}}.

    c) How many more farmers rear animals than grow crops?

  11. Calculate the number of farmers that rear animals.

    From part a), the fraction rearing animals is 35\frac{3}{5}.

    Number of farmers rearing animals=35×50\text{Number of farmers rearing animals} = \frac{3}{5} \times 50 Number of farmers rearing animals=3×505\text{Number of farmers rearing animals} = 3 \times \frac{50}{5} Number of farmers rearing animals=3×10\text{Number of farmers rearing animals} = 3 \times 10 Number of farmers rearing animals=30\text{Number of farmers rearing animals} = 30
  12. Find the difference between the number of farmers rearing animals and the number of farmers growing crops.

    Difference=Farmers rearing animals−Farmers growing crops\text{Difference} = \text{Farmers rearing animals} - \text{Farmers growing crops} Difference=30−20\text{Difference} = 30 - 20 Difference=10\text{Difference} = 10
    • The final answer is 10 farmers\boxed{10 \text{ farmers}}.
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