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Tree A grew by 15 rac12 centimetres in 2012. Tree B grew 17 rac34 centimetres more than Tree A in the same year. What was the increase in the height of Tree B?

Mathematics

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4. Tree A grew by 151215\frac{1}{2} centimetres in 2012. Tree B grew 173417\frac{3}{4} centimetres more than Tree A in the same year. What was the increase in the height of Tree B?

  1. Identify the growth of Tree A and the additional growth of Tree B.

    Tree A's growth: 151215\frac{1}{2} cm
    Tree B's additional growth: 173417\frac{3}{4} cm

  2. Add Tree A's growth to the additional growth to find the total increase in height for Tree B.

    1512+173415\frac{1}{2} + 17\frac{3}{4}
  3. Add the whole numbers.

    15+17=3215 + 17 = 32
  4. Find a common denominator for the fractions 12\frac{1}{2} and 34\frac{3}{4}.

    The least common denominator is 4.

    12=1×22×2=24\frac{1}{2} = \frac{1 \times 2}{2 \times 2} = \frac{2}{4}
  5. Add the fractions.

    24+34=2+34=54\frac{2}{4} + \frac{3}{4} = \frac{2+3}{4} = \frac{5}{4}
  6. Convert the improper fraction to a mixed number.

    54=114\frac{5}{4} = 1\frac{1}{4}
  7. Add this to the sum of the whole numbers from Step 3.

    32+114=331432 + 1\frac{1}{4} = 33\frac{1}{4}

    The increase in the height of Tree B was \boxed{\text{33\frac{1}{4} cm}}.

    5. At a 121212\frac{1}{2} kilometres running competition, Jamie ran for 37103\frac{7}{10} kilometres before he stopped for a rest. After the rest, he completed the running competition. How far did Jamie run after the rest?

  8. Identify the total distance and the distance run before the rest.

    Total distance: 121212\frac{1}{2} km
    Distance before rest: 37103\frac{7}{10} km

  9. Subtract the distance run before the rest from the total distance to find the distance run after the rest.

    1212−371012\frac{1}{2} - 3\frac{7}{10}
  10. Convert the mixed numbers to improper fractions.

    1212=(12×2)+12=25212\frac{1}{2} = \frac{(12 \times 2) + 1}{2} = \frac{25}{2} 3710=(3×10)+710=37103\frac{7}{10} = \frac{(3 \times 10) + 7}{10} = \frac{37}{10}
  11. Find a common denominator for the fractions 252\frac{25}{2} and 3710\frac{37}{10}.

    The least common denominator is 10.

    252=25×52×5=12510\frac{25}{2} = \frac{25 \times 5}{2 \times 5} = \frac{125}{10}
  12. Perform the subtraction.

    12510−3710=125−3710=8810\frac{125}{10} - \frac{37}{10} = \frac{125 - 37}{10} = \frac{88}{10}
  13. Convert the improper fraction to a mixed number and simplify.

    8810=8810=88÷210÷2=845\frac{88}{10} = 8\frac{8}{10} = 8\frac{8 \div 2}{10 \div 2} = 8\frac{4}{5}

    Jamie ran \boxed{\text{8\frac{4}{5} km}} after the rest.

    6. The total mass of two bags was 281428\frac{1}{4} kilograms. One of the bags was 1411014\frac{1}{10} kilograms. What was the mass of the other bag?

  14. Identify the total mass and the mass of one bag.

    Total mass: 281428\frac{1}{4} kg
    Mass of one bag: 1411014\frac{1}{10} kg

  15. Subtract the mass of one bag from the total mass to find the mass of the other bag.

    2814−1411028\frac{1}{4} - 14\frac{1}{10}
  16. Subtract the whole numbers.

    28−14=1428 - 14 = 14
  17. Find a common denominator for the fractions 14\frac{1}{4} and 110\frac{1}{10}.

    The least common multiple of 4 and 10 is 20.

    14=1×54×5=520\frac{1}{4} = \frac{1 \times 5}{4 \times 5} = \frac{5}{20} 110=1×210×2=220\frac{1}{10} = \frac{1 \times 2}{10 \times 2} = \frac{2}{20}
  18. Subtract the fractions.

    520−220=5−220=320\frac{5}{20} - \frac{2}{20} = \frac{5-2}{20} = \frac{3}{20}
  19. Combine the whole number from Step 3 and the fraction from Step 5.

    14+320=1432014 + \frac{3}{20} = 14\frac{3}{20}

    The mass of the other bag was \boxed{\text{14\frac{3}{20} kg}}.

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