Word Problems With Fractions

Step-by-Step Solutions and Problem-Solving Guide

Word problems involving fractions are presented along with their detailed solutions for guidance.

Word Problems and Detailed Solutions

  1. Problem 1: A farmer wants to sell his corn harvest to 3 buyers. The first would buy \( \dfrac{2}{7} \) of it, the second would buy \( \dfrac{2}{5} \) of it, and the third would buy one-third of it. Can he satisfy all the buyers?

    2. Solution

    To determine if he can satisfy all buyers, we need to find out if the sum of these fractions is less than or equal to 1 (the total harvest).
    First, let's add the fractions by finding a common denominator. The common denominator for 7, 5, and 3 is 105.

    \[ \dfrac{2}{7} = \dfrac{2 \times 15}{7 \times 15} = \dfrac{30}{105} \] \[ \dfrac{2}{5} = \dfrac{2 \times 21}{5 \times 21} = \dfrac{42}{105} \] \[ \dfrac{1}{3} = \dfrac{1 \times 35}{3 \times 35} = \dfrac{35}{105} \]

    Now, add the fractions:

    \[ \dfrac{30}{105} + \dfrac{42}{105} + \dfrac{35}{105} = \dfrac{107}{105} \]

    Since \( \dfrac{107}{105} \) is greater than 1, the farmer cannot satisfy all the buyers.

  2. Problem 2: To make a rabbit pâté, half of the rabbit meat's weight is mixed with chicken liver, and a quarter of its weight is mixed with veal. Calculate the ratio of the rabbit meat's weight to the weight of the pâté.

    3. Solution

    Let the weight of the rabbit meat be \( x \).
    The total weight of the pâté is:

    \[ \text{Total weight} = x + \dfrac{1}{2}x + \dfrac{1}{4}x \]

    Add the terms:

    \[ \text{Total weight} = x\left(1 + \dfrac{1}{2} + \dfrac{1}{4}\right) = x\left(\dfrac{4}{4} + \dfrac{2}{4} + \dfrac{1}{4}\right) = x\left(\dfrac{7}{4}\right) \]

    The ratio of rabbit meat to the total weight of the pâté:

    \[ \text{Ratio} = \dfrac{x}{\dfrac{7}{4}x} = \dfrac{4}{7} \]

    So, the ratio of the rabbit meat's weight to the weight of the pâté is \( 4:7 \).

  3. Problem 3: Three people buy a barrel of juice together. The first takes \( \dfrac{2}{5} \), and the second takes \( \dfrac{3}{10} \). What will be the share of the third person?

    4. Solution

    First, let's find the combined share of the first two people:

    \[ \dfrac{2}{5} + \dfrac{3}{10} \]

    Convert \( \dfrac{2}{5} \) to have a denominator of 10:

    \[ \dfrac{2}{5} = \dfrac{4}{10} \]

    Now, add:

    \[ \dfrac{4}{10} + \dfrac{3}{10} = \dfrac{7}{10} \]

    The remaining share for the third person:

    \[ 1 - \dfrac{7}{10} = \dfrac{10}{10} - \dfrac{7}{10} = \dfrac{3}{10} \]

    So, the third person gets \( \dfrac{3}{10} \) of the barrel.

  4. Problem 4: Rapeseed contains about 48% of its weight in oil, but only \( \dfrac{2}{3} \) of this oil can be extracted by pressing. What fraction of the total weight of the rapeseed does the extracted oil represent?

    5. Solution

    The fraction of the total weight of the rapeseed represented by the extracted oil:

    \[ \text{Fraction} = 48\% \times \dfrac{2}{3} = \dfrac{48}{100} \times \dfrac{2}{3} = \dfrac{48 \times 2}{100 \times 3} = \dfrac{96}{300} = \dfrac{32}{100} = \dfrac{8}{25} \]

    So, the extracted oil represents \( \dfrac{32}{100} \) or \( 32\% \) (or \( \dfrac{8}{25} \)) of the total weight of the rapeseed.

  5. Problem 5: \( \dfrac{2}{9} \) of a post is painted white, and \( \dfrac{4}{7} \) of the remaining part is painted red. What fraction of the post is left unpainted?

    6. Solution

    First, find the remaining part after the white paint:

    \[ \text{Remaining after white paint} = 1 - \dfrac{2}{9} = \dfrac{9}{9} - \dfrac{2}{9} = \dfrac{7}{9} \]

    Now, find the part painted red:

    \[ \text{Red painted part} = \dfrac{4}{7} \times \dfrac{7}{9} = \dfrac{28}{63} = \dfrac{4}{9} \]

    The unpainted part:

    \[ \text{Unpainted part} = \dfrac{7}{9} - \dfrac{4}{9} = \dfrac{3}{9} = \dfrac{1}{3} \]

    So, \( \dfrac{1}{3} \) of the post is left unpainted.

  6. Problem 6: A farmer takes three days to plow his field. The first day, he plows \( \dfrac{1}{3} \) of it; the second day, he plows \( \dfrac{3}{8} \) of the remainder. What fraction of the field's area remains to be plowed?

    7. Solution

    First, calculate the remaining part after the first day:

    \[ \text{Remaining after the first day} = 1 - \dfrac{1}{3} = \dfrac{3}{3} - \dfrac{1}{3} = \dfrac{2}{3} \]

    Now, calculate the part plowed on the second day:

    \[ \text{Second day's plowed part} = \dfrac{3}{8} \times \dfrac{2}{3} = \dfrac{6}{24} = \dfrac{1}{4} \]

    Finally, calculate the remaining part of the field:

    \[ \text{Remaining part} = \dfrac{2}{3} - \dfrac{1}{4} \]

    Find a common denominator (12):

    \[ \dfrac{2}{3} = \dfrac{8}{12}, \quad \dfrac{1}{4} = \dfrac{3}{12} \]

    Subtract:

    \[ \text{Remaining part} = \dfrac{8}{12} - \dfrac{3}{12} = \dfrac{5}{12} \]

    So, \( \dfrac{5}{12} \) of the field is left to plow.

  7. Problem 7: A wood merchant first sells one-third of his merchandise. Then, he sells \( \dfrac{2}{5} \) of the remaining amount. What fraction of his merchandise does he have left?

    8. Solution

    First, calculate the remaining part after the first sale:

    \[ \text{Remaining after first sale} = 1 - \dfrac{1}{3} = \dfrac{2}{3} \]

    Now, calculate the part sold in the second sale:

    \[ \text{Second sale part} = \dfrac{2}{5} \times \dfrac{2}{3} = \dfrac{4}{15} \]

    Finally, calculate the remaining part of the merchandise:

    \[ \text{Remaining part} = \dfrac{2}{3} - \dfrac{4}{15} \]

    Find a common denominator (15):

    \[ \dfrac{2}{3} = \dfrac{10}{15} \]

    Subtract:

    \[ \text{Remaining part} = \dfrac{10}{15} - \dfrac{4}{15} = \dfrac{6}{15} = \dfrac{2}{5} \]

    So, \( \dfrac{2}{5} \) of the merchandise is left.

  8. Problem 8: A plane makes a 6.5-hour journey. What fraction of the journey has it completed after 1.5 hours of flight?

    9. Solution

    The fraction of the journey completed:

    \[ \text{Fraction} = \dfrac{1.5}{6.5} = \dfrac{15}{65} = \dfrac{3}{13} \]

    So, the plane has completed \( \dfrac{3}{13} \) of the journey.

  9. Problem 9: What fraction of the day has passed when it is 8 o'clock in the morning?

    10. Solution

    A day has 24 hours, and 8 hours have passed by 8 AM.
    The fraction of the day that has passed:

    \[ \text{Fraction} = \dfrac{8}{24} = \dfrac{1}{3} \]

    So, \( \dfrac{1}{3} \) of the day has passed by 8 AM.

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