Word Problems With Fractions
Word problems involvong fractions are presented along with their detailed solutions.
problems
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A farmer wants to sell his corn harvest to 3 buyers. The first would buy 2/7 of it, the second would buy 2/5 of it, and the third would buy one-third of it. Can he satisfy all the buyers?
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To make a rabbit pate, 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 pate.
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Three people buy a barrel of juics together. The first takes 2/5, and the second takes 3/10. What will be the share of the third person?
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Rapeseed contains about 48% of its weight in oil, but only 2/3 of this oil can be extracted by pressing. What FRACTION of the total weight of the rapeseed does the extracted oil represent?
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2/9 of a post is painted white, and 4/7 of the remaining part is painted red. What FRACTION of the post is left unpainted?
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A farmer takes three days to plow his field. The first day, he plows 1/3 of it; the second day, he plows 3/8 of the remainder. What FRACTION of the field's area remains to be plowed?
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A wood merchant first sells one-third of his merchandise. Then, he sells 2/5 of the remaining amount. What FRACTION of his merchandise does he have left?
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A plane makes a 6.5-hour journey. What FRACTION of the journey has it completed after 1.5 hours of flight?
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What FRACTION of the day has passed when it is 8 o'clock in the morning?
Solutions
Let's work through each problem step by step.
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Can the farmer satisfy all the buyers?
The farmer wants to sell his corn harvest to three buyers:
The first buyer would purchase \( \dfrac{2}{7} \) of the harvest.
The second buyer would purchase \( \dfrac{2}{5} \) of the harvest.
The third buyer would purchase \( \dfrac{1}{3} \) of the harvest.
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. To do so, we'll find a common denominator.
The common denominator for 7, 5, and 3 is 105.
Convert each fraction to have this common denominator:
\( \dfrac{2}{7} = \dfrac{2 \times 15}{7 \times 15} = \dfrac{30}{105} \)
\( \dfrac{2}{5} = \dfrac{2 \times 21}{5 \times 21} = \dfrac{42}{105} \)
\( \dfrac{1}{3} = \dfrac{1 \times 35}{3 \times 35} = \dfrac{35}{105} \)
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.
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To make a rabbit pate:
Half of the rabbit meat's weight is mixed with chicken liver.
A quarter of the rabbit meat's weight is mixed with veal.
Let the weight of the rabbit meat be \( x \).
The total weight of the pate:
\[ \text{Total weight} = x + \dfrac{1}{2}x + \dfrac{1}{4}x \]
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 pate:
\[ \text{Ratio} = \dfrac{x}{\dfrac{7}{4}x} = \dfrac{4}{7} \]
So, the ratio of the rabbit meat's weight to the weight of the pate is \( 4:7 \).
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Three people buy a barrel of juics together:
The first person takes \( \dfrac{2}{5} \).
The second person takes \( \dfrac{3}{10} \).
We need to find the share of the third person.
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.
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Fraction of total weight of rapeseed oil obtained after pressing
Rapeseed contains 48% oil by weight, but only \( \dfrac{2}{3} \) of this oil can be extracted.
The fraction of the total weight of the rapeseed represented by the extracted oil:
\[ \text{Fraction} = 48\% \times \dfrac{2}{3} = \dfrac{48}{100} \times \dfrac{2}{3} = \dfrac{48 \times 2}{100 \times 3} = \dfrac{96}{300} = \dfrac{32}{100} = 0.32 \]
So, the extracted oil represents \( \dfrac{32}{100} \) or \(32\% \) of the total weight of the rapeseed.
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\( \dfrac{2}{9} \) of the post is painted white.
\( \dfrac{4}{7} \) of the remaining part is painted red.
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.
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Fraction of the field left to plow
- The farmer plows \( \dfrac{1}{3} \) of the field on the first day.
- On the second day, he plows \( \dfrac{3}{8} \) of the remaining part.
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.
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Fraction of the merchant's merchandise left
- The merchant first sells \( \dfrac{1}{3} \) of his merchandise.
- Then he sells \( \dfrac{2}{5} \) of the remaining part.
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.
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Fraction of the journey completed
- The total journey time is 6.5 hours.
- The plane has flown for 1.5 hours.
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.
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Fraction of the day elapsed by 8 AM
- A day has 24 hours.
- 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.
More References and links
- Fractions Questions and Problems with Solutions .
- Fractions and Mixed Numbers- Grade 7 Maths Questions and Problems With Answers .
- Fractions and Mixed Numbers- Grade 6 Math Questions and Problems With Answers .
- Fractions - Grade 5 Maths Questions With Solutions .
- Fractions - Grade 4 Maths Questions With Answers .