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### Problem 1: Baker's Flour
A baker bought [tex]\(5 \frac{1}{4}\)[/tex] kilos of flour. He used [tex]\(\frac{3}{8}\)[/tex] of a kilo for hotenkes and [tex]\(\frac{1}{4}\)[/tex] of a kilo for cookies. How much flour had he left?
#### Step-by-Step Solution:
1. Convert [tex]\(5 \frac{1}{4}\)[/tex] kilos to an improper fraction:
[tex]\[ 5 \frac{1}{4} = \frac{21}{4} \][/tex]
2. Add the amounts of flour used for hotenkes and cookies:
[tex]\[ \frac{3}{8} + \frac{1}{4} = \frac{3}{8} + \frac{2}{8} = \frac{5}{8} \][/tex]
3. Subtract the used flour from the total flour:
[tex]\[ \frac{21}{4} - \frac{5}{8} = \frac{42}{8} - \frac{5}{8} = \frac{37}{8} \][/tex]
Therefore, the amount of flour left is [tex]\(\frac{37}{8}\)[/tex] kilos.
### Problem 2: Mrs. Acosta's Chickens
Mrs. Acosta had 2 chickens for dinner. One weighed [tex]\(3 \frac{1}{4}\)[/tex] kilos and the other weighed [tex]\(2 \frac{1}{8}\)[/tex] kilos. Find the weight of the two chickens. Find the difference in their weight.
#### Step-by-Step Solution:
1. Convert [tex]\(3 \frac{1}{4}\)[/tex] kilos to an improper fraction:
[tex]\[ 3 \frac{1}{4} = \frac{13}{4} \][/tex]
2. Convert [tex]\(2 \frac{1}{8}\)[/tex] kilos to an improper fraction:
[tex]\[ 2 \frac{1}{8} = \frac{17}{8} \][/tex]
3. Find the total weight of the two chickens:
[tex]\[ \frac{13}{4} + \frac{17}{8} = \frac{26}{8} + \frac{17}{8} = \frac{43}{8} \][/tex]
4. Find the difference in their weight:
[tex]\[ \left|\frac{13}{4} - \frac{17}{8}\right| = \left|\frac{26}{8} - \frac{17}{8}\right| = \left|\frac{9}{8}\right| = \frac{9}{8} \][/tex]
Therefore, the total weight of the two chickens is [tex]\(\frac{43}{8}\)[/tex] kilos, and the difference in their weight is [tex]\(\frac{9}{8}\)[/tex] kilos.
### Problem 3: Watermelon Fractions
Mother brought home a big watermelon. She sliced the watermelon into 5 equivalent parts. She gave Mylene [tex]\(\frac{1}{6}\)[/tex] and Jerry [tex]\(\frac{1}{2}\)[/tex]. What fraction of the watermelon had Mother left?
#### Step-by-Step Solution:
1. The total watermelon represented as a whole is [tex]\(1\)[/tex].
2. Add the fractions given to Mylene and Jerry:
[tex]\[ \frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3} \][/tex]
3. Subtract the given fractions from the whole:
[tex]\[ 1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3} \][/tex]
Therefore, the fraction of the watermelon left is [tex]\(\frac{1}{3}\)[/tex].
### Problem 4: Dressmaker's Lace
A dressmaker had 12 meters of lace. After cutting [tex]\(1 \frac{1}{4}\)[/tex] meters, [tex]\(2 \frac{3}{8}\)[/tex] meters, and [tex]\(3 \frac{1}{3}\)[/tex] meters, how much lace did she have left?
#### Step-by-Step Solution:
1. Convert the cuts to improper fractions:
[tex]\[ 1 \frac{1}{4} = \frac{5}{4}, \quad 2 \frac{3}{8} = \frac{19}{8}, \quad 3 \frac{1}{3} = \frac{10}{3} \][/tex]
2. Find a common denominator and add the fractions:
[tex]\[ \frac{5}{4} + \frac{19}{8} + \frac{10}{3} = \frac{10}{8} + \frac{19}{8} + \frac{80}{24} = \frac{10}{8} + \frac{19}{8} + \frac{40}{12} = \frac{40}{24} + \frac{57}{24} = \frac{121}{24} \][/tex]
3. Subtract the total cut from the initial length of lace (converted to an improper fraction):
[tex]\[ 12 - \frac{121}{24} = \frac{288}{24} - \frac{121}{24} = \frac{167}{24} \][/tex]
Therefore, the length of the lace left is [tex]\(\frac{167}{24}\)[/tex] meters.
### Problem 5: Pizza Pieces
Some pizzas were sold at the school canteen. One pizza was cut into 8 pieces of equivalent sizes. Earl bought [tex]\(\frac{1}{8}\)[/tex] of it and his classmate bought [tex]\(\frac{3}{8}\)[/tex]. What part of the pizza was left?
#### Step-by-Step Solution:
1. The whole pizza is represented as [tex]\(1\)[/tex].
2. Add the fractions bought by Earl and his classmate:
[tex]\[ \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2} \][/tex]
3. Subtract the bought fraction from the whole:
[tex]\[ 1 - \frac{1}{2} = \frac{1}{2} \][/tex]
Therefore, the fraction of the pizza left is [tex]\(\frac{1}{2}\)[/tex].
A baker bought [tex]\(5 \frac{1}{4}\)[/tex] kilos of flour. He used [tex]\(\frac{3}{8}\)[/tex] of a kilo for hotenkes and [tex]\(\frac{1}{4}\)[/tex] of a kilo for cookies. How much flour had he left?
#### Step-by-Step Solution:
1. Convert [tex]\(5 \frac{1}{4}\)[/tex] kilos to an improper fraction:
[tex]\[ 5 \frac{1}{4} = \frac{21}{4} \][/tex]
2. Add the amounts of flour used for hotenkes and cookies:
[tex]\[ \frac{3}{8} + \frac{1}{4} = \frac{3}{8} + \frac{2}{8} = \frac{5}{8} \][/tex]
3. Subtract the used flour from the total flour:
[tex]\[ \frac{21}{4} - \frac{5}{8} = \frac{42}{8} - \frac{5}{8} = \frac{37}{8} \][/tex]
Therefore, the amount of flour left is [tex]\(\frac{37}{8}\)[/tex] kilos.
### Problem 2: Mrs. Acosta's Chickens
Mrs. Acosta had 2 chickens for dinner. One weighed [tex]\(3 \frac{1}{4}\)[/tex] kilos and the other weighed [tex]\(2 \frac{1}{8}\)[/tex] kilos. Find the weight of the two chickens. Find the difference in their weight.
#### Step-by-Step Solution:
1. Convert [tex]\(3 \frac{1}{4}\)[/tex] kilos to an improper fraction:
[tex]\[ 3 \frac{1}{4} = \frac{13}{4} \][/tex]
2. Convert [tex]\(2 \frac{1}{8}\)[/tex] kilos to an improper fraction:
[tex]\[ 2 \frac{1}{8} = \frac{17}{8} \][/tex]
3. Find the total weight of the two chickens:
[tex]\[ \frac{13}{4} + \frac{17}{8} = \frac{26}{8} + \frac{17}{8} = \frac{43}{8} \][/tex]
4. Find the difference in their weight:
[tex]\[ \left|\frac{13}{4} - \frac{17}{8}\right| = \left|\frac{26}{8} - \frac{17}{8}\right| = \left|\frac{9}{8}\right| = \frac{9}{8} \][/tex]
Therefore, the total weight of the two chickens is [tex]\(\frac{43}{8}\)[/tex] kilos, and the difference in their weight is [tex]\(\frac{9}{8}\)[/tex] kilos.
### Problem 3: Watermelon Fractions
Mother brought home a big watermelon. She sliced the watermelon into 5 equivalent parts. She gave Mylene [tex]\(\frac{1}{6}\)[/tex] and Jerry [tex]\(\frac{1}{2}\)[/tex]. What fraction of the watermelon had Mother left?
#### Step-by-Step Solution:
1. The total watermelon represented as a whole is [tex]\(1\)[/tex].
2. Add the fractions given to Mylene and Jerry:
[tex]\[ \frac{1}{6} + \frac{1}{2} = \frac{1}{6} + \frac{3}{6} = \frac{4}{6} = \frac{2}{3} \][/tex]
3. Subtract the given fractions from the whole:
[tex]\[ 1 - \frac{2}{3} = \frac{3}{3} - \frac{2}{3} = \frac{1}{3} \][/tex]
Therefore, the fraction of the watermelon left is [tex]\(\frac{1}{3}\)[/tex].
### Problem 4: Dressmaker's Lace
A dressmaker had 12 meters of lace. After cutting [tex]\(1 \frac{1}{4}\)[/tex] meters, [tex]\(2 \frac{3}{8}\)[/tex] meters, and [tex]\(3 \frac{1}{3}\)[/tex] meters, how much lace did she have left?
#### Step-by-Step Solution:
1. Convert the cuts to improper fractions:
[tex]\[ 1 \frac{1}{4} = \frac{5}{4}, \quad 2 \frac{3}{8} = \frac{19}{8}, \quad 3 \frac{1}{3} = \frac{10}{3} \][/tex]
2. Find a common denominator and add the fractions:
[tex]\[ \frac{5}{4} + \frac{19}{8} + \frac{10}{3} = \frac{10}{8} + \frac{19}{8} + \frac{80}{24} = \frac{10}{8} + \frac{19}{8} + \frac{40}{12} = \frac{40}{24} + \frac{57}{24} = \frac{121}{24} \][/tex]
3. Subtract the total cut from the initial length of lace (converted to an improper fraction):
[tex]\[ 12 - \frac{121}{24} = \frac{288}{24} - \frac{121}{24} = \frac{167}{24} \][/tex]
Therefore, the length of the lace left is [tex]\(\frac{167}{24}\)[/tex] meters.
### Problem 5: Pizza Pieces
Some pizzas were sold at the school canteen. One pizza was cut into 8 pieces of equivalent sizes. Earl bought [tex]\(\frac{1}{8}\)[/tex] of it and his classmate bought [tex]\(\frac{3}{8}\)[/tex]. What part of the pizza was left?
#### Step-by-Step Solution:
1. The whole pizza is represented as [tex]\(1\)[/tex].
2. Add the fractions bought by Earl and his classmate:
[tex]\[ \frac{1}{8} + \frac{3}{8} = \frac{4}{8} = \frac{1}{2} \][/tex]
3. Subtract the bought fraction from the whole:
[tex]\[ 1 - \frac{1}{2} = \frac{1}{2} \][/tex]
Therefore, the fraction of the pizza left is [tex]\(\frac{1}{2}\)[/tex].
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