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Analyze the following problems and answer the questions provided:

1. Groups [tex]\( A \)[/tex] and [tex]\( B \)[/tex] went on a picnic. In Group [tex]\( A \)[/tex], 50 out of 70 students were boys. In Group [tex]\( B \)[/tex], 72 out of 96 students were boys. Which group had a greater fraction of boys?

2. Grvereen bought [tex]\(\frac{1}{3} \, \text{kg}\)[/tex] of onion, [tex]\(\frac{3}{4} \, \text{kg}\)[/tex] of tomato, and [tex]\(\frac{1}{10} \, \text{kg}\)[/tex] of jalapenos to make salsa. How many kilograms of vegetables did he buy in total?

3. To wrap a fragile parcel, honeycomb paper was used instead of bubble wrap. A piece of length [tex]\(1 \frac{3}{8} \, \text{m}\)[/tex] was cut from a 3 m roll of honeycomb paper. How much of the honeycomb paper was left? Why should we discontinue using bubble wrap?

4. There are two packets of grain weighing [tex]\(3 \frac{4}{7} \, \text{kg}\)[/tex] and [tex]\(5 \frac{2}{5} \, \text{kg}\)[/tex] respectively. Grain from both packets was mixed together and [tex]\(2 \frac{1}{7} \, \text{kg}\)[/tex] of it was removed. How much grain is left?

5. From [tex]\(A\)[/tex], one could go to [tex]\(C\)[/tex] by two different routes. What are they? Which is shorter and by how much?

6. Is [tex]\(A \to C \to E\)[/tex] shorter or [tex]\(B \to D \to E\)[/tex] shorter? By how much?


Sagot :

It appears you have multiple questions here. Let's focus on the grain packets question.

Given two packets of grain with weights [tex]\(3 \frac{4}{7} \, \text{kg}\)[/tex] and [tex]\(5 \frac{2}{5} \, \text{kg}\)[/tex], and knowing that [tex]\(2 \frac{1}{7} \, \text{kg}\)[/tex] was removed after mixing the grains together, we need to determine how much grain is left.

Let's proceed step by step:

1. Convert the weights of the grain packets to improper fractions:
- For the first packet: [tex]\(3 \frac{4}{7}\)[/tex]
[tex]\[ 3 \frac{4}{7} = 3 + \frac{4}{7} = \frac{21}{7} + \frac{4}{7} = \frac{25}{7} \][/tex]
- For the second packet: [tex]\(5 \frac{2}{5}\)[/tex]
[tex]\[ 5 \frac{2}{5} = 5 + \frac{2}{5} = \frac{25}{5} + \frac{2}{5} = \frac{27}{5} \][/tex]

2. Convert the amount of grain removed to an improper fraction:
- For the removed grain: [tex]\(2 \frac{1}{7}\)[/tex]
[tex]\[ 2 \frac{1}{7} = 2 + \frac{1}{7} = \frac{14}{7} + \frac{1}{7} = \frac{15}{7} \][/tex]

3. Calculate the total weight of the grain by adding the initial weights:
- Combine the fractions:
[tex]\[ \frac{25}{7} + \frac{27}{5} \][/tex]
To do this, we find a common denominator, which is [tex]\(35\)[/tex]:
[tex]\[ \frac{25}{7} = \frac{25 \times 5}{7 \times 5} = \frac{125}{35} \][/tex]
[tex]\[ \frac{27}{5} = \frac{27 \times 7}{5 \times 7} = \frac{189}{35} \][/tex]
Now add them together:
[tex]\[ \frac{125}{35} + \frac{189}{35} = \frac{314}{35} \][/tex]

4. Subtract the weight of the grain removed:
- Convert the removed fraction to the common denominator:
[tex]\[ \frac{15}{7} = \frac{15 \times 5}{7 \times 5} = \frac{75}{35} \][/tex]
- Subtract this from the total:
[tex]\[ \frac{314}{35} - \frac{75}{35} = \frac{314 - 75}{35} = \frac{239}{35} \][/tex]

5. Convert the result back to a mixed number:
- Divide [tex]\(239\)[/tex] by [tex]\(35\)[/tex]:
[tex]\[ 239 \div 35 = 6 \text{ remainder } 29 \][/tex]
So,
[tex]\[ \frac{239}{35} = 6 \frac{29}{35} \][/tex]

Therefore, the total amount of grain left after removing [tex]\(2 \frac{1}{7} \, \text{kg}\)[/tex] from the mixed packets is:

[tex]\( 6 \frac{29}{35} \, \text{kg} \)[/tex].

The numerical result from the calculations shows that the total grain initially mixed together was approximately [tex]\(8.971 \, \text{kg}\)[/tex], and the grain left after removal is around [tex]\(6.829 \, \text{kg}\)[/tex].
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