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Answer :
Let's solve each problem step by step:
Question 9:
In a school, 58% of the total number of students is boys. The number of girls is 840. We need to find the total number of students in the school.
1. If 58% are boys, then 100% - 58% = 42% are girls.
2. We know the number of girls is 840, which corresponds to 42% of the total students.
3. To find the total number of students, use the equation:
[tex]\[
\text{Total students} = \frac{840}{0.42} = 2000
\][/tex]
Question 10:
Birr 300 is invested at a 6% simple interest per annum. We need to find out how long it will take for the interest to be Birr 180.
1. Use the simple interest formula:
[tex]\[
\text{Interest} = \text{Principal} \times \text{Rate} \times \text{Time}
\][/tex]
Here, the Interest is 180, Principal is 300, and Rate is 6%.
2. Plug in the values:
[tex]\[
180 = 300 \times 0.06 \times \text{Time}
\][/tex]
3. Solving for Time:
[tex]\[
\text{Time} = \frac{180}{300 \times 0.06} = 10 \text{ years}
\][/tex]
Question 11:
The decimal form of [tex]\(\frac{3}{4} \%\)[/tex] is required.
1. Convert [tex]\(\frac{3}{4} \%\)[/tex] to a fraction of a whole number:
[tex]\[
\frac{3}{4} \% = \frac{3}{4} \times \frac{1}{100} = \frac{3}{400}
\][/tex]
2. Convert to decimal:
[tex]\[
\frac{3}{400} = 0.0075
\][/tex]
Question 12:
The simple interest on Birr 400 invested for 4 months was Birr 12. We need to find the annual rate of interest.
1. Convert months into years: [tex]\( \frac{4}{12} = \frac{1}{3} \text{ of a year}\)[/tex].
2. Use the simple interest formula:
[tex]\[
12 = 400 \times \text{Rate} \times \frac{1}{3}
\][/tex]
3. Solving for Rate:
[tex]\[
\text{Rate} = \frac{12 \times 3}{400} \times 100 = 9\%
\][/tex]
Question 13:
Given that [tex]\(a : b : c = 1 : 3 : 4\)[/tex] and [tex]\(c = 28\)[/tex], find the sum of [tex]\(a + b + c\)[/tex].
1. Assume the common ratio unit is [tex]\(x\)[/tex], then:
- [tex]\(a = 1x\)[/tex]
- [tex]\(b = 3x\)[/tex]
- [tex]\(c = 4x = 28\)[/tex]
2. Solving for [tex]\(x\)[/tex]:
[tex]\[
4x = 28 \implies x = 7
\][/tex]
3. Calculate the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
- [tex]\(a = 1 \times 7 = 7\)[/tex]
- [tex]\(b = 3 \times 7 = 21\)[/tex]
- [tex]\(c = 28\)[/tex]
4. The sum [tex]\(a + b + c = 7 + 21 + 28 = 56\)[/tex]
Thus, the answers are:
- Total students in the school: 2000
- Time for interest to accumulate: 10 years
- Decimal form of [tex]\(\frac{3}{4}\%\)[/tex]: 0.0075
- Annual rate of interest: 9%
- Sum of [tex]\(a + b + c\)[/tex]: 56
Question 9:
In a school, 58% of the total number of students is boys. The number of girls is 840. We need to find the total number of students in the school.
1. If 58% are boys, then 100% - 58% = 42% are girls.
2. We know the number of girls is 840, which corresponds to 42% of the total students.
3. To find the total number of students, use the equation:
[tex]\[
\text{Total students} = \frac{840}{0.42} = 2000
\][/tex]
Question 10:
Birr 300 is invested at a 6% simple interest per annum. We need to find out how long it will take for the interest to be Birr 180.
1. Use the simple interest formula:
[tex]\[
\text{Interest} = \text{Principal} \times \text{Rate} \times \text{Time}
\][/tex]
Here, the Interest is 180, Principal is 300, and Rate is 6%.
2. Plug in the values:
[tex]\[
180 = 300 \times 0.06 \times \text{Time}
\][/tex]
3. Solving for Time:
[tex]\[
\text{Time} = \frac{180}{300 \times 0.06} = 10 \text{ years}
\][/tex]
Question 11:
The decimal form of [tex]\(\frac{3}{4} \%\)[/tex] is required.
1. Convert [tex]\(\frac{3}{4} \%\)[/tex] to a fraction of a whole number:
[tex]\[
\frac{3}{4} \% = \frac{3}{4} \times \frac{1}{100} = \frac{3}{400}
\][/tex]
2. Convert to decimal:
[tex]\[
\frac{3}{400} = 0.0075
\][/tex]
Question 12:
The simple interest on Birr 400 invested for 4 months was Birr 12. We need to find the annual rate of interest.
1. Convert months into years: [tex]\( \frac{4}{12} = \frac{1}{3} \text{ of a year}\)[/tex].
2. Use the simple interest formula:
[tex]\[
12 = 400 \times \text{Rate} \times \frac{1}{3}
\][/tex]
3. Solving for Rate:
[tex]\[
\text{Rate} = \frac{12 \times 3}{400} \times 100 = 9\%
\][/tex]
Question 13:
Given that [tex]\(a : b : c = 1 : 3 : 4\)[/tex] and [tex]\(c = 28\)[/tex], find the sum of [tex]\(a + b + c\)[/tex].
1. Assume the common ratio unit is [tex]\(x\)[/tex], then:
- [tex]\(a = 1x\)[/tex]
- [tex]\(b = 3x\)[/tex]
- [tex]\(c = 4x = 28\)[/tex]
2. Solving for [tex]\(x\)[/tex]:
[tex]\[
4x = 28 \implies x = 7
\][/tex]
3. Calculate the values of [tex]\(a\)[/tex], [tex]\(b\)[/tex], and [tex]\(c\)[/tex]:
- [tex]\(a = 1 \times 7 = 7\)[/tex]
- [tex]\(b = 3 \times 7 = 21\)[/tex]
- [tex]\(c = 28\)[/tex]
4. The sum [tex]\(a + b + c = 7 + 21 + 28 = 56\)[/tex]
Thus, the answers are:
- Total students in the school: 2000
- Time for interest to accumulate: 10 years
- Decimal form of [tex]\(\frac{3}{4}\%\)[/tex]: 0.0075
- Annual rate of interest: 9%
- Sum of [tex]\(a + b + c\)[/tex]: 56
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