Question

In a school there are 1800 students. Last day except \(4 \%\) of the boys all the students were present in the school. Today except \(5 \%\) of the girls all the students are present in the school, but in both the days no. of students present in the school, were same. The no. of girls in the school is : (a) 1200 (b) 800 (c) 1000 (d) 600

Short answer

Solution: Using the given information, we first set up equations for total students and present students on both days. After solving the equations, we find that there are 800 girls in the school.

Step-by-step solution

Write the equation for total students

B + G = 1800 On the first day, \(4 \%\) of the boys were absent, meaning \(96 \%\) of the boys were present. Therefore, the total number of students present on the first day can be represented as:

Write the equation for present students on day one

0.96B + G = Present\, students\, on\, day\, one On the second day, \(5 \%\) of the girls were absent, meaning \(95 \%\) of the girls were present. The number of students present on the second day is the same as that of the first day. So,

Write the equation for present students on day two

B + 0.95G = Present\, students\, on\, day\, two Now we have two equations with present students on both days being the same:

Set the two present students equations equal to each other

0.96B + G = B + 0.95G To solve for G, we can first isolate the number of boys in terms of the number of girls.

Solve for G

0.04B = 0.05G => G = \frac{4B}{5} Now substitute this expression for G in the total students equation.

Substitute G in the total students equation

B + \frac{4B}{5} = 1800 Now solve for B.

Solve for B

\frac{9B}{5} = 1800 => B = 1000 Now that we have the value for B, we can determine the value of G using the expression we found in step 5.

Find the value of G

G = \frac{4B}{5} = \frac{4(1000)}{5} = 800 So there are 800 girls in the school. The correct option is: (b) 800

Key concepts

Word Problems

Word problems are a common feature in algebra and involve scenarios where you translate a real-world situation into a mathematical problem. The goal is to comprehend the problem narrative and then apply algebraic concepts to find a solution.
In this exercise, you need to find the number of girls in a school given certain attendance conditions over two days. Understanding the problem's context is essential: the number of students present on both days is the same, despite different absence rates for boys and girls on each day. This involves reading the problem carefully and identifying key pieces of information, such as percentages, total students, and the condition that the total present remains constant on both days.
To effectively approach word problems:

• Read the problem thoroughly.
• Identify what you're solving for—in this case, the number of girls.
• Extract and organize given information into manageable steps.
• Formulate one or more equations based on the problem's conditions.

As you work through the problem, ensure each condition is represented in your equations. This clarifies what relationship or calculation is necessary to find the solution.

Equations

Equations are mathematical statements that express the equality between two expressions. They are fundamental tools in algebra for solving word problems. In this exercise, we use equations to express the conditions given by the problem.
Firstly, we establish the total number of students with the equation:

• \( B + G = 1800 \)

where \( B \) represents the number of boys and \( G \) represents the number of girls.
Next, we create equations based on the attendance conditions:

• On day one, with 4% of boys absent: \( 0.96B + G \)
• On day two, with 5% of girls absent: \( B + 0.95G \)

Both expressions equal the number of students present on both days. This allows us to set up another equation:

• \( 0.96B + G = B + 0.95G \)

These equations are the key to solving for either boys or girls. By manipulating them, you can rearrange and substitute values, leading ultimately to the solution. Understanding how to set up and solve these equations is crucial when dealing with algebraic word problems.

Percentage Calculations

Percentage calculations are integral in solving problems that involve parts of a whole expressed as fractions of 100. This exercise uses percentage calculations to describe boys' and girls' attendance.
On the first day, 4% of boys are absent, meaning 96% are present. This is expressed as:

• Presence of boys: \( 0.96B \)

On the second day, 5% of girls are absent, meaning 95% are present, represented by:

• Presence of girls: \( 0.95G \)

These calculations transform percentage figures into decimal multipliers that are used in equations. By understanding the conversion between percentages and decimals:

• 4% becomes 0.04 and 96% becomes 0.96.
• 5% becomes 0.05 and 95% becomes 0.95.

This helps in setting up accurate equations from the given conditions, allowing you to find unknown values by solving the equations. Understanding how to perform and interpret percentage calculations is a fundamental skill in solving real-world algebraic problems.