Students of a class are made to sit in rows of equal number of chairs. If number of students is increased by 2 in each row, then the number of rows decrease by 3. If number of students is increased by 4 in each row, then the number of rows decreases by 5. What is the number of students in the class?
120
Understanding the Student and Row Problem
This problem describes a scenario involving students sitting in rows, where the number of students per row and the number of rows change under different conditions, but the total number of students remains constant. We need to find the original total number of students in the class.
Let's define some variables to represent the initial situation:
- Let r be the original number of rows.
- Let c be the original number of chairs (students) in each row.
- The total number of students in the class is the product of the number of rows and the number of chairs per row: Total Students = r × c.
Setting Up Equations from the Conditions
The problem gives us two different scenarios describing how the number of rows and students per row change. We can translate these scenarios into algebraic equations.
Condition 1: Increased Students per Row by 2, Rows Decrease by 3
If the number of students in each row is increased by 2, the new number of students per row becomes \(c + 2\). The number of rows decreases by 3, making the new number of rows \(r - 3\). The total number of students remains the same.
So, the total students in this case is \((r - 3)(c + 2)\). This must equal the original total number of students, \(rc\).
\( (r - 3)(c + 2) = rc \)
Let's expand the left side of the equation:
\( rc + 2r - 3c - 6 = rc \)
Subtract \(rc\) from both sides:
\( 2r - 3c - 6 = 0 \)
Rearranging this, we get our first equation:
\( 2r - 3c = 6 \) (Equation 1)
Condition 2: Increased Students per Row by 4, Rows Decrease by 5
If the number of students in each row is increased by 4, the new number of students per row becomes \(c + 4\). The number of rows decreases by 5, making the new number of rows \(r - 5\). The total number of students is still the same.
So, the total students in this case is \((r - 5)(c + 4)\). This must also equal the original total number of students, \(rc\).
\( (r - 5)(c + 4) = rc \)
Let's expand the left side of this equation:
\( rc + 4r - 5c - 20 = rc \)
Subtract \(rc\) from both sides:
\( 4r - 5c - 20 = 0 \)
Rearranging this, we get our second equation:
\( 4r - 5c = 20 \) (Equation 2)
Solving the System of Linear Equations
Now we have a system of two linear equations with two variables (\(r\) and \(c\)):
- \( 2r - 3c = 6 \)
- \( 4r - 5c = 20 \)
We can solve this system using methods like substitution or elimination. Let's use the elimination method. We can multiply Equation 1 by 2 to make the coefficient of \(r\) the same as in Equation 2.
Multiply Equation 1 by 2:
\( 2 \times (2r - 3c) = 2 \times 6 \)
\( 4r - 6c = 12 \) (Equation 3)
Now we subtract Equation 3 from Equation 2:
\( (4r - 5c) - (4r - 6c) = 20 - 12 \)
\( 4r - 5c - 4r + 6c = 8 \)
\( c = 8 \)
So, the original number of chairs (students) per row was 8.
Now substitute the value of \(c = 8\) into either Equation 1 or Equation 2 to find \(r\). Let's use Equation 1:
\( 2r - 3c = 6 \)
\( 2r - 3(8) = 6 \)
\( 2r - 24 = 6 \)
\( 2r = 6 + 24 \)
\( 2r = 30 \)
\( r = \frac{30}{2} \)
\( r = 15 \)
So, the original number of rows was 15.
Calculating the Total Number of Students
The total number of students is the original number of rows multiplied by the original number of students per row:
\( \text{Total Students} = r \times c \)
\( \text{Total Students} = 15 \times 8 \)
\( \text{Total Students} = 120 \)
Therefore, the number of students in the class is 120.
Verification of the Solution
Let's check if our values \(r=15\) and \(c=8\) satisfy the original conditions:
- Original: 15 rows, 8 students/row. Total = \(15 \times 8 = 120\) students.
- Condition 1: Rows decrease by 3 (\(15-3=12\) rows), students per row increase by 2 (\(8+2=10\) students/row). Total = \(12 \times 10 = 120\) students. This matches.
- Condition 2: Rows decrease by 5 (\(15-5=10\) rows), students per row increase by 4 (\(8+4=12\) students/row). Total = \(10 \times 12 = 120\) students. This also matches.
The solution is consistent with the problem statements.
Revision Table: Student-Row Problem
| Description | Original State | Condition 1 State | Condition 2 State |
|---|---|---|---|
| Number of Rows | \(r\) (15) | \(r - 3\) (12) | \(r - 5\) (10) |
| Students per Row | \(c\) (8) | \(c + 2\) (10) | \(c + 4\) (12) |
| Total Students | \(r \times c\) (120) | \((r - 3)(c + 2)\) (120) | \((r - 5)(c + 4)\) (120) |
Additional Information: Solving Word Problems
Solving word problems like this involves translating the given information into mathematical equations. Here are some general steps:
- Read Carefully: Understand the scenario and what is being asked.
- Define Variables: Assign letters to represent the unknown quantities. Be specific about what each variable represents.
- Translate to Equations: Convert the verbal descriptions of relationships between quantities into algebraic equations. Look for keywords like "is," "equals," "sum," "difference," "product," "ratio," etc.
- Solve the Equations: Use appropriate algebraic techniques (like substitution, elimination, or graphing) to solve the system of equations for the unknown variables.
- Answer the Question: Make sure you answer what was asked in the problem. If you found variables, calculate the final value requested (e.g., total students).
- Verify the Solution: Check if your answer makes sense in the context of the original problem and satisfies all the given conditions.
This problem specifically used a system of two linear equations, which is a common technique for problems involving two related unknown quantities and two given conditions.
Similar Questions
The values of x and y from the equations x - y = 6 and x/3 + y/2 = 12 are:
What is the value of x, 2x/3 + y/ 2 = 4 and x/3 - y/2 = 1?
How many integral values of x and y satisfy the equation 5x + 9y = 7, where -500 < x < 500 and -500 < y < 500?
A person carries Rs. 500 and wants to buy apples and oranges out of it. If the cost of one apple is Rs. 5 and the cost of one orange is Rs. 7 then what is the number of ways in which a person can buy both apples and oranges using total amount?
How many pairs of natural numbers are there such that the difference of their squares is 35?
What is the maximum value of the expression \(\frac{1}{{{x^2}\; + \;5x\; + \;10}}?\)
A man who recently died left a sum of Rs. 3,90,000 to be divided among his wife, five sons and four daughters. He directed that each son should receive 3 times as much as each daughter receives and that each daughter should receive twice as much as their mother receives. What was the wife’s share?
At present the average of the ages of a father and a son is 25 years. After seven years, the son will be 17 years old. What will be the age of father after 10 years?
The age of a woman is two-digit integer. On reversing this integer, the new integer is the age of her husband who is elder to her. The difference between their ages is one eleventh of their sum. What is the difference between their ages?
A quadratic polynomial ax 2+ bx + c = 0 is such that when it is divided by x, (x - 1) and (x + 1), the remainders are 3, 6 and 4 respectively. What is the value of (a + b)?
Important Questions from Linear Equation in 2 Variable
If 2 x + 3 y = 17;
2 x+2 - 3 y+1 = 5
then the values of x and y are:
The solution of pair of linear equations \(\dfrac{1}{2}x+\dfrac{2}{3}y=-1,x-\dfrac{1}{3}y=3\) by the elimination method, is:
Kumar tried his skill at shooting at a fun fair. He has to hit the target and if he hits the target he gets 1 Rs. and if he misses he has to pay 50 paise. He attempted 25 shots and won 10 Rs. In how many did he hit the target?
The sum of two numbers is 66 and their difference is 22. What is the ratio of the two numbers?
Shyam spent half of his money and was left with as many as he had rupees before, but with half as many rupees as he had paise before. Which of the following is a possible amount of money he is left with?
