Problem 8
Question
Use the five-step strategy for solving word problems to find the number or numbers described in Exercises. \(70 \%\) of what number is \(252 ?\)
Step-by-Step Solution
Verified Answer
The number is 360.
1Step 1: Understanding the problem
The problem is trying to find a number that is such that 70% of this number equals 252. In order to solve this, we must set up an equation to represent this problem.
2Step 2: Setting up the equation
Let's let the unknown number be 'x'. We translate the problem into a mathematical equation by writing '70% of x equals 252'. In mathematical terms, this is \(0.70 * x = 252\). This equation represents our problem.
3Step 3: Solving the equation
To find out what x is, we should divide both sides of the equation by 0.70. This gives us \(x = 252 / 0.70\).
4Step 4: Calculate the value of x
The solution to the equation \(x = 252 / 0.70\) gives \(x = 360\). Therefore, the number, such that 70% of it equals to 252, is 360.
Key Concepts
Percentages in AlgebraSetting Up EquationsAlgebraic Problem-Solving Techniques
Percentages in Algebra
Working with percentages in algebra involves converting the given percentage to a decimal to facilitate easier calculations. For instance, the question, 'what number is 70% of 252?' requires understanding that the term 'percent' means 'per hundred'. So, 70% can be expressed as the decimal 0.70. This conversion is crucial as it simplifies the algebraic manipulation needed to solve equations.
To translate a percentage problem into an algebraic equation, we multiply the decimal form of the percentage by the variable representing the unknown number. In mathematical terms, this relationship is expressed as '0.70 times x equals 252'. This step is integral to setting up a workable equation that can lead to finding the solution.
To translate a percentage problem into an algebraic equation, we multiply the decimal form of the percentage by the variable representing the unknown number. In mathematical terms, this relationship is expressed as '0.70 times x equals 252'. This step is integral to setting up a workable equation that can lead to finding the solution.
Setting Up Equations
Setting up equations is a fundamental step in algebraic problem-solving. To solve word problems, first, we identify what we are asked to find and let a variable represent this unknown. In our case, we are trying to find a number where 70% of it is 252, hence we let 'x' represent this unknown number.
The next step is to translate the word problem into an equation using algebraic symbols. The phrase '70% of x equals 252' becomes the equation \(0.70 * x = 252\). Ensuring that the equation accurately represents the problem is key. Any errors at this stage will lead to incorrect solutions. Once the equation is set up correctly, solving for 'x' involves inverse operations that undo the multiplication or division present in the equation.
The next step is to translate the word problem into an equation using algebraic symbols. The phrase '70% of x equals 252' becomes the equation \(0.70 * x = 252\). Ensuring that the equation accurately represents the problem is key. Any errors at this stage will lead to incorrect solutions. Once the equation is set up correctly, solving for 'x' involves inverse operations that undo the multiplication or division present in the equation.
Algebraic Problem-Solving Techniques
Algebraic problem-solving utilizes a variety of techniques to find solutions to equations. Once we have set up the equation, the next step is to solve for the unknown variable. In our example, the equation \(0.70 * x = 252\) can be solved by performing the inverse operation - which in this case is division. By dividing both sides of the equation by 0.70, we isolate 'x' on one side to get \(x = 252 / 0.70\).
Utilizing this technique is simple and efficient, especially when dealing with linear equations. Double-checking our work can prevent common mistakes, such as mishandling decimals or inverting operations. By following these algebraic techniques methodically, you can solve a variety of word problems in algebra with ease.
Utilizing this technique is simple and efficient, especially when dealing with linear equations. Double-checking our work can prevent common mistakes, such as mishandling decimals or inverting operations. By following these algebraic techniques methodically, you can solve a variety of word problems in algebra with ease.
Other exercises in this chapter
Problem 8
Express each interval in set-builder notation and graph the interval on a number line. $$(3, \infty)$$
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By factoring and then using the zero-product principle. $$ 9 y^{3}+8=4 y+18 y^{2} $$
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Solve and check linear equation. \(13 x+14=12 x-5\)
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In Exercises \(1-12\), plot the given point in a rectangular coordinate system. $$ (3,-2) $$
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