Do I have a problem? Basic proportion problem.

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In summary, it seems like you are concerned that pursuing a math-related career could limit your ability to solve simple problems due to your experience with more complex problems in high school. However, it is important to remember that solving linear problems is still possible and that you can use concepts from more complex problems to solve linear problems.
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Hello, this is an odd question but when I compared how I solved it compared to how the book solved it I kinda got worried. It was a very basic problem I been going back and touching up on my basics before I start pre-cal and trig in college.

The problem was: After completing 7/10 of his math homework assignment, Josh has 15 more questions complete. What is the total number of questions on his assignment? The answer was 50.

I got that by

[itex]\frac{7}{10}[/itex]=[itex]\frac{(x-15)}{x}[/itex] Followed by cross multiplying and dividing.

However the book does it simply by doing the subtraction 10-7 = 3 then setting it up as

[itex]\frac{3}{10}[/itex]=[itex]\frac{15}{x}[/itex] Then cross multiply and divide.

I know someone will say "It works so you can do what you are doing". But is there a possibility this kind of over complication( haha not hardly complicated!) Can hurt me later on? The way I performed it seemed much more intuitive... I have a high school record of up to ap calc and ap chem and I'm worried if I pursue engineering in college I'll have a hard time solving simple things because of my use to complexity?
 
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  • #2
(3/10)x=15
x=50
 
  • #3
Life-Like said:
Hello, this is an odd question but when I compared how I solved it compared to how the book solved it I kinda got worried. It was a very basic problem I been going back and touching up on my basics before I start pre-cal and trig in college.

The problem was: After completing 7/10 of his math homework assignment, Josh has 15 more questions complete.
Do you meant "15 more questions to complete"? What you wrote could be interpreted as meaning has completed 15 questions. If he has already completed 7/10 of his assignment, he still has 3/10 left. If he has 15 questions more to complete, letting "x" be the total number of problems, (3/10)x= 15 so x= (10/3)(15)= 150/3= 50.

What is the total number of questions on his assignment? The answer was 50.

I got that by

[itex]\frac{7}{10}[/itex]=[itex]\frac{(x-15)}{x}[/itex] Followed by cross multiplying and dividing.

However the book does it simply by doing the subtraction 10-7 = 3
Okay, that is, they are calculating 1- 7/10= 10/10- 7/10= (10- 7)/10= 3/10 as I did.

then setting it up as

[itex]\frac{3}{10}[/itex]=[itex]\frac{15}{x}[/itex] Then cross multiply and divide.

I know someone will say "It works so you can do what you are doing". But is there a possibility this kind of over complication( haha not hardly complicated!) Can hurt me later on? The way I performed it seemed much more intuitive... I have a high school record of up to ap calc and ap chem and I'm worried if I pursue engineering in college I'll have a hard time solving simple things because of my use to complexity?
As long as you have linear problems, you can set them up as "proportions" and get the correct answer.

 
  • #4
Thanks guys
 
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FAQ: Do I have a problem? Basic proportion problem.

What is a basic proportion problem?

A basic proportion problem is a mathematical problem that involves comparing two ratios or fractions to determine if they are equal or not.

How do I know if I have a basic proportion problem?

If you are given two ratios or fractions and are asked to compare them, then you have a basic proportion problem.

What is the formula for solving a basic proportion problem?

The formula for solving a basic proportion problem is: a/b = c/d, where a and b are the first ratio, and c and d are the second ratio.

How do I solve a basic proportion problem?

To solve a basic proportion problem, you can use the cross-multiplication method, where you multiply the numerator of one ratio by the denominator of the other, and then set the two resulting products equal to each other.

Can basic proportion problems be used in real-life situations?

Yes, basic proportion problems can be used in real-life situations, such as calculating ingredient ratios for a recipe, determining the cost of an item based on its weight, or finding the amount of time it takes to travel a certain distance at a given speed.

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