How do I calculate percentage increase for ratios in math homework?

  • Thread starter Thread starter asmalik12
  • Start date Start date
  • Tags Tags
    Ratio
AI Thread Summary
To calculate the percentage increase for ratios, the first example shows that a photocopier's paper increase from a ratio of 4:5 results in a 25% increase. For the second example involving an alloy with a ratio of copper, alloy, and lead at 5:7:3, the initial calculation of 57.14% for the alloy to lead is incorrect. The correct approach is to compare the parts of alloy (7) to lead (3), resulting in a 133.33% increase. Clarification is needed regarding the presentation of the problem, as it may cause confusion in understanding the ratios. Accurate calculations and clear definitions of parts are essential for solving ratio problems effectively.
asmalik12
Messages
8
Reaction score
0

Homework Statement


1. A photocpier increases the paper with the ratio 4:5, what % is the paper increased
2. An alloay has a copper, alloy and lead with the ratio 5:7:3, what % is the alloy to lead


2. The attempt at a solution

1. Difference in equation = 5 - 4 = 1
= 1/4 * 100
= 25 %

2. Difference between the ratio of alloy and lead = 7 - 3 = 4
= 4/7 * 100
= 57.14 %

Am I doing it rightly?
 
Physics news on Phys.org


asmalik12 said:

Homework Statement


1. A photocpier increases the paper with the ratio 4:5, what % is the paper increased
2. An alloay has a copper, alloy and lead with the ratio 5:7:3, what % is the alloy to lead


2. The attempt at a solution

1. Difference in equation = 5 - 4 = 1
= 1/4 * 100
= 25 %
Yes, this is correct.
asmalik12 said:
2. Difference between the ratio of alloy and lead = 7 - 3 = 4
= 4/7 * 100
= 57.14 %

Am I doing it rightly?
No. There are 7 parts alloy to 3 parts lead, so there is about 2.33 times as much alloy as lead.
 


For Question No. 2

Difference in the ratio of alloy to lead = 7 - 3 = 4
= 4/3 * 100
= 133.33 %

Now ?
 


For question number two - are there 3 parts that you are comparing the 1 part to? Why does the problem refer to alloy as both the whole and a part? (I'm wondering if part of the confusion is in the presentation of the problem)

If the ratio is: ALLOY = Copper:OtherALLOY:Lead then I second Mark44's solution.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top