Weight/height Problem

This is a discussion on Weight/height Problem within the C++ Programming forums, part of the General Programming Boards category; Well the whole problem goes as such: Write a program that asks for the user's height, weight, and age, and ...

  1. #1
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    Weight/height Problem

    Well the whole problem goes as such:

    Write a program that asks for the user's height, weight, and age, and then computes clothing sizes according to the forumlas:

    Hat size = weight in pounds divided by height in inches and all that multiplied by 2.9

    Jacket size (chest in inches) = height times weight divided by 288 then adjusted by 1/8 of an inch for each 10 years over age 30 ( Note that the adjustment only takes place after a full 10 years. So there is no adjustment for ages 30 through 39, but 1/8 of an inch is added for age 40).

    Waist in inches = weight divided by 5.7 and then adjusted by adding 1/10 of an inch for each 2 years over age 28. (Note that the adjustment only takes place after a full 2 years. So, there is no adjustment for age 29, but 1/10 of an inch is added for age 30)

    Use functions for each calculation. Your program should allow the user to repeat this calculation as often as the user wishes.
    Ok so I have the following for the hatsize method:

    Code:
    double HatSize(double weight, double height)
    {
    
    double result = 2.9*(weight/height);
    
    return result;
    
    }
    What I need help is with the JacketSize and Waist in inches, maybe if you show me the jacketsize method I can figure out the waist in inches.


    Thanks for all help!!
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  2. #2
    Deathray Engineer MacGyver's Avatar
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    Dude, they give you the answer right there... Don't be so lazy!

    Jacket size (chest in inches) = height times weight divided by 288 then adjusted by 1/8 of an inch for each 10 years over age 30 ( Note that the adjustment only takes place after a full 10 years. So there is no adjustment for ages 30 through 39, but 1/8 of an inch is added for age 40).
    (height * weight / 288).... then for every 10 years over 30, add 1/8 of an inch. Hint: (Age - 30) / 10 (integer division btw) is the amount of times to add 1/8 of an inch I believe (provided the person is over 30), but I'm not paying careful attention because obviously you barely put any effort into this.

    Seriously, the instructions are right there in front of you. This assignment hardly takes any effort in terms of actual coding. Don't be lazy and expect people to do your assignments.

    I already told someone off for the same kind of thing....
    Last edited by MacGyver; 03-24-2007 at 06:52 PM.

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    how would I get 1/8 of a inch though?

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    Registered User Noir's Avatar
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    how would I get 1/8 of a inch though?
    Did you try dividing 1 by 8?

  5. #5
    Deathray Engineer MacGyver's Avatar
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    Sometimes I wonder why people sign up for programming courses....

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