How do you have a logical operator of true and false as your type but 0 and 1 as your value?

Hi. I am having trouble with this homework question: Write a function called eligible that helps the admission officer of the Graduate School of Vanderbilt University decide whether the applicant is eligible for admission based on GRE scores. The function takes two positive scalars called v and q as input. They represent the percentiles of the verbal and quantitative portions of the GRE respectively. You do not need to check the input. The applicant is eligible if the average percentile is at least 92% and both of the individual percentiles are over 88%. The function returns the logical true or false. What I have attempted is below:

function [true, false] = eligible(v, q)
    if mean(v, q)>= 92 && v>88 && q>88 
           fprintf('true\n');
    elseif mean(v, q)<92 
        fprintf('false\n');
    elseif v<=88 
          fprintf('false\n');
    elseif q<=88
         fprintf('false\n');

I think my problem is that I need the false to appear as the type but 0 to appear as my value and true to appear as the type but 1 to appear as my value.

[Pranav nair]

function admit = eligible(v,q)
  c = (v+q)/2;
  if  (c>= 92 && v>88 && q>88)
    admit = true;
  else (v < 88 && q < 88)
    admit = false;
  end

[DGM]

See also:

https://www.mathworks.com/matlabcentral/answers/471264-write-a-function-called-eligible-that-helps-the-admission-officer-of-the-graduate-school-decide-whet

https://www.mathworks.com/matlabcentral/answers/525704-how-can-i-overcome-the-error-variable-admit-must-be-of-data-type-logical-instead-of-double-i-have

Answer 1

[Image Analyst]

You used mean() incorrectly. Look at this:

>> mean(10,30) % WRONG!
ans =
    10
>> mean([10,30]) % Right way uses brackets.
ans =
    20

So instead of

      if mean(v, q)>= 92 && v>88 && q>88 
           fprintf('true\n');
      elseif mean(v, q)<92 
           fprintf('false\n');
      elseif v<=88 
           fprintf('false\n');
      elseif q<=88
          fprintf('false\n');

use

      if mean([v, q]) >= 92 && v > 88 && q > 88 
           fprintf('true\n');
           isEligible = true;
      else
           fprintf('false\n');
           isEligible = false;
      end

[mallavarapu hema saikumar]

function a=eligible(v,q)
c=(v+q)/2;
if (c>=92)&&(v>88&&q>88)
a=true;
else
a=false;
end

[Walter Roberson]

function a=eligible(v,q)
c=(v+q)/2;
a = (c>=92)&&(v>88&&q>88);

[the cyclist]

Or, if you are playing Cody ...

function ans = eligible(v,q)
((v+q)>=184)&&(v>88&&q>88);

[This is not a serious comment. Please do not take it as such!]

Answer 2

[the cyclist]

You are on the right track. Here is a hint ...

Instead of outputting two variables named true and false, I think you'd want to just output one variable -- maybe call it isEligible -- that takes on the values true/false.

It looks like you know how to calculate the value of isEligible, because you have used it to determine what to display to the screen.

Answer 3

[Jorge Briceño]

Hi Alexandra,

Maybe another alternative would be:

 function [isEligible]=eligible( v,q )
    Your_Output_Name= v>88 && q>88 && ((q+v)/2)>=92
 end

Cheers, Jorge

[Walter Roberson]

Your_Output_Name would have to be isEligible for this to work.

[Jorge Briceño]

Ok, I had a silly typo. Thanks, Walter.

   function [Your_Output_Name]=eligible( v,q )
    Your_Output_Name= v>88 && q>88 && ((q+v)/2)>=92
 end

Answer 4

[ledinh lam]

I think it will be :

function el=eligible(v,q)
    if mean([v,q]) >= 92 && v>88 && q >88 
        el=true; 
    else
        el=false; 
    end
end

Answer 5

[Duddela Sai Prashanth]

function [out] = eligible(v, q)
if v > 88 && q > 88 && (v+q)/2 >= 92
    out = true;
else
    out = false;
end

[Walter Roberson]

What essential difference is there between this and https://www.mathworks.com/matlabcentral/answers/298895-how-do-you-have-a-logical-operator-of-true-and-false-as-your-type-but-0-and-1-as-your-value#comment_586846 ?

Answer 6

[Yamen Al-Jajan]

function admit=eligible(v,q)
if v>=88 & q>=88
    ave=(v+q)/2;
    if ave>=92
        admit=true;
    else
        admit=false;
    end
else
    admit=false;
end

Answer 7

[Rahul Krishna]

function admit = eligible(v,q)
if (q +v)/2 >=92 && (v>88 && q>88)
    admit = true
   fprintf(' the candidate is eligible \n')
else 
    admit = false
    fprintf(' the candidate is not eligible \n')
end

Answer 8

[MALK adil]

function out = eligible(v,q)
out = (v+q)./2 >= 92  && v > 88 && q > 88;
if out==1
    out= true;
else  out= false;
end

[Image Analyst]

@MALK adil your if block is totally unnecessary.

out is already a boolean (true or false) by how you define it. You do not need to check if it's a 1 (double) and then assign it to true (which it already is). I believe that when you compare a logical to a double (like bln == 7) it converts the logical to a double (false converts to 0, and true converts to 1) and then it compares that double to the number (7) and returns a double.

out = true

out = logical
   1

result = out == 7 % Convert out to double then compare the two doubles and return a logical

result = logical
   0

And if it's not 1 (the only other choice is false) there is no need to set it to false. It's already false!

So you could simply have done

function out = eligible(v,q)
out = (v+q)./2 >= 92  && v > 88 && q > 88;