Program to find the common ratio of three numbers
Given a:b and b:c. The task is to write a program to find ratio a:b:c
Examples:
Input: a:b = 2:3, b:c = 3:4
Output: 2:3:4
Input: a:b = 3:4, b:c = 8:9
Output: 6:8:9
Approach: The trick is to make the common term ‘b’ equal in both ratios. Therefore, multiply the first ratio by b2 (b term of second ratio) and the second ratio by b1.
Given: a:b1 and b2:c
Solution: a:b:c = (a*b2):(b1*b2):(c*b1)
For example:
If a : b = 5 : 9 and b : c = 7 : 4, then find a : b : c.
Solution:
Here, Make the common term ‘b’ equal in both ratios.
Therefore, multiply the first ratio by 7 and the second ratio by 9.
So, a : b = 35 : 63 and b : c = 63 : 36
Thus, a : b : c = 35 : 63 : 36
Below is the implementation of the above approach:
C++
#include <bits/stdc++.h>
using namespace std;
void solveProportion( int a, int b1, int b2, int c)
{
int A = a * b2;
int B = b1 * b2;
int C = b1 * c;
int gcd = __gcd(__gcd(A, B), C);
cout << A / gcd << ":"
<< B / gcd << ":"
<< C / gcd;
}
int main()
{
int a, b1, b2, c;
a = 3;
b1 = 4;
b2 = 8;
c = 9;
solveProportion(a, b1, b2, c);
return 0;
}
|
Java
import java.util.*;
import java.lang.*;
import java.io.*;
class GFG{
static int __gcd( int a, int b){
return b== 0 ? a : __gcd(b, a%b);
}
static void solveProportion( int a, int b1, int b2, int c)
{
int A = a * b2;
int B = b1 * b2;
int C = b1 * c;
int gcd = __gcd(__gcd(A, B), C);
System.out.print( A / gcd + ":"
+ B / gcd + ":"
+ C / gcd);
}
public static void main(String args[])
{
int a, b1, b2, c;
a = 3 ;
b1 = 4 ;
b2 = 8 ;
c = 9 ;
solveProportion(a, b1, b2, c);
}
}
|
Python 3
import math
def solveProportion(a, b1, b2, c):
A = a * b2
B = b1 * b2
C = b1 * c
gcd1 = math.gcd(math.gcd(A, B), C)
print ( str (A / / gcd1) + ":" +
str (B / / gcd1) + ":" +
str (C / / gcd1))
if __name__ = = "__main__" :
a = 3
b1 = 4
b2 = 8
c = 9
solveProportion(a, b1, b2, c)
|
C#
using System;
class GFG
{
static int __gcd( int a, int b)
{
return b == 0 ? a : __gcd(b, a % b);
}
static void solveProportion( int a, int b1,
int b2, int c)
{
int A = a * b2;
int B = b1 * b2;
int C = b1 * c;
int gcd = __gcd(__gcd(A, B), C);
Console.Write( A / gcd + ":" +
B / gcd + ":" +
C / gcd);
}
public static void Main()
{
int a, b1, b2, c;
a = 3;
b1 = 4;
b2 = 8;
c = 9;
solveProportion(a, b1, b2, c);
}
}
|
PHP
<?php
function __gcd( $a , $b )
{
return $b == 0 ? $a : __gcd( $b , $a % $b );
}
function solveProportion( $a , $b1 , $b2 , $c )
{
$A = $a * $b2 ;
$B = $b1 * $b2 ;
$C = $b1 * $c ;
$gcd = __gcd(__gcd( $A , $B ), $C );
echo ( $A / $gcd ) . ":" .
( $B / $gcd ) . ":" . ( $C / $gcd );
}
$a = 3;
$b1 = 4;
$b2 = 8;
$c = 9;
solveProportion( $a , $b1 , $b2 , $c );
?>
|
Javascript
<script>
function __gcd(a, b)
{
return b == 0 ? a : __gcd(b, a % b);
}
function solveProportion(a, b1, b2, c)
{
let A = a * b2;
let B = b1 * b2;
let C = b1 * c;
let gcd = __gcd(__gcd(A, B), C);
document.write( A / gcd + ":" + B / gcd + ":" + C / gcd);
}
let a, b1, b2, c;
a = 3;
b1 = 4;
b2 = 8;
c = 9;
solveProportion(a, b1, b2, c);
</script>
|
Time Complexity : O(log(A+B)) ,where A=a*b2 and B = b1*b2
Space Complexity : O(1), since no extra space has been taken.
Last Updated :
20 Aug, 2022
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