Question

Three numbers are in the ratio 6:3:1, and the sum of these numbers is 420. If the first number is reduced by 50%, and the second number is increased by 42, and the sum of the numbers remain the same, find the resulting ratio of the numbers .

Answer 1

`x:y:z\ \ \ \Leftrightarrow\ \ \ 6:3:1\\ \\ (x)/(z) =(6)/(1)\ \ \ \Rightarrow\ \ \ x=6z\ \ \ \ \ \ \ and\ \ \ \ \ \ \ (y)/(z) =(3)/(1)\ \ \ \Rightarrow\ \ \ y=3z\\ \\x+y+z=420\ \ \ \Rightarrow\ \ \ 6z+3z+z=420\ \ \ \ \Rightarrow\ \ \ \ 10z=420\ /:10\\ \\z=42\ \ \ \Rightarrow\ \ \ \ x=6\cdot42=252\ \ \ \ and\ \ \ \ y=3\cdot42=126\\ \\ \\a=x-50\%x=50\%x= (1)/(2) \cdot252=126=3\cdot42\\ \\b=y+42=126+42=168=4\cdot42\\ \\c=420-126-168=126=3\cdot42\\ \\a:b:c\ \ \ \Leftrightarrow\ \ \ 3:4:3`