Let the mass of #"60 % Mn" = xcolor(white)(l) "kg"#
Look at the problem another way.
You are mixing 1000 kg of metal with #xcolor(white)(l) "kg of 60 % Mn"# and getting #"(1000 +x) kg of 50 % Mn"#.
#"Mass of Mn in 60 % = mass of Mn in 50 %"#
#x color(red)(cancel(color(black)("kg"))) × 60 color(red)(cancel(color(black)("% Mn"))) = (1000 +x) color(red)(cancel(color(black)("kg"))) × 50 color(red)(cancel(color(black)("% Mn")))#
#60x = 50(1000 + x) = "50 000" + 50x#
#10x = "50 000"#
#x = "50 000"/10 = 5000#
Thus, you would add #"5000 kg of 60 % Mn"# to 1000 kg of metal, and you would get #"6000 kg of 50 % Mn"#.
Check:
#"5000 kg of 60 % Mn"# contains #"3000 kg of Mn"#, and #"6000 kg of 50 % Mn"# also contains #"3000 kg of Mn"#.
