Let the %age of honey and milk in the two solutions be (a%, b%) and (c%, d%) respectively. According to the question:
(a−10)
(10−c)
=
(b−16)
(16−d)
(It does not make a difference whether a is greater than c or whether b is greater than d, because taking -1 common from both the numerator and the denominator will nullify the effect.) Solving, we get, 16a−ad−160+10d=10b−160−bc+16c So, ad−bc=16a−10b−16c+10d ...(1) Similarly from the second and third proportions we can say that ad−bc=12a−12b−12c+12d. ...(2) and ad−bc=xa−16b−xc+16d ...(3) From (1) and (2), we get