In two types of stainless steel the ratio of chromium and steel are 2 : 11 and 5 : 21 respectively. In what proportion should the two types be mixed so that the ratio of chromium to steel in the mixed type becomes 7 : 32 = ?

In two types of stainless steel the ratio of chromium and steel are 2 : 11 and 5 : 21 respectively. In what proportion should the two types be mixed so that the ratio of chromium to steel in the mixed type becomes 7 : 32 = ? Correct Answer 1 : 2

Given:
⇒ Let the alloys be mixed in the ratio of $$x:y$$   (Assumption)
⇒ In 1st alloy, chromium = $$\frac{2x}{13}$$
⇒ In 1st alloy, steel = $$\frac{11x}{13}$$
⇒ In 2nd alloy, chromium = $$\frac{5y}{26}$$
⇒ In 2nd alloy, steel = $$\frac{21y}{26}$$

⇒ The ratio in which 2 alloys must be mixed to get a new alloy with a ratio of chromium and steel be 7 : 32 =?

Now we have,
$$\eqalign{ & \left( {\frac{{2x}}{{13}} + \frac{{5y}}{{26}}} \right) : \left( {\frac{{11x}}{{13}} + \frac{{21y}}{{26}}} \right) = 7 : 32 \cr & \Rightarrow \frac{{\left( {\frac{{2x}}{{13}} + \frac{{5y}}{{26}}} \right)}}{{ \left( {\frac{{11x}}{{13}} + \frac{{21y}}{{26}}} \right) }} = \frac{7}{{32}} \cr & \Rightarrow \frac{{\frac{{4x + 5y}}{{26}}}}{{ \frac{{22x + 21y}}{{26}} }} = \frac{7}{{32}} \cr & \Rightarrow \frac{{4x + 5y}}{{22x + 21y}} = \frac{7}{{32}} \cr & \Rightarrow 128x + 160y = 154x + 147y \cr & \Rightarrow 154x - 128x = 160y - 147y \cr & \Rightarrow 26x = 13y \cr & \Rightarrow \frac{x}{y} = \frac{{13}}{{26}} \cr & \therefore \frac{x}{y} = \frac{1}{2} \cr} $$

∴ Ratio in which 2 alloys must be mixed to get a new alloy with a ratio of chromium and Steel to be 7 : 32 is 1 : 2