Question

Nickel and iron are mixed in the ratio of 4:5. And 4:11 to make two allys namely,x and y, respectively.suppose we want to make the third alloy,z,by mixing equal quantities of x and y.what will be the ratio of nickel and iron present in the alloy z?

Answer 1

Given :-

• Ratio of Nickel and iron in x = 4 : 5
• Ratio of Nickel and iron in y = 4 : 11
• Equal quantities of x and y are mixed to make third alloy z .

To Find :-

• what will be the ratio of nickel and iron present in the alloy z ?

Solution :-

→ Nickel and iron in x = 4 : 5 = 9 Total .

→ Nickel and iron in y = 4 : 11 = 15 Total .

As, we need Equal quantities of both x and y , we will Multiply x by 5 and y by 3 . ( 9 * 5 = 15 * 3)

So ,

→ Nickel and iron in x = 5(4 : 5) = 20 : 25 = 45 Total .

→ Nickel and iron in y = 3(4 : 11) = 12 : 33 = 45 Total .

Therefore,

→ Total Nickel in z now = (Nickel in x + Nickel in y) = 20 + 12 = 32 .

→ Total iron in z now = (iron in x + iron in y) = 25 + 33 = 58 .

Hence,

→ Ratio of Nickel and iron present in the alloy z = 32 : 58 = 16 : 29 (Ans.)