In two alloys the ratio of iron and copper is 4 ratio 3 and 6 ratio 1

1.	In two alloys the ratio of iron and copper is 4 ratio 3 and 6 ratio 1 respectively if 14 kg of the first Roy and 42 kg of the second all is mixed together to form a new law and what will be the ratio of iron to copper in the new alloy
Answer» Solution: In 14 KG of first Alloy, Ratio of iron and copper=4 :3 Let ratio of Iron and copper in first alloy be , 4 x and 3 x. 4 x + 3 x= 14 7 x =14 Dividing both sides by, 2 gives x=2 So, Amount of Iron = 4 × 2= 8 Kg Amount of Copper = 3 × 2= 6 Kg In 14 Kg of Second Alloy, Ratio of iron and copper=6:1 Let ratio of Iron and copper in first alloy be , 6 x and 1 x. 6 x + 1 x= 42 7 x =42 Dividing both sides by, 7 gives x=6 So, Amount of Iron = 6 × 6= 36 Kg Amount of Copper = 6 × 1= 6 Kg When two Alloys are Mixed total amount of Alloy=42 Kg + 14 Kg =56 Kg Total Amount of Iron in COMBINATION of two Alloy = 36 +8=44 Kg Total Amount of Copper in Combination of two Alloy = 6 +6=12 Kg Ratio of Iron to copper in the new alloy $=\frac{44}{12}\\\\=\frac{11}{3}$ =11 : 3

In two alloys the ratio of iron and copper is 4 ratio 3 and 6 ratio 1 respectively if 14 kg of the first Roy and 42 kg of the second all is mixed together to form a new law and what will be the ratio of iron to copper in the new alloy

Answer»

Solution:

In 14 KG of first Alloy, Ratio of iron and copper=4 :3

Let ratio of Iron and copper in first alloy be , 4 x and 3 x.

4 x + 3 x= 14

7 x =14

Dividing both sides by, 2 gives

x=2

So, Amount of Iron = 4 × 2= 8 Kg

Amount of Copper = 3 × 2= 6 Kg

In 14 Kg of Second Alloy, Ratio of iron and copper=6:1

Let ratio of Iron and copper in first alloy be , 6 x and 1 x.

6 x + 1 x= 42

7 x =42

Dividing both sides by, 7 gives

x=6

So, Amount of Iron = 6 × 6= 36 Kg

Amount of Copper = 6 × 1= 6 Kg

When two Alloys are Mixed total amount of Alloy=42 Kg + 14 Kg

=56 Kg

Total Amount of Iron in COMBINATION of two Alloy = 36 +8=44 Kg

Total Amount of Copper in Combination of two Alloy = 6 +6=12 Kg

Ratio of Iron to copper in the new alloy

$=\frac{44}{12}\\\\=\frac{11}{3}$

=11 : 3