Question

a mixture contains three constituents a, b and c in the ratio 5 : 9 : 28 by volume. the solution is altered such that the ratio of a to b is halved and the ratio of b to c is quadrupled. what is the volume of c present in the new mixture if 10 cubic cm of a is present in it?

Answer 1

Answer:

5x=10 cm³,x= 2 cm²

So c=40*2=80 cm³

Step-by-step explanation:

mixture contains three constituents a, b and c in the ratio 5 : 9 : 28 by volume

Let the volumes are

a=5x,b=9x  and c=28x cm³

Now the solution is altered such that the ratio of a to b is halved

and the ratio of b to c is quadrupled

In new solution:

a=5x and b=2a=10x cm³

c=4b=4*10=40 x cm³

Now in new mixture a=10 cm³

5x=10 cm³,x= 2 cm²

So c=40*2=80 cm³