<PRE><B>Question 145304</B>
n alloy of tin and copper contains 14 pounds of tin and 26 pounds of copper.
a second alloy of tin and copper contains 8 pounds of tin and 24 pounds of copper. How many pounds of each alloy must be taken to form a third alloy
containing 15 pounds of tin and 35 pounds of copper.
:
Solve this using the decimal value of %tin:
:
Find the per cent tin in the 1st alloy:
{{{14/((14+26))}}} = .35
:
Find the per cent tin in the 2nd alloy
{{{8/((8+24))}}} = .25
:
Find the per cent tin in the resulting alloy
{{{15/((15+35))}}} = .30
:
Let x = amt of the 1st alloy required
:
The resulting total amt is (15 + 35) = 50 lb, therefore:
(50-x) = amt of the 2nd equation
:
The per cent tin equation:
.35x + .25(50-x) = .3(50)
:
.35x + 12.5 - .25x = 15
:
.35x - .25x = 15 - 12.5
:
.10x = 2.5
x = {{{2.5/.1}}}
x = 25 lb of alloy #1
and
50 - 25 = 25 lb of alloy #2
:
:
Check solution using %tin equation
.35(25) + .25(25) = .30(50)
8.75 + 6.25 =15</PRE>