There are two types of ore. First ore contains 68% iron , second ore contains 63% iron. Find the amount of each that will contain 65% of iron in 100 ton of ore?????
Answers
Define x and y:
Let the amount of the first (68% iron) ore be x
The amount of second (63% iron) ore be y
Form the equations:
Total ore is 100 tons
⇒ x + y = 100 --------------------------------- [ 1 ]
Total percentage of iron is 65%
0.68x + 0.63y = 0.65(100)
⇒ 0.68x + 0.63y = 65 --------------------- [ 2 ]
Solve the equations:
x + y = 100 -------------------------------- [ 1 ]
0.68x + 0.63y = 65 --------------------- [ 2 ]
[ 1 ] x 0.68:
0.68 x + 0.68y = 68 --------------------- [ 3 ]
FInd y:
[ 3 ] - [ 2 ] :
0.05y = 3
y = 3 ÷ 0.05 = 60 tons
Find x:
From [ 1 ]:
x + y = 100
sub y = 60 into the equation:
x + 60 = 100
x = 100 - 60
x = 40 tons
Find the amount of each of type of ore:
First (68% iron) ore = x = 40 tons
Second (63% iron) ore = y = 60 tons
Answer: There are 40 tons of the 68% iron ore and 60 tons of the 63% iron ore.