Math, asked by NishantBabloo, 1 year ago

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

Answered by TooFree
4

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.


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