One alloy of iron is 49% iron and another 28% iron. How many tons of each should be to make 120 tons of 42% iron alloy?
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An alloy of iron is 49% iron and other is 28% of iron.How many kilogram of each should be used to get 120 kg of 42%iron alloy?
Let x kg of 49% iron alloy is used
and y kg of 28% iron alloy is used.
therefore, acc. to the question,
x + y = 120. - - - - - - - (1)

●on multiplying (i) by 4 and subtracting from (ii) , we get,
3x= 240
==> x= 80
put in (i), we get,
80+y=120
y=12080
==> y=40
●thus, 80 kg of 49% iron alloy is used and 40 kg of 28% iron alloy is used.
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# NIKKY
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GOOD NIGHT :))
_____________________
An alloy of iron is 49% iron and other is 28% of iron.How many kilogram of each should be used to get 120 kg of 42%iron alloy?
Let x kg of 49% iron alloy is used
and y kg of 28% iron alloy is used.
therefore, acc. to the question,
x + y = 120. - - - - - - - (1)
●on multiplying (i) by 4 and subtracting from (ii) , we get,
3x= 240
==> x= 80
put in (i), we get,
80+y=120
y=12080
==> y=40
●thus, 80 kg of 49% iron alloy is used and 40 kg of 28% iron alloy is used.
________________________
HOPE IT HELPS U DEAR! !!!
# NIKKY
☺☺
GOOD NIGHT :))
VijayaLaxmiMehra1:
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Answered by
4
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