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Mr lee had some grains. He sold 380 kg of the grains. By feeding the chicken in his farm with the same amount of grains each day, he would use up all his remaining grains in 50 days. If in the first 30 days, the chicken were fed 1/8 of the original amount of the grains, how many kg of grains did the farmer feed the chicken on the 1st 30 days?

 Feb 1, 2016

Best Answer 

 #1
avatar+118667 
+5

Mr lee had some grains. He sold 380 kg of the grains. By feeding the chicken in his farm with the same amount of grains each day, he would use up all his remaining grains in 50 days. If in the first 30 days, the chicken were fed 1/8 of the original amount of the grains, how many kg of grains did the farmer feed the chicken on the 1st 30 days?

 

originally he had      380+x     kg

 

 

50d=x                         (1)

30d=1/8*(380+x)        (2)

30*8d = 380+x

240d   = 380+x

sub in (1)

240d =380 +50d

190d = 380

d=2

so each day the chickens eat 2 kg

So in the first 30 days the farmer fed the chickens  60Kg

 Feb 1, 2016
 #1
avatar+118667 
+5
Best Answer

Mr lee had some grains. He sold 380 kg of the grains. By feeding the chicken in his farm with the same amount of grains each day, he would use up all his remaining grains in 50 days. If in the first 30 days, the chicken were fed 1/8 of the original amount of the grains, how many kg of grains did the farmer feed the chicken on the 1st 30 days?

 

originally he had      380+x     kg

 

 

50d=x                         (1)

30d=1/8*(380+x)        (2)

30*8d = 380+x

240d   = 380+x

sub in (1)

240d =380 +50d

190d = 380

d=2

so each day the chickens eat 2 kg

So in the first 30 days the farmer fed the chickens  60Kg

Melody Feb 1, 2016
 #2
avatar
+5

Thanks melody I understood it but I changed the equations from algebra to normal math lolz

 Feb 1, 2016
 #3
avatar+33660 
0

I'm puzzled (I often am!).  What do you mean by "normal" math here?  In what way doesn't this involve algebra?

 Feb 1, 2016

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