Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

Mely who owns a coffee store in Divisoria Market wishes to mix first-class coffee that sells 200 per kg with 25 kg of imported cofee that sells 150 per kg. How many kgs of first-class coffee must be mixed with the imported coffee to obtain a mixture of worth 175 per kg? How much of the mixture will be produced?

OpenStudy (anonymous):

25 kg, 50 kg

OpenStudy (anonymous):

but how?

OpenStudy (anonymous):

1:1 ratio.. simple

OpenStudy (anonymous):

kindly show your solution

OpenStudy (anonymous):

wait for sometime.. will be right back

OpenStudy (anonymous):

say, first class coffee is x kg and imported coffee is y kg in amount in the mixture so, 200x+150y=175 (x+y) u can get, x=y y is given 25 kg so x is also 25 kg

OpenStudy (anonymous):

i'm sorry but it is clear in the given that: mix first-class coffee that sells 200 per kg with 25 kg of imported cofee that sells 150 per kg. meaning --- 1 kg: 25 kg

OpenStudy (anonymous):

no, u got it wrong

OpenStudy (anonymous):

mix first class coffee that sells 200 per kg that means the price of first class coffee costs 200 we have to mix them in correct ratio

OpenStudy (anonymous):

then, how many kgs of first-class coffee must be mixed with the imported coffee to obtain a mixture of worth 175 per kg?

OpenStudy (anonymous):

1:1ratio.. i showed the work

OpenStudy (dumbcow):

X*200 + 25(150) = 175(X+25) its like a weighted average, since 175 is right in the middle between 150 and 200 then the amounts are equal. X = 25 solve the equation to verify

OpenStudy (anonymous):

yup.. when u face this kind of problems, to solve these fast.. take the difference of two given values from the required value then take the reciprocal of the ratio of differences that will be the required ratio

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!