Question

P,Q,R are three cups having capacities of 120,180 and 200 cc respectively which are

completely filled with tea of three different varieties. They are all mixed in a separate vessel of

capacity more than that of P,Q,R taken together, and then the mixture is poured successively

into P,Q,R Then how much (in cc) of Pand Q’s tea will be there in R?


ratio of those three cups in cc is
120:180:200
= 6:9:10
quantity of 2nd and 3rd mixtures in 3rd cup is
200×6/25=48
200×9/25=72​

Answer 1

Given :- P,Q,R are three cups having capacities of 120,180 and 200 cc respectively which are completely filled with tea of three different varieties. They are all mixed in a separate vessel of capacity more than that of P,Q,R taken together, and then the mixture is poured successively into P,Q,R Then how much (in cc) of Pand Q’s tea will be there in R ?

Answer :-

→ In P tea = 120 cc

→ In Q tea = 180 cc

→ In R tea = 200 cc

so,

→ In fourth vessel quantity of P : Q : R = 120 : 180 : 200 = 6 : 9 : 10 .

now, when this mixture is poured again into P, Q and R .

then,

→ Capacity of R = 200 cc .

therefore,

→ Quantity of P in cup R = 200 * [6/(6 + 9 + 10)] = 200 * (6/25) = 48 cc .

→ Quantity of Q in cup R = 200 * [9/(6 + 9 + 10)] = 200 * (9/25) = 72 cc .

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