Question

54. A vessel contains mixture of tin and copper in the ratio of 2:3.
Some amount of mixture is taken out and 28 gm copper is added
to the remaining mixture so that amount of copper becomes
66(2/3)% in the new mixture. If 12(1/2)% of initial mixture is
22.5 gm then, find what amount of tin was taken out from the
initial mixture?​

Answer 1

Answer:

tin amount is 4 gm taken out

Step-by-step explanation:

given that 12(1/2)% of initial mixture is 22.5gm

so 100% is (22.5/(25/2))×100= 180 gm

so the initial amount of tin and copper are in the ratio 2:3

so tin is (2/5)×180=72 gm

and copper is (3/5)×180= 108 gm

as the tin is remove the copper amount in the new mixture is 108+28=136 gm which is 66(2/3)% so total mixture is 100 % = (136/(200/3))×100

= 136×3×100/200= 204 gm

as let the amount of tin remove is x so we can have this 180gm (initail) +28 gm (copper)- x gm tin= 204 gm (new mixture)

=>. 208-x = 204

=>. x = 4 gm