Question

A vessel is full of a mixture of kerosene and petrol in which there is 18% kerosene.Eight litres are drawn off and then the vessel is filled with petrol.If the kerosene is now 15% how much does the vessel hold?

a)40 lit b)32 lit c)36lit d)48lit

Answer 1

initially
kerosene=18%
so petrol =82%
after
kerosene =15%
petrol=85%

so 3% =8 lit
so total kerosene in the mixture =(8x18/3) =48 lit

Answer 2

somebody please explain the above solution

Answer 3

@amistre64

Answer 4

from yesterday

k+p = initial amount

k/(k+p) = .18
p/(k+p) = .82

(k+p) - 8 + Lp is the new amount

k/(k+p) - 8 + Lp = .15
p/(k+p) - 8 + Lp = .85

Answer 5

forogt a set ofo () ...

Answer 6

k/.18 = p/.82

k = 9p/41

(9p/41)/((9p/41+p) - 8 + Lp) = .15
p/((9p/41+p) - 8 + Lp) = .85

hmm, the wolf says no solution exists

Answer 7

what is the error in my process?

Answer 8

i think when 8 litres of kerosene is taken ,then 8 litres of petrol is filled

Answer 9

hmm, i cant see that in the instructions, but its not a bad assumption

Answer 10

removing 8L doesnt change our percentage, only adding in the new p does

Answer 11

can u explain the solution given in first comment,

Answer 12

how can we say that 3% = 8L?

Answer 13

because when 8 litres is taken kerosene $ reduces from 18 to 15

Answer 14

also, the question seems to ask how much does the vessel hold, and not how much kerosene is in the vessel. the amount of kerosene hasnt changed any

Answer 15

if we have k and p in the same container, and remove 8L we havent changed the ratio by removing some of the contents

Answer 16

its only in adding new p that the ratio gets changed

Answer 17

in the 8 litres removed there can be petrol and kerosene

but petrol is filled after that and kersosene % is now 15

so it can be 8 L= 3 %

Answer 18

if we start with 100L that has a ratio of 18k to 82p

then removing 8L still leaves us with 92L with a ratio of 18k to 82p

Answer 19

adding p to this mixture will give us 92 + L, at a ratio of 15k to 85p

Answer 20

removing 8L means that we removed: 8(.18)k and 8(.82)p

the ratio remains the same, but the quantity reduces

Answer 21

k/(k+p) = .18
p/(k+p) = .82

k/(k-8(.18)k + p-8(.82)p + Lp) = .15
p/(k-8(.18)k + p-8(.82)p + Lp) = .85

Answer 22

i found this equation somewhere

18[1-(8/x)] = 15

Answer 23

also found this but i cant understand

Kerosene initial = 0.18 of Capacity of Vessel...
When 8 litres are withdrawn,Total Kerosene withdrawn=0.18*8=1.44 litres and total Petrol withdrawn = 8-1.44 = 6.56 litres..
After this percentage of kerosene = .15 of Capacity of Vessel...

Thus 1.44 litres --> 3 %
Hence 1% --> .48 litres
hence 100% --> 48 litres..

Answer 24

i can only explain my own ideas ..

lets try this new setup with k=9p/41 since we have reduced the original amount properly while retaining the same initial construction

Answer 25

wolf hates me ... ugh. im at a loss for it and im not going to be able to explain someone elses approach.

Answer 26

the first part seems to be the same conclusion i came to, removing 8 removes proportionate parts

Answer 27

i dont follow the reasoning afterwards tho .. somewhat, but it seems to be making an assumption to me is all

Answer 28

1.44L of k was removed. but not sure why that is 3% of the new mixture