Question

If there are 6 litres of a drink and 80% of it is the special mixture, and 20% of it is just water, how many litres of water must be added so that the drink has 50% of it's special mixture

PS - Thanks math helper for the other question, but i dont get one part of the question - check my previous message

Answer 1

just saw your reply - thanks!

Answer 2

This is the same concept as your last one.
After reading my previous response, and realizing that in this case
the amount of the "special mixture" does not change, let me know what
you get

Answer 3

ok so this is what i devised out -

0.80(6) + 0x = 0.50(6 + x)

Hence, i get
x = 3.6

I dont know if this is right, for the answer sheet gives
x = 2
thanks

Answer 4

please reply

:sobbing face:

Answer 5

That's what I got also.
Here is why 2 L is not correct:

In the original 6 L there was 80% of the sp (special mixture)
or 4.8 L of sp
This cannot change
so new amount would be 8 L, 1/2 of that is supposed to be sp
but that would be 4 L instead of 4.8 L

my answer of 3.6 L
original amount of sp = .8(6) = 4.8 L
new volume = 6 L + 3.6 L = 9.6 L , where the 3.6 L does NOT contain any additional sp, so .....
amount of sp in that is 1/2 of 9.6 L or 4.8 L
Their answer key is wrong!