mixture problem. please hep with simple solution

This topic has expert replies
User avatar
Newbie | Next Rank: 10 Posts
Posts: 3
Joined: Tue Oct 31, 2017 8:20 am
There are two containers A and B filled with mentha oil with different prices and with volumes 140 and 60 liters respectively. Equal quantities are drawn from both A and B in such a manner that the oil drawn from A is poured in into B and oil drawn from B is poured into A. After doing so, the price per liter becomes equal in both containers. What is the (equal) quantity that was drawn?


what is the concept behind this problm ??

User avatar
Master | Next Rank: 500 Posts
Posts: 134
Joined: Fri Jan 23, 2009 1:23 pm
Location: California
Thanked: 59 times
Followed by:15 members

by dabral » Wed Nov 01, 2017 9:40 pm
Hi shaparage,

With these type of problems, the key is to be able to set up the problem and translate it into an equation. One cannot just memorize the approach to these problems. The variations are too many. In this case, we can say that x liters of solution are moved from A to B, and vice versa. Let the price of oil be p1 $/liter in container A and p2 $/liter in container B. The total cost of oil in container A will be: (140-x)p1 + (x)p2

Similarly, the cost of oil in container B will be (60-x)p2 + (x)p1. Set the price per liter in the two containers to be equal and you will find that the terms involving p1 and p2 drop out and x turns out to be 42 liters.

Cheers,
Dabral
Free Video Explanations: 2021 GMAT OFFICIAL GUIDE.

GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

by Matt@VeritasPrep » Thu Nov 09, 2017 5:55 pm
Say that each liter in A costs $a and each liter in B costs $b. Then the price of the barrels is $140a and $60b, respectively.

After we switch c liters from each barrel to the other, we've got $(140 - c)*a + cb in one barrel and $(60 - c)*b + c*a in the other. Since these create an equal price per liter, we have

((140 - c)a + bc)/140 = ((60 - c)b + ac)/60

60 * (140a - ac + bc) = 140 * (60b - bc + ac)

8400a - 60ac + 60bc = 8400b - 140bc + 140ac

8400a - 200ac = 8400b - 200bc

42a - ac = 42b - bc

42a - 42b - ac + bc = 0

(42 - c) * (a - b) = 0

So either c = 42 or a = b. We're told (in the stem) that the prices are different, so a ≠ b and c = 42 is our only solution.

GMAT Instructor
Posts: 2630
Joined: Wed Sep 12, 2012 3:32 pm
Location: East Bay all the way
Thanked: 625 times
Followed by:119 members
GMAT Score:780

error in subject

by Matt@VeritasPrep » Thu Nov 09, 2017 5:59 pm
By the way, that is not a trivial GMAT problem! It takes more than the normal amount of of setup and legwork to sort out, and feels a bit more like an algebra homework problem than anything you'd encounter on this test.