Problem Solving

Initially Jar A has 8 pounds of water and Jar B has x pounds of grain. Then 4 pounds of water are transferred from A to B, thoroughly mixed with the grain, and 10 pounds is transferred from B to A. If the final mixture in A has 6 pounds of water then x equals

(a) 10
(b) 25/2
(c) 40/3
(d) 16
e) 18

Answer is D

dtweah wrote:Initially Jar A has 8 pounds of water and Jar B has x pounds of grain. Then 4 pounds of water are transferred from A to B, thoroughly mixed with the grain, and 10 pounds is transferred from B to A. If the final mixture in A has 6 pounds of water then x equals

(a) 10
(b) 25/2
(c) 40/3
(d) 16
e) 18

Jar A v/s Jar B
------ ------
8 x
4 x+4
14 (x+4)-6 => x-6

Given, final solution in Jar A contains 6 pounds of water
=> Out of 10 pounds that were transfered, 8 pounds were that of grain alongwith two pounds of water,

Since the mixture ration of grain and water is not specified in Jar B,

I assume it to be of equally proportianate, and conclude that the remaining in Jar B are 8 pounds of grain + 2 pounds of water.

=> originally x must be 16.

Thus D.

Please correct if wrong.

dtweah wrote:Initially Jar A has 8 pounds of water and Jar B has x pounds of grain. Then 4 pounds of water are transferred from A to B, thoroughly mixed with the grain, and 10 pounds is transferred from B to A. If the final mixture in A has 6 pounds of water then x equals

(a) 10
(b) 25/2
(c) 40/3
(d) 16
e) 18

Since there are 6 pounds of water in A finally, 10 pounds of mixture transferred from B to A contains 2 pounds of water. (remember that jar A contains 4 pounds of water before these 10 pounds from jar B are transferred)

So 10 pounds from jar B contains 2 pounds of water. So the ratio of water and grain in jar B is 1:4.

So obviously after transferring 8 pounds of water from A to B, the ratio of water and grain in B is 1:4.

Hence initially B contains 16 pounds of grain. (since the ratio of water and grain is 1:4 and water is 8 pounds)

All I have done is the backward calculation