let suppose Y has water=a and Z has water =b in the begining,,,,,,,, after ist step the water in Z becomes 2 times i.e=b+b=2b and water in Y becomes b less a i.e=a-b ,,,,,,,, after 2nd step water in Y becomes 2 times i.e (a-b)+(a-b)=2a-2b and water in Z becomes 2b-(a-b)=3b-a,,,,,,,,in 3rd step water in Z becomes 2 times i.e 3b-a+3b-a= 6b-2a and finally in Y becomes (2a-2b)-( 3b-a)=3a-5b,,,,,,, finally water becomes 64 in each i.e Y=3a-5b=-64 and Z= -2a+6b=64 ,,,, by solving simultaneously ans comes Y=88 and Z=40

Since it was a computer science problem, I just wrote this and saved myself a whole lot of braincells.

#include<stdio.h>

main(){
    int z=64, y=64;
    z/=2;y+=z;
    y/=2;z+=y;
    z/=2;y+=z;
    printf("y=%d  z=%d\n",y,z);
}

Therfore giving me

Y=88 and Z=40

$$ The total final water in barrel $Y$ and barrel $Z$ is $64+64=128\text{ gallons}$ , therefore the total initial water in $Y$ and $Z$ must also $128\text{ gallons}$ . From the options given, only one option that has a total sum $128\text{ gallons}$ , which is $88\text{ gallons}$ in $Y$ , $40\text{ gallons}$ in $Z$ .

$\text{\# }\mathbb{Q.E.D.}\text{ \#}$

there are some calculations for the solutions of the problem. But most simple is by inspecting the answer: 64+64=128; 88+40=128.

Solve from reverse y= 64 z=64 y= 64+32=96 Z= 64-32=32 y= 96-48 =48 z= 32+48 =80 y= 48+40 = 88 Z= 80-40 =40

Just see the answers. The only answer that has the same amount as the end is 88 and 40.

Y=88 and Z=40 , Because at the end both contain 64 L ie 128 collectively and only this answer confirms to this condition.

There is a shortcut for this: 64+64 = 128

88 + 40 is the only one that adds up to 128. So, there you go

water in both the barrels is 64 gallons so at the end water is 64+64=128 gallons similarly at the begining water should be 128 gallons. From the options only 3rd option gives 88+40=124. Hence answer is 88+40

at the end, both barrels contained 64+64=128. so, at start both should also contain the total 128. so, in 1st option, y=88 & z=40. by adding them we get, 128. so I think, this is correct.

You don't actually need to solve it, the given possible solutions give it away since ony 1 option sum to 128 gallons of water

Both barrels contain 64 gallons. The only solution that has the same is 88 and 40

Total volume of water at the end os 64x2. Only option 2 has total volume of water of 128 ;D

The total volume will remain the same then the total amount = 64 ×2 =128 Only one option accomplish that 88 , 40

Bad options. Water is neither created nor destroyed in the problem. So by conservation law the net water remains a constant in time. Thus only one option satisfies.

first we should add y+z in all options then 88+40=128 then yz/2=128/2=64

the choices are not so intelligent i think. anything that will add up to 128 will be the answer. simple as that.

Back Tracking y=64, z=64 Step 2 y=96 and z=32, Step 3 y=48 and z=80 step 4 y=88 and z=40

starting from backward divide big figure by 2 and add the same value to small figure. Go it 3 time, and you will get answer

The following actions of pouring water from Y to Z , Z to Y blah blah blah is just an illusion to the statement. Focus on the final given that both barrels contain 64 gallons each therefore the total gallons of water is 128 gallons. Among the choices , Letter A is the correct answer in which the equivalent total amount of water is 128 gallons.

simple just look at the ans and add them it must be same as double of 64

let in Y =a+b, Z=a, 1.moved from y so Y=b,Z=2a 2.moved from Z so Y=2b,Z=2a-b 3.moved from Y so Y=2b-2a+b ,Z=4a-2b 3b-2a=64,4a-2b=64 so y=a+b=88,z=a=40

Y Z

1: y-z , 2z;

2:2(y-z),2z-(y-z);

3:2y-2z-(3z-y),2(3z-y);

¦¦¦ 3y-5z=64 - 2y+6z=64,,,,,,y=88;z=40