matches problem

Number Theory Level 3

You have 48 matches that were divided into 3 groups.

From the first group, you removed a number of matches equal to those in the second group, and added them to the second group.
Then from the second group, you removed a number of matches equal to those in the third group, and added them to the third group.
Then, from the third group, you removed a number of matches equal to those in the first group, and added them to the first group.
Lo an behold, there are an equal number of matches in each group.

How many matches were there in the second group originally?

The answer is 14.

3 solutions

Milind Joshi
Jun 4, 2014

(1.)x..............y...................48-x-y (2.)x-y..........2y.................48-x-y (3.)x-y.........3y+x-48.......96-2x-2y (4.)2x-2y.....3y+x-48.......96-3x-y After this step....all are equal...solving that eq.....we get y=14

at the final stage, each pile has same no of matches(16), 2a-b=2b-c=2c-a=16, solving this we get 16 as answer, pls tell me where i am wrong?

Balagopal Indiradevi - 6 years, 10 months ago

your first assumption is wrong if you consider that a , b and c are the main no. of matches for each group , so , the right assumption must be as follow : (a-b)*2 = 2b-c = 2c-a = 16

because you take "b" from "a" group and then add the same number at the last step ... that means (a-b)*2 = 16

try this and you will get the right result ..

Ahmed Abdelbasit - 6 years, 9 months ago

that is a good analogy .. that is all right

Ahmed Abdelbasit - 7 years ago

Aslam Hossain
Jul 30, 2014

let 1st grp=x, 2nd=y,3rd=z structuring the works as equation, x-y+(x-y)=16.......1; y+y-z=16...........2; z+z-(x-y)=16......3; as they are equal after rearranging. so 48/3=16; for equation 1; 2x-2y=16; x-y=8; x=8-y; Now input the value of x in equation 3; 2z-(8-y-y)=16; 2z-8=16z; z=12; now putting the value of z into equation 2; 2*12-(x-y)=16; 24-x+y=16; 24-8+2y= 16 (as,x=8-y); 2y=28;
y=14 (ans)

Excellent .. Good thinking

Ahmed Abdelbasit - 6 years, 9 months ago

Nick Turtle
Jan 12, 2018

I'm too lazy to get paper and pencil to use algebra to solve this problem, so I'm going to try working out the problem in my head by solving it backward.

At the end, there is an equal number of matches in each group. Since the total is $48$ , each group has $16$ .

Before that, we removed a number of matches equal to those in the first group from the third group and added them to the first group. This means that at this step, the number of matches in the first group doubled. So, before this step, group 1 had $\frac{16}{2}=8$ matches, group 2 stayed at $16$ , and group 3 (which previously had the $8$ matches given to group 1) at $16+8=24$ matches.

Before that, we removed a number of matches equal to those in the third group from the second group and added them to the third group. This means that at this step, the number of matches in the third group doubled. So, before this step, group 3 had $\frac{24}{2}=12$ matches, group 2 (which previously had the $12$ matches given to group 3) at $16+12=28$ , and group 1 stayed at $8$ .

Before that, we removed a number of matches equal to those in the second group from the first group and added them to the second group. This means that at this step, the number of matches in the second group doubled. So, before this step, group 2 had $\frac{28}{2}=14$ matches, group 1 (which previously had the $14$ matches given to group 1) at $8+14=22$ , and group 3 remained with $12$ .

So, the answer is group two had $14$ matches originally.

matches problem

The answer is 14.

3 solutions

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