Numbers - Please help

[franco6571]

Dawn, Ann and Megan had 504 beads altogether. Ann gave some of her beads to Megan and the the number of beads Megan had doubled. Megan then gave some of her beads to Dawn and the number of beads Dawn had also doubled. In the end, all of them had the same number of beads. How many beads did Ann have at first?


Note: I have the answer for this question in the book but I still do not understand the working and how they got the answer. Maybe someone here can show a different way to get the answer which I can understand.

 

[HallsofIvy]

Let a be the number of beads Ann had, initially, let b be the number of beads Dawn had, initially, and let c be the number of beads Megan had, initially.

"Dawn, Ann, and Megan had 504 beads altogether". So a+ b+ c= 504.

"Ann gave some of her beads to Megan". Let that number of beads be x, so Ann now has a- x beads and Megan has c+ x. "The number of beads Megan had doubled". So c+ x= 2c and, subtracting c from both sides, x= c. In order to double the number of beads Megan had, Ann must have given Megan a number of beads equal to what she had.

At this point Ann has a- x= a- c beads and Megan has c+ x= 2c beads.

"Megan then gave some of her beads to Dawn and the number of beads Dawn had also doubled."
Let the number of beads Megan gives to Dawn be y. Now Megan has 2c- y beads and Dawn has b+ y beads. Since that is double what Dawn had, b+ y= 2b so y= b.

At this point, Ann has a- c beads, Megan has 2c- b beads and Dawn has 2b beads.

"In the end, all of them had the same number of beads."
So a-c = 2c- b and 2c- b= 2b. We also have, from the first, a+ b+ c= 504. We have three linear equations to solve for a, b, and c.

We can reduce a- c= 2c- b to a+ b- 3c= 0 so a= 3c- b.
We can reduce 2c- b= 2b to 2c- 3b= 0 so c= (3/2)b and then a= 3c- b= (9/2)b- b= (7/2)b.

Replace a by (7/2)b and c by (3/2)b in a+ b+ c= 504 to get
(7/2)b+ b+ (3/2)b= 6b= 504.

Solve that for b and then, since the problem asks specifically for the number of Ann's beads at first, use a= (7/2)b.

 

[Deleted member 4993]

Dawn, Ann and Megan had 504 beads altogether. Ann gave some of her beads to Megan and the the number of beads Megan had doubled. Megan then gave some of her beads to Dawn and the number of beads Dawn had also doubled. In the end, all of them had the same number of beads. How many beads did Ann have at first?


Note: I have the answer for this question in the book but I still do not understand the working and how they got the answer. Maybe someone here can show a different way to get the answer which I can understand.

Since you posted this problem in arithmetic, HoI's method may be "hard" for you. You can solve this by elementary arithmetic also;

Since after the end all three had equal number beads - they each had (504/3=) 168 bead

Since Dawn had doubled her bead count - she must have started with (168/2 =) 84 beads

Meghan must have given her 84 beads. So before giving beads to Dawn, Meghan must have had (168+84 = ) 252 beads. Meghan must have started with (252/2 = )126 beads.

Now you calculate the number beads that Ann started with at first.

Make sure you check your answer.......

 

[Dr.Peterson]

Dawn, Ann and Megan had 504 beads altogether. Ann gave some of her beads to Megan and the the number of beads Megan had doubled. Megan then gave some of her beads to Dawn and the number of beads Dawn had also doubled. In the end, all of them had the same number of beads. How many beads did Ann have at first?

Note: I have the answer for this question in the book but I still do not understand the working and how they got the answer. Maybe someone here can show a different way to get the answer which I can understand.

When what you don't understand is someone else's work, it's a good idea to show that work -- even coming from a book, it could be wrong! And if it's right, showing it and telling us what part you don't follow could save the helpers a lot of time.

I am guessing SK's "working backward" approach is likely what the book showed; but it can be hard to explain such thinking briefly in words, so it may be we just have to help you with a bit of their work that was too brief.

 

[franco6571]

This is the answer from the book. Please check to see if the answer is right or wrong. Thanks

 

[franco6571]

Let a be the number of beads Ann had, initially, let b be the number of beads Dawn had, initially, and let c be the number of beads Megan had, initially.

"Dawn, Ann, and Megan had 504 beads altogether". So a+ b+ c= 504.

"Ann gave some of her beads to Megan". Let that number of beads be x, so Ann now has a- x beads and Megan has c+ x. "The number of beads Megan had doubled". So c+ x= 2c and, subtracting c from both sides, x= c. In order to double the number of beads Megan had, Ann must have given Megan a number of beads equal to what she had.

At this point Ann has a- x= a- c beads and Megan has c+ x= 2c beads.

"Megan then gave some of her beads to Dawn and the number of beads Dawn had also doubled."
Let the number of beads Megan gives to Dawn be y. Now Megan has 2c- y beads and Dawn has b+ y beads. Since that is double what Dawn had, b+ y= 2b so y= b.

At this point, Ann has a- c beads, Megan has 2c- b beads and Dawn has 2b beads.

"In the end, all of them had the same number of beads."
So a-c = 2c- b and 2c- b= 2b. We also have, from the first, a+ b+ c= 504. We have three linear equations to solve for a, b, and c.

We can reduce a- c= 2c- b to a+ b- 3c= 0 so a= 3c- b.
We can reduce 2c- b= 2b to 2c- 3b= 0 so c= (3/2)b and then a= 3c- b= (9/2)b- b= (7/2)b.

Replace a by (7/2)b and c by (3/2)b in a+ b+ c= 504 to get
(7/2)b+ b+ (3/2)b= 6b= 504.

Solve that for b and then, since the problem asks specifically for the number of Ann's beads at first, use a= (7/2)b.

++++++++++++++
Good Day Sir
Thank you for taking the time to answer my question. I have posted the answer from the book. Please check to see if the book is right or wrong.
Thank You again.

Regards
Francis A,

 

[franco6571]

Since you posted this problem in arithmetic, HoI's method may be "hard" for you. You can solve this by elementary arithmetic also;

Since after the end all three had equal number beads - they each had (504/3=) 168 bead

Since Dawn had doubled her bead count - she must have started with (168/2 =) 84 beads

Meghan must have given her 84 beads. So before giving beads to Dawn, Meghan must have had (168+84 = ) 252 beads. Meghan must have started with (252/2 = )126 beads.

Now you calculate the number beads that Ann started with at first.

Make sure you check your answer.......

++++++++++++++++++++++++++++++
Good Day Sir
Thank you for taking the time to answer my question. I have posted the answer from the book. Please check to see if the book is right or wrong.
Thank You again.

Regards
Francis A,

 

[franco6571]

When what you don't understand is someone else's work, it's a good idea to show that work -- even coming from a book, it could be wrong! And if it's right, showing it and telling us what part you don't follow could save the helpers a lot of time.

I am guessing SK's "working backward" approach is likely what the book showed; but it can be hard to explain such thinking briefly in words, so it may be we just have to help you with a bit of their work that was too brief.

+++++++++===
Good Day Sir
Thank you for taking the time to answer my question. I have posted the answer from the book. Please check to see if the book is right or wrong.
Thank You again.

Regards
Francis A,

 

[Dr.Peterson]

Dawn, Ann and Megan had 504 beads altogether. Ann gave some of her beads to Megan and the the number of beads Megan had doubled. Megan then gave some of her beads to Dawn and the number of beads Dawn had also doubled. In the end, all of them had the same number of beads. How many beads did Ann have at first?

I would not describe what was written as an explanation of how to obtain the answer, but only of the final part of that process. It doesn't tell us how to draw the picture.

If I were trying to use this method, I might first sketch the three strips without the little squares, just showing a dotted part of A moving to B, and a shaded part of that moving to D. Then, since the results for M and D are equal, I would see that the shaded parts and the original D are equal, so each must be equal to the original D. At that point I might add a vertical line in M matching the original D.

Then I'd realize that the entire strip for M is divided into thirds, while the original M is 1/2 of that whole. So I could divide the original M into three boxes, and use that same unit to divide the others. At that point I'd have the drawing they show, and I could start the thinking they explain.

Algebra is easier! In particular, it's far easier to explain. To really do so using the method shown, I'd need to take the time to draw pictures for you.