Easy math puzzles

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October 15th, 2024 at 3:15:46 PM permalink

...and see that your Step 3 corresponds to "Two Adjacent". I think you need to go through "Two opposite" before you can guarantee a win, thus this could take two moves. Also I can't see how you can avoid the possibility of having to use the alternative route (even if starting with two adjacent rather than two opposite, as the logic is similar). Thus I can only see the solution in five moves.

(The "alternative route" means for the second step, if you find a red-green or green-green, then set them to red-red for a win; otherwise you find red-red and can set them to green-green to ensure the other two are red-green.)

Wizard
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October 16th, 2024 at 2:29:12 AM permalink

Quote: unJon
I may be lost but don�t see how your step 4 works.
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You're right. I'm just spent an hour on a revised solution and see that it fails.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

Gialmere

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October 16th, 2024 at 9:29:45 AM permalink

Quote: Wizard
Quote: Gialmere

Above are six pieces of rope. Imagine that you grasp the two ends of the rope and pull until the rope is straight.

Which of the six ropes will end up with a knot in them, and which ones just form loops that can be pulled out?
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A -- No loop
B -- No loop
C -- Loops
D -- Loop (figure 8 knot)
E -- No loop
F -- No loop

I must admit I got a piece of rope, mimicked these diagrams and pulled on the ends. It's possible I didn't construct some correctly.

I'd be interested to know the method to solve such puzzles mentally. I have a feeling it's some kind of odd/even thing to do with the rope going over/under itself.
link to original post

Hmm, I'm not sure what your answer is saying but...

Only C and D form knots.

Apologies for the delay. I'm having terrible computer issues and can only use my phone these days.
-----------------------------------------

Have you tried 22 tonight? I said 22.

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October 17th, 2024 at 10:02:17 AM permalink

Update 1:35PM 10/17/24 -- Please ignore this post -- It has an error

Here is my revised solution to the four switch puzzle.

Key:
U = Switch in up position
D = Switch in down position.

Since this one is rather stale, I'm not putting it in spoiler tags.

1. For turn 1, choose opposite switches. If the same, flip both switches. If opposite, flip the D switch.
2. For turn 2, choose opposite switches.
3. If the turn 1 switches were originally in the same position, and the turn 2 switches are in the same position, then flip either switch on turn 2. This will result in either the puzzle being solved or switches in an UUDD configuration.
4. If the turn 1 switches were originally both up, and the turn 2 switches are in opposite positions, then flip the up switch. This will result in the puzzle being solved in a DDDD configuration.
5. If the turn 1 switches were originally both down, and the turn 2 switches are in opposite positions, then flip the down switch. This will result in the puzzle being solved in the UUUU configuration.
6. If the turn 1 switches were opposite, and the turn 2 switches are in the same position, then flip either switch. This will either result in a solved puzzle or an UUDD configuration.
7. If the turn 1 switches were opposite, and the turn 2 switches are in opposite positions, then flip the D switch. This will either result in a solved puzzle in the UUUU configuration.
8. For turn 3, if the puzzle isn't solved yet, it must be in a UUDD configuration. Pick adjacent switches and flip both. This will result in the puzzle being solved or a UDUD configuration.
9. For turn 4, pick opposite switches and flip both. This will solve the puzzle.

Last edited by: Wizard on Oct 17, 2024

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

charliepatrick

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October 17th, 2024 at 1:01:20 PM permalink

I'm not sure about 0) then 6) (this is where I couldn't do it in four moves).
0) This is where you finid opposites so U?D? is turned to U?U? (all you now know at this stage is that ?? cannot be UU).
6) You're starting with U?U? and picking opposites
(a) obviously if you see any D, then you're looking at ?? and can win by turning them both to UU.
(b) however if you see UU then you must have picked the same two again. Even if you turn them to UD you're left with U?D?; you have no further information. So my proposed solution is to turn them to D?D?. You know ?? is not UU and now you know ?? is not DD; hence you know ?? = UD so must have UDDD. (On the next move you can turn UDDD into UUDD by flipping opposite DD to UD.)
8) I agree on "Turn 3" if you're in a UUDD configuration you can force a win in two moves by picking adjacent switches. If you see UU turn them to DD for DDDD; if you see DD turn them to UU for UUUU; if you see UD (or DU) swap them around to get UDUD.

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October 17th, 2024 at 1:35:19 PM permalink

Quote: charliepatrick
I'm not sure about 0) then 6) (this is where I couldn't do it in four moves).
0) This is where you finid opposites so U?D? is turned to U?U? (all you now know at this stage is that ?? cannot be UU).
6) You're starting with U?U? and picking opposites
(a) obviously if you see any D, then you're looking at ?? and can win by turning them both to UU.
(b) however if you see UU then you must have picked the same two again. Even if you turn them to UD you're left with U?D?; you have no further information. So my proposed solution is to turn them to D?D?. You know ?? is not UU and now you know ?? is not DD; hence you know ?? = UD so must have UDDD. (On the next move you can turn UDDD into UUDD by flipping opposite DD to UD.)
8) I agree on "Turn 3" if you're in a UUDD configuration you can force a win in two moves by picking adjacent switches. If you see UU turn them to DD for DDDD; if you see DD turn them to UU for UUUU; if you see UD (or DU) swap them around to get UDUD.
link to original post

Thank you for your comments. I agree I screwed up, again.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

charliepatrick

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October 19th, 2024 at 2:24:27 PM permalink

Here is an analysis of the puzzle which may help. I suspect the logic is similar if you start with two adjacent picks, the aim is to get onto the b-c-d path.

You can only have a handful of positions as many are rotations or mirror of the others.
(a) 4-0 : UUUU or DDDD (these are winners)
(b) 3-1 : UDDD or DUUU
(c) 2-2 adjacent : UUDD (which is the same as DDUU, DUUD, UDDU)
(d) 2-2 opposite : UDUD (which is the same as DUDU)

The starting position cannot be (a)
Let�s assume your first move is to pick opposite

1A U?U? : you change this to D?D?
(i) uuud/uduu goes to dudd/dddu = (b)
(ii) udud goes to dddd = win
(iii) uudd : this cannot create the case where you see U?U?.
In this case you know you have (b) (or have won)

1B U?D? : you change this to D?D?
(i) uddd goes to dddd = win
(ii) uudu goes to dudu = (d)
(iii) uddu/uudd goes to dddu/dudd = (b)
The problem is that after 1B, you don�t know whether you are (b) or (d), see (e) below.

(b) (assuming you are starting with DDDU, pick opposite)
(i) D?U? : since you have found the only U, change it to D and win
(ii) D?D? : if you change it to d?u? then you must have duud or dduu = (c)

(c) (starting with DDUU or equivalent, pick adjacent)
(i) DD?? : since the other two are Us, flip DD to UU to win.
(ii) UU?? : since the other two are Ds, flip UU to DD to win.
(iii) DU : since you have DU from DUud, swap them to create UDUD =(d)

(d) (starting with DUDU or equivalent, pick opposite)
(i) D?D? : since the other two are Us, flip the Ds to Us to win
(ii) U?U? : since the other two are Ds, flip the Us to Ds to win

Now 1A U?U? can go via (b) (c) and (d) to win in four moves or fewer
The problem with 1B is you don�t know whether you have (d) or (b).

(I suspect this is where we still need to find a four move solution, but I can�t see it!)
(e) (starting with 1B U?D? and converting it to D?D, not pick opposite)
One way is to pick opposite
(x) U?U? : since you know there are two Ds, you change the two Us to Ds (D?D?)
(i) udud goes to dddd = win
(y) D?D? : you don�t know the other two, this is equivalent to 1A, so change to U?U?
(i) dudu goes to uuuu = win
(ii) dudd goes to uuud = (b)
(z) U?D? : this can only happen if you had uddd
(i) uddd goes to dddd = win
So 1B can either win or go to (b) (which via (c) and (d) can be solved, but it takes five turns).

Summary : 1A goes to (b) (c) and (d) (or wins earlier) and 1B goes to (e) before (b) (c) and (d) (or wins earlier).

charliepatrick

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October 26th, 2024 at 10:10:31 AM permalink

I stumbled across this puzzle on facebook today (so have taken screenshots). The first is how they posed the puzzle and the second (within spoiler tags) shows the construction one might make. My suggestion is try to solve it and then, if it helps, you can use the reference diagram (if you used the same method as they did to solve the puzzle). Note the hidden diagram does not show the solution, just the approach.

ThatDonGuy

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October 26th, 2024 at 12:35:53 PM permalink

Using the diagram, let P be the point where the line segment connecting the two circles' centers intersects the circles.
Let x be the length of the side of the square.
OP = OB = OF = 5; EF = x, so OE = 5 - x
Note that BE = AB / 2 = 5; since AB / 2 = x / 2, BE = 5 - x / 2
Pythagorean Theorem on OEB: (5 - x)^2 + (5 - x/2)^2 = 25
25 - 10 x + x^2 + 25 - x^2 + x^2 / 4 = 25
5/4 x^2 - 15 x + 50 = 25
x^2 - 12 x + 20 = 0
x = 2 or 10, but x < 5 as otherwise BC > OF.
The area of the square = x^2 = 4.

Check: BE = 5 - 1 = 4 and OE = 5 - 2 = 3; OEB is a 3-4-5 right triangle.

charliepatrick

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October 26th, 2024 at 1:38:42 PM permalink

^ Well done - what's nice about the puzzle is how, as you've seen, [your last line] confirms it's the correct answer.

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November 8th, 2024 at 4:01:50 PM permalink

Here is my solution if the radius of the circles were 1:

(1-2x)^2 + (1-x)^2 = 1
5x^5 - 6x + 1 = 0
x = (3+/-2)/5

The only one that makes sense is x=1/5.

So, if the radius of the circles were 1, the area of the square would be (2/5)^2 = 4/25.

Then multiply by 5^2 = 25 if the circles have radius 5: (4/25)*25 = 4

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

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November 8th, 2024 at 4:03:54 PM permalink

I hate to ask a question I can't solve, but this is one of those times. I wrote it in my notes for future "ask the wizard" questions. I put down the answer, but forgot how I solved it. I have a feeling there is some trig substation involved. This one beer worth for an answer in pure form with solution.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

AutomaticMonkey

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November 8th, 2024 at 4:39:01 PM permalink

Quote: Wizard

I hate to ask a question I can't solve, but this is one of those times. I wrote it in my notes for future "ask the wizard" questions. I put down the answer, but forgot how I solved it. I have a feeling there is some trig substation involved. This one beer worth for an answer in pure form with solution.
link to original post

I can solve it. But only by that method I used in high school algebra class, that would earn me a disdainful frown from the teacher and to be called lazy and a waste of great potential.

(He was only half right. I have worked very hard at being a waste of great potential.)

ThatDonGuy

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November 8th, 2024 at 4:52:01 PM permalink

Quote: AutomaticMonkey
Quote: Wizard

I hate to ask a question I can't solve, but this is one of those times. I wrote it in my notes for future "ask the wizard" questions. I put down the answer, but forgot how I solved it. I have a feeling there is some trig substation involved. This one beer worth for an answer in pure form with solution.
link to original post

I can solve it. But only by that method I used in high school algebra class, that would earn me a disdainful frown from the teacher and to be called lazy and a waste of great potential.

(He was only half right. I have worked very hard at being a waste of great potential.)
link to original post

Square both sides: 3 - x = 9 - 6 x^2 + x^4, limited to x <= 3 (as otherwise the square root is imaginary)
x^4 - 6 x^2 + x + 6 = 0
(x + 1) (x - 2) (x^2 + x - 3) = 0
This has solutions -1, 2, -1/2 - sqrt(13)/2, and -1/2 + sqrt(13)/2, but I think only x = -1 is valid for the original equation

ChesterDog

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November 8th, 2024 at 4:52:30 PM permalink

Wizard
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November 8th, 2024 at 5:08:49 PM permalink

Well done, Don and CD. However how did you easily factor:

x^4 - 6 x^2 + x + 6 = (x + 1) (x - 2) (x^2 + x - 3)?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

ChesterDog

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November 8th, 2024 at 5:28:44 PM permalink

Quote: Wizard
Well done, Don and CD. However how did you easily factor:

x^4 - 6 x^2 + x + 6 = (x + 1) (x - 2) (x^2 + x - 3)?
link to original post

I should have mentioned that if x^4 - 6 x^2 + x + 6 has rational solutions, they must be either -6, -3, -2, -1, 1, 2, 3, or 6.

I used "synthetic division" to find that x = -1 is a solution. So, x + 1 is a factor of x^4 - 6 x^2 + x + 6.

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November 8th, 2024 at 5:40:53 PM permalink

Thank you to both of you!

I forgot about the rational root formula. I probably haven't used it since high school.

A well-earned beer to both of you!

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

camapl

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November 11th, 2024 at 5:56:01 PM permalink

Quote: Wizard
Thank you to both of you!

I forgot about the rational root formula. I probably haven't used it since high school.

A well-earned beer to both of you!

link to original post

Tomorrow is my 11th birthday in sobriety� I might have to rethink that now - Oktoberfest, here I come! (I know, it�s already over!)

It�s a dog eat dog world. �Or maybe it�s the other way around!

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November 12th, 2024 at 7:07:14 AM permalink

You have four identical cups and six distinct balls. How many distinguishable ways can you place the balls in the cups?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

camapl

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November 12th, 2024 at 9:32:28 AM permalink

Quote: Wizard
You have four identical cups and six distinct balls. How many distinguishable ways can you place the balls in the cups?
link to original post

6,498

Edit to add� I misread the OP. My optic nerves must�ve been on a break! My calculations consider distinct cups AND balls�

Last edited by: camapl on Nov 12, 2024

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charliepatrick

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November 12th, 2024 at 10:05:27 AM permalink

I've used a spreadsheet to confirm it's 187.

Four identical cups, six unique items: how many arrangements
For arguments sake put (1) into a cup, and see how many are in the main cup.
1 (i) 6-0 Only one way of having six in the main cup
5 (ii) 5-1 Five ways to pick the one that isn't in main cup
10 (iii) 4 - 2 (10 ways to pick the two that will not be in the main cup)
10 (iv) 4 - 1 - 1 (10 ways to pick the two that will not be in the main cup)
10 (v) 3 - 3 (10 ways to pick the two to accompany the main cup)
30 (vi) 3 - 2 - 1 (10 ways to pick the two for the main cup, then 3 ways to pick the solo)
10 (vii) 3 - 1 - 1 - 1 (10 ways to pick the two for the main cup)
5 (vii) 2 - 4 (5 ways to pick the ball to accompany the main cup)
20 (viii) 2 - 3 - 1 (5 ways to pick the ball to accompany the main cup, 4 ways to pick the odd ball)
15 (ix) 2 - 2 - 2 (5 ways to pick the ball to accompany the main cup, 3 ways to organise the rest into pairs)
30 (x) 2 - 2 - 1 - 1 (5 ways to pick the ball to accompany the main cup, 6 ways to pick the pair)
1 (xi) 1 -5 (Only one way of having the others in another cup)
5 (xii) 1 - 4 -1 (5 ways to pick the solo, leaving the other four to go into a cup)
10 (xiii) 1 - 3 - 2 (10 ways to pick the pair, leaving the triple to go into a cup)
10 (xiv) 1 - 3 - 1 - 1 (10 ways to pick the two solos, similar to picking the pair)
15 (xv) 1 - 2 - 2 - 1 (5 ways to pick the solo, 3 ways to organise the rest into pairs)
0 (xvi) 0 - ? (by definition the main cup must have the (1) ball, so impossible.
187

Last edited by: charliepatrick on Nov 12, 2024

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November 12th, 2024 at 12:51:58 PM permalink

I could be the one in error, but I agree with neither answer above.

Quote:
5 (ii) 5-1 Five ways to pick the one that isn't in main cup

Aren't there six ways to pick the singleton ball?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

charliepatrick

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November 12th, 2024 at 2:49:32 PM permalink

^

You may recall I assumed ball #1 was placed in a cup and then considered how the other five could be distributed. Starting with where the #1 ball is (I've called that the main cup) you could have a total of five balls in the main cup, hence #2,#3,#4,#5 or #6 could be the odd ball (5 ways). This is the 5-1 option. The other way is that #1 is alone and the other five #2,#3,#4,#5 and #6 are in one cup, this is (1-5) and has 1 way. This gives a total of 6 ways for a 5:1 distribution.
I've got another approach that I'll be looking at, so hopefully can post that later.

charliepatrick

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November 12th, 2024 at 3:31:55 PM permalink

Let's consider how we can put three balls into the cups.
Then for each of these possibilities look at how the other three might fit in.
(a) All three balls can go into one cup (1 way) (ABC) (empty) (empty) (empty)
(b) Two balls in one cup, and one in another (3 ways) (AB) (C) (empty) (empty) (or AC B, BC A)
(c) One ball in each of three cups (1 way) (A) (B) (C) (empty)

Now for each of these options see how the other three balls might fit in.
(i) All three balls into one cup (1 way) DEF
(ii) Two balls in one cup, one in another (3 ways) DE F
(iii) One ball in each of three cups (1 way) D E F

(a)(i) Same cup (ABC+DEF) or different cup (ABC,DEF) (2 ways) total = 2 ways
(a)(ii) ABC+DE,F / ABC+F,DE / ABC,DE,F (3 ways) total = 9 ways
(a)(iii) ABC+D,E,F / ABC+E,D,F / ABC+F,D,E / ABC,D,E,F (4 ways) total = 4 ways
(b)(i) AB+DEF,C / AB,C+DEF / AB,C,DEF (3 ways) total = 9 ways
(b)(ii) AB+DE,C+F / AB+DE,C,F / AB+F,C+DE / AB+F,C,DE / AB,C+DE,F / AB,C+F,DE / AB,C,DE,F (7 ways) total = 63 ways
(b)(iii) AB+D,C+E,F / AB+D,C+F,E / AB+D,C,E,F / AB+E,C+D,F / AB+E,C+F,D / AB+E,C,D,F / AB+F,C+D,E / AB+F,C+E,D / AB+F,C,D,E / AB,C+D,E,F / AB,C+E,D,F / AB,C+F,D,E (12 ways) total = 36 ways
(c)(i) A+DEF,B,C / A,B+DEF,C / A,B,C+DEF / A,B,C,DEF (4 ways) total = 4 ways
(c)(ii) A+DE,B+F,C / A+DE,B,C+F / A+DE,B,C,F / A+F,B+DE,C / A+F,B,C+DE / A+F,B,C,DE / A,B+DE,C+F / A,B+DE,C,F / A,B+F,C+DE / A,B+F,C,DE / A,B,C+DE,F / A,B,C+F,DE (12 ways) total = 36 ways
(c)(iii) Ax,Bx,Cx (6 ways) / Ax,Bx,C,x (6 ways) / Ax,B,Cx,x (6 ways) / A,Bx,Cx,x (6 ways) total = 24 ways

2+9+4+9+63+36+4+36+24 = 187

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November 12th, 2024 at 3:34:37 PM permalink

Quote: Wizard
You have four identical cups and six distinct balls. How many distinguishable ways can you place the balls in the cups?
link to original post

The problem is not worded clearly. What is a "distinguishable way," especially if the cups are identical?
Since the cups are identical, does any "order" in which they are arranged important?
Also, does there have to be a ball in every cup?
Assuming (a) order is unimportant - e.g. placing "red, blue, green, and yellow balls" in cups is a single "distinguishable way" regardless of which balls go into which cups, and (b) you can leave some, or even all, of the cups empty:

There is 1 way to leave all of the cups empty
There are C(6,1) = 6 ways to use 1 ball
There are C(6,2) = 15 ways to use 2 balls
There are C(6,3) = 20 ways to use 3 balls
There are C(6,4) = 15 ways to use 4 balls
The total is 57, or 56 if you have to put at least one ball underneath a cup

charliepatrick

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November 12th, 2024 at 3:54:13 PM permalink

^^ If the balls were A B C D E and F, then I have assumed that AB CD E F is the same as F AB E CD (as you could rearrange the cups to make it look like AB CD E F) but not the same as AC BD E F. Similarly ABCDE F is different from ABCDF E or A BCDEF. Thus I have assumed that A goes in cup 1, B either goes with A or into cup 2, etc.
^ I have assumed you have to place each of the six balls into a cup, but that cups can land up unused and empty. The reply above addresses a different puzzle where a cup can contain a maximum of one ball and asks how many ways the cups could land up from 6 different balls. Perhaps this is why I have a different answer.

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November 13th, 2024 at 4:45:51 AM permalink

Quote: ThatDonGuy
Since the cups are identical, does any "order" in which they are arranged important?

No. The order does not matter.

Quote:
Also, does there have to be a ball in every cup?

No.

Also, every ball must be used.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

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November 13th, 2024 at 4:48:38 AM permalink

Quote: charliepatrick
^^ If the balls were A B C D E and F, then I have assumed that AB CD E F is the same as F AB E CD

This is correct.

Quote:
Similarly ABCDE F is different from ABCDF E or A BCDEF.

That is correct.

Quote:
I have assumed you have to place each of the six balls into a cup, but that cups can land up unused and empty.

That is true.

Quote:
The reply above addresses a different puzzle where a cup can contain a maximum of one ball and asks how many ways the cups could land up from 6 different balls. Perhaps this is why I have a different answer.

Each cup can contain any number of balls, from 0 to 6.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

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November 13th, 2024 at 12:24:47 PM permalink

I now agree with Charlie's answer. All due apologies for the incorrectly disagreeing before.

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

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November 16th, 2024 at 1:54:42 PM permalink

Quote: Wizard
You have four identical cups and six distinct balls. How many distinguishable ways can you place the balls in the cups?
link to original post

I got complaints this puzzle wasn't worded clearly. How would you have worded it?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

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November 16th, 2024 at 1:55:36 PM permalink

Nathan buys 100 pounds of beans. The beans consist of 99% water by weight. She leaves them in the sun to let them dry out. The next day, she finds they consist of 98% water. What is their new weight?

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

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November 16th, 2024 at 2:32:29 PM permalink

Quote: Wizard
Quote: Wizard
You have four identical cups and six distinct balls. How many distinguishable ways can you place the balls in the cups?
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I got complaints this puzzle wasn't worded clearly. How would you have worded it?
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You have four identical cups and six distinct balls. How many distinguishable ways can you place the balls in the cups, such that each ball is under a cup?
You may put more than one ball under a particular cup, and one or cups may be empty.
Two ways are "distinguishable" if they cannot be made the same merely by moving the cups around.

Quote: Wizard
Nathan buys 100 pounds of beans. The beans consist of 99% water by weight. She leaves them in the sun to let them dry out. The next day, she finds they consist of 98% water. What is their new weight?
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Originally, the beans consist of 99 pounds of water and 1 pound of "bean."
Assuming the amount of actual bean is unaffected by the drying, there is still 1 pound of bean, so there must be 49 pounds of water.
The new weight is 50 pounds.

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November 16th, 2024 at 2:38:56 PM permalink

Quote: ThatDonGuy
You have four identical cups and six distinct balls. How many distinguishable ways can you place the balls in the cups, such that each ball is under a cup?
You may put more than one ball under a particular cup, and one or cups may be empty.

Thank you. That seems rather wordy. How about: "You have four identical cups and six distinct balls. How many ways can all six balls be placed in the cups?

Note I removed the word "distinguishable" as that seemed to cause more confusion than clarity.

Quote:

Originally, the beans consist of 99 pounds of water and 1 pound of "bean."
Assuming the amount of actual bean is unaffected by the drying, there is still 1 pound of bean, so there must be 49 pounds of water.
The new weight is 50 pounds.

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Correct!

"For with much wisdom comes much sorrow." -- Ecclesiastes 1:18 (NIV)

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November 16th, 2024 at 4:23:25 PM permalink

Quote: Wizard
Nathan buys 100 pounds of beans. The beans consist of 99% water by weight. She leaves them in the sun to let them dry out. The next day, she finds they consist of 98% water. What is their new weight?
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Is the answer 198 pounds? 🤔💡

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