# These are the Medium Puzzle Grid Solutions

WELCOME TO PUZZLES-ONLINE-NICHE

This page contains the MEDIUM puzzle grid solutions to all of the PDF logic puzzle grids on our web site.

You may consider this page a sort of cheat or crib sheet for our logic puzzles.

You can expand each solution by clicking on the title, and some solutions are available in

PDF printable format, (click "PDF PRINTABLE VERSION" link).

Solutions banner

MEDIUM :

## MEDIUM :

 1
AT THE PET SHOP(2)-SOLUTION

## AT THE PET SHOP(2) (show / hide)

Bill -Monkey-Toto
Jen -Snake-Wobbles
Roger -Dog-Spot
Susan -Cat-Spike

#### step-by-step:( show / hide)

• Lets look at the first clue "Roger's pets name was not Wobbles."

• >Locate the row labeled 'Roger' in the chart and find the column with 'Wobbles'
Now click the grid square[Roger-Wobbles] until the 'red xx' appears.

• Lets look at the next clue "The girl who named her pet Spike is afraid of snakes."

• The first part of the clue actually tells us that 'no boy' (Bill, Roger) would have named his
pet 'Spike', So locate grid squares [Bill-Spike, Roger-Spike] and place a 'red xx' in both.

As for the second part of the clue ("Spike...is afraid of snakes"), we could infer that she would not have named her pet Spike.
(So locate the grid square [Spike-Snake] and place a 'red xx' there).

• Lets look at the next clue "Jan did not name her pet Toto who is not a Dog."

• The first part here is a simple elimination, so locate grid square[Jan-Toto] and place a 'red xx' there.

Again for the second part("Toto...is not a dog"), we have the grid square elimination [Toto-Dog] as well.

• The next clue "Bill purchased a monkey."

• Not difficult at all, simply locate [Bill-Monkey] and place a 'green box' there.

*IMPORTANT:*
(However 'DO NOT' forget to make the following grid square eliminations) :

[Bill-Cat,Bill-Dog, Bill-Snake], as well as
[Jan-Monkey,Roger-Monkey, Susan-Monkey] and throw in grid square
[Spike-Monkey]-- because we know from the previous clue, Bill did not name his pet Spike.

• Our next clue is "No pets name begins with the same letter as its species."

• Now the only grid square that meets this requirement is "Spot-Snake".
So find [Spot-Snake] and place a 'red xx' in that box.

• The very next clue we have is"The Cat was named either Spike or Wobbles."

• This merely implies that the cat , was 'NOT' also named either Spot or Toto ,

leading us to the elimination of the grid squares [Spot-Cat, Toto-Cat].

• Lets proceed to the last clue "Jan likes the name Wobbles and gave that name to the pet Snake she purchased."

• So locate the grid squares [Jan-Snake, Jan-Wobbles, Wobbles-Snake] and fill them all with 'green' boxes.

We also start our grid square eliminations as follows:
Locate (and eliminate) [Jan-Cat, Jan-Dog, Jan-Spike, Jan-Spot],
as well as [Bill-Wobbles,Susan-Wobbles,Roger-Snake, Susan-Snake, Snake-Toto],
and don't forget to eliminate the lower grid squares [Wobbles-Cat,Wobbles-Dog, Wobbles-Monkey].

*NOTE : * (Since we have no remaining clues)
Lets search our grid for more possible 'eliminations-solutions' ,
The first one is "Susan-Spike" ,(it is the only remaining grid square in the 'Spike' column).

So after the grid square [Susan-Spike] is filled with a 'green box',
Make the following grid square eliminations : [Susan-Spot, Susan-Toto].

Looking at the bottom grid squares we see in the row labeled 'Toto' there is only one
possible solution remaining: 'Toto-Monkey' , (which eliminates 'Spot-Monkey'), leaving
only 'Spot-Dog' (as a possible solution in the 'Spot' row) , ( which in turn eliminates 'Spike-Dog'),
and leaves 'Cat-Spike' as the last 'pet-name' combo.

Now knowing Susan's pet was named 'Spike' ( and that 'Spike is a Cat')
We can conclude : Susan's pet is a 'Cat' ,
(and after filling the grid square[Susan-Cat] with a 'green box') , we eliminate [Roger-Cat]
(and select the solution[Roger-Dog]) , and because the dog is named 'Spot' ,
it must also be true that Roger (must have) purchased the dog,
So select grid square [Roger-Spot and highlight it 'green' (which eliminates [Roger-Toto] ), and reveals
the final grid for the puzzle: [Bill-Toto], (and eliminates [Bill-Spot]).

• Congratulations! Puzzle solved. To summarize:

Bill chose the monkey and named it Toto.
Jan chose the snake and named it Wobbles.
Roger chose the dog and named it Spot.
Susan chose the cat and named it Spike.
###### PDF PRINTABLE VERSION

2.

AIRPORT ARR/DEP-SOLUTION

## AIRPORT -SOLUTION (show / hide)

Mary-10am-1:15pm
Kathy-9am-2pm
Susan 12-1:45pm

#### step-by-step:( show / hide)

• Lets look at the first clue "The person who arrived earliest also left latest."

• Locate the row labeled '2:00pm' in the chart and find the column with '9am'
Now click the grid square[2:00pm-9am] until the 'green box' appears.

You can also eliminate [2:00pm-10am, 2:00pm-1130am, 2:00pm-12noon, 1:15pm-9am, 1:30pm-9am, 1:45-9am] by placing a 'red xx' in each square.

• The next clue begins:"Max arrived sometime after Kathy , but before Susan.."

• The fact that 'Max arrived after Kathy' means he could not have arrived at 9am( nor left at 2pm).
So click the grid square[Max-9am, Max-2:00pm] until the 'red xx' appears.

• Lets look at the next clue "Susan departed before Kathy, who departed last.."

• The key to this clue is 'Kathy, who departed last'

To solve this clue : First locate[Kathy-2:00pm] and fill that square with a 'green box',
that Kathy must be this 'person', go ahead and click square [Kathy-9am] until a green box appears there.

*IMPORTANT:*
(However 'DO NOT' forget to make the following grid square eliminations) :

[Kathy-10am,Kathy-1130am, Kathy-12noon,Kathy-1:15pm, Kathy-1:30pm, Kathy-1:45pm], as well as

• Our next clue is "Adam's arrival time was exactly 2 hours before his departure time."

• This clue requires us to use a bit of math, adding 2 to each arrival time to match a departure time.
So after trying 10am +2 =12, 1130am+2= 1:30pm , we see the only logical solution is 1130am-1:30pm
Therefore we locate (and fill with a 'green box') grid squares [Adam-1130am,Adam-1:30pm], while eliminating these grid squares
and [1:15pm-1130am, 1:45pm-1130am], and then we can fill in the square [1:30pm-1130am],
and finally eliminate these squares : [1:30pm-10am, 1:30pm-12noon].

• Lets proceed to the last clue "Max left exactly 15 min prior to Adam."

• Since Adam left at 1:30pm, Max must have left at 1:15pm, therefore go ahead and fill grid square[Max-1:15pm]
with a 'green box' and eliminate [Susan-1:15pm, Max-1:45pm], ( which means [Susan-1:45pm] must be a 'green').

To solve the remaining grid squares, we rely on a piece of a previous clue ("....Max arrived....before Susan")
Therefore Max arrived at 10am and not 12noon(which is Susan's arrival time).

• Congratulations! Puzzle solved. To summarize:

Max 10am and 1:15pm.
Kathy 9am and 2:00pm.
Susan 12noon and 1:45pm.
###### PDF PRINTABLE VERSION

3.

INSURANCE CO.-SOLUTION

## INSURANCE CO.-SOLUTION (show / hide)

Bill -Home-Jane.
Joan-Life-Mary.
Roger-Auto-Stan.
Susan-Theft-Alex.

#### step-by-step:( show / hide)

• Lets look at the first clue "Roger did not sell a policy to a woman."

• Locate the row labeled 'Roger' and find the columns with both women customers( 'Jane', 'Mary')
Now click the grid squares[Roger-Jane, Roger-Mary] until the 'red xx' appears.

• Our next clue "Mary did not buy her policy from Susan."

• Click the grid squares[Susan-Mary] until the 'red xx' appears.

• Reading the next clue"Jane purchased neither the auto or the life policy."

• Click the grid squares[Jane-Life, Jane-Auto] until the 'red xx' appears.

• The next clue is "A woman client purchased a new life insurance policy with a woman agent."

• Locate the row with women clients ('Mary, Jane') and the column with 'LIFE', notice that
Jane-Life has been eliminated, therefore Mary has to be the "...woman client " of this clue.
Therefore, click on grid square [Mary-Life] until the 'green box' appears.
*IMPORTANT:*
(However 'DO NOT' forget to make the following grid square eliminations) :
[ Mary-Home, Mary-Auto, Mary-Theft, Alex-Life, Stan-Life], and since neither roger or susan has sold a policy to Mary, eliminate these grid squares:[Susan-Life, Roger-Life]

Now from the last part of this clue ( "...woman agent ") , we can make
the following elimination: [Bill-Mary] ,
(By doing so we can see that the 'woman agent' must be 'Joan' !).

So click on the grid squares [Joan-Life, Joan-Mary] until the 'green box' appears for both.

*IMPORTANT:* Lets make our grid square eliminations for the following:
[Joan-Home, Joan-Auto, Joan-Theft, Bill-Life,Roger-Life, Susan-Life] as well as
[Joan-Alex, Joan-Stan, Joan-Jane,Roger-Mary, Susan-Mary].

• The next clue "Stan bought an auto policy from either bill or roger . "

• Locate and click on grid square[Stan-Auto] until this box is 'green'.

Make your grid eliminations: [Stan-Home, Stan-Theft, Alex-Auto].
The second part of this clue ("...from either bill or roger" ), allows us to make another elimination
of the grid square [Susan-Auto].

• This next clue"Both women agents did not sell policies to both women (neither did the men agents) ."

• To understand this clue we take into account that agent 'Joan has sold a policy to Mary', therefore,

Susan could not have sold a policy to Jane. Locate [Susan-Jane] and place a red xx in that square.
This leads to the conclusion that Bill must have sold the policy to Jane (locate grid square
[Bill-Jane] and click until the 'green box' appears there- this of course leads to the
elimination of [Bill-Alex, Bill-Stan, Bill-Auto] ).

• Now our last clue is"Susan did not sell the home policy".

• First locate [Susan-Home] and after adding the 'red xx', we see that Susan must have
sold the "Theft" policy.

Select grid [Susan-Theft] and add a 'green box' there, making the eliminations:
[Roger-Theft, Bill-Theft].

This results in the conclusion that Bill must have sold the "HOME" policy
So find [Bill-Home, Jane-Home] and add a 'green box' to each
( which means[Roger-Home, Alex-Home] are eliminated in turn),

This leads to the following conclusions: Roger sells the auto policy to Stan!
(grid square[Roger-Auto,Roger-Stan] to 'green), (eliminating
[Roger-Alex,Susan-Stan], leading to the final grid squares:
[Susan-Alex, Alex-Theft] to 'green'.

• Congratulations! Puzzle solved. To summarize:

Bill sold the Home policy to Jane.
Joan sold the Life policy to Mary.
Roger sold the Auto policy to Stan.
Susan sold the Theft policy to Alex.
###### PDF PRINTABLE VERSION

4.

ICE CREAM TREATS-SOLUTION

## ICE CREAM TREATS (show / hide)

Ben -Lotso choc- 2pm.
Jerry-Rocky nut-3pm.
Reba-Alamo fudge-1pm.

#### step-by-step:( show / hide)

• Lets look at the first clue "Sadie did not come in First."

• Locate the row labeled 'Sadie' and find the column '1pm'
Now click the grid square[Sadie-1pm] until the 'red xx' appears.

• Our next clue "Ben came in before at least one other person he, like Reba did not order Rocky nut."

• Lets evaluate the second part first, ( "...he, like Reba did not order Rocky nut").
Locate and eliminate the grid squares [Ben-Rocky Nut, Reba-Rocky nut].
Now the clue"...before at least one other person", translates to , he didnt come in last, so eliminate grid square[Ben-4pm].

• The next clue is"The woman who ordered Alamo fudge came in before Jerry , (who didn't come in before 2 pm) ."

• First since only a woman ordered Alamo fudge, we can eliminate the grid squares
[Ben-Alamo-Fudge, Jerry-Alamo-Fudge].
This woman also came in before Jerry, who did not come in prior to 2pm.
Lets eliminate the following grid squares :[Jerry-1pm, Jerry-2pm], and [Alamo-Fudge-4pm].

• Our next clue"The one who ordered Lot so Choco came in at 2 pm".

• Simple. Locate grid square [2pm-Lotso-Choco] and click until the 'green box' appears. This will lead to
the following eliminations:[2pm-Pink-Jub,2pm-Rocky nut, 2pm-Alamo fudge] as well as
[1pm-Lotso Choco,3pm-Lotso Choco,4pm-Lotso choco].

• The clue"The two women in some order, entered the shop at least 3 hours apart".

• Requires a bit of logical math, first we know that Sadie didnt come in at 1pm, therefore lets assume Sadie came in at 2pm, that means Reba would have had to come in at 11am or 5pm(at least 3hours apart), which is not possible.
The only possible way that this can be true is if Reba came in at 1pm, then Sadie comes in at 4pm.
Locate the grid squares [Reba-1pm, Sadie- 4pm] and fill them in with 'green boxes'.

This will result in the following grid square eliminations:
Which results in the grid square [Ben-2pm] becoming the only solution, for 2pm, and thus it follows that grid square
[Jerry-3pm] as a remaining solution- highlight both squares green.
We can also make some more logical conclusions-
Since the one who ordered lotso choco came in at 2pm, then Ben must have ordered lotso choco!
( find[Ben-Lotso choco] fill it in with a 'green box'),which in turn results in the following
eliminations: [Ben-Pink Jub, Reba-Lotso, Sadie-Choco, Jerry-Lotso].

• Lets look at our last clue"Rocky nut was ordered somewhere between 2 and 3 pm."

• With a previous clue ( "...Lotso Choco....2pm" )already claiming the 2pm time slot, then Rocky nut could have only
been ordered at 3pm.
Find grid square [Rocky nut- 3pm] and click until the 'green box' appears.

Lets make the following eliminations [Pink Jub-3pm, Alamo fudge-3pm, 1pm-Rocky nut, 4pm-Rocky nut]
This reveals the following solution squares [4pm-Pink jub, 1pm-Alamo fudge ], which can be made 'green'.

The rest of the grid-solution-combos can be solved without additional clues, by employing some logical deduction

Reba came in at 1pm and Alamo Fudge was purchased at 1pm, thus Reba purchased Alamo Fudge.
( If Reba purchased Alamo fudge we eliminate grid square [Reba-Pink jub, Sadie-Alamo fudge ] )
The one who came in at 3pm purchased Rocky Nut, therefore Jerry came in at 3pm and purchased Rocky nut!
If jerry purchased Rocky nut, he 'DID NOT' purchase Pink Jub(eliminate grid square [Jerry-Pink jub],

• Congratulations! Puzzle solved. To summarize:

Ben purchased Lotso Choco at 2pm.
Reba purchased Alamo fudge at 1pm.
Sadie purchased Pink jub at 4pm.
Jerry purchased Rocky nut at 3pm.
###### PDF PRINTABLE VERSION

5.

PLANET LOGIC GRID PUZZLE-SOLUTION

## PLANET LOGIC GRID PUZZLE-SOLUTION (show / hide)

Bob -Margo-Monday.
Carl-Bargo-Thursday.
Edie-Argo-Tuesday.
Igor-Sargo-Friday.
Max-Fargo-Wednesday.

#### step-by-step:( show / hide)

• Lets look at the first clue "Carl and Igor discovered their planets Sargo and Bargo (in some order) , on consecutive days but not starting Monday ."

• First find 'Carl' and Igor in the grid, we are given that between the two scientists they discovered -Sargo & Bargo,
to the exclusion of the other scientists, thus we thus we can make the following grid square eliminations:
[Carl-Argo, Carl-Fargo, Carl-Margo, Igor-Argo, Igor-Fargo, Igor-Margo] as well as
[Bob-Bargo,Bob-Sargo, Edie-Bargo,Edie-Sargo, Max-Bargo, Max-Sargo].

For the second part (".... not starting monday ), we locate the grid squares
[Carl-Monday, Igor-Monday] and place 'red xx' in those squares.

• "Bob did not discover Fargo which was discovered sometime later in the week after Bob's discovery".

• To begin simply eliminate grid square [Bob-Fargo] .
Next locate and eliminate [Monday-Fargo], because "....Fargo was discovered ....later in the week"
It is also possible to eliminate [Bob-Friday], because " ......Fargo was discovered...after Bob's discovery."

• "Argo was discovered before Bargo , but after Margo."

• Eliminate [Argo-Friday] because it was 'before Bargo' and
Eliminate [Argo-Monday] because it was 'after Margo'. Also
Eliminate [Margo-Friday] because it was 'before Argo'. Lastly
Eliminate [Bargo-Monday, Bargo-Tuesday] because it was after BOTH Argo & Margo.

• "Edie discovered her planet the very next day after Bob's discovery , she did not find Margo ."

• Eliminate [Edie-Monday] because it was 'after Bob's discovery'.
Eliminate [Edie-Margo] because she did not find Margo.

• "Sargo was discovered last ."

• Locate grid square [Fri-Sargo] as 'last', can only mean 'Friday'. Also make these eliminations:
[Fri-Bargo, Fri-Fargo] as well as [Mon-Sargo, Tue-Sargo, Wed-Sargo, Thu-Sargo ]
This also means Margo was discovered first, locate and fill in a 'green box' for [Mon-Margo], and
eliminate [ Tue-Margo, Wed-Margo, Thu-Margo ]

• "Carl discovered his planet 3 days after Bob ."

• From this clue we deduce that Bob discovered his planet either on monday ( Carl-Thursday) or
tuesday(Carl-Friday).
Make the following grid square eliminations:
[Bob-Wed, Bob-Thu, Bob-Fri] and [Carl-Mon, Carl-Tue, Carl-Wed ].
It is also true since Bob's discovery was either monday or tuesday, edie's discovery must be
tuesday or wednesday, therefore eliminate [Edie-Thu, Edie-Fri].

• "The day of Max's discovery Margo and Argo had already been found."

• This clue is really telling us that 'Max' discovered NEITHER of these planets.
So locate grid squares [Max-Argo, Max-Margo] and fill with 'red xx'.
Max could have only discovered Fargo, therefore set grid [Max-Fargo] to 'green' (which in turn eliminates [Edie-Fargo] ),
Which in turn means [Edie-Argo] is highlighted to 'green', from which it follows that the grid square
[Bob-Argo] is eliminated, leaving only [Bob-Margo] as the corrrect solution for Bob's row.
Of the possible days for Bob-Margo, only Monday is correct, and this also means Edie's day was 'tuesday'("the very next day after Bob's discovery") therefore these grid squares
[Bob-Mon, Mon-Margo,Edie-Tue] are highlighted 'green' (while eliminating Edie-Wed,Bob-Tue, Igor-Tue, Max-Mon, Max-Tue]
(Now that we know Bob's discovery day we also know Carl's "....three days after Bob")
So highlight in 'green' grid square[Carl-Thu], while eliminating
[Carl-Fri, Igor-Thu,Max-thu]
We now have more eliminations including [Carl-Sargo], because remember "Sargo was discovered Last" (friday).
If Carl did not discover Sargo, he could have only discovered Bargo( "Carl and Igor discovered their planets , Sargo and Bargo")
Leaving Igor to discover Sargo, so lets fill in the grid squares [Carl-Bargo, Igor-Sargo], and make
further eliminations including:[Igor-Bargo, Igor-Wed], Leaving grid square [Igor-Fri]
to be highlighted 'green'( while eliminating [ Max-Fri], and
and eliminate [Wed-Bargo, Thu-Argo, Thu-Fargo].
Also Eliminate [Max-Fri], leaving grid square: [Max-Wed] to be highlighted 'green, from which
the remaining grid squares[Tue-Argo, Wed-Argo, Tue-Fargo, Wed-Fargo] can be deduced.

• Congratulations! Puzzle solved. To summarize:

Bob discovered the planet Margo on Monday.
Carl discovered the planet Bargo on Thursday.
Edie discovered the planet Argo on Tuesday.
Igor discovered the planet Sargo on Friday.
Max discovered the planet Fargoon Wednesday.
###### PDF PRINTABLE VERSION

6.

MIXED DRINKS-SOLUTION

## MIXED DRINKS-SOLUTION (show / hide)

David-Soda-Vanilla.
Mary-Juice-Strawberry.
Wilma-Tea-Lemon.

#### step-by-step:( show / hide)

• Lets look at the first clue "Wilma did not drink water mixed with her lemon ."

• Find grid square [Wilma-lemon] and click until it is 'green'. We also have the following
elimination squares [Wilma-water, Wilma-Lime,Wilma-s.berry, Wilma-Vanilla] as well as
[Adam-Lemon, David-Lemon, Mary-Lemon, Lemon-Water] so add 'red xx' to all squares now.

• "Only the men drank soda or water. "

• Both men(Adam, David) either had soda or water, therefore select these grid squares for elimination:
Eliminate [Mary-Soda,Mary-Water, Wilma-Soda].

• "Mary was not the one who had water & lime. "

• Lets eliminate [Mary-Lime] straight away.
We are also given a drink-mix combo, so locate [Lime-Water] and click the square until a'green box' appears there,
(which leads to further eliminations) [Lime-Soda, Lime-Tea, Lime-Juice] as well as
[S.berry-Water, Vanilla-Water].

• "David did not mix his drink with lime. "

• If this is true, then David did not drink the water, either.
Lets eliminate [David-Lime, David-Water] to begin with.
Furthermore, if he did not drink water(well then he must have had Soda)
Find and highlight 'green'[David-Soda], which eliminates [Adam-Soda],

• "The woman who had strawberry did not have tea. "

• Since David is not a woman, lets eliminate [David-Strawberry] which leads to the following logical conclusions:
David had Vanilla , and Mary was the woman who had strawberry.
Furthermore Mary did not have tea(since she had the strawberry), thus she could have only had the Juice.
To conclude (if Mary had the juice), then 'Wilma must have had the Tea(with Lemon)' ,and
'Mary had the juice mixed with strawberry' , while 'David had the Soda mixed with Vanilla'.

• Congratulations! Puzzle solved. To summarize:

David had the Soda mixed with Vanilla.
Mary had the Juice mixed with Strawberry.
Wilma had the Tea mixed with Lemon.
###### PDF PRINTABLE VERSION

7.

TEST SCORES-SOLUTION

## TEST SCORES - show / hide-

Amber-75-85.
Marc-90-60.
Marie-65-95.
Matt-85-75 .
Stacy-100-90.

#### step-by-step:( show / hide)

• This first clue "No child received the same score on both quizzes."

• We are already given the two sets of scores, so the eliminations are as follows:

for the following Rows 75 - 75 , 85 - 85 , and 90 - 90.

• Our next clue is " The girl who received a perfect score on the first quiz, wound up with an average of 95 from both quizzes."

• Taking the first part of our clue , we eliminate (all the boys) from the
Column 100 - Marc, Matt.

The last part of the clue requires a bit of math, because we know the first score was 100, we are then told the Average after the second quiz is 95, and using the formula for average = (Quiz 1 +Quiz 2)/2 = 95, which can be
re-written as (Quiz 1 +Quiz 2)= 95 x 2, and since we know Quiz 1= 100, by substitution we have,
100 + Quiz 2 = 190, (or solving for Quiz 2 ), Quiz 2 = 190 - 100 = 90. Therefore our first solution is 100 - 90.

We can now eliminate first in Column 100 - 60, 75, 85, 95. While in Row 90 - 65, 75, 85 and in Column 90 - Marc, and Matt.

• Our next clue is "Marc had the lowest overall average of 75( he received a score of 60 on one of the quizzes)."

• Lets take the second part of the clue and fill-in the solution Marc - 60, from which we can make the
elimination in
Row Marc - 75, 85, 95 (for the second set of quizzes) , and then the eliminations in Column 60 -
Amber, Marie, Matt, and Stacy.

We are now prepared to find Marc's first score based on the 75 average , using the formula
average = (Quiz 1 + Quiz 2)/2 , ( and making the appropriate substitutions for average = 75, and Quiz 2 = 60 ), and solving for
Quiz 1 = 150 - 60 = 90, so we have a solution for Marc's set of scores : 90 - 60 ,
leading to the following eliminations :

Row Marc - 65, 75, 85 , and in Column 90 - Amber, Marie, Matt , Stacy, 75, 85, 95, then
Row 60 - 65, 75, 85.

• Our next clue states " Amber scored higher on her second quiz."

• We can make the elimination from the Row Amber - 100.

• Our next clue states " On the second quiz Marie improved her score by more than 20 points."

• We must first look at the available Quiz scores for Marie, first from
Quiz 1 : 65, 75, 85, 100 and
Quiz 2 : 75, 85, 90, 95 ,

The only possible scores she could have received are on Quiz 1 65, and on Quiz 2 a 95( which is 'more than 20 points' higher than Quiz 1 ).

Therefore, we have the solutions : Marie - 65 and Marie - 95 ,
( from which we make the following eliminations) in Row Marie - 75, 85, 100 ( from Quiz 1 ) , and Marie - 75, 85, 90( from Quiz 2).

( From which we derive more eliminations) in Column 65 - Amber, Matt, Stacy, 75, 85 and Row 9 5- 75, 85, and
returning to Column95 - Amber, Matt, Stacy.

Which leads to the following solutions:75 - 85 and 85-75.

• Our last clue is " Matt's second score was 10 points lower than his first quiz score."

• Quiz 1 : 65, 75, 85, and Quiz 2 : 75, 85, 95 ,

which means the only logical combination would be 85-75 ,

so filling-in these solutions Matt - 85 (Quiz 1) and Matt - 75 (Quiz 2) ,

results in the following eliminations : first, for Row Matt - 65, 75 (Quiz set 1) , and Matt - 85, 95 (Quiz set 2),

then for Columns 85 - Amber, Stacy and 75 - Amber and for Row 75 - 65 .

• Congratulations! Puzzle solved. To summarize:
Amber-75-85.
Marc-90-60.
Marie-65-95.
Matt-85-75 .
Stacy-100-90.
###### PDF PRINTABLE VERSION

8.

THE STANDINGS-SOLUTION

## THE STANDINGS - show / hide-

Bears 17 wins- 3 losses.
Indians 4 wins- 16 losses.
Giants 15 wins- 5 losses.
Rangers 2 wins- 18 losses .
Angels 12 wins - 8 losses.

#### step-by-step:( show / hide)

• This first clue "The Indians lost more games than both the Angels and Giants."

• This clue yields the following eliminations:
Row Indians - 3, 5 and Angels - 18 and Giants - 18.

• This next clue is "The Bears had more wins than losses."

• Simply locate the wins & losses Columns which include
Wins - 17, 15, 12, 4, 2. & Losses - 3, 5, 8, 16, and 18.

The Bears must have had one of these possible win-loss combinations ( 17-3, 15-5 , or 12-8 ) , which leads to
the following eliminations in the Row Bears - 4, 2, 16, 18.

• This next clue "The Rangers lost more games than at least three other teams."

• Is a straightforward one, they must have lost either8, 16, or 18 games.
So make the following elimination in Row Rangers - 3, 5.

• The next clue states "The Angels had twice as many losses as Indians had wins."

• To determine the Angels total losses we look at the possible wins column and double the numbers, to reveal
34, 30, 24, 8, or 4. The only two possibilities are that the Indians had either 2 Wins ( in which case the
or the Indians had 4 wins ( in which case the Angels had 8 losses ).

(But no team had 4 losses ), therefore the solutions are as follows:
Indians - 4 (wins) , and Angels - 8 (losses) ,
( and also because each team only played 20 games ), the following solutions are also true

Indians - 16 (losses) , and Angels - 12 (wins).

We can now also make some eliminations

For Row Indians - 17, 15, 12, and 2( wins) - 8, 18 ( losses).

For Row Angels - 17, 15, 4, and 2( wins) - 3, 5, or 16 ( losses). Which leads to eliminations in
Columns 4- Giants, Rangers, 16 - Giants, Rangers, and 8 - Bears, Giants, Rangers ,

( which uncovers the solution ) Rangers - 18 , ( from which it follows

the Rangers could only have had 2 wins) ,

therefore we can fill-in the solution
Rangers - 2 , and back to eliminations of Rangers - 17, 15, 12.

• The last clue states "The Giants placed second in the standings."

• The second amount of wins is 15, therefore we have the solution :
Giants - 15 and as a consequence Giants - 5 , which leads to the eliminations
in Row Bears - 15, which leads to the solution :
Bears - 17

• Congratulations! Puzzle solved. To summarize:
Bears 17 wins- 3 losses.
Indians 4 wins- 16 losses.
Giants 15 wins- 5 losses.
Rangers 2 wins- 18 losses .
Angels 12 wins - 8 losses.
###### PDF PRINTABLE VERSION

9.

TEACHER'S CLASSES-SOLUTION

## TEACHER'S CLASSES - show / hide-

Miss Jones-Art-80.
Mr. Smith-History-95.
Mrs. Taylor-Spanish-90.
Miss Wilson-Music-85.

#### step-by-step:( show / hide)

Miss Jones taught either Art or History
• This first clue " Miss Jones taught either Art or History."

• This clue yields the following eliminations:
Row Jones - Music, History

• The next clue " Mrs. Taylor's students averaged above 85 for the entire semester."

• Another simple elimination clue
RowTaylor - 85, 80.

• The next clue states " Mr. Smith did not teach either Spanish or Art(one of these classes finished with an 80 average)."

• Which leads to an immediate elimination in
Row Smith - Spanish, Art, then Row 80 - Music, History.

• The next clue states " Miss Wilson taught Music.."

• Our first direct solution clue reveals Wilson - Music, which also allows for the following
eliminations starting in row Wilson - Art, History, Spanish, 80 , and for Column Music - Smith, Taylor.

Which reveals a solution Smith - History.

• The last clue states "Mr. Smith's class averaged higher than any of the women teachers classes."

• The highest average is 95, therefore fill-in the solutions Smith - 95 and History - 95, with the attendant
eliminations - for Row Smith - 90, 85, 80, and Column History - 90, and 85, , as well as, Column 95 - Jones, Taylor and Wilson , and Row 95 - Music, Art, Spanish, , and we now have some revealed solutions:
Taylor - Spanish, and Taylor - 90 as well as 90 - Spanish, allowing us to eliminate
Row Taylor - Art and Column90 - Jones, Smith , Wilson , from which we can make these grid solutions:
Wilson - 85 and , Jones - Art , and the eliminations Jones - 85, which gives the solutions :
Jones - 80 and , 80 - Art with the eliminations: Music - 90, reveals the solution: Music - 85.

• Congratulations! Puzzle solved. To summarize:
Miss Jones-Art-80.
Mr. Smith-History-95.
Mrs. Taylor-Spanish-90.
Miss Wilson-Music-85.

10.

## STUDENT'S GRADES - show / hide-

Ashley - C / Home Economics.
Ben - A / Science.
Paul - B+ / English.
Mary - A+ / Math.
Rachel - B / Geography.

#### step-by-step:( show / hide)

• This first clue "The student who received the lowest grade (in HOME EC.) is not a male."

• This clue results in the following solution : Row Home - C and corresponding eliminations
in Row Home - A+, A, B+, and B , while further eliminations can be made in
in Column C - Math, English, Science, and Geography.
Locate Column Home - Ben, and Paul .

(These two eliminations are made from the last part of the clue ( "... is not a male. " ) ).

• This next clue "The highest grade was not given in either Science or Geography."

• Which leads to two simple eliminations :
Locate Column A+ - Science , and Geography .

• The next clue is "Mary received either an A+ or an A in Math."

• Begin with recording the solution Mary - Math then proceed with the following eliminations :
Row Mary - B+, B, C , English, Science, Geography and Home EC.
Then for Column Math - Ashley, Ben, Paul , and Rachel. Finally, for Row Math - B+, and B.

• The next clue states : "Rachel received a higher grade than Ashley, but a lower grade than Paul."

• This of course means Rachel did not get the C, and Ashley did not get the highest grade( A+ ).
therefore make the eliminations Rachel - C and Ashley - A+ .

Now from the second part of the clue ( " ... a lower grade than Paul. " ), means that Rachel is also
eliminated from the highest grade ( A+ ), remove ( Rachel - A+ ), and since

Paul had to have scored higher than BOTH Rachel & Ashley,
we can in turn, eliminate from Row Paul - B, C.

• The next clue is : "The student who was in the Geography class scored lower than the student in the English class , who scored lower than the one in the Math class."

• Again we begin with some logical deductions:

( If the one in the Geography class scored lower than the one in English ),this means
we can eliminate Geography - A+ and also since

" ... the one in the English class scored lower than the one in the Math class. " , we can then
also eliminate Geoghraphy - A , as well as English - A + , which now reveals that :

the highest grade was received by the one in the Math class ( allowing us to fill the solution
Math - A+ ) and eliminate Math - A,

(which also by logical deduction, yields the solution) :

Mary - A+ , ( since we know Mary was the one who took the Math course ), from which

these eliminations follow in Column A+ - Ben, Paul, , and in Row Mary - A .

• The next clue is : "The male student who who took the Science class received a higher grade than the one in the English class( who was not Ben )."

• Begin with the obvious elimination, ( since none of the girls could have taken the Science class ),
we eliminate from Column Science - Ashley, and Rachel.

The second part of this clue tells us that the two boys took either the Science or English classes,

( But be aware the last part also tells us : " ... the one in the English class( who was not Ben )." ).
Thus our solution must be (for the boys) :
Ben - Science and Paul - English ,

( From which we can make further eliminations ) :
in Columns English - Ben, Ashley, and Rachel and Science - Paul, also from Row Paul - Geography.

Before we can end this clue we can make another logical deduction :

since we know now that Paul (in Science) received a higher grade than Ben (in English) , ( and
checking the availability of the grades for these boys ), Ben could only have who received the A, and Paul, the B+.

( Fill-in these two solutions : Ben - A and Paul - B+ ) ,

from which we can then make our eliminations:

in Column A - Ashley, Rachel, English and
Column B+ - Ashley, Rachel, Science, and Geography, and from Column B - English, Science ,
which leads to the solution : Geography - B .

• We have our optional last clue : " Ben received a grade that was exactly one grade higher than Rachel."

• Since we already know Ben received the A , then Rachel received the B, and the solution for Ashley is :

Ashley - C and Ashley - Home EC. , leaving the solution : Rachel - Geography.

• Congratulations! Puzzle solved. To summarize:
Ashley - C / Home Economics.
Ben - A / Science.
Paul - B+ / English.
Mary - A+ / Math.
Rachel - B / Geography.
###### PDF PRINTABLE VERSION

11.

JUMPING BEANS-SOLUTION

## JUMPING BEANS - show / hide-

Jane-Green-5.
Jasper-Yellow-11.
Jeff-Blue-3.
Jim-Orange-7.
Julie-Red-9.

#### step-by-step:( show / hide)

• This first clue "Jim's set has more than the one with the green-colored beans, but less than that of the red set."

• Here we are given two immediate eliminations for Jim:
Row Jim - Green, and Red.

Additionally , since Jim " ...has more than the green... " and " ... less than the red set. ", we can make these eliminations ( by logical deduction)
Row Jim - 3, and 11.

• By our next clue : "Jane's set of five beans is colored neither blue or yellow."

• We start with a solution here Jane - 5 , and making the eliminations for :
Row Jane - 3, 7, 9, 11 , as well as Column 5 - Jasper, Jim, Jeff and Julie.

Focusing now on the last part of the clue ( " ... colored neither blue or yellow. " ),
we can make these eliminations , starting in Row Jane - Blue, Yellow, followed by the same for
Row 5 - Blue, Yellow

• Now our next clue states : "Julie's set contains more beans than either Jane's or Jeff's (combined)."

• Which taken the given ( Jane = 5 ) , we make note of the available beans ( 3, 7, 9, or 11 ) and if we add
5 to each we get ( 8, 12, 14, and 16 ) of which, the totals above 12 (including 14 and 16) , are just not possible,
therefore the only solution can be that Jane's + Jeff's ( beans ) = 8 , ( which is less than both 9 and 11 ),

therefore we have the solution : Jeff - 3 , (and since we now know that Julie has either 9 or 11 beans ) ,
we can then
eliminate from Row Julie - 3, 5, and 7 , which allows for further eliminations :

in Row Jeff - 5, 7, 9, 11 and Column 3 - Jasper .

• The next clue is : " The orange-colored set either contains 5 or 7 total."

• Simply eliminate the impossible, namely, find ( and remove ) from the Column Orange - 3, 9, 11.

**NOTE** An additional elimination can be made because we know Julie can only have 9 or 11 beans
and the orange set " ... either contains 5 or 7 total " , it is not possible for Julie to have the orange beans, thus
eliminate in Row Julie - Orange ,

and after applying this same principle for Jeff ( who has "exactly 3 beans ") ,
we have the elimination in Row Jeff- Orange.

• The next clue reads : " The yellow-colored set (which is not Julie's), contains not less than 9 beans."

• Let's start with the obvious elimination in Row Julie - Yellow, and then for
Column Yellow - 3, 5 and 7.

• The next to last clue states : "Jeff's beans are blue."

• This is a solution clue , Jeff - Blue , which ( since we know Jeff has 3 beans ) means the following
must also be the solution : 3 - Blue , from which we make some
further eliminations :
Row Jeff - Green, Red, and Yellow , as well as
Row Blue - Jasper, Jim, Julie, 5, 7, 9, and 11 , and
Row 3 - Green, and Red.

• The last clue reveals : " There are less beans in Julie's red-colored set than the yellow set."

• Begin with the solution Julie - Red , then make eliminations in
Column Red - Jasper, Jim, 5, and 7. as well as (back to ) Row Julie - Green.

Lastly we are given : ( "... less beans in Julie's...than the yellow set " ) , which can only lead to a

valid solution (or two) Julie - 9 , Red -9 and Yellow - 11,

( From which we make our eliminations as follows ) :
Row Julie - 11, , and Row 11 - Green, then from Column Red - 11, then from Row 9 - Green, Yellow

from Column 9 - Jasper and Jim, which reveals this solution Jasper - 11 ,
which means : Jim must be 7 , so fill in solution Jim - 7, and make the eliminations:

Row Jim - Yellow , leads to the solution Jim - Orange ] , from which this solution

directly follows: 7 - Orange ,

which triggers eliminations : first, in Column Orange - Jane and Jasper , followed by
Row 5 - Orange ,( which provides the solution ) 5 - Green, from which the solution :
Jane - Green logically follows.

• Congratulations! Puzzle solved. To summarize:
Jane-Green-5.
Jasper-Yellow-11.
Jeff-Blue-3.
Jim-Orange-7.
Julie-Red-9.
###### PDF PRINTABLE VERSION

12.

MATTER AFTER MATTER-SOLUTION

## MATTER AFTER MATTER - show / hide-

Alvin 5-P-49
Louie-7-G-27.
Margaret-3-T-15.
Grace-6-A-12.
Charlie-9-O-36.

#### step-by-step:( show / hide)

• This first clue "Grace doubled her initial particles after using Compound A."

• Here we are given an immediate solution : Grace - A. From which we can make our eliminations
Row Grace - O, T , P, and G , and in Column A - Alvin, Louie, Margaret, and Charlie.

If Grace doubled her initial particles , she began with : 6 and increased it to 12 , so we have the solutions

Grace - 6 and Grace - 12 , from which we have the following eliminations

Row Grace - 3, 5 , 7, 9 , 15, 27, 36, and 49 and Column 12 - Alvin, Louie, Margaret, and Charlie , and

Column 6 - Alvin, Louie, Margaret, Charlie , 15, 27, 36, 49, O, T , P, and G , then

Row 12 - 3, 5 , 7, 9 , O, T , P, and G .

Row A - 3, 5 , 7, and 9 , and back to Column A - 15, 27, 36, and 49.

• This next clue is " Using Compound P allowed one scientist to finish with the most amount of particles."

• From which we can make the solution P - 49 and eliminations from Row49 - O, T, and G , and
in Column P - 15, 27, and 36.

• This next clue is " After applying his compound Charlie's sample increased by a factor of 4."

• To determine the possible solution we must multiply each initial sample ( 3,5,7,9 ) by 4 and try to match it to one of the final samples( 15, 27, 36, or 49).

( The resulting pairs would be 3 - 12, 5 - 20 , 7 - 28 , or 9 - 36.
From which it follows that only 9-36 fufills the clue.

Therefore we have the solutions : Charlie - 9 , and Charlie - 36 , from which we make these additional eliminations

first in Row Charlie - 3, 5, 7, 15, 27, and 49 , followed by
Column 9 - Alvin, Louie, Margaret, 15, 27, and 49 and Column 36 - Alvin, Louie, Margaret , and Row 36 - 3, 5, and 7.

• This next clue states "Compound G was used by the one whose initial sample was 7, and and increased his/her final particle count by at least 20."

• First we fill-in the given solution G - 7, with the attendant eliminations in Row G - 3, 5, and 9, and also
Column 7 - O, T, and P.

Now the last part " ... increased his/her final particle count by at least 20 " , means we can limit our

combination of initial - final pairs to 7 - 27, or 7 - 49 , ( but since the one who used Compound P finished with 49 particles ),

the only solution here MUST be 7 - 27.

This allows the following eliminations : in Column 7 - 15, 49 , and in
Row 27 - 3 , 5, O, and T, and gives the solution : 27 - G , followed by eliminations in
Column G - Charlie, 15, and 36.

• Next clue is "Margaret was able to increase her initial sample by a factor of 5, she didn't use Compound O."

• Lets first make the elimination Margaret - O.

Now to solve the first part of the clue we take the remaining available initial samples for Margaret ( 3, 5, or 7 ) and multiply
each by 5, with the resulting pairs being 3 - 15 , 5 - 25 , or 7 - 35 , from which the only solution can be 3 - 15, and therefore
we also have solutions Margaret - 3 , and Margaret - 15, from which we can make our grid eliminations:

Starting with Row Margaret - 5, 7, 27, and 49, and Column 15 - Alvin and Louie, and Row 15 - 5 ,
and Column 3 - Alvin, Louie, and 49 , which leads to the solutions

49 - 5 and 5 - P.

Next, eliminating from Column 5 - O, and T , leads to
eliminations in Row P - 3 , and 9 .

**NOTE** We can make some logical deductions for Margaret by observing the relationship between, first Margaret - 3 - 15 ,
and the available compounds : First we eliminate from Column 3 - O , which reveals the solution :

3 - T and ( from which it follows ) Margaret - T , and 15 - T,

and leads to the following eliminations:
Row Margaret - P, and G, Column T - Alvin, Louie, Charlie, and 36 , , and Row T - 9.

Which leads to a new set of solutions : ( Starting with )

O - 9 , 36 - O , Charlie - O,
with these eliminations : Row Charlie - P , Row 15 - O , then Column O - Alvin, and Louie .

• The last clue reveals : "Alvin's initial particle sample was lower than at least 3 others, his final particle count exceeded at least three of the others."

• To begin with , if Alvin's initial sample was lower, this means he must have had either 3, 5, or 6, but only 5 remains, so we have the solution for Alvin:
Alvin - 5 and therefore Alvin - P and Alvin - 49
From which we can complete( eliminate ) from his Row Alvin - 7, G, and 27, and likewise for Louie :

Louie - 7 and therefore Louie - G and Louie - 27
From which we can complete( eliminate ) from his Row Louie - 5, P and 49.

• Congratulations! Puzzle solved. To summarize:
Alvin 5-P-49
Louie-7-G-27.
Margaret-3-T-15.
Grace-6-A-12.
Charlie-9-O-36.

13.

## VOTES ARE IN - show / hide-

Alex-90-545pm.
Billy-70-600pm.
Cathy-60-5pm.
Robin-50-530pm.
Susie-40-630pm.
Tanya-20-515pm.

#### step-by-step:( show / hide)

• This first clue "The one who arrived latest did not receive the fewest votes."

• We are given the two variables The one who arrived latest (630pm), and the fewest votes (20), therefore
we can eliminate the grid square (in column) 20 - 630pm.

• Our next clue is "Susie (the one who received 40 votes), also (arrived at least 30 minutes) after the one who received 70 votes did".

• We are given a solution right away Susie - 40, from which we can make eliminations, beginning with
Row Susie - 20, 50, 60, 70, 90, and
Column 40 - Alex, Billy, Cathy, Robin, and Tanya.

But we are not done with the clue, the last part ( "... arrived at least 30 minutes) after the one who received 70 votes did." ),
which means Susie could not have arrived at 500pm or 515pm, and the one who received 70 votes could not have arrived later than 600pm.
We can eliminate from Row Susie - 500pm, 515pm and from Column 70- 630pm.

• The next clue is "Alex received more votes than at least 3 others, and also arrived after three others".

• The first part means Alex did not receive 20, 40 or 50 votes, which leads to the elimination in
Row Alex - 20, 40, and 50.

And the last part refers to the time of Alex's arrival, if he arrived "after 3 others",
he could only have arrived at 545pm, and we are now able to make some eliminations in
Row Alex - 5, 515, 530, 600, 630 , followed by eliminations in
Column 545 - Billy, Cathy, Robin, Susie, and Tanya. As a consequence of this result we can also
apply the following eliminations in
Row 545pm - 20, 40, and 50.

• The next clue states "Billy arrived at least 1/2 hour before Susie did and received exactly 30 more votes than she did".

• Begin by eliminating in Row Billy - 515, 630.

The second part of the clue yields a solution namely, Billy - 70 (which is "... exactly 30 more votes than she ...")
(which will lead to the following eliminations ) , first in Row Billy - 20, 50, 60, and 90 and for Column 70 - Alex, Cathy, Robin , Tanya, 515, and 545.

• Another clue begins : "Cathy who arrived precisely at 5pm , received more than 50 votes.".

• Immediately gives the solution Cathy - 500pm, which allows us to begin our eliminations in
Column 500pm - Billy, Robin, Tanya

( **NOTE** the elimination of Billy - 500, also results in the elimination of 70 - 500 and Susie - 530. )

Now taking the last part of the clue ( "...more than 50 votes." ), we can now eliminate from
Row Cathy - 20, 50, 515, 530, 600 and 630.

• The Clue : "The one who arrived at exactly 545pm received the most votes".

• Well this one, we know is Alex, so we can fill in the solution Alex - 90, make the eliminations
from Column 90 - Cathy, Robin, Tanya, 5pm, 515, 530, 600, and 630pm, then from
Row Alex - 60 and 545-60.

It then reveals the solution Cathy-60, from which we make the eliminations in
Column 60 - Robin, Tanya, go to Row 5pm complete the solution 5 pm - 60,
make the eliminations 5 pm - 20, 40, and 50, and returning to
Column 60 - 515, 530, 600, 630.

• The Last clue " Neither Robin (who arrived exactly one hour before Susie and received more votes than her), nor Tanya receive more than 50 votes.."

• We look at our remaining available times for Robin, which are 515, 530, 600, and 630, but she arrived
BEFORE Susie, so this eliminates 'ALL' possibilities but 515 and 530, and since there is not a 615 arrival,
Robin could have only arrived at 530pm, making Susie's arrival 630pm,
we can now plot these solutions in our grid ( Robin - 530 and Susie - 630 ) , which leads to the solution for Billy (which was 1/2 hour BEFORE Susie), so
Billy - 600 and these solutions follow by logical deduction : 70 - 600,40 - 630, and Tanya - 515.

To determine the final vote count we note that :

" Robin ... who arrived... before Susie ... received more vote than her... ",

• Congratulations! Puzzle solved. To summarize:
Alex-90-545pm.
Billy-70-600pm.
Cathy-60-5pm.
Robin-50-530pm.
Susie-40-630pm.
Tanya-20-515pm.
###### PDF PRINTABLE VERSION

14.

TRANSPORTATION-SOLUTION

## "T" for TRANSPORTATION - show / hide-

Tony-8am-Taxi.
Timmy-10am-Tour Bus.
Tammy-7am-Trolley.
Tanya-noon-Train.
Tabitha-9am-Tugboat.

#### step-by-step:( show / hide)

• This first clue "The earliest available transport is the Trolley (which is not taken by any of the men, or Tanya)."

• A simple solution clue is provided in the first part of the clue: Trolley - 7 ,
( from which we can then make the eliminations)
beginning in Column7 - Train, Tour Bus, Taxi, Tugboat followed by
eliminations in RowTrolley - 8, 9, 10 , noon.

Now from the last part of the clue ( ".... Trolley (which is not taken by any of the men, or Tanya)" ) ,
we make further eliminations in Column7 AM - Tony, Timmy, and Tanya.

• The next clue "Nobody departs at noon from either the Taxi or the Tour bus."

• This is an elimination clue, so locate the following Column 7AM - Taxi, Tour Bus and make the eliminations.

• The next clue is " Tabitha either rides the train or the tugboat."

• We must eliminate the transports Tabitha doesn't take (Trolley, Tour Bus, Taxi), therefore locate
( and eliminate ) from RowTabitha - Trolley, Tour Bus, and Taxi.

**NOTE** We can also eliminate [Tabitha-7], because this is the Trolley departure time, (which Tabitha does not
take). Which leads to a solution: 7 AM - Tammy,( the only remaining person in Column 7AM ), from which
we can make further eliminations,
So locate Row ( and eliminate ) Tammy - 8, 9, 10, noon, Train, Tour Bus, Taxi, and Tugboat ,
and for Column Trolley - Tony, Timmy, and Tanya.

• The next clue "The Taxi is taken by Tony every morning sometime before 9am."

• Begins with a given solution Tony - Taxi, so make the eliminations in
Row Tony - Train, Tour Bus, and Tugboat, followed by eliminations in
Column Taxi - Timmy, and Tanya.

Now from the last part of the clue ( "... sometime before 9am." ), we can now make eliminations in
RowTony - 9, 10, and noon.

( Which means we have the solution Tony - Taxi - 8AM), which means we can eliminate from
Column 8AM - Timmy, Tanya, Tabitha, Train, Tour Bus, and Tugboat , which leads to more eliminations in
Row Taxi - 9, and 10AM.

• The next clue "The Tugboat is not the last transport of the day."

• Which means it does not depart at noon, therefore we can eliminate from
Column Noon - Tugboat, (which reveals the solution Noon - Train), resulting in more eliminations
from row Train - 9, 10Am.

• Our last clue states : "Tanya leaves 2 hours after Timmy."

• Looking at the grid Timmy's available departure times are 9, 10 or noon , and since Tanya leaves
2 hours later, then
Tanya's departures could only be 11, 12 or 2, from which only one combination can be true:

therefore we have the solutions Timmy - 10AM and Tanya - Noon, from which it follows that the
solution for the final person-time slot is Tabitha - 9AM.

So now that we have "ALL" of the times in place, we can solve this puzzle by matching the final pieces as follows:
Tanya takes the train at noon, which means Timmy takes the Tour Bus at 10, and Tabitha takes a Tugboat at 9am.

• Congratulations! Puzzle solved. To summarize:
Tony-8am-Taxi.
Timmy-10am-Tour Bus.
Tammy-7am-Trolley.
Tanya-noon-Train.
Tabitha-9am-Tugboat.
###### PDF PRINTABLE VERSION

From Medium Puzzle Grid Solutions to Logic Puzzle Grid

From Medium Puzzle Grid Solutions to Puzzle Games Online