Below is your guide to the Detailed Easy Puzzle Grid Solutions



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Below are the Details to the Easy puzzle Grid Solutions.

This page contains the Easy 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).

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SOLUTIONS :

EASY :

EASY :

1.

AT THE PET SHOP(1)-SOLUTION

AT THE PET SHOP(1) (show / hide)

Bill --- snake.
Jen --- dog.
Roger --- monkey.
Susan --- cat.

step-by-step:( show / hide)

  • Lets look at the first clue "Nobody chose a pet which started with the same first letter as their name."

  • Locate Susan in the chart and find the column with Snake

    Now click the grid square(susan-snake) until the 'red xx' appears.

  • Lets look at the next clue "Bill already has a dog."

  • Locate Bill in the chart and find the column with Dog

    Now click the grid square(bill-dog) until the 'red xx' appears.

  • Lets look at the next clue "Jenis afraid of snakes."

  • Locate Jen in the chart and find the column with snake

    Now click the grid square(jen-snake) until the 'red xx' appears.

  • Lets look at the next clue "Roger did not choose the snake."

  • Locate Roger in the chart and find the column with snake

    Now click the grid square(Roger-snake) until the 'red xx' appears.

  • *IMPORTANT:*
    Look at the column under 'SNAKE'. We have now eliminated 3 of the possible 4 squares in that column.

  • Therefore the square bill-snake must be the correct solution

    So click the grid square(bill-snake) until the 'green box' appears.
    At the same time place 'red xx' in all other squares in the bill-row(bill-cat, bill-monkey).

  • Lets look at the next clue "Neither Jen nor Susan like monkeys."

  • Locate Jen and Susan in the chart and find the column with monkey

    Now click the grid square(jen-monkey, susan-monkey) until the 'red xx' appears.

  • *IMPORTANT:*
    Look at the column under 'MONKEY'. We have now eliminated 3 of the possible 4 squares in that column.

  • Therefore the square roger-monkey must be the correct solution

    So click the grid square(roger-monkey) until the 'green box' appears.
    At the same time place 'red xx' in all other squares in the roger-row(roger-cat, roger-dog).

  • Lets look at the last clue "Jen did not choose the cat."

  • Locate Jen in the chart and find the column with cat

    Now click the grid square(jen-cat) until the 'red xx' appears.

    This logically will conclude our puzzle as follows:
    Since jen did not choose the cat, she could have only chosen the dog,
    (jen-dog square should be highlighted with a 'green-box',while jen-cat is filled with a 'red xx').

    The opposite is selected for susan, since she could have only selected the cat.

  • Congratulations! Puzzle solved. To summarize:
    Bill chose the snake.
    Jen chose the dog.
    Roger chose the monkey.
    Susan chose the cat.
PDF PRINTABLE VERSION

2.

A MATTER OF DOLLS-SOLUTION

A MATTER OF DOLLS (show / hide)

Ann:yellow-cries
Mary: pink-sings
Stacy: white-talks

step-by-step:( show / hide)

  • Lets look at the first clue "The doll named Ann is not wearing the pink or white dress."

  • Locate Ann in the chart and find the columns with pink & white

    Now click the grid square(ann-pink,ann-white) until the 'red xx' appears.
    *IMPORTANT:*--Please note by eliminating the other two choices for Ann's dress color, the obvious is revealed , Ann's dress color CAN ONLY be "yellow".

  • Therefore, locate the grid square[ann-yellow] and click until the 'green box' appears.
    Also note : locate the grid squares [mary-yellow, Stacy-yellow] and click until the 'red xx' appears.

  • Lets look at the next clue "Grandma's white dress matches the doll who talks ,which is not named mary."


    • We are actually looking at two clues here :
    1. ... white dress matches the doll who talks
      2. and,
    2. ... which is not named mary.
  • For clue 1. look in the chart and find the column with talks and white

    Now click the grid square[talks-white] until the 'green box' appears.

    • *NOTE:*We can also click on the following grid squares[cries-white,sings-white] as well as [talks-pink, talks-yellow]until the 'red xx' appears for all of these grid squares.

  • For clue 2. Find [Mary-talks,Mary-white] and click until the 'red xx' appears.


    • *NOTE:*By using visual logic we can see the square [Mary-pink] is the only remaining option , so click on this square until the 'green box' appears.

    ( We can now eliminate [Stacy-pink], by placing a red xx in that box, leaving only [Stacy-white] and because the doll with the white dress=talks, we locate the square[Stacy-talks] and place 'green boxes' in both these squares, while eliminating these three squares[Stacy-cries,Stacy-sings,Ann-talks]).

  • Lets look at the final clue : "The doll that sings is dressed in a pink outfit. "

  • Now click the grid square[sings-pink] until the 'green box' appears.
    • *NOTE:*We can also click on the following grid squares[sings-yellow, cries-pink]until the 'red xx' appears for these grid squares.

    ( As a consequence we can see that [cries-yellow] should now be filled with a 'green box' --
    which NOW leads us to the final solution after locating [ann-cries], which gets a 'green box',
    while the grid squares[Mary-cries, Ann-sings] both get 'red xx' boxes)
    which in turn ,leaves only [    Mary-sings] which of course is filled with the final 'green box'.

  • Congratulations! Puzzle solved. To summarize:
    Ann:yellow-cries
    Mary: pink-sings
    Stacy: white-talks
PDF PRINTABLE VERSION

3.

BACKPACKS-SOLUTION

BACKPACKS(show / hide)

SAM: green-5th
SAL: blue-3rd
SEAN: purple-lunch

step-by-step:( show / hide)

  • Lets look at the first clue "Sam did not lose his backpack before lunch."
  • Locate SAM in the chart and find the columns with 3rd

    Now click the grid square[Sam-3rd] until the 'red xx' appears.
    (Because from this clue, we only know he had it before lunch, and nothing else).

  • Lets look at the next clue "Sal's backpack which was not green was lost during 3rd period. ."


    • We are actually looking at two clues here :
    • ... which was not green
      • and,
    • ... lost during 3rd period.


  • For clue 1 look in the chart and find the column with Sal and green

    Now click the grid square[sal-green] until the 'red xx' appears.

  • For clue 2 Find [Sal-3rd] and click until the 'green box' appears.

    *NOTE:*(This clue is actually quite useful because it allows us to eliminate some other grid-squares as follows.)

    Locate the following [Sal-Lunch,Sal-5th,Sean-3rd, green-3rd]--eliminate these squares by placing the 'red xx' in their respective boxes.
  • Lets look at the final clue : "Sean lost sight of his purple backpack during lunch. "

  • We should immediately fill the grid squares[Sean-purple, Sean-Lunch, lunch-purple] with 'green boxes'.
  • *NOTE:*This is by far our most valuable clue yet, because it will allow us to complete Sean's row, the purple column, and the Lunch row as follows:

    Locate the following grid squares[Sean-blue,Sean-green, Sean-5th] and [Sam-purple,Sal-purple,3rd-purple,5th-purple] as well as the Lunch grid squares[Lunch-blue,Lunch-green, Sam-Lunch]

    fill ALL of them with 'red xx' now.

  • This in turn leads to other logic conclusions : (in no particular order)
    • [Sam-fifth], the only remaining period in which a backpack can be lost, so go ahead and fill in that sqaure with a 'green box'.

    • [Sal-blue,3rd-blue],the only remaining possible combinations for Sal's backpack.
      (Which in turn eliminates [Sam-blue,5th-blue], and leaves only two possible combinations to solve the entire grid: [Sam-green,5th-green] once those squares are filled with green boxes, we have our solution
  • Congratulations! Puzzle solved.
    To summarize:
    • SAM: lost his green packpack sometime during 5th period.

    • SAL: lost his blue backpack sometime during 3rd period.

    • SEAN: lost his purple backpack sometime during the lunch period.
PDF PRINTABLE VERSION

4.

Pet SHOP Days-pg71-SOLUTION

Pet SHOP Days(show / hide)

Bill: Monkey
Jane: Snake
Roger : Dog
Susan: Cat

step-by-step:( show / hide)

  • Lets look at the first clue "No one bought a pet that began with the first letter of their name."
  • Locate Susan in the chart and find the columns with Snake
    (as this is the only grid square which satisfies the clue)
    Now click the grid square[Susan-Snake] until the 'red xx' appears.
  • Lets look at the next clue "The woman who bought the snake is not Susan.
  • A very careful reading of this clue reveals that if there are two women, and we already know Susan did not buy the snake (from the first clue), then Jane is the only woman left -so locate the grid square [Jane-Snake] and click until the 'green box' appears.

  • This clue now allows us to make some wholesale eliminations of the following grid squares:
    [Jane-Cat, Jane-Dog, Jane-Monkey] and [Bill-Snake, Roger-Snake]- all filled with 'red xx' now.

  • The next clue is "Roger is allergic to cats."
    A pretty straight-forward clue, if Roger is 'allergic to cats', it would be logical to assume he would not purchase one as a pet, therefore lets find the grid square[Roger-Cat] and place a 'red xx' there.


  • Finally lets turn our attention to the last clue:"Bill did not purchase a dog or a cat.

    Again a straightforward clue, lets go ahead and eliminate those two grid squares
    [Bill-Dog, Bill-Cat].

    Well, right away it is clear the only possible remaining pet Bill could have bought is
    the monkey. (So click on [Bill-Monkey] and fill the square with a'green box'.)


  • This will allow us to now eliminate two additional grid squares
    ( [Roger-Monkey, Susan-Monkey],

    and if we have followed our logic correctly , this reveals another singular possible pet selection, namely 'Roger-Dog',

    so that if we select this grid square and make it green, we will then see that the final person-pet combination is 'Susan-Cat'.

    • One could at this point complete the grid, but since the puzzle is solved, it is not a requirement!
  • Congratulations! Puzzle solved.
    To summarize:

    • Bill: purchased a monkey as a pet.

    • Jane: purchased a Snake as a pet.

    • Roger: purchased a Dog as a pet.

    • Susan: purchased a Cat as a pet.

    		•
    PDF PRINTABLE VERSION

    5.

    Pet SHOP Days-2-pg71-SOLUTION

    Pet SHOP Days-2(show / hide)

    Bill: Dog
    Jane: Monkey
    Roger : Cat
    Susan: Snake

    step-by-step:( show / hide)

    • Lets look at the first clue "One person bought a pet that began with the first letter of their name."
    • Locate Susan in the chart and find the columns with Snake
      (as this is the only grid square which satisfies the clue)
      Now find the grid square[Susan-Snake] and fill with a 'green box' .

      • *NOTE* :At this point do not forget to add your elimination grid squares as follows:
      [Susan-Cat, Susan-Dog, Susan-Monkey] as well as those for the column 'Snake':
      [Bill-Snake, Jane-Snake, Roger-Snake] should all now be filled with 'red xx'.
    • Lets look at the next clue " Roger already has a pet monkey and did not buy another one."
    • A very simple clue -so locate the grid square [Roger-Monkey] and fill with a 'red xx' .

    • Finally lets turn our attention to the last clue:"Bill bought a puppy after deciding not to purchase a cat."

      A straight forward clue, lets go ahead and fill in a 'green box' for the grid square[Bill-Dog] and eliminate these
      grid squares[Bill-Cat, Bill-Monkey], as well as those in the column 'Dog':[Jane-Dog, Roger-Dog].

    • Now if we have followed the logic correctly, we should be able to see in the 'Monkey' column , that there is
      only one open grid square remaining,namely [Jane-Monkey], which means after eliminating grid square
      [Jane-Cat], there is only one grid square remaining (in the column titled 'Cat'), [Roger-Cat].
  • Congratulations! Puzzle solved.

  • To summarize:

    • Bill: purchased a Dog as a pet.

    • Jane: purchased a Monkey as a pet.

    • Roger: purchased a Cat as a pet.

    • Susan: purchased a Snake as a pet.

    		•
    PDF PRINTABLE VERSION

    6.

    AUTO REPAIR INTERACTIVE-SOLUTION

    AUTO REPAIR INTERACTIVE -show / hide-

    Bill -Engine-Willy
    Jim -Check/up-Sam
    Roger Oil-Adam
    Susan -Tire-Tom

    step-by-step:( show / hide)


    • Lets look at the first clue "Susan didn't have the oil change , nor did she see mechanic named Willy."

      • This provides us with two immediate elimination clues: [Susan-Oil, Susan-Willy] - fill both with 'red xx'.

    • Lets look at the next clue: "Roger who didn't have engine problems or a check up , saw neither Tom nor Sam."

      • This clue simply provides us with several more elimination grid squares :
      [Roger-Engine, Roger-Check,Roger-Sam,Roger-Tom].

    • Lets look at our next clue:"The woman who had the tire change , saw either Tom or Adam."

      • (Cheeky monkey eh?) The 'woman' , which allows us to eliminate the grid squares
      [Bill-Tire,Jim-Tire,Roger-Tire], Why? Because none of these people are women!

      Lets now take a moment to update our grid squares as follows :

      First locate[Susan-Tire] and fill it in with a 'green box'.

      Next , for the grid squares [Susan-Engine, Susan-Check] choose 'red xx' for both.

      Lastly, (Since we now know she ....."either saw Tom or Adam"),

      we can eliminate the following grid squares:[Susan-Sam], as well as [Sam-Tire, Willy-Tire].

    • Lets proceed to our next clue:"Jim who had a routine check-up , didn't see Willy."
      • Easy enough, locate grid square[Jim-Check] and turn it into a'green box',
      Then for grid square [Jim-Willy] a 'red xx'.

      *NOTE:* (This clue will also allow us to uncover another),
      To do so : First eliminate [Bill-Check, Willy-Check], then look at the 'ENGINE' column,
      (and by using the process of elimination), you will see there is now only one possible (logical) solution here:

      'Bill-Engine' ,so go ahead and mark the grid square [Bill-Engine] with a 'green box'.

      Now, by eliminating the grid square[Bill-Oil], reveals yet another solution :
      'Roger-OIL'!

      Now since we have 'Roger-OIL', we can eliminate[Sam-Oil,Tom-Oil], by mutual exclusion,
      or in other words, they(Sam, Tom) have already been eliminated from Roger's row, they are eliminated from the 'Oil' Column.

    • Finally, we have our last clue : "Adam performed an oil change for one motorist, who is not Bill."
      • Well, since we already know who the motorist is who had the oil change ('Roger'- (see previous clue) ),

      Lets fill in grid square [Roger-Adam] with a 'green box', then make the (appropriate) eliminations of these grid squares:

      [Roger-Willy, Bill-Adam, Jim-Adam, Susan-Adam],

      which now reveals Susan's mechanic (remember, from clue 3 : "....it was either 'Tom or Adam' ) so that the grid square [Susan-Tom] may now be highlighted 'green'.
      Doing so results in the elimination of the grid squares: [Bill-Tom, Jim-Tom],
      which in turn, now reveals our final solution (by looking at column 'Willy'), namely,

      'Bill-Willy' (again by the process of elimination), and after we add the 'green box' in grid square
      [Bill-Willy], it logically follows that our last combination of motorist-mechanic can only be:

      'Jim-Sam'!

      (You may fill in the lower grid squares if you like, as an additional check on the logical eliminations!)

    • Congratulations! Puzzle solved.

      To summarize:
      • Bill: who had Engine problems saw the mechanic named Willy.

      • Jim: who had a routine Check-up saw the mechanic named Sam.

      • Roger: who had an Oil change saw the mechanic named Adam.

      • Susan: who had a flat Tire saw the mechanic named Tom.
    PDF PRINTABLE VERSION

    7.

    MY LITTLE PONIES-SOLUTION

    MY LITTLE PONIES (show / hide)

    Pam: 2nd-place-Giggles
    Penny: 3rd-show-Sierra
    Polly: 1st-win-Patches

    step-by-step:( show / hide)

    • Lets look at the first clue "Pam's pony was not named Sierra , the pony that did not win."

      • So to begin locate the grid square[Pam-Sierra] and place a 'red xx' in that square.
      Next we have "....Sierra....did not win" , so locate grid square[Sierra-Win] and place a 'red xx' in that square.

    • The next clue "Giggles won the second prize (place)."


      • Simply locate grid square[Giggles-Place] and add a 'green box' in that square.

      *IMPORTANT:* We must also eliminate the following grid squares :

      [Giggles-Win,Giggles-Show], as well as [Patches-Place, Sierra-Place].

      We can now see that in the "WIN" column there is only one empty square--
      'Patches-Win', so mark that grid square with a 'green box' and add a 'red xx' for 'Patches-Show'
      Which allows us to add a'green box' to ['Sierra-Show'], and since we know from clue 1 that "...Pam's pony was not Sierra" and Sierra was third ,
      then we can eliminate ['Pam-Show'] by adding a 'red xx' .

    • The next clue "Polly's pony was the winner."

      • Start by filling the grid square['Poly-Win'] with a 'green-box' and place the 'red xx' in the
      corresponding elimination squares[Polly-Place, Polly-Show, Pam-Win, Penny-Win].

      Since "Polly was the winner" she must have ridden Patches, find grid-square
      [Poly-Patches] add a 'green box', and a'red xx' for
      [Poly-Giggles,Polly-Sierra] and [Pam-Patches, Penny-Patches].

      We can see from our grid that the only remaining grid square in "SIERRA" column is ['Penny-Sierra'], which leaves 'Pam-Giggles' as the only remaining grid square
      possibility. (Go ahead and fill both of those squares with 'green boxes').

      *IMPORTANT:* At this point we can now fill out the rest of the grid squares with the information at hand! That is since Pam rides Giggles, and Giggles placed, then [Pam-Giggles] can only be 'green', and thus we eliminate grid square [Penny-Place], revealing the final grid square [Penny-Show] as a 'green box', which makes sense logically, because we already know Sierra placed last, and Penny was on Sierra.

      The grid is now complete and we did not even get to our last clue , which is :

      "Pam's pony finished ahead of Penny's."

      Though we did not use this clue to fill up the grid, we can use it as a solution check, and by viewing the grid, we can see this clue reinforces our logic conclusions from the previous step/clue!!

    • Congratulations! Puzzle solved. To summarize:


      Pam: 2nd-place-Giggles
      Penny: 3rd-show-Sierra
      Polly: 1st-win-Patches
    PDF PRINTABLE VERSION

    8.

    THREE-LEGGED RACE-SOLUTION

    THREE-LEGGED RACE(show / hide)

    1ST: STAN-SALLY
    2ND: MARY-FRED
    3RD: Ann-Martin

    step-by-step:( show / hide)

    • Lets look at the first clue "All partners were mixed boy-girl."

    • This will allow us to eliminate the following grid squares[Ann-Sally,Mary-Sally,Stan-Martin,Stan-Fred] by placing 'red xx' in those squares.
      This also leads us to our first team pairing 'Stan-Sally', so go ahead and locate the grid square[Stan-Sally] and click until the 'green box' is highlighted.

    • Lets look at the next clue "Fred did not finish 1ST ,nor was Ann his partner. ."

      We are actually looking at two clues here :

      1. ... Fred did not finish 1ST


      2. ... nor was Ann his partner.


    • For clue 1 look in the chart and find the column with Fred and 1st

      Now click the grid square[Fred-1st] until the 'red xx' appears.

    • For clue 2 Find [Ann-Fred] and click until the 'red xx' appears.

      *NOTE:*(This clue is actually quite useful because it allows us to see that the only possible team-mate
      for Ann must be Martin, so for the grid square [Ann-Martin] add a 'green box'.

      *NOTE:*At this point we can eliminate the grid square[Mary-Martin] and reveal the final pair
      of team mates 'Mary-Fred'. (click grid square[Mary-Fred] until the 'green box' appears.)

      *NOTE:*You may also now exclude Mary from 1st--(since Fred did not finish first, neither can Mary).
      Locate grid square[Mary-1st] and click until the 'red xx' appears there.


    • Lets look at the final clue "Mary and her partner finished 2nd ,behind Stan and his partner. ."

      Lets go to the column labeled '2nd', (and keep in mind that what applies to Mary also applies to Fred),

      locate the grid squares[Mary-2nd, Fred-2nd] and ensure both are filled with 'green boxes'.

      • *NOTE:*Be sure to eliminate all other grid squares in the '2nd' column

      [Ann-2nd, Stan-2nd,Sally-2nd, Martin-2nd]) by adding 'red xx' to all appropriate squares.

    • Lastly we have "....2nd behind Stan and his partner."
      Which means Stan and his partner, (which we now know is Sally), actually won the race,
      leaving the team of 'Ann-Martin' as the third place contestants.

      • You can now complete the remaining grid squares(if you desire).

    • Congratulations! Puzzle solved.

      To summarize:

    • 1ST Place: Stan with his partner, Sally.
    • 2ND Place: Mary with her partner Fred.
    • 3RD Place: Ann with her partner Martin
    PDF PRINTABLE VERSION

    9.

    FLEA MARKET-SOLUTION

    FLEA MARKET - show / hide-

    Bill-12-Ties.
    Jane-14-Belts.
    Kelly-6-Dresses.
    Mary-18 Figures .
    Tom-7-Stuffed animals.

    step-by-step:( show / hide)

    • This first clue "Tom sold one-half as many items as Jane.."


      • Requires a bit of math. We are given the number of items involved ( 6, 7, 12, 14, or 18), therefore we need to determine what is possible with this set of numbers.
      Since "Tom sold 1/2 as many....", lets assume Jane sold 18, than Tom sold 9 (not possible), but

      if Jane sold 14, then Tom sold 7, and if Jane sold 12, then Tom sold 6, are good possibles
      Lets make the following eliminations for these two person's rows as follows:
      Jane - 6, 7, 18. and Tom - 12, 14, and 18.

    • This next clue : "Bill sold more ties than at least 2 others."


      • Here we are given one solution, (Bill-Ties), which allows us to make these eliminations in Bill's row:
      Bill- Belts, Dresses, Figures, and Animals. Followed by eliminations in the Ties Column:
      Ties - Jane, Kelly, Mary , and Tom.

      The last part of the clue("...more ties than at least 2 others.") Allows us to make the additional eliminations in Bill's and Ties rows as follows:
      Bill - 6, 7.
      Ties - 6, 7.

    • The next clue states : "Mary sold at least 3 times as many items as Kelly."


      • We need to use our math skills once again, and realize that Mary must have sold 18 items(because 18 divided by 3 = 6, which is the only mathematical solution that fits within this given set of numbers), therefore we now have two solutions:
      Kelly - 6 , and Mary - 18, which allows us to make these eliminations in the following rows:
      Kelly - 7, 12, 14, 18 and Mary - 6, 7, 12, and 14. Which leads to additional logical eliminations:
      Tom - 6 , (which leads to the solutions Tom-7 (and consequently) Jane-14), which leads to further eliminations:
      Jane - 12, and Bill - 14 (which leads to the solution Bill-12 (and consequently) Ties-12), which leads to the eliminations:
      Ties - 14, 18 and (for the Column 12):
      12 - Belts, Dresses, Figures, and Animals.

    • This final clue is "More figures were sold than animals , and more Belts (which was neither 12 or 18 in quantity) than ties, and less Dresses than Animals ."


      • Lets take each part separately, beginning with : "More figures were sold than animals....", which
      leads to two eliminations: Figures - 6, and Animals-18.

      Next we have "... more Belts (which was neither 12 or 18 in quantity) than Ties..."

      Which means if there were more belts than ties and (since Ties =12), and the Belts do not equal 12, or 18,
      then Belts must equal 14, so make the following eliminations:
      Belts - 6, 7, 18 and (for column 14) 14- Dresses, Figures, Animals , which leads to the solution Jane-14-Belts
      and the eliminations Jane - Dresses, Figures, and Animals , as well as (for Column Belts)
      Belts - Kelly, Mary, and Tom , and Kelly - Figures.

      Our final part of the clue is ".....less dresses than Animals..."
      For this to be true it means we have to eliminate Dress - 18 and Animals - 6.

      (Which in turn leads to the solutions) :
      Figures - 18 , Dresses - 6 and Animals - 7.

      From which it follows:

      Kelly is the one who sold 6 dresses.
      Mary is the one who sold 18 Figures and,
      Tom is the one who sold 7 stuffed Animals.

    • Congratulations! Puzzle solved. To summarize:
      Bill-12-Ties.
      Jane-14-Belts.
      Kelly-6-Dresses.
      Mary-18 Figures .
      Tom-7-Stuffed animals.
    PDF PRINTABLE VERSION

    10.

    SNO CONES-SOLUTION

    SNO CONES - show / hide-

    Valerie 5 -Vanilla.
    Sue 4 -Candy.
    Sam 3- Strawberry.
    Vance 2- Chocolate .

    step-by-step:( show / hide)

    • This first clue "Valerie either had 5 vanilla scoops or 3 strawberry scoops."


      • This clue is giving us the information to allow us to make the following eliminations from Row:
      Valerie - 2, 4, Chocolate, and Candy.
    • This next clue : "Vance was not the one who had 4-candy flavored scoops."


      • Here we are given one solution, (4-Candy), which allows us to make these eliminations in 4's Column:
      4 - Vance, Vanilla, Chocolate, and Strawberry, Followed by eliminations in the Candy row:
      Candy - 2, 3, 5.

    • Next we are given the clue : "Sue had one more scoop than did Sam(who did not have the fewest scoops)."


      • Exploring the last part of this clue first ("Sam, who did not have the fewest"), which allows us to eliminate
      Sam-2 as well as Sue - 2, and 3. (Which reveals the only solution for Column 2, namely Vance-2), which of course allows for the elimination of these grid squares in Vance's row :
      Vance - 3, , 5 and Candy.

      .
    • The last clue states : "A boy chose the strawberry flavored sno-cone."


      • Which implies neither Sue nor Valerie purchased strawberry, which allows the elimination of
      Val-S.Berry and Sue-S.Berry, which in turn leads to the solution Val-Vanilla.

      (With the subsequent eliminations for Vanilla's Column) :
      Vanilla - Sue, Sam, and Vance.

      Now recalling our initial clue "Valerie either had ... " allows us to solve Val-5 , while
      eliminating 5 - Sue, Sam, Chocolate, and S.berry,

      (which in turn reveals the solutions Sue-4 , and Sam-3 ) ,

      (and since the person who had 3-scoops also had S.berry), then this person can only be Sam , and
      Vance had 2 scoops of chocolate!

    • Congratulations! Puzzle solved. To summarize:
      Valerie 5 -Vanilla.
      Sue 4 -Candy.
      Sam 3- Strawberry.
      Vance 2- Chocolate .
    PDF PRINTABLE VERSION

    11.

    AT THE STATION-SOLUTION

    AT THE STATION - show / hide-

    A-Express-810.
    B-Red-eye -801.
    C-Direct-800.
    D-Bullet-815.
    E-Metro-805.

    step-by-step:( show / hide)

    • This first clue "The earliest arrival is on platform C but is not the Bullet, or the Express."


      • The earliest arrival would be 800, thus we have our first solution, namely, C - 800

      , and we are also given eliminations ("... not the Bullet, or Express.").

      Locate row C and make the eliminations: C - Express, Bullet, 801, 805, 810, 815.

      Then for Column 800- A, B, D, E

      and Row 800 - Bullet, Express.



    • This next clue : "The Metro (not arriving at 8:15) arrives sometime after the Direct which arrives before the Red-eye."


      • Lets make the first elimination in Column Metro as follows : Metro - 815

      The next part of the Clue ("... sometime after the Direct..." ) will allow us to make

      more eliminations in column Metro - 800, Direct - 800, 815.

      The last part of the clue ("... Direct arrives before Red Eye" ) , Leads to the logical elimination Metro - 801
      and (more importantly) in column Red-Eye - 800, 815 , which leads to the solution Direct - 800.

      From which we make the following eliminations for Column
      Direct - 801, 805, 810, 815.

      Finally because we have (from the first clue) C - 800, it follows that
      Direct - C must also be true, and we can make the following

      eliminations(by consequence) :

      C - Metro, Red-Eye (and) in column Direct - A, B, D, and E.

    • The next clue : "The Train that arrives on platform D, is at least 10 minutes later than the Train scheduled to arrive on platform B."


      • This means that the train on Plat. D cannot arrive before 815, ( since the one on plat. B can only arrive at 801, or 805,)
      and by the clue ["... at least 10 minutes later ..."], limits our selection for the train on platform D
      (to the exclusion of all others), to the solution D - 815.

      (This also means we make the following eliminations in rows B- 810, 815 and D- 801, 805, 810),
      and for column 815 - A, B, E ,

      Then in row D- Metro, Red-Eye (because from previous clues we know neither of these arrives at 815).

    • The very next clue : "The Red-eye (arriving before at least 3 other trains) is not scheduled to arrive on platform A."


      • Lets make the first obvious elimination: [Red-eye - A.]

      and since we have " ... arriving before at least 3 other..." this means
      Red Eye could not arrive at 805, 810, or 815, so by the logic of elimination,
      it could only have arrived at 801. Therefore make the eliminations in
      Column Red-eye - 805, 810, 815, followed by eliminations in
      Row
      801- Express, Bullet.

    • Now our next clue states : "The Metro is scheduled for platform E ( sometime after 801)."


      • Here, we are simply given a solution Metro - E , which leads to the eliminations starting with
      Column
      Metro- A , B, and D, and then in
      Row E - Express, Bullet, Red-eye, and 801.

    • Finally the last clue is "The Express(arriving on platform A), arrives before only one other Train."



      • Which yields a solution Express-A, ( allowing us to eliminate A-Bullet ) ,and in
      Column Express - B, and D.

      (Now revealing other solutions), beginning with

      D - Bullet and (by consequence) B - Red-Eye.

      Lastly, we are told the Express is arriving ( "... before only one other Train. "),
      which, of course, means Express could only have arrived at 810 , (which is before only one other train!)

      (Which leads to the only remaining solution: Metro - 805 )

    • Congratulations! Puzzle solved. To summarize:
      A-Express-810.
      B-Red-eye -801.
      C-Direct-800.
      D-Bullet-815.
      E-Metro-805.
    PDF PRINTABLE VERSION

    12.

    BUNGLING BURGLARS-SOLUTION

    BUNGLING BURGLARS - show / hide-

    Sticky-920 Mulberry-Smith.
    Lefty-837 Mulberry-Baker.
    Shifty-901 Mulberry-Wilkes.
    Slappy-833 Mulberry-Rogers.
    Sharky-820 Mulberry-Jones.

    step-by-step:( show / hide)

    • This first clue " Neither the Jones' nor the Rogers' live at 920 Mulberry."


      • A simple elimination clue is provided, therefore locate the following grid squares (and eliminate them):
      Row 920 - Jones, Rogers.

    • This next clue : "The Smiths live 19 house numbers from the Wilkes family home, (which Shifty was caught burglarizing)."


      • We have the following house numbers : 820, 833, 837, 901, and 920, therefore we must find the number combos
      that are exactly nineteen apart. Going over the list the only pair that fits this clue is 901 - 920,

      and we can make the following eliminations in the :
      Columns Smith - 820, 833, and 837. Likewise for the column Wilkes- 820, 833, and 837, while eliminating squares in these Rows 901 - Jones, Bakers, Rogers, (as well as) 920 - Jones, Bakers, and Rogers.

      Don't forget we have from the last portion of this clue : ( "...the Wilkes family home, which Shifty
      was caught burglarizing.")       Therefore, we are given the solution Shifty-Wilkes,

      (From which we can then make these eliminations) :
      Row Shifty - Jones, Smith, Bakers, Rogers, 820, 833,and 837.

      Next, we can eliminate from the Column Wilkes - Sticky, Lefty, Slappy, and Sharky.

    • The next clue "The two closest homes involved belonged to the Bakers and the Rogers (in some order)."


      • From our list of house numbers , the closest two are 833 and 837 , therefore we can make the following eliminations:

      In Rows 833 - Jones (Followed by row) 837- Jones , which leads to the solution 820-Jones.

    • The following clue "Lefty tried to break into a house that was 17 house numbers away from the Jones'."


      • Is telling us the house Lefty tried to burglarize was 17 away from the Jones(which we just revealed as 820 Mulberry),

      therefore the house referred to here is 837, and we have the solution Lefty - 837 , which leads to the eliminations
      from Row Lefty - Jones, Smith, 820, 833, 901 and 920, then from

      Column 837 - Sticky, Slappy, and Sharky6.

    • The next clue "Slappy was caught at 833 Mulberry (the Rogers' home)."


      • Here we are provided with a multiple solution Slappy - 833 - Rogers, from which
      We can make our eliminations as follows, begin with

      Row Slappy - Jones, Smith, Bakers, 820, 901 and 920.

      Then from Column 833 - Sticky, Sharky Followed by Column Rogers - Sticky, Lefty, Sharky, 837, ,

      and from Row 833 - Bakers, which leads directly to the solution 837 - Bakers, (which is the house Lefty
      was caught at), so the solution Bakers-Lefty must also be true, which allows us to complete the Bakers column with these eliminations:

      Bakers - Sticky, Sharky .

    • Our final clue states: "The Smiths' house is at 920 Mulberry which is 100 house numbers away from the house that Sharky broke into.."


      • Begin with applying the solution Smith - 920 to the grid, (which eliminates Wilkes-920 (and as a consequence)

      reveals the solution Wilkes - 901,and (consequently) Shifty - 901).

      Finally we have the last part of the clue ("...100 house numbers away from the house that Sharky broke into").

      Which means Sharky broke into the house at 820 ( not the Smith's), therefore, Sticky is the one who

      broke into the Smith's house at 920 , and Sharky broke into the Jones' home at 820 ( which is 100 houses from the Smith's).

    • Congratulations! Puzzle solved. To summarize:
      Sticky-920 Mulberry-Smith.
      Lefty-837 Mulberry-Baker.
      Shifty-901 Mulberry-Wilkes.
      Slappy-833 Mulberry-Rogers.
      Sharky-820 Mulberry-Jones.
    PDF PRINTABLE VERSION




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