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).
1.
AT THE PET SHOP(1) (show / hide)
Bill  snake.
Jen  dog. Roger  monkey. Susan  cat. stepbystep:( show / hide)

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2.
A MATTER OF DOLLS (show / hide)
Ann:yellowcries
Mary: pinksings Stacy: whitetalks stepbystep:( show / hide)
We are actually looking at two clues here : *NOTE:*By using visual logic we can see the square [Marypink] is the only remaining option , so click on this square until the 'green box' appears. ( We can now eliminate [Stacypink], by placing a red xx in that box, leaving only [Stacywhite] and because the doll with the white dress=talks, we locate the square[Stacytalks] and place 'green boxes' in both these squares, while eliminating these three squares[Stacycries,Stacysings,Anntalks]). ( As a consequence we can see that [criesyellow] should now be filled with a 'green box'  which NOW leads us to the final solution after locating [anncries], which gets a 'green box', while the grid squares[Marycries, Annsings] both get 'red xx' boxes) which in turn ,leaves only [Marysings] which of course is filled with the final 'green box'. 
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3.
BACKPACKS(show / hide)
SAM: green5th
SAL: blue3rd SEAN: purplelunch stepbystep:( show / hide)
We are actually looking at two clues here : 
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4.
Pet SHOP Days(show / hide)
Bill: Monkey
Jane: Snake Roger : Dog Susan: Cat stepbystep:( show / hide)
One could at this point complete the grid, but since the puzzle is solved, it is not a requirement! To summarize:

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5.
Pet SHOP Days2(show / hide)
Bill: Dog
Jane: Monkey Roger : Cat Susan: Snake stepbystep:( show / hide)
[SusanCat, SusanDog, SusanMonkey] as well as those for the column 'Snake': [BillSnake, JaneSnake, RogerSnake] should all now be filled with 'red xx'.

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6.
AUTO REPAIR INTERACTIVE show / hide
Bill EngineWilly
Jim Check/upSam Roger OilAdam Susan TireTom stepbystep:( show / hide)
This provides us with two immediate elimination clues: [SusanOil, SusanWilly]  fill both with 'red xx'. This clue simply provides us with several more elimination grid squares : [RogerEngine, RogerCheck,RogerSam,RogerTom]. (Cheeky monkey eh?) The 'woman' , which allows us to eliminate the grid squares [BillTire,JimTire,RogerTire], Why? Because none of these people are women! Lets now take a moment to update our grid squares as follows : First locate[SusanTire] and fill it in with a 'green box'. Next , for the grid squares [SusanEngine, SusanCheck] choose 'red xx' for both. Lastly, (Since we now know she ....."either saw Tom or Adam"), we can eliminate the following grid squares:[SusanSam], as well as [SamTire, WillyTire]. Then for grid square [JimWilly] a 'red xx'. *NOTE:* (This clue will also allow us to uncover another), To do so : First eliminate [BillCheck, WillyCheck], 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: 'BillEngine' ,so go ahead and mark the grid square [BillEngine] with a 'green box'. Now, by eliminating the grid square[BillOil], reveals yet another solution : 'RogerOIL'! Now since we have 'RogerOIL', we can eliminate[SamOil,TomOil], 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. Lets fill in grid square [RogerAdam] with a 'green box', then make the (appropriate) eliminations of these grid squares: [RogerWilly, BillAdam, JimAdam, SusanAdam], which now reveals Susan's mechanic (remember, from clue 3 : "....it was either 'Tom or Adam' ) so that the grid square [SusanTom] may now be highlighted 'green'. Doing so results in the elimination of the grid squares: [BillTom, JimTom], which in turn, now reveals our final solution (by looking at column 'Willy'), namely, 'BillWilly' (again by the process of elimination), and after we add the 'green box' in grid square [BillWilly], it logically follows that our last combination of motoristmechanic can only be: 'JimSam'! (You may fill in the lower grid squares if you like, as an additional check on the logical eliminations!) 
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7.
MY LITTLE PONIES (show / hide)
Pam: 2ndplaceGiggles
Penny: 3rdshowSierra Polly: 1stwinPatches stepbystep:( show / hide)
So to begin locate the grid square[PamSierra] and place a 'red xx' in that square. Next we have "....Sierra....did not win" , so locate grid square[SierraWin] and place a 'red xx' in that square. Simply locate grid square[GigglesPlace] and add a 'green box' in that square. *IMPORTANT:* We must also eliminate the following grid squares : [GigglesWin,GigglesShow], as well as [PatchesPlace, SierraPlace]. We can now see that in the "WIN" column there is only one empty square 'PatchesWin', so mark that grid square with a 'green box' and add a 'red xx' for 'PatchesShow' Which allows us to add a'green box' to ['SierraShow'], and since we know from clue 1 that "...Pam's pony was not Sierra" and Sierra was third , then we can eliminate ['PamShow'] by adding a 'red xx' . Start by filling the grid square['PolyWin'] with a 'greenbox' and place the 'red xx' in the corresponding elimination squares[PollyPlace, PollyShow, PamWin, PennyWin]. Since "Polly was the winner" she must have ridden Patches, find gridsquare [PolyPatches] add a 'green box', and a'red xx' for [PolyGiggles,PollySierra] and [PamPatches, PennyPatches]. We can see from our grid that the only remaining grid square in "SIERRA" column is ['PennySierra'], which leaves 'PamGiggles' 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 [PamGiggles] can only be 'green', and thus we eliminate grid square [PennyPlace], revealing the final grid square [PennyShow] 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!! 
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8.
THREELEGGED RACE(show / hide)
1ST: STANSALLY
2ND: MARYFRED 3RD: AnnMartin stepbystep:( show / hide)
[Ann2nd, Stan2nd,Sally2nd, Martin2nd]) by adding 'red xx' to all appropriate squares. 
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9.
FLEA MARKET  show / hide
Bill12Ties.
Jane14Belts. Kelly6Dresses. Mary18 Figures . Tom7Stuffed animals. stepbystep:( show / hide)
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. Here we are given one solution, (BillTies), 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. 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 Tom7 (and consequently) Jane14), which leads to further eliminations: Jane  12, and Bill  14 (which leads to the solution Bill12 (and consequently) Ties12), which leads to the eliminations: Ties  14, 18 and (for the Column 12): 12  Belts, Dresses, Figures, and Animals. Lets take each part separately, beginning with : "More figures were sold than animals....", which leads to two eliminations: Figures  6, and Animals18. 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 Jane14Belts 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. 
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10.
SNO CONES  show / hide
Valerie 5 Vanilla.
Sue 4 Candy. Sam 3 Strawberry. Vance 2 Chocolate . stepbystep:( show / hide)
This clue is giving us the information to allow us to make the following eliminations from Row: Valerie  2, 4, Chocolate, and Candy. Here we are given one solution, (4Candy), 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. Exploring the last part of this clue first ("Sam, who did not have the fewest"), which allows us to eliminate Sam2 as well as Sue  2, and 3. (Which reveals the only solution for Column 2, namely Vance2), which of course allows for the elimination of these grid squares in Vance's row : Vance  3, , 5 and Candy. . Which implies neither Sue nor Valerie purchased strawberry, which allows the elimination of ValS.Berry and SueS.Berry, which in turn leads to the solution ValVanilla. (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 Val5 , while eliminating 5  Sue, Sam, Chocolate, and S.berry, (which in turn reveals the solutions Sue4 , and Sam3 ) , (and since the person who had 3scoops also had S.berry), then this person can only be Sam , and Vance had 2 scoops of chocolate! 
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11.
AT THE STATION  show / hide
AExpress810.
BRedeye 801. CDirect800. DBullet815. EMetro805. stepbystep:( show / hide)
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. 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 RedEye  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, RedEye (and) in column Direct  A, B, D, and E. 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, RedEye (because from previous clues we know neither of these arrives at 815). Lets make the first obvious elimination: [Redeye  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 Redeye  805, 810, 815, followed by eliminations in Row 801 Express, Bullet. 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, Redeye, and 801. Which yields a solution ExpressA, ( allowing us to eliminate ABullet ) ,and in Column Express  B, and D. (Now revealing other solutions), beginning with D  Bullet and (by consequence) B  RedEye. 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 ) 
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12.
BUNGLING BURGLARS  show / hide
Sticky920 MulberrySmith.
Lefty837 MulberryBaker. Shifty901 MulberryWilkes. Slappy833 MulberryRogers. Sharky820 MulberryJones. stepbystep:( show / hide)
A simple elimination clue is provided, therefore locate the following grid squares (and eliminate them): Row 920  Jones, Rogers. 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 ShiftyWilkes, (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. 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 820Jones. 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. 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 BakersLefty must also be true, which allows us to complete the Bakers column with these eliminations: Bakers  Sticky, Sharky . Begin with applying the solution Smith  920 to the grid, (which eliminates Wilkes920 (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). 
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