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).
1. |
AT THE PET SHOP(2) (show / hide)
Bill -Monkey-Toto
Jen -Snake-Wobbles Roger -Dog-Spot Susan -Cat-Spike step-by-step:( show / hide)
>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. 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). 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. 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. Now the only grid square that meets this requirement is "Spot-Snake". So find [Spot-Snake] and place a 'red xx' in that box. 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]. 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]). |
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2.
AIRPORT -SOLUTION (show / hide)
Adam -11:30am-1:30pm
Mary-10am-1:15pm Kathy-9am-2pm Susan 12-1:45pm step-by-step:( show / hide)
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 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. 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 [Adam-9am ,Adam-2:000pm, Susan-9am ,Susan-2:000pm] 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 [Adam-10am,Adam-12noon, Adam-1:15pm, Adam-1:45pm] as well as [Max-1130am, Max-1:30pm, Susan-1130am, Susan-1:30pm] 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]. 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). |
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3.
INSURANCE CO.-SOLUTION (show / hide)
Bill -Home-Jane.
Joan-Life-Mary. Roger-Auto-Stan. Susan-Theft-Alex. step-by-step:( show / hide)
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. Click the grid squares[Susan-Mary] until the 'red xx' appears. Click the grid squares[Jane-Life, Jane-Auto] until the 'red xx' appears. 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]. 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]. 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] ). 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'. |
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4.
ICE CREAM TREATS (show / hide)
Ben -Lotso choc- 2pm.
Jerry-Rocky nut-3pm. Reba-Alamo fudge-1pm. Sadie Pink jubilee-4pm step-by-step:( show / hide)
Locate the row labeled 'Sadie' and find the column '1pm' Now click the grid square[Sadie-1pm] until the 'red xx' appears. 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]. 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]. 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]. 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: [Reba-2pm, Reba-3pm, Reba-4pm, Sadie-2pm, Sadie-3pm, Ben-1pm, Jerry-4pm]. 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]. 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], Sadie'MUST' have purchased it! |
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5.
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)
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. 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." 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. Eliminate [Edie-Monday] because it was 'after Bob's discovery'. Eliminate [Edie-Margo] because she did not find Margo. 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 ] 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]. 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. |
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6.
MIXED DRINKS-SOLUTION (show / hide)
Adam-Water-Lime.
David-Soda-Vanilla. Mary-Juice-Strawberry. Wilma-Tea-Lemon. step-by-step:( show / hide)
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. Both men(Adam, David) either had soda or water, therefore select these grid squares for elimination: [Adam-Tea,Adam-Juice, David-Tea, David-Juice, Lemon-Water] and for the women(Mary, Wilma) Eliminate [Mary-Soda,Mary-Water, Wilma-Soda]. 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]. 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], which means Adam drank water & Lime, so select Adam-Water and Adam-Lime, which in turn, eliminates [Adam-S.berry, Adam-Vanilla]. 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'. |
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7.
TEST SCORES - show / hide-
Amber-75-85.
Marc-90-60. Marie-65-95. Matt-85-75 . Stacy-100-90. step-by-step:( show / hide)
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. 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. 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. We can make the elimination from the Row Amber - 100. 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. 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 . |
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8.
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 clue yields the following eliminations: Row Indians - 3, 5 and Angels - 18 and Giants - 18. 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. Is a straightforward one, they must have lost either8, 16, or 18 games. So make the following elimination in Row Rangers - 3, 5. 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 Angels had 4 losses) 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 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 |
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9.
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)
This clue yields the following eliminations: Row Jones - Music, History Another simple elimination clue RowTaylor - 85, 80. Which leads to an immediate elimination in Row Smith - Spanish, Art, then Row 80 - Music, History. 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 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. |
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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 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. " ) ). Which leads to two simple eliminations : Locate Column A+ - Science , and Geography . 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. 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. 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 . 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 . 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. |
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11.
JUMPING BEANS - show / hide-
Jane-Green-5.
Jasper-Yellow-11. Jeff-Blue-3. Jim-Orange-7. Julie-Red-9. step-by-step:( show / hide)
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. 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 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 . 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. Let's start with the obvious elimination in Row Julie - Yellow, and then for Column Yellow - 3, 5 and 7. 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. 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. |
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12.
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)
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. 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. 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. 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. 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 . 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. |
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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)
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. 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 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. 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. 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. 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. 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... ", which means Robin received 50 votes and therefore Tanya could only have received 20. |
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14.
"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)
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. This is an elimination clue, so locate the following Column 7AM - Taxi, Tour Bus and make the eliminations. 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. 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. 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. 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. |
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