Logic Game Challenge #35: The Toy Maker Game (Easy)

[wingedfeetxc]

The Toy Maker Game (Easy) will be live in a few.

Discuss the game here.

[wsjstockstar]

Toy Maker Game (Easy)

6 possible colors (F,G,H,I,J,K)
3 different types of toys: toy a, toy rc, and toy t
2 colors assigned to each toy, each color assigned once

Rules:
If G(t), then K(t)
Color I requires assignment with either G or J
Color H cannot be with Color J
Either Color F or Color G must be assigned to rc, but not both

Question Analysis and Answers:
Question 1
a) J and H cannot be assigned to the same toy. In this setup, J and H are assigned to toy a.
b) This arrangement satisfies the conditions. NOTE: During a real LSAT exam, it would be in the test takers best interest to stop here. There is no need to pursue other answer choices because we have identified an arrangement that is satisfactory. This is all we need, and in order to use time as efficiently as possible, we could proceed to Question 2. In the interests of the post, though, we can explore Question 1 c, d, & e.
c) Colors F and G are both assigned to toy rc. This is not acceptable, because either Color F or Color G must be assigned to toy rc, but not both.
d) Notice that Color G is assigned to toy t. If G(t), then K(t). We cannot have G and H on toy t due to the conditions. Therefore, this is not an acceptable arrangement.
e) Notice that Color I is assigned to toy a. If Color I is assigned to a toy, it must be assigned with either Color G or Color J. Color I cannot be assigned to a toy and then paired with Color H. Therefore, this is an unacceptable arrangement.

CORRECT ANSWER FOR QUESTION 1 is B


Question 2
a) If Goldenrod is used on the airplane, this means that Fuchsia must be used on the race car. Now there are choices available for the other colors. Indigo could be with Jonquil, or Indigo could be with Goldenrod. We can stop at this point because the colors for the toys cannot be determined if Goldenrod is used on the airplane.

b) If Goldenrod is used on the race car, we know that fuchsia cannot be used on the race car. Given this amount of little information, we can see that we have not been able to determine much. Therefore, this prompt would not determine the colors for all three models.

c) If Goldenrod is used on the truck, khaki must be used on the truck. Since the race car requires either goldenrod or fuchsia, and goldenrod has already been used, the race car will use fuchsia. Since indigo must be paired with either goldenrod or indigo, and goldenrod has already been paired, indigo will be paired with jonquil. Since indigo and jonquil must be a pair in this circumstance, indigo and jonquil will be assigned to the airplane. This leaves only one space left, the second color for the race car. Since the race car already has fuchsia, the race car will use fuchsia and heliotrope. We can see that given this answer choice, we can determine the colors for all three models. Again, if this were a real LSAT exam, it would be best to stop here and move on to question 3, so that time would not be wasted examining incorrect answer choices. However, for the purposes of this explanation, we can examine Question 2 d & 2 e.

d) If fuchsia is used on the airplane, we know that goldenrod will be assigned to the race car. It's important to note that there is a lot we don't know, and we could creatively fill in the blanks. If we could creatively fill in the blanks, then the order has not been determined.

e) If fuchsia is used on the race car, we know that goldenrod will not be used on the race car. However, goldenrod could be used on the truck, or goldenrod could be used on the airplane. There is no indication of the placement of other colors, thus, there is very little information to help us determine the colors for all three models.

CORRECT ANSWER FOR QUESTION 2 is C


Question 3
If goldenrod and khaki are used together on one model, we can set up three customized, nifty diagrams (see link below).

https://sites.google.com/site/appaloosaappaloosadual/

We can see that there is a pattern in our diagrams which is that fuchsia and helitrope are used on one model. Now, the question is, is this pairing necessary? Must fuchsia and helitrope be used on one model? Yes! Yes, they must. Therefore, we must find the answer choice that reflects this "must-have" pairing.

CORRECT ANSWER CHOICE FOR QUESTION 3 is E


Question 4
If fuchsia and jonquil are used on the race car, we know several facts. First, indigo cannot be paired with jonquil. Second, goldenrod cannot be used on the race car.

Since indigo needs to be paired with either goldenrod or jonquil, indigo must be paired with goldenrod, for jonquil has already been assigned. We know that goldenrod cannot be assigned to the truck, for if goldenrod were assigned to the truck, khaki would also be assigned to the truck. But indigo must be paired with goldenrod. So, if goldenrod cannot be assigned to the truck, this means that goldenrod (paired with indigo) must be assigned to the airplane. This leaves us with two colors, khaki and heliotrope. There are two spaces left (under the truck) and there are two colors left. So, therefore, there is only on pair of colors that can be used on the truck.

CORRECT ANSWER CHOICE FOR QUESTION 4 is A


Question 5
a) If goldenrod is used on the airplane and heliotrope is used on the airplane, fuchsia is used on the race car, and indigo is paired with jonquil on the truck.
b) If goldenrod is used on the airplane and khaki is used on the airplane, fuchsia is used on the race car, and indigo is paired with jonquil on the truck.
c) If goldenrod is used on the airplane and jonquil is used on the race car, fuchsia is used on the race car, and indigo is used on the airplane.
d) If goldenrod is used on the airplane and indigo is used on the race car, we have a problem. See, fuchsia must be used on the race car if goldenrod is used on the airplane, and this is a problem for indigo. Indigo must be paired with either goldenrod or jonquil, but since this answer choice is forcing indigo to be paired with fuchsia, this answer choice is a logical impossibility. If this were a real LSAT exam, it would be best to stop here and move on to the last question. For the purposes of this explanation, we can examine 5e.

e) If goldenrod is used on the airplane and heliotrope is used on the truck, fuchsia is used on the race car, indigo is used on the airplane, and the other colors can fall into their desired places.

CORRECT ANSWER CHOICE FOR QUESTION 5 is D


Question 6
If indigo is used on the race car, we could have either fuchsia or goldenrod on the race car, right? No. We could only have indigo paired with goldenrod. Therefore, fuchsia, heliotrope, jonquil and khaki must find suitable arrangements on the airplane and on the truck. We could have jonquil and khaki on the truck, or heliotrope and khaki on the truck, or jonquil and fuchsia on the truck, or heliotrope and fuchsia on the truck. But, we cannot have heliotrope and jonquil on the truck -- this pairing is forced by the conditions in answer choice C.

CORRECT ANSWER CHOICE FOR QUESTION 6 is C

[canarykb]

Is this contest still going? I wanna winnn it.

Starting out
I draw out a little chart if at all possible, so the first thing I did when reading the prompt was to draw this:
A R T
_ _ _
_ _ _
(Recognizing that the order of the colors on each toy does not matter.)

We also learn the following rules:
1. HxJ
2. R has F OR G
3. IG or IJ
4. G only on T if K on T
With these type of problems where the diagram is pretty clearly laid out, and most of them aren’t if/then statement I try to avoid writing out the logic. I find it muddles it for me, I just write the rules by the diagram in a way that makes things clear.

1. WHICH OF THE FOLLOWING COULD REPRESENT THE COLORS USED FOR EACH TOY?

My strategy for these type of "find a possible answer" question is to just go rule by rule and find all the ones that violate each one in turn and cross them out. So we start with all options : ABCDE

1. A violates rule 1 so now we only have options BCDE
2. C violates rule 2 so now we have options BDE
3. E violates rule 3 b/c I is with H, so now we have options BD
4. In (D) G=T, but K =/ T, violating rule 4

SO WE’RE LEFT WITH B

2. WHICH OF THE FOLLOWING WOULD DETERMINE THE COLORS FOR ALL THREE MODELS?

For these I look first at the answers dealing with variables that have a lot of rules about them, because their placement will determine the most. Once I find and answer that works, I just move on. So looking through our rules I know we know a lot about G. We also know a lot about G in relation to truck specifically, so I tried out C first.

(C) Goldenrod is used on the truck.

The initial rule means out diagram looks like this:
A R T
_ _ G
_ _ _
And apply rule 4, we know that K must = T, if G = TA R T
_ _ G
_ _ K

Another thing we know about G is in rule 3, that F or G must = R. Because G is no longer available, F = R
A R T
_ F G
_ _ K
The Final thing we know about G is if G isn’t with I, J is with I (Rule 3). So if J & I must be together, they only have one option left!
A R T
I F G
J H K

I filled H in in the final spot. To double check my work, I made sure that it doesn’t violate rule 1, which it doesn’t.

SO I KNOW C DETERMINES THE DIAGRAM AND IS THE CORRECT ANSWER

At this point I WOULD MOVE ON. (Other people have explored the other possibilities, so you can read those)

3. IF GOLDENROD AND KHAKI ARE USED TOGETHER ON ONE MODEL, WHICH OF THE FOLLOWING MUST BE TRUE?

Looking at the answers, we see that there are two ways to go, either determining the potential placements of individual colors and where things MUST be placed(Answers A, B, C) OR determining the potential pairings and which things MUST be paired (Answers D, E).
Potential pairings is quick to figure out, so I looked at that first:
So we know there must be three pairings (one for each toy). G and K are paired which leaves IFHJ to be paired. Rule 3 says I has to be with either G or J, it can’t be with G, so it must be with J.

So the pairings MUST be: GK; IJ; FH

ANSWER (E) STATES THAT FUCHSIA AND HELIOTROPE MUST BE USED ON ONE MODEL. WHICH WE KNOW MUST BE TRUE.

So at this point, i would move on, since i found the correct one. (I think you can look in the forum to go over other possibilities)

4. IF FUCHSIA AND JONQUIL ARE USED ON THE RACE CAR, HOW MANY DIFFERENT PAIRS OF COLORS CAN BE USED ON THE TRUCK?

This is another question where pairing is important, so I figured out the pairs first. So we know that F is with J. Applying rule 3, we know that I must be with J or G, and it cant be with J, so we know I is with G. Which leaves us with the following colors that MUST be paired:

FJ; IG; KH

Placing them in the diagram, we’ve been told FJ is on the racecar. So:
A R T
_ F _
_ J _

So IG or KH are left to be in on the Truck. Applying rule 4, we know that if G is on the Truck, K is on the truck. But, since they cannot be paired on the same Toy, G cannot be a on the Truck. So KH must be paired on the truck.

SO THE ANSWER IS (A) 1, BECAUSE WE ONLY HAVE ONE OPTION.

5. IF GOLDENROD IS USED ON THE AIRPLANE, EACH OF THE FOLLOWING COULD BE TRUE EXCEPT:
Let’s diagram it:
(applying rule 2....)
A R T
G F _
_ _ _

We don’t get much, but applying rule 3, we know that either I is with G on A, or it must be with J.
So that splits out 2 options:
OPTION1:
A R T
G F K
I _ _
(Because H cannot be with J(rule1), we know that K must be on the Truck here)
OPTION 2:
A R T
G F I
_ _ J

So now we check them against the answers, and find the one that is NOT possible on EITHER diagram
A) Heliotrope is used on the airplane. _ could be on Opt 2
(B) Khaki is used on the airplane. - could be on Opt 2
(C) Jonquil is used on the race car. - could be on Opt 1 or 2
(D) Indigo is used on the race car. - NOPE
(E) Heliotrope is used on the truck. - could be on Opt 2

SO WE KNOW D IS THE CORRECT ANSWER

6. IF INDIGO IS USED ON THE RACE CAR, WHICH OF THE FOLLOWING CANNOT BE THE PAIR OF COLORS USED ON THE AIRPLANE?
Another pairing Question! I figured out the pairs first:
If I is on R, and F or G must be on R (rule 2), but I can only be with J or G (rule 3) the only option for the racecar is IG. We don’t know much else except H cannot be with J. So the possible pairs are:

OPTION 1: IG (on R); HK (on A or T); JF (on A or T)
OPTION 2: IG (on R); HF (on A or T); JK (on A or T)

Now that we’ve got our options, we check against the answers:
(A) fuchsia and heliotrope _ could be on opt 2
(B) fuchsia and jonquil _ could be on opt 1
(C) fuchsia and khaki _ NOPE
(D) heliotrope and khaki _ could be on opt 1
(E) jonquil and khaki_ could be on opt 2

SO C MUST BE THE CORRECT ANSWER

[sahilsalunke.21]

The Toy Maker Game (Easy) will be live in a few.

Discuss the game here.

As we have to answer 6 questions, we will not use a question wise approach to setups. Instead we can easily invest 5 minutes of time to build the full setup of the game as follows:-

Out of all the constraints, only two are critical to build the setup.
-We can only have F or G with the race car. So that limits the placeholder options.
- IG or IJ pair must exist. (Therefore GJ is not possible)

We will use the first rule to divide the possibilities into: race car using F and race car using G. The second rule will be used in conjunction so as to find the remaining pairs.

Race car using F
Suppose we use F for the race car, then we can only pair it with H, J and K. This gives us the following 5 possibilities using all the rules of the game:
Case Aircraft Race car Truck
1 IG FH KJ
2 IJ FH KG
3 KG FH IJ
4 IG FJ KH
5 HG FK IJ

Race car using G
Now if we use G for the race car, then we can only pair it with I, H and K. (Note: GJ pair is not possible as it will result in I being left without any pair)
Listing the possibilities using all the rules of the game we have:
Case Aircraft Race car Truck
6 FH IG KJ
7 FJ IG KH
8 KJ IG FH
9 KH IG FJ
10 IJ GH FK
11 FK GH IJ
12 IJ GK FH
13 FH GK IJ
--------------------------------------------------------------------
Now for the questions,
Q1.
Option (A) HJ on airplane. As we cant find HJ pair in first column of any of the 13 cases option (A) is wrong.
Option (B) KH on airplane. We identify it in case 9. It is easily seen that other columns match to that in option (B)

There is NO need to check further options.

So answer is (B)


Q2.
We need an option which gives us only one situation/case.
Option(A) G on the airplane. We look for G in the first column. We spot cases 1,3,4 and 5. Clearly more than one, so (A) is wrong.
Option(B) G on the race car. We look for G in the second column. We spot cases 6 to 13. Clearly more than one, so (A) is wrong.
Option(C) G on the truck. We look for G in the third column. As it results in only one case i.e. case 2.

So (C) is our answer.

Q3.
GK are on one model so we have cases 2,3,12 and 13. In all these case we have fixed remaining pairs FH and IJ. So option (E).
Alternately, see if options apply to ALL of 2,3,12 and 13 cases.

So (E) is our answer.

Q4.
FJ is used on the race car. Look for FJ in second column.
We see only Case 4. So, only one possibility.

(A) is our answer.

Q5.
G is used on aircraft. So look for G in column 1. Cases 1,3,4 and 5. Check if options apply to any of cases 1,3,4 and 5.
Clearly, (D) is not true to any of these cases.

(D) is our answer.

Q6.
I is used on the race car. Look for I in column 2. Cases 6,7,8 and 9.
So only FH,FJ,KJ and KH can be used on the aircraft. So (C) is incorrect as FK is not on the aircraft.

(C) is our answer.

---------------------------------------------------------------

[ohthatpatrick]

Those were WONDERFUL submissions! Nice work, everyone.

wsjstockstar - Your explanation was definitely the most comprehensive. And I loved the way you mimicked a test-taker's thought process, reacting to the question stem or answer choice by reflecting on how certain stimuli would trigger us to ask ourselves certain questions or analyze certain rules.

canarykb - Your explanations were definitely the best formatted for screen. You did a great job showing diagrams and made great use of boldface. Most importantly, though, I thought you provided some invaluable tips about how to approach each question most efficiently.

sahilsalunke.21 - You get the award for "most original" response. Your take on itemizing every permutation of the game before diving into the questions was very interesting, and it obviously made short-work of working through the questions.

That type of explanation raises an important conversation about the tradeoffs involved in sketching out all the possibilities ahead of time.

Pros:
- We can consolidate our thinking into two different phases: 1. Listing out the possibilities 2. Using that data to answer questions

It could be argued that listing out possibilities can move very quickly and efficiently, since the rules stay fresh in your mind without the distractions of specific questions giving temporary constraints.

Cons:
- This approach will not work for many (if not most) games. The poster did a great job of contextualizing what it was about this game's setup and rules that led him/her to see that there was an approachably finite set of possibilities. However, games with more uncertainty would have so many permutations that it would be untenable to try listing them all. Since we want our approach to games to be as consistent as possible (so that we feel comfortable and familiar each time we attack a new one), it might be better to just attack the questions, as we'll have to do that in most games.

- By definition, doing an exhaustive list of possibilities is doing more work than you need to. The questions that go with a game are never going to test every possible scenario, so by listing out all possibilities we're doing more work than the game demands. Since time is of the essence, there seems to be something inherently flawed in that strategy.

(as I mentioned earlier, the counterargument here is to say that one is quicker listing out all possibilities than one would be dealing with the uncertainty involved in each question -- although there are probably certain games and certain test-takers for which this could be true, I think these would be the exception to the rule)

The one question I'd like to briefly mention is Q5. It tells us that G is assigned to A (which seems to only lead to the inference that F must thereby be assigned to R). It is tempting to start sketching out scenarios to fill in the blanks that remain, but if we were to first take a quick pass through the answers, we would probably see that (D) must be false. After all, we can't have IF together.

I think this problem gives us a valuable reminder that it can be fruitful to take a quick pass through a MBT or MBF question's answer choices to see if anything 'clicks', before we undertake to sketch out the possibilities in more detail.

Overall, great work everyone! Thanks for your participation!