Logical Reasoning & DI (LRDI) Test - 2

We know that the slack in pipes A-B, A-C and A-D is zero. So, the volume of oil that flows in these three pipes is 8 million litres. B’s capacity is 3 million litres. So, the remaining 5 million litres is distributed into D and E equally. So, 2.5 million litres flow in both B-E and B-D pipes.
Similarly, the oil that flows into C is 8 million litres. C’s capacity is 2.5 million litres. So, the remaining 5.5 litres is distributed into D and F equally. So, 2.75 litres flow in both C-D and C-F pipes.
So, a total of 8 + 2.5 + 2.75 = 13.25 million litres has flown into D. D’s capacity is 4.5 million litres. The remaining 8.75 million litres is distributed into D-E, D-G and D-F pipes equally. So, 8.75/3 = 2.917 million litres flows in the D-G pipeline.

We know that the slack in pipes A-B, A-C and A-D is zero. So, the volume of oil that flows in these three pipes is 8 million litres. B’s capacity is 3 million litres. So, the remaining 5 million litres is distributed into D and E equally. So, 2.5 million litres flow in both B-E and B-D pipes.
Similarly, the oil that flows into C is 8 million litres. C’s capacity is 2.5 million litres. So, the remaining 5.5 litres is distributed into D and F equally. So, 2.75 litres flow in both C-D and C-F pipes.
So, a total of 8 + 2.5 + 2.75 = 13.25 million litres has flown into D. D’s capacity is 4.5 million litres. The remaining 8.75 million litres is distributed into D-E, D-G and D-F pipes equally. So, 8.75/3 = 2.917 million litres flows in the D-F pipeline.
So a total of 2.75 + 2.917 = 5.667 million litres flow into the F oil well. The capacity of F is 4 million litres. So, 5.667 - 4 = 1.667 million litres flow into the F-G pipeline.
The flow in D-E pipeline is 2.917 million litres and B-E pipeline is 2.5 million litres. So, a total of 5.417 million litres flows into E. The capacity of E is 2 million litres. So, the extra 3.417 million litres flows into E-G pipe.
So, a total of 3.417 + 2.917 + 1.667 = 8 million litres flows into G, which is distributed into 20 reservoirs. So, each reservoir has 8million/20 = 0.4 million litres.
Alternative approach:
The total flow into G cannot also be found by subtracting the requirement in each of the oil wells from the total outflow from A.
Total outflow from A = 3*8 = 24 million litres
Total requirement of B, C, D, E and F = 3 + 2.5 + 4.5 + 2 + 4 = 16 million litres
Total flow into G = 24 - 16 = 8 million litres, which is distributed into 20 reservoirs.
⇒ Each reservoir has 8 million/20 = 0.4 million litres.

We know that the slack in pipes A-B, A-C and A-D is zero. So, the volume of oil that flows in these three pipes is 8 million litres. B’s capacity is 3 million litres. So, the remaining 5 million litres is distributed into D and E equally. So, 2.5 million litres flow in both B-E and B-D pipes.
8 million litres come into D from A and 2.5 million litres come into D from B. Nothing comes from C because the C-D pipeline is closed. Hence the total volume into D is 10.5 million litres. The capacity of D is 4.5 million litres. So, 10.5 - 4.5 = 6 million litres is distributed into D-E, D-F and D-G pipelines equally. So, the flow in each of these pipelines is equal to 6/3 = 2 million litres. So, the slack in D-G pipeline is equal to 8 - 2 = 6 million litres.

It is known that the total number of students is 100. We also know that the number of students who opted for Marketing is more than the number of students who opted for Finance which is more than the number of students who opted for Operations.
We have to maximize the number of students who chose both Finance and Operations as their specialization. We know that the number of students who chose Marketing will be greater than the number of students who opted for Finance and Operations as their specialization. This can be done as shown in the following Venn diagram.

Thus, the maximum number of students who opted for Finance and Operations but not Marketing = 48

We know that the game does not end in the first round, and at the end of the first round 3 white coins remain. This means that, for the game to not end, the minimum number of black coins that must be left on the board is 1.
Since D can pocket one white coin to end his turn, (A,B,C) in their turns pocket a total of 5 white coins and 0 black coins. This can be done using many combinations like (A,B,C) =  (3,0,2), (2,1,2) etc.
We now have to maximise the number of coins that D should pocket. 
This means that D has to pocket all the black coins on the board except one, which has to be left on the board as the game does not end.
Therefore D should have pocketed 8 black coins, and also should also have pocketed the red coin during his turn. 
Also, D should have ended his turn by pocketing a white coin at the end.
Therefore, the total number of coins that D could have pocketed is 8 black + 1 red + 1 white = 10 coins.

We consider all scores possible for Team A where the opponent is Team B.
Case 1: Red is pocketed by Team B
In such a case where the opponent has pocketed the red, then he must have taken at least one of his coins. Thus, the opposition team can have from 0 coins to 8 coins on board. 
Therefore, all the possible scores of Team A are 0,2,4,6,......16. 
Sum of this series = 2*(1+2+.. 8)= 2*36 = 72 points
Case 2: Red is pocketed by Team A
In this case, 0 to 9 coins of Team B can be on the board.
As the red is pocketed, then 5 points are added to each of the above scores. Hence, now, the possible scores are 5+2*0, 5+2*1 ... 5+2*9 = 5,7,9,......23 points. Sum of this series = 5*10 + 2*(1+2+...+9) = 50+90 = 140
The total is 72 + 140 = 212 points.

If white coin is the last coin to be pocketed, this means that there are 0 white coins left on the board and at least one black coin on the board. In such a case, the points of black can be 5 or 0, depending on whether black has pocketed the red coin. As it is given that black wins, we can infer that the black team has pocketed the red coin. Hence, Black team points are 5.
Suppose there is 1 black coin left on the board, then, White's points would be 2.
Suppose there are 2 black coins left on the board, White would have 4 points. When 3 black coins exist , White would have 6 points. But this case would not exist as then White would win. 
Hence, the point differences of Black - White can be 5-2 and 5-4.
The maximum number of values that can exist is 2.

We know that the slack in pipes A-B, A-C and A-D is zero. So, the volume of oil that flows in these three pipes is 8 million litres. B’s capacity is 3 million litres. So, the remaining 5 million litres is distributed into D and E equally. So, 2.5 million litres flow in both B-E and B-D pipes.
Similarly, the oil that flows into C is 8 million litres. C’s capacity is 2.5 million litres. So, the remaining 5.5 litres is distributed into D and F equally. So, 2.75 litres flow in both C-D and C-F pipes.
So, a total of 8 + 2.5 + 2.75 = 13.25 million litres has flown into D. D’s capacity is 4.5 million litres. The remaining 8.75 million litres is distributed into D-E, D-G and D-F pipes equally. So, 8.75/3 = 2.917 million litres flows in the D-F pipeline.
So a total of 2.75 + 2.917 = 5.667 million litres flow into the F oil well. The capacity of F is 4 million litres. So, 5.667 - 4 = 1.667 million litres flow into the F-G pipeline. So, the slack of F-G pipeline = 8 - 1.667 = 6.333 million litres.

Let us use J, K, L, M and N as codes for Jack, Kiara, Liam, Mike and Nico respectively. 
Using statements 2, 4 and 6, we can have the following table to start with:
Statement 2 gives that the person sitting in second position copied his third answer from the person before him. Hence person sitting in the first position must answer B for question 3.
Statement 6 mentions that the person sitting in the middle did not copy his answer.Since he answered C for question 2. The person sitting in front must not have answered C for question 2. Hence the person in position 2 for question can answer either A/B.

Given J sits one place in front of L. Since L cannot sit at position 2 and 5 . J can sit only at 2,3. If J sits at 2 or 3 L must sit at 3 or 4.
Given in statement 4 M does not not sit position 5 nor answer option B for question 3. Hence he can only sit at either 3 or 4.
If J sits at 3 and L sits at 4 M does not have a place to sit. Hence J must sit at 2, L at 3 and M at 4.
Also, statement 3 says that J marked C twice and hence A and B were marked exactly once by him. Hence C must be answered for Q1 and Q4.
Statement 1 says that B was marked twice in question 1 hence it must be answered for by L and M because the person sitting at first and final position cannot answer B and person at position has already answered Q1 with C hence B must be answered for Q1 by persons L and M and since they are consecutively seated they must have copied. After placing as per the mentioned conditions the table looks like this :

Since the a set of persons must copy only one answer. The person sitting at position 1 cannot place C for Q1 because if he does that then there will be two copied answers between 1st and 2nd seated people. Hence he must answer A for question 1.
Note that in statement 3 it is mentioned that Nico marked the answer as option C only for Q1. Hence Nico is at position 5 and Kiara at position 1.
Also, exactly 2 persons marked B as answers to Q1. So,

Now, we have copied answers for pairs (3,4) and (1,2) and for questions 1 and 3.
So, pairs (2,3) and (4,5) must have copied answers for Q2 and Q4.
And since pair (2,3) did not copy answer for Q2, they must have done it for Q4. And the pair (4,5) must have copied for Q2.

Statement 4 tells that person sitting at the last position answers B for question 2. Since (4,5) copied Q2 person at position 4 must answer B for Q4
Now since, each of the 5 person must have at least one A, B and C each as the answers. We further deduce the following:
For M one of Q3 and Q4 must be answered with A and C. Since Q4 cannot be answered C by M as it would mean that he had copied this from L this is a false case. Hence he answers A for Q3 and C for Q4.
For L since we must have at least  one answer to be A we must Q3 with A.
For K we must have at least 1 answer to be C and we cannot answer Q4 with C we must answer Q2 with C.
The remaining Q4 for K can be answered with A/B but not C because J has answered Q4 with C.
The last person can answer Q3 with A/B because A/B because C cannot be placed as M answered Q3 with C similarly for Q4 only B/C can be used.
Also, the last person marked B as the answer for Q2. So

Now, Nico only answered C for Q1. If he answers B for Q3 . He cannot answer A for any of the Questions hence he must answer Q3 with option A. Since he only answers one question with option C he must answer Q4 with B.
The final table looks like :

Let us use J, K, L, M and N as codes for Jack, Kiara, Liam, Mike and Nico respectively.
Using statements 2, 4 and 6, we can have the following table to start with:
Statement 2 gives that the person sitting in second position copied his third answer from the person before him. Hence person sitting in the first position must answer B for question 3.
Statement 6 mentions that the person sitting in the middle did not copy his answer.Since he answered C for question 2. The person sitting in front must not have answered C for question 2. Hence the person in position 2 for question can answer either A/B.

Given J sits one place in front of L. Since L cannot sit at position 2 and 5 . J can sit only at 2,3. If J sits at 2 or 3 L must sit at 3 or 4.
Given in statement 4 M does not not sit position 5 nor answer option B for question 3. Hence he can only sit at either 3 or 4.
If J sits at 3 and L sits at 4 M does not have a place to sit. Hence J must sit at 2, L at 3 and M at 4.
Also, statement 3 says that J marked C twice and hence A and B were marked exactly once by him. Hence C must be answered for Q1 and Q4.
Statement 1 says that B was marked twice in question 1 hence it must be answered for by L and M because the person sitting at first and final position cannot answer B and person at position has already answered Q1 with C hence B must be answered for Q1 by persons L and M and since they are consecutively seated they must have copied. After placing as per the mentioned conditions the table looks like this :

Since the a set of persons must copy only one answer. The person sitting at position 1 cannot place C for Q1 because if he does that then there will be two copied answers between 1st and 2nd seated people. Hence he must answer A for question 1.
Note that in statement 3 it is mentioned that Nico marked the answer as option C only for Q1. Hence Nico is at position 5 and Kiara at position 1.
Also, exactly 2 persons marked B as answers to Q1. So,

Now, we have copied answers for pairs (3,4) and (1,2) and for questions 1 and 3.
So, pairs (2,3) and (4,5) must have copied answers for Q2 and Q4.
And since pair (2,3) did not copy answer for Q2, they must have done it for Q4. And the pair (4,5) must have copied for Q2.

Statement 4 tells that person sitting at the last position answers B for question 2. Since (4,5) copied Q2 person at position 4 must answer B for Q4
Now since, each of the 5 person must have at least one A, B and C each as the answers. We further deduce the following:
For M one of Q3 and Q4 must be answered with A and C. Since Q4 cannot be answered C by M as it would mean that he had copied this from L this is a false case. Hence he answers A for Q3 and C for Q4.
For L since we must have at least one answer to be A we must Q3 with A.
For K we must have at least 1 answer to be C and we cannot answer Q4 with C we must answer Q2 with C.
The remaining Q4 for K can be answered with A/B but not C because J has answered Q4 with C.
The last person can answer Q3 with A/B because A/B because C cannot be placed as M answered Q3 with C similarly for Q4 only B/C can be used.
Also, the last person marked B as the answer for Q2. So

Now, Nico only answered C for Q1. If he answers B for Q3 . He cannot answer A for any of the Questions hence he must answer Q3 with option A. Since he only answers one question with option C he must answer Q4 with B.
The final table looks like :

Since Kiara is the first person among the 5 friends, she did not copy any answer from her friends.

Let us use J, K, L, M and N as codes for Jack, Kiara, Liam, Mike and Nico respectively.
Using statements 2, 4 and 6, we can have the following table to start with:
Statement 2 gives that the person sitting in second position copied his third answer from the person before him. Hence person sitting in the first position must answer B for question 3.
Statement 6 mentions that the person sitting in the middle did not copy his answer.Since he answered C for question 2. The person sitting in front must not have answered C for question 2. Hence the person in position 2 for question can answer either A/B.

Given J sits one place in front of L. Since L cannot sit at position 2 and 5 . J can sit only at 2,3. If J sits at 2 or 3 L must sit at 3 or 4.
Given in statement 4 M does not not sit position 5 nor answer option B for question 3. Hence he can only sit at either 3 or 4.
If J sits at 3 and L sits at 4 M does not have a place to sit. Hence J must sit at 2, L at 3 and M at 4.
Also, statement 3 says that J marked C twice and hence A and B were marked exactly once by him. Hence C must be answered for Q1 and Q4.
Statement 1 says that B was marked twice in question 1 hence it must be answered for by L and M because the person sitting at first and final position cannot answer B and person at position has already answered Q1 with C hence B must be answered for Q1 by persons L and M and since they are consecutively seated they must have copied. After placing as per the mentioned conditions the table looks like this :

Since the a set of persons must copy only one answer. The person sitting at position 1 cannot place C for Q1 because if he does that then there will be two copied answers between 1st and 2nd seated people. Hence he must answer A for question 1.
Note that in statement 3 it is mentioned that Nico marked the answer as option C only for Q1. Hence Nico is at position 5 and Kiara at position 1.
Also, exactly 2 persons marked B as answers to Q1. So,

Now, we have copied answers for pairs (3,4) and (1,2) and for questions 1 and 3.
So, pairs (2,3) and (4,5) must have copied answers for Q2 and Q4.
And since pair (2,3) did not copy answer for Q2, they must have done it for Q4. And the pair (4,5) must have copied for Q2.

Statement 4 tells that person sitting at the last position answers B for question 2. Since (4,5) copied Q2 person at position 4 must answer B for Q4
Now since, each of the 5 person must have at least one A, B and C each as the answers. We further deduce the following:
For M one of Q3 and Q4 must be answered with A and C. Since Q4 cannot be answered C by M as it would mean that he had copied this from L this is a false case. Hence he answers A for Q3 and C for Q4.
For L since we must have at least one answer to be A we must Q3 with A.
For K we must have at least 1 answer to be C and we cannot answer Q4 with C we must answer Q2 with C.
The remaining Q4 for K can be answered with A/B but not C because J has answered Q4 with C.
The last person can answer Q3 with A/B because A/B because C cannot be placed as M answered Q3 with C similarly for Q4 only B/C can be used.
Also, the last person marked B as the answer for Q2. So

Now, Nico only answered C for Q1. If he answers B for Q3 . He cannot answer A for any of the Questions hence he must answer Q3 with option A. Since he only answers one question with option C he must answer Q4 with B.
The final table looks like :

Liam and Kiara has same answers for only Q2.

Let us use J, K, L, M and N as codes for Jack, Kiara, Liam, Mike and Nico respectively.
Using statements 2, 4 and 6, we can have the following table to start with:
Statement 2 gives that the person sitting in second position copied his third answer from the person before him. Hence person sitting in the first position must answer B for question 3.
Statement 6 mentions that the person sitting in the middle did not copy his answer.Since he answered C for question 2. The person sitting in front must not have answered C for question 2. Hence the person in position 2 for question can answer either A/B.

Given J sits one place in front of L. Since L cannot sit at position 2 and 5 . J can sit only at 2,3. If J sits at 2 or 3 L must sit at 3 or 4.
Given in statement 4 M does not not sit position 5 nor answer option B for question 3. Hence he can only sit at either 3 or 4.
If J sits at 3 and L sits at 4 M does not have a place to sit. Hence J must sit at 2, L at 3 and M at 4.
Also, statement 3 says that J marked C twice and hence A and B were marked exactly once by him. Hence C must be answered for Q1 and Q4.
Statement 1 says that B was marked twice in question 1 hence it must be answered for by L and M because the person sitting at first and final position cannot answer B and person at position has already answered Q1 with C hence B must be answered for Q1 by persons L and M and since they are consecutively seated they must have copied. After placing as per the mentioned conditions the table looks like this :

Since the a set of persons must copy only one answer. The person sitting at position 1 cannot place C for Q1 because if he does that then there will be two copied answers between 1st and 2nd seated people. Hence he must answer A for question 1.
Note that in statement 3 it is mentioned that Nico marked the answer as option C only for Q1. Hence Nico is at position 5 and Kiara at position 1.
Also, exactly 2 persons marked B as answers to Q1. So,

Now, we have copied answers for pairs (3,4) and (1,2) and for questions 1 and 3.
So, pairs (2,3) and (4,5) must have copied answers for Q2 and Q4.
And since pair (2,3) did not copy answer for Q2, they must have done it for Q4. And the pair (4,5) must have copied for Q2.

Statement 4 tells that person sitting at the last position answers B for question 2. Since (4,5) copied Q2 person at position 4 must answer B for Q4
Now since, each of the 5 person must have at least one A, B and C each as the answers. We further deduce the following:
For M one of Q3 and Q4 must be answered with A and C. Since Q4 cannot be answered C by M as it would mean that he had copied this from L this is a false case. Hence he answers A for Q3 and C for Q4.
For L since we must have at least one answer to be A we must Q3 with A.
For K we must have at least 1 answer to be C and we cannot answer Q4 with C we must answer Q2 with C.
The remaining Q4 for K can be answered with A/B but not C because J has answered Q4 with C.
The last person can answer Q3 with A/B because A/B because C cannot be placed as M answered Q3 with C similarly for Q4 only B/C can be used.
Also, the last person marked B as the answer for Q2. So

Now, Nico only answered C for Q1. If he answers B for Q3 . He cannot answer A for any of the Questions hence he must answer Q3 with option A. Since he only answers one question with option C he must answer Q4 with B.
The final table looks like :

Option A was chosen 5/6 times.
Option B was chosen 8/7 times.
Option C was chosen 7 times.
So, even if Kiara marked A as the answer for Q4, A would still be the least chosen option.

The amount of money (in thousands) spent by each of the friends in six cities as per the chart provided is:

After spending in Hyderabad, each of the six friends are left with same amount of money, say 'x'. 
By then Aman, Billu, Dinesh, Ekansh, Ganesh and Hritesh have spent a total of 50, 75, 100, 50, 70 and 35 thousand respectively.
Since, they are left with the same amount after this, all of the six friends must have started the tour with (x+50), (x+75), (x+100), (x+50), (x+70) and (x+35) thousand respectively.
The following table shows the amount left with each of the friends after visiting the particular cities:

Hritesh had the least amount before the start of the tour.

The amount of money (in thousands) spent by each of the friends in six cities as per the chart provided is:

After spending in Hyderabad, each of the six friends are left with same amount of money, say 'x'.
By then Aman, Billu, Dinesh, Ekansh, Ganesh and Hritesh have spent a total of 50, 75, 100, 50, 70 and 35 thousand respectively.
Since, they are left with the same amount after this, all of the six friends must have started the tour with (x+50), (x+75), (x+100), (x+50), (x+70) and (x+35) thousand respectively.
The following table shows the amount left with each of the friends after visiting the particular cities:

After Bangalore's trip, Dinesh is left with the highest amount.

The amount of money (in thousands) spent by each of the friends in six cities as per the chart provided is:

Each spent at least 50% of money 
Now Maximum amount will be left if each spent as less as possible which means each spent 50% of total money with him.
At the beginning of the trip they can have a maximum total of :
Aman  =220
Billu - 220
Dinesh -250
Ekansh- 240
Ganesh- 280
Hritesh - 190
But considering the additional condition that all of them have the same equal amounts by the time they reach Delhi :
Let initially their amounts be a, b, c, d, e, f
Since by the time they reach Delhi all of them must be equal :
By the time they reach Delhi
For Aman : a - 50000 
If Aman has 220000 he will end up with 220000- 50000 = 170000 by the time he reaches Delhi. This is the maximum limit.
For Billu : b - 75000
If Billu has 220000 he will end up with 220000- 75000 = 145000 by the time he reaches Delhi. This is the maximum limit.
For Dinesh : c - 100000
If Dinesh has 250000 he will end up with 250000- 100000 = 150000 by the time he reaches Delhi. This is the maximum limit.
For Ekansh : d - 50000
If Ekansh has 240000 he will end up with 240000- 50000 = 190000 by the time he reaches Delhi. This is the maximum limit.
For Ganesh : e - 70000

If Ganesh has 280000 he will end up with 280000- 70000 = 210000 by the time he reaches Delhi. This is the maximum limit.
For Hritesh : f - 35000
If Hritesh has 190000 he will end up with 190000- 35000 = 155000 by the time he reaches Delhi. This is the maximum limit.
The lowest among the maximum for all the people was for Billu which was 145000.
Hence everyone must have Rs 145000 by the time they reach Delhi.
Before the beginning of Mumbai trip :
Aman will have 145000 + 10000 + 15000 =  Rs 170000
Billu will have 145000 + 30000 + 25000 = Rs 200000
Dinesh will have 145000 + 40000 + 35000 = Rs 220000
Ekansh will have 145000 + 15000 + 5000 = Rs 165000
Ganesh will have 145000 + 15000 + 20000 = Rs 180000
Hritesh will have 145000 + 20000 + 5000 = Rs 170000
Among these Dinesh has the highest of Rs 220000 before the Mumbai trip.

The bottom and top faces of the cube have been painted using green colour. The numbers assigned to the cubes on the bottom face are 1 to 25 while the numbers assigned to the cubes on the top face are 101 to 125.
Therefore, the required sum = 25 x 26/2  + 101 + 125/2  x  25 = 3150
Hence, [4].
Alternatively,
Each tube of the top face is numbered 100 more the cube which is in its position at the bottom face. If sum of cubes of bottom face is x, sum of cubes of the top face = x + 25 × 100.
The required sum is more than 2500. 3150 is the only option greater than 2500.

The amount of money (in thousands) spent by each of the friends in six cities as per the chart provided is:

After spending in Hyderabad, each of the six friends are left with same amount of money, say 'x'.
By then Aman, Billu, Dinesh, Ekansh, Ganesh and Hritesh have spent a total of 50, 75, 100, 50, 70 and 35 thousand respectively.
Since, they are left with the same amount after this, all of the six friends must have started the tour with (x+50), (x+75), (x+100), (x+50), (x+70) and (x+35) thousand respectively.
The following table shows the amount left with each of the friends after visiting the particular cities:

Ganesh is left with Rs. (x-70) thousand after the trip ends.
So, x=Rs. 100,000
Amount left with Billu after his trip to Mumbai= Rs. (x+30) thousand= Rs. 130,000

Let x, y, z be the number of sovereigns received by the 3 applicants. Then x≥3,y≥3,z≥3 and x+y+z=16.
Let u=x-3,v=y-3and w=z-3,then u≥0,v≥0,w≥0
So, u+3+v+3+w+3=16
Or, u+v+w=7
The total number of the solutions of the given equation is ^7+3-1C_3-1 = ⁹C₂ = 36.

It is given that the sum triples in 4 years. If it has become 9 times itself, it was tripled twice. Let the amount be P, and it gets tripled in 4 years at R% p.a.
So P(1 + (r/100))⁴ = 3P
Or, (1 + (R/100))⁴ = 3
Let it take n years to become 9 times.
Or, P(1 + (R/100))ⁿ = 9P
Or, (1 + R/100)ⁿ = 9
Or, [(1 + (R/100))⁴]^n/4 = 32
Or, 3^n/4 = 3² or, n/4 = 2
Or n = 8.

For n cm sized cube the number of cubes with 3 faces painted = 8
Number of cubes with 2 faces painted = 12x(n-2)
Number of cubes with 1 face painted = 6(n-2)²
Number of cubes with 0 face painted = (n-2)³
Given 6(n-2)² - 12(n-2) = 90
So (n-2)(n-4) = 15
Or n = 7
Therefore number of cubes with 0 face painted = (7-2)³ = 125