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Hi, Puys CAT 2011 is finally over :: :: Lets start the Preparation for CAT 2012..please use this thread for all kind of LR-DI discussion,posting questions,submitting answers etc. Here are few links for previous years DI-LR thread ...

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@techsurge said:from P and Q one is definitley a liar or alternator....... looking at R lets assume Q is a truth teller => R is an alternator => P is a liarso Q is a doctor

Thnaks bro for your instant reply.Please clarify me this doubt:Firstly,I assume P is truth teller=>Then both Q & R is liar,so not possible.Now let's assume Q to be truth teller=>then P is liar & R is alternator.Also if we assume R to be Truth-teller=>then P is liar & Q is alternator. Am I right here??In both cases Q comes out to be a doctor but my doubt is that are both assumptions right??Because it mismatches with Time's solution.Thanks!

GUYS HELP ME OUT IN THIS QUESTION:

===========================

The people of a kavya tribe always make 2 statements while replying.P,Q,R belong to this.One is truth teller,one is liar,one is alternator in any order.Question askd them is :Who among you is doctor?

P:I am the doctor.Q is liar.

Q:I am the doctor.R is liar.

R:Q is the doctor.P is liar.

Then who is doctor.Only One is doctor among them.

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is there a link from where we can get some insight on truth-lie questions also?..

For all the MATCHSTICKS ( pick from a no. to win ) and CUBES( some faces painted and so on) kind of problems....

http://www.pagalguy.com/news/cubes-matchsticks-logical-reasoning-tricks-cat-2011-a-18681

guys check out the following link of Handa Sir for

->>MATCHSTICKS/COINS ( how many to be picked so that you win and so on...) and ->>CUBES (no. of cubes when it is cut .... painted faces....so on...)

http://www.pagalguy.com/news/cubes-matchsticks-logical-reasoning-tricks-cat-2011-a-18681

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Hi Friends, Can you please give me tips on how to solve complex ratios,like the one I mentioned below??

(3845.74*39.45-45.43*2893.67)/(27.41*1771)

It is stopping me every time I attempt a question like this.

Please guide me on it.

Thanks in advance.

Can we find out how many teams can score maximum points at pool stage???

---- Can anyone please tell the approach for these type of questions----

16 teams participated in a tournament.

Teams are divided into tow groups of 8 .

First round each team plays exactly one match against the other of the same group.

In each match two points are awarded to the winning and no points awarded to the losing and 1 point is awarded to each team in case of a draw.

Top two teams from each group will go to the semifinals and winners will go to the finals.

1 -> What could be the MAXIMUM points scored by a team that failed to reach the semi-finals?

2 -> What could be tje MAXIMUM points scored by a team that finished LAST in the pool at the end of pool stage?

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1) what i s the no of cubes needed to completely cover a cube of size 5*5*5 which is kept in a corner of a room such that it's 2 edges are along the 2 adjacent walls?

2) if year 2005 starts with a saturday, then find the no of Sundays in those months which have 30 days.

3) which days of the week cant be the 1st day of a century?

P.S: plz give solutions/approach also

@sonamaries7 said:@scrabbler : help help!Three different faces of a cube are painted in three different colours(adjacent pair of faces) - red, green and blue. This cube is now cut into 216 smaller but identical cubes.(The adjacent faces are in the same colour: red, red, then next adjacent pair in green and so on) 1). What is the number of the smaller cubes that will have exactly three faces painted ?2). How many of the smaller cubes have exactly two faces painted ?3). What is the number of small cubes that have exactly one face painted ?4). What is number of small cubes that have exactly one face painted red and no other face painted ?5). What is the number of small cubes that have one face painted green and one face blue and no other face painted ?6). What is the numbers of cubes that have no face painted at all ?Plz share the approach also..is there any approach other than drawing and visualizing the cube?

1) 8= fixed in all cases

2) 12(n-2) = 12( 6-2) = 48

3) 6(n-2)^2= 96

4) 2(n-2)^2= 32

5) 4(n-2) = 16

6) (n-2)^3= 64

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ďťż@sonamaries7 said:@scrabbler : help help!Three different faces of a cube are painted in three different colours(adjacent pair of faces) - red, green and blue. This cube is now cut into 216 smaller but identical cubes.(The adjacent faces are in the same colour: red, red, then next adjacent pair in green and so on) 1). What is the number of the smaller cubes that will have exactly three faces painted ?2). How many of the smaller cubes have exactly two faces painted ?3). What is the number of small cubes that have exactly one face painted ?4). What is number of small cubes that have exactly one face painted red and no other face painted ?5). What is the number of small cubes that have one face painted green and one face blue and no other face painted ?6). What is the numbers of cubes that have no face painted at all ?Plz share the approach also..is there any approach other than drawing and visualizing the cube?

Would you rather byheart a formula than just draw and analyze?

Number of identical cubes = 216 = 6^3

Number of cubes per side = 6

1) To have three faces painted, the cubes will be at the corners. There are 8 corners.

2) To have exactly two faces painted, they must be at the sides but not the corners. Per side, there are (6-2) = 4 such cubes. There are 12 such sides. Hence, 48 cubes.

3) To have one face painted, the cubes must be landlocked. ie not at the sides. Thus they are located as locked squares of (6-2) dimension. There are 6 sides with 1 square each. So there are 6*4*4 = 96

4) To have only one face painted, and that too as red, it will be a third of (3), since there will be two (adjacent) sides painted red. Hence, there are 32 cubes.

5) To have two faces painted, one green and the other blue, the cubes will be located along three sides but not at those corners. ie (6-2)*3 = 12 cubes

6) To have no face painted at all, the cubes will lie within a lanlocked cube excluding the outer surface of the original. Thus cube of dimension (6-2) ie 4^3 = 64 cubes

ďťż
Number of identical cubes = 216 = 6^3

Number of cubes per side = 6

1) To have three faces painted, the cubes will be at the corners. There are 8 corners.

2) To have exactly two faces painted, they must be at the sides but not the corners. Per side, there are (6-2) = 4 such cubes. There are 12 such sides. Hence, 48 cubes.

3) To have one face painted, the cubes must be landlocked. ie not at the sides. Thus they are located as locked squares of (6-2) dimension. There are 6 sides with 1 square each. So there are 6*4*4 = 96

4) To have only one face painted, and that too as red, it will be a third of (3), since there will be two (adjacent) sides painted red. Hence, there are 32 cubes.

5) To have two faces painted, one green and the other blue, the cubes will be located along three sides but not at those corners. ie (6-2)*3 = 12 cubes

6) To have no face painted at all, the cubes will lie within a lanlocked cube excluding the outer surface of the original. Thus cube of dimension (6-2) ie 4^3 = 64 cubes

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Three different faces of a cube are painted in three different colours(adjacent pair of faces) - red, green and blue. This cube is now cut into 216 smaller but identical cubes.(The adjacent faces are in the same colour: red, red, then next adjacent pair in green and so on)

1). What is the number of the smaller cubes that will have exactly three faces painted ?

2). How many of the smaller cubes have exactly two faces painted ?

3). What is the number of small cubes that have exactly one face painted ?

4). What is number of small cubes that have exactly one face painted red and no other face painted ?

5). What is the number of small cubes that have one face painted green and one face blue and no other face painted ?

6). What is the numbers of cubes that have no face painted at all ?

Plz share the approach also..is there any approach other than drawing and visualizing the cube?

@smashed said:lets say u leave 'x' coins at the end for the opponent, so the opponent will lose the game if you leave 1,2 or 3 coins at the end. so 'x' can take 1,2 or 3 values.now to be sure of winning you can match your number of coins with the opposition, so that the total is min + max (3 + 6 in this case). So if the opposition takes out 3 coins u take out 6 coins, for 4 coins take out 5 coins and so on.so you are going to win, if you leave x+ 9k coins the end. As the total coins in the Q is 50 assume 9k to be 45.so you win the game if u leave 45+3/2/1 coins at the end i.e - 48/47/46 coins.as you cannot take out 2 coins, the ans is 3 or 4 coins.

ANY1 WHO HAS OTHER THEN ABOVE MENTIONED APPROACH PLEASE POST

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