### Discussion :: Dice - Dice 2 (Q.No.1)

Six dice with upper faces erased are as shows.

The sum of the numbers of dots on the opposite face is 7.

Amit said: (Sep 14, 2010) | |

Please tell the rules of dice type II. |

Prudhvi said: (Sep 30, 2010) | |

You told that even numbered dice have even number of dots on their top faces if we set it as four it should be 12 but you take it as 6. Can we take 4 or not? |

Pragya said: (Nov 10, 2010) | |

You told that 2, 4 & 6 dice have even number of dots on their top faces then here should be one option with value 6 b'cz 2 is also even number & here can be answer 12 also. If your answer is right then explain it that why are you taking 6 dots in one dice?. |

Isaac said: (Jan 13, 2011) | |

When you observe carefully you see only 4, 5, 2, 3 will on the sides and the remaining two nos 6, 1 are left for us to deal with and amongh these two only 6 is even that explains every ones question. |

Deepak Varshney said: (Jan 19, 2011) | |

I think so, ans is 10. No. Of dots on dice II = 2. No. Of dots on dice IV = 6 (as 4 and 2 at adjacent). No. Of dots on dice VI = 2. |

Anand Kumar Gone said: (Aug 8, 2011) | |

Yeah I got the answer..... Let us read the question very carefully.... 1) dices are not different kind 2)the given condition is sum of number of dots on bottom faces(opposite to erased upper faces) is 7. According to above second point we have only one possibility i.e. 1+1+1+1+2+1=7 Because the fifth dice have number 1 dot on forward face so the bottom face value should be 2. Hence bottom face number of 1st dice=1 2nd dice=1 3rd dice=1 4th dice=1 5th dice=2 6th dice=1 And coming to problem adjacent faces of number 4 are 1,2,5,6(remember that number 1 is at bottom). So number 4 is opposite to number 3. Now consider fourth dice,we have the following numbers on the forward face=2 right face=4 left face=3 (since 4 is opposite to 3) bottom face=1 (which we got earlier from given condition) Now we left with top and backward faces. From given data even numbered dices have only even number on their top faces. So,we left only with one even number 6. Therefore top face=6 backward face=5. and 1 is opposite to 6 Second,fourth,sixth dices have 1 on bottom face.so they all have number 6 on their top. Hence total number of dots on thier top faces is 6+6+6 = 18. |

Venky said: (Aug 24, 2011) | |

Anand kumar your great. |

Sameeksha said: (Aug 28, 2011) | |

Yup great explanation anand:). |

Nisha said: (Sep 5, 2011) | |

Well done Anand. :-) |

Vadivelu said: (Sep 6, 2011) | |

- |

Rohit Dakave said: (Jan 7, 2012) | |

Consider a dice,opposite faces are always in pairs(1,6)(2,5)(3,4) ->"No need to look at the condition" 1)In 1st ques we want even numbered dice i.e 2,4 and 6th dice top faces with even no. from fig(2) we can see that 4 and 5 are visible means we have 3 opposite of 4 and 2 opposite of 5.so now we are left with 1 and 6. now we want 6 at the top(even)so 1 wiil be at the bottom face so from 2nd dice we got 6 Similarly in 4th dice we can see 2 and 4, means 5 will be opposite of 2 and 3 will opposite of 4 left with 1 and 6 we want 6(even) at the top so 1 wiil be at the bottom face so from 4th dice we get 6 Similarly in 6th dice we can see 4 and 5, means 3 will be opposite of 4 and 2 will opposite of 5 left with 1 and 6 we want 6(even) at the top so 1 wiil be at the bottom face so from 6th dice we get 6 Adding 6+6+6=18. Using above procedure rest of the problems can be solved easily. |

Angel said: (Mar 18, 2012) | |

Rohit best explanation! keep it up ! |

Sujith(Spt) said: (Apr 17, 2013) | |

Keep in mind that the sum of the numbers of dots on the opposite face is 7. take 4 dotted face : here from figure the adjacent faces are [5,3,2] and the remaining are {1,3} so opp. to 4 is 3 (4+3=7). Take 5 dotted face : here from figure the adjacent faces are [4,1] and the remaining are {2,3,6} so opp. to 5 is 2 (5+2=7). Take 6 dotted face : here from figure the adjacent faces are [3,5] and the remaining are {1,2,4} so opp. to 6 is 1 (6+1=7). Keep in mind that here even numbers can only be 2,4,and 6. Now consider (II) : here 4 is there,and opp. to 5 is 2 then only possibility is 6 on top. Consider (IV): here 2 and 4 is there then only possibility is 6 on top. Consider (VI): same as (II). so 6 on top. So adding 6+6+6 = 18 is the answer. |

Rahul said: (May 13, 2014) | |

Nicely explained but too long answer I don't think this will come for nata exam. |

Arun said: (Nov 4, 2014) | |

I think 2 4 6 dice contain 9, 6, 9 dots so answer will be 24 I think once 0 through this. |

Tanay said: (Aug 3, 2016) | |

But by the figures V and VI, 1 is opposite to 4. |

Tanay said: (Aug 3, 2016) | |

Well done Rohit now I understood the question very well. Thanks a lot. |

Jeetsingh said: (Mar 8, 2017) | |

Thanks @Anand. |

Santhosh said: (Apr 18, 2017) | |

As explained above all 6, except 5th are having 1 at the bottom. And all should have 6 at the top, but 1st dice is having 6 adjacent to 1. |

Harsh said: (Jul 18, 2017) | |

These are standerd dice and each nearest faces sum is not equal to 7 that's why we taking 6. |

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