Verbal Reasoning - Dice - Discussion

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

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

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.

Rohit best explanation! keep it up !

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Well done Anand. :-)

Yup great explanation anand:).

Anand kumar your great.

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.

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.