Verbal Reasoning - Dice - Discussion

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 !

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.

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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.

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.