Verbal Reasoning - Dice - Discussion

Here even number dice are=2,4,6.

(Note:The sum of the numbers of dots on the opposite face is 7.)
if we take dice2...in there 5and 4dots;
So;
5 + 2 = 7.
4 + 3 = 7.
Then 2dice have 2 + 3 = 5.

If we take dice4; in there 2 and 4 dots.
So;
2 + 5 = 7
4 + 3 = 7.
Then 4dice have 5 + 3 = 8

**If we take dice6 in there 4and 5
So.
4+3=7
5+2=7.
Then dice6 have 3+2=5

No add three even numbers (2,4,6) dice
We get 5+8+5=18
Hence proved.

Guys here's a Straightforward explanation!

Key points(Rules) from the question:

1) Opposite sides sums up to 7 [{1,6}, {2,5}, {4,3}]
2) Select Even index dice [ dice2, dice4, dice6 ] & if their top side face have only even number
then how many dots are total in the top face(which are even) of even index dice?
3) Total means summing
4) No of dots on a face = Numeric representation of those dots.


Take dice2
Use rule (1):

5<->2 -> {5 is in front then opposite is 2 which will be on back}
4<->3 -> {4 is in the right then the opposite is 3 which will be on the left}
1<->6 ----> only top face for even index dice = 6 {question said even is always on top so remaining even number is 6}

select 6

take dice4
use rule (1):

5<->2 {2 is on face then opposite 5 will be on back}
4<->3 {4 is on the side then on another side, there will be 3}
1<->6 ----> only top face for even idex dice = 6 {question said only even will be on top & remaining choice for even is 6}

select 6
take dice4
use rule (1):

5<->2
4<->3
1<->6 ----> only top face for even index dice = 6.

select 6 -> same reason as above;

Perform total:
Total = 6 + 6 + 6.

OKAY First of all 6 dices are given.

And according to question, we must take even number of dices i.e. fig 2, 4, 6.

Given, the opposite face of dices will have a Sum of 7.

So in fig 2:we have 5 front so its opposite will be 2 (their sum goes like this 5+2=7) and in 4 in right, so opposite will be 3 (4+3=7) remaining 6 and 1 (6+1=7) as it will be opposite with each other. According to the question, even the number of dots will be in the top so obviously in between remaining number 6 and 1.

6 will be on top in fig 2. And the same trick goes to rest fig 4 and 6.

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.

Sum of opp. phases is =7.
Then,

Even number on the tip phase;
so,
6+1=7.
so,
I think that he takes 6.

@ALL. Even number of dices are 2, 4, 6. It's clear that 2, 4, 5 are adjacent to each other. And it's said that top numbers should be even. Among 1 to 6, 2, 4, 6 are even, but 2 and 4 are adjacent. So, 6 is the answer.

Well done, Thanks @Anand.

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.

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

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