Verbal Reasoning - Dice - Discussion

Even numbered dice 1, 2, and 4 now, the top palace should even dots like;
dice 2 top place: 2 or 6
dice 2 top place: 6 [because 2 and 4 are already present]
and dice 4 top place : 2 or 6.

Now compare 2 and 4th dice if the 2nd dice top is 6 then 2's opposite 5 but the 4th dice top can't be 2 because 5 is adjacent so it's top will be 6 also
and if the 2nd dice top is 2 then 5's opposite 6, now 4th top can't be 6 because 5 is adjacent so it will be 2 also;
Now;
total of even numbers dice may be 6+6+6 or 2+6+2 so 18 or 10
So, the answer will be 18.

Now in 4th dice.

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.

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.

1 Opposite 6, 2 opposite 5, 3 opposite 4.

So Then the Odd Number Of Dots on their Faces are 5, 1, 5, 1, 3, 1 Next.

Even the Number of Dots on their Faces are 2, 6, 2, 6, 4, 6.

Then Fill the Blanks and you will get the answers.

Sum of opp. phases is =7.
Then,

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

Why 6 dots? Explain.

Well done, Thanks @Anand.

Nice explanation @Anand Kumar.

@All.

Consider the first two dices. They have a common down face which is 1. According to Rule No 3, 5 and 4 should be the opposite faces of 3 and 6 respectively. Then, 2 should be face-up. But it seems contradict to the answer. Can anyone tell me to get the correct answer?