Verbal Reasoning - Dice - Discussion

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.

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.

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.

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.

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.

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.

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.

@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?

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?.

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.