# Some more maths logic questions?

1. In a row of five persons, A is next to B. E is on right side of A who has D on his left side. C and D do not sit together. Who are sitting at the two ends of the row?

2. A person has nine coins, all looking alike but one of them slightly lighter than the others. Using a balance without weights, can he discover the lighter coin in just two weighings? If the number of coins is 12, how many minimum weighings will be required to find the lighter coin?

3. You have 10 stacks of coins with 10 coins in each stack. One stack of coins are all counterfeit and weigh one gram more than genuine ones. A genuine coin weighs 13 grams. Using a pointer scale (which shows the actual weight), what is the smallest number of weighings necessary to determine which stack is counterfeit?

4. A three centimeter cube has been painted red on all its sides. It is cut into one centimeter cubes. How many small cubes will be there in all? How many of these will be there with only

-one side painted red

-two sides painted red

-three sides painted red

-no side painted red?

no one has got all correct till now

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1 C and D are on the ends

2 yes, for 9 coins he can do it in 2 weighings

Make 3 stacks of coins, each with 3 coins. Put 1 stack on one side and another stack on the other side of the balance and leave the 3rd stack off.

- If one side is lighter, he knows the odd coin is in that stack of 3 coins.

- If they balance, he knows the odd coin must in the 3rd stack.

Once he knows which stack has the odd coin, he puts 1 coin on 1 side, 1 on the other side.

- If they balance, he knows the 1 he didn’t weigh is the odd one.

- If one side is lighter, he knows that coin is the odd coin

For 12 coins, the minimum is 3 weighings. He splits the coins into 2 stacks of 6 coins each. He follows the same method as above.

3. 3 weighings. First put 5 stacks on each side. The heavier side contains the stack with the counterfeit coins. Follow as above weighing 2 stacks then 1 stack.

4. There are 27 small cubes. (Think of a Rubik’s cube!)

There are 6 cubes with 1 side painted red (the center of each face of the cube)

There are 12 cubes with 2 sides painted red (the middle edge of each face)

There are 8 cubes with 3 sides painted red (the corner pieces of each face)

There is 1 cube with no sides painted red (the center cube)

1. C & D?

2. ?

3. ?

4. 24

4. A. 4 w/ 1 side painted red

4. B. 8 w/ 2 sides painted red

4. C. 8 w/ 3 sides painted red

4. D. 1 w/ no side painted red?

4. ? I first came up w/ 24, then recounted again and came up w/ 25.

Since I can’t figure these out, here’s a star.

1) D & C.

2) 3

12 is divided into two the lighter of the two is taken

6 is divided into two the lighter of the two is taken

2 coins are measured the lighter is going to be the light coin, but if equal the coin that wasn’t measured is the light coin

3) least possible 3, but 4 is possible as well

Has basically the same procedure as 2

10 is divided into two, the heavier stacks are taken

Get 4 stacks from the 5 stacks and put two stacks on each side

The heavier is taken, if equal then the stack that isn’t measured is the counterfeit, but if one is heavier take it and split into one stack at each side and the heavier is counterfeit

4. 24

4. A. 4 w/ 1 side painted red

4. B. 8 w/ 2 sides painted red

4. C. 8 w/ 3 sides painted red

4. D. 1 w/ no side painted red?

(taken from the person above…soz)

1- d, c

2- no, 6

3- 5

4- 24 , 1) 5 2) 6 3) 12 4) 1