Hi,there is a trap in that question. the answer 60 is correct ONLY if "IDENTICAL cubes" is mentioned. This is not the case here. The answer is 27. You need to take cubes of different sizes to get this figure. I am still looking for the detailed answer. I have come across this qn before and remember the answer to be 27. Will write to you once i have the details.

I have attemted answering couple of other qns as well.

rakesh kumar <vicky_4147@...> wrote:

Dear Garima,

I could solve one of the problems.

The minimum number of cubes to fill a cuboid 20*16*12

For the cube to fit in the space given its side must be a multiple of 20,16,12

For minimum number this should be the highest commom multiple of 20,16,12 that is the h.c.f. = 4

The number will be = (20*16*12) / (4*4*4) = 60 .

Bye,

vickygarima mittal <mgarima_20@...> wrote:

HI EVERYONE,

PLEASE HELP ME SOLVE THESE.Q1. IN HOW MANY WAYS CAN 30 IDENTICAL BALLS BE DIVIDED IN GROUPS OF 3.

1)146 2)75 3)130 4)91.

Since the balls are identical we can proceed as below.

First Group 1 ball : The remaining 29 can be split in 14 ways [1,28 : 2,27: 3,26 etc .. 14,15]First Group 2 balls : The remaining 28 can be split in 13 ways [2,27 : 2,26: 3,25 etc .. 14,14]

Here note that we start at 2,27: 1,28 has been taken care of in step 1.First Group 3 balls : The remaining 27 can be split in 11 ways [3,24 : 2,26: 3,25 etc .. 13,14]

and so on. the final tally would look like this

Balls in first group - Num of ways1 - 14

2 - 13

3 - 11

4 - 10

5 - 8

6 - 7

7 - 5

8 - 4

9 - 2

10 - 1Total of 14, 13 .... 1 is 75 thats the answer ,75.

Q2. HOW MANY +VE INTEGRAL SOLUTIONS (X,Y,Z)ARE THERE FOR X+(-1)^Z *Y = 2Z , AND X,Y,Z<=10.

1)3 2)37 3) 35 4)38Q3.HOW MANY RIGHT ANGLED TRIANGLES ARE POSSIBLE GIVEN THAT OE OF TH SIDES IS 15 CM LONG.

1)2 2)3 3)5 4)6.

Possible Pythagorean triplets9,12,15

15,36,39

8,15,17

15,20,25cant think of more.

Something wrong. It could be my approach, the question, the answer choices. ;). Will surely write if i get more info on this.

Q4. WHAT IS THE MINIMUM NUMBER OF CUBES WITH WHICH ONE CAN CONSTRUCT A CUBOID OF DIMENSIONS 20*16*12.

1) 60 2)27 3) 20 4)9.

27

Do you Yahoo!?

Yahoo! SiteBuilder - Free web site building tool.**Try it!**- hi,Regarding the number of cubes: the answer is 20. yes, we have to consider the cubes of different sides.I solved the problem like this:cuboid is of size--- 20*16*12first try to fit the maximum possible cube...in this case a cube of side 12.........one cube used.then considered the remaining part as different cuboids..we are left with cuboid of 12*4*12 , 8*16*12.consider 12*4*12again try to fit maximum possible cube ..... it will be of side 4 ....which requires 3*1*3 = 9 cubes.consider 2*16*12maximum cube size is 8 .... no of cubes is 2.we are again left with 4*8*16for this a cube of side 4 is required .... no:of cubes is 8.therefore total number of cubes required is 20 .... 27 is the wrong answer.

IT LOOKS RIGOUROUS BUT WHEN U START DOING IT GETS SIMPLER.byeyuvawrote:*Shiv Keskar <shivkeskar@...>*

Hi,there is a trap in that question. the answer 60 is correct ONLY if "IDENTICAL cubes" is mentioned. This is not the case here. The answer is 27. You need to take cubes of different sizes to get this figure. I am still looking for the detailed answer. I have come across this qn before and remember the answer to be 27. Will write to you once i have the details.

I have attemted answering couple of other qns as well.

rakesh kumar <vicky_4147@...> wrote:

Dear Garima,

I could solve one of the problems.

The minimum number of cubes to fill a cuboid 20*16*12

For the cube to fit in the space given its side must be a multiple of 20,16,12

For minimum number this should be the highest commom multiple of 20,16,12 that is the h.c.f. = 4

The number will be = (20*16*12) / (4*4*4) = 60 .

Bye,

vickygarima mittal <mgarima_20@...> wrote:

HI EVERYONE,

PLEASE HELP ME SOLVE THESE.Q1. IN HOW MANY WAYS CAN 30 IDENTICAL BALLS BE DIVIDED IN GROUPS OF 3.

1)146 2)75 3)130 4)91.

Since the balls are identical we can proceed as below.

First Group 1 ball : The remaining 29 can be split in 14 ways [1,28 : 2,27: 3,26 etc .. 14,15]First Group 2 balls : The remaining 28 can be split in 13 ways [2,27 : 2,26: 3,25 etc .. 14,14]

Here note that we start at 2,27: 1,28 has been taken care of in step 1.First Group 3 balls : The remaining 27 can be split in 11 ways [3,24 : 2,26: 3,25 etc .. 13,14]

and so on. the final tally would look like this

Balls in first group - Num of ways1 - 14

2 - 13

3 - 11

4 - 10

5 - 8

6 - 7

7 - 5

8 - 4

9 - 2

10 - 1Total of 14, 13 .... 1 is 75 thats the answer ,75.

Q2. HOW MANY +VE INTEGRAL SOLUTIONS (X,Y,Z)ARE THERE FOR X+(-1)^Z *Y = 2Z , AND X,Y,Z<=10.

1)3 2)37 3) 35 4)38Q3.HOW MANY RIGHT ANGLED TRIANGLES ARE POSSIBLE GIVEN THAT OE OF TH SIDES IS 15 CM LONG.

1)2 2)3 3)5 4)6.

Possible Pythagorean triplets9,12,15

15,36,39

8,15,17

15,20,25cant think of more.

Something wrong. It could be my approach, the question, the answer choices. ;). Will surely write if i get more info on this.

Q4. WHAT IS THE MINIMUM NUMBER OF CUBES WITH WHICH ONE CAN CONSTRUCT A CUBOID OF DIMENSIONS 20*16*12.

1) 60 2)27 3) 20 4)9.

27

Do you Yahoo!?

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