The Official CAT 2007 Quant Thread

[horrible]

Originally Posted by jaybind

the ans for 2nd ques is 60.here is the solution:
case1:take unit place=5 i.e 1 way
then 10th place can be filled in 4 ways,100th place in 3 ways,1000th palce in 2 ways and 10000th place in 1 way.so no of ways=1*4*3*2*1=24
case2:unit place=4,10th place=5 i.e.1 way
100th place can be filled in 3 ways and 1000th in 2 ways and 10000th place in 1 way...
no of ways=6
case3:unit place=4,10th place=any one of (1,2,3)i.e 3 ways
100th place in 2 ways,1000th in 2 and 10000th place in 1 way.
no of ways=12
case 4:unit place=3,10th place=any one of (1,2) i.e 2 ways
100th place in 1 way and 1000th in 2 way and 10000th place in 1 way ...no of ways=4
case5:unit place=3,10th palce=any one of (4,5) i.e 2 ways
100th palce in 2 way,1000th in 2 way and 10000th place in 1 way
no of ways=8
case6:unit place=2
10th place in 3 ways,100th palce in 1 way,1000th in 2 way 10000th place in 1 way
so no of ways=6
so ans=24+6+12+4+8+6=60
i know its quite difficult to understand but then problem is quite complex.any clarifications guys??

5 digits 1,2,3,4,5
Total number of numbers = 1 × 2 × 3 × 4 × 5 = 120

In these numbers, half of them will have hundred's digit > unit's digit AND
half of them will have hundred's digit < unit's digit, because all digits are distinct.

Thus, answer = 120/2 = 60.

CAT is all about keeping the things simple...

[Krack_CAT]

Originally Posted by karthikeyantcs

Mohit, Horrible and I have already come to the conclusion that 9 bags is the answer. I think you mistook our posts.

But 9 is not there as one of the options.. That's what confused me....

[horrible]

Originally Posted by mohit.pasrija

How many numbers can be formed by using 8,7,5,3,2 such that they are divisible by

125 ? Digits to be used exactly once.

(1) 20 (2) 4 (3) 2 (4) 1

Sol...for a number to be divisible by 125 its last three digits should ve divisible by 125.
so from the above given numbers.
we could have 4 possible numbers i.e.
1.32875
2.23875
3.28375
4.82375

tell me if i m wrong...

I think u r right. Here goes the method ...

Divisible by 125 means last 3 digits divisible by 125
125 or 250 or 375 or 500 or 625 or 750 or 875 or 000.

From the available digits, only 375 and 875 are possible.
For case I i.e. (--375), the first 2 digits could be 28 or 82
Similarly for case II (--875), the first 2 digits could be 23 or 32.

Hence, 4 cases.

[jaybind]

Originally Posted by horrible

5 digits 1,2,3,4,5
Total number of numbers = 1 × 2 × 3 × 4 × 5 = 120

In these numbers, half of them will have hundred's digit > unit's digit AND
half of them will have hundred's digit < unit's digit, because all digits are distinct.

Thus, answer = 120/2 = 60.

CAT is all about keeping the things simple...

hey buddy could u elaborate the highlighted part a bit further...dint get it...
and yes i agree that cat is all about keeping the things simple but then dont get surprised if it turns out to be complex...

[horrible]

Originally Posted by Krack_CAT

But 9 is not there as one of the options.. That's what confused me....

CAT now has 5 options .. so assume the 5th option as 9 and mark it as ur answer

[horrible]

Originally Posted by jaybind

hey buddy could u elaborate the highlighted part a bit further...dint get it...
and yes i agree that cat is all about keeping the things simple but then dont get surprised if it turns out to be complex...

Oh there's nothing to be surprised about that sir.. we all are here to learn. It incidently happened that this funda struck me upon seeing this qs. May be there is an even simpler method to it. May be you know some other thing which I would do in a complex manner. Let us learn from each other...


Let's take hundred's digit to be x and unit's digit to be y
Now, three universal cases exist for x and y.
x > y
x < y
x = y

The third case doesn't hold for our problem since all digits are distinct (1,2,3,4,5).

Since there is no restriction as such and in this situation, all the digits (1,2,3,4,5) are equally likely to get at any position, we can safely assume that there would be no distinction in the values that could be assumed by the variables x and y... ... (hard to explain... someone pls elaborate...)

Thus, x > y and x < y would happen equal number of times in 120 combinations i.e. 60-60 times.

Hence, answer should be 60 (this is how I do it in a fast manner... though it has its own limitations. Faculty told me not to apply it when one of the digits is 0... since 0 cannot be at leftmost place... kinda weird.)

I am awaiting along with you for a more lucid explanation...

[Krack_CAT]

Originally Posted by horrible

CAT now has 5 options .. so assume the 5th option as 9 and mark it as ur answer

HAHAHAHA

But i still have a gut feeling that 9 is not the correct answer.... will try and get the "correct" answer and detailed solution

[mohit_353]

Originally Posted by Krack_CAT

But 9 is not there as one of the options.. That's what confused me....

see...that was also one of my doubts...so if we need to bring the solution thank we can bring 11 or 15

in case of 11 ...we can consider another 3 bags having random no of coins like 90,30, 7...it will give 11 as answer

or we can bring 15 by considering 1,2,4,8,16,32,64........in 7 bags...it will bring 15...
wht to consider if same question appear in CAT-07 ...the minimum one....

wht. to say.....

[mohit_353]

Originally Posted by horrible

Oh there's nothing to be surprised about that sir.. we all are here to learn. It incidently happened that this funda struck me upon seeing this qs. May be there is an even simpler method to it. May be you know some other thing which I would do in a complex manner. Let us learn from each other...


Let's take hundred's digit to be x and unit's digit to be y
Now, three universal cases exist for x and y.
x > y
x < y
x = y

The third case doesn't hold for our problem since all digits are distinct (1,2,3,4,5).

Since there is no restriction as such and in this situation, all the digits (1,2,3,4,5) are equally likely to get at any position, we can safely assume that there would be no distinction in the values that could be assumed by the variables x and y... ... (hard to explain... someone pls elaborate...)

Thus, x > y and x < y would happen equal number of times in 120 combinations i.e. 60-60 times.

Hence, answer should be 60 (this is how I do it in a fast manner... though it has its own limitations. Faculty told me not to apply it when one of the digits is 0... since 0 cannot be at leftmost place... kinda weird.)

I am awaiting along with you for a more lucid explanation...

so this means that if we change the question like if ten's digit is always greater than units digit or unit's digit is greater than ten's digit ...the answer will be same means 60.... .....

wht to say....

[mohit_353]

another sitter....

1.)121212.......upto 300 digits is divided by 99, what is the

remainder ?
a)15
b)18
c)12
d)22

acc. to me ans. should be 18...not completely sure.....

wht to say???

2.)3) How many rectangles can one make from a regular chess board ?
a)1296
b)1332
c)1242
d)1221

happy solving.....

guys come on ...show some entuz....




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