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...

hey ...sory 4 wht..we r here to learn from each other..i must be thankful to u guys for taking interest ........

some more stuff......



4. A number when divided by 3 and 5 gives remainders 1 and 3. What will be the

remainder when the same number is divided by 15?


Happy solving.........

any other methods for solving this???

hey ...sory 4 wht..we r here to learn from each other..i must be thankful to u guys for taking interest ........

some more stuff......





4. A number when divided by 3 and 5 gives remainders 1 and 3. What will be the

remainder when the same number is divided by 15?

Happy solving.........

the ans for this will be 13.
the reqired nos will be like 13,28,43 and so on...
which when divided by 15 will give a remainder of 13...
am i right???

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...

hey that looks pretty ok...sohuld be correct..

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...

hey ...sory 4 wht..we r here to learn from each other..i must be thankful to u guys for taking interest ........

some more stuff......


2.How many 5 digit numbers can be made using 1, 2, 3, 4, 5 such that the digit in the unit

place is greater than the digit in the hundred's place ?

(1) 40 (2) 120 (3) 60 (4) 720





Happy solving.........

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??

Hey Guys Can Ny1 Tell Me Which Buks Should Be Referred For Preparing Maths, Di, Lr.
As I M Not Able To Join Ny1 Institute...

Plzz Guide Me....for Cat07

hey u can refer arun sharma for quant...its available for di as well...also the career launcher publication book is quite good.u can have a look at these.
will be very helpful:)

What i meant was, that, only 9 bags are required to get any combination of 382

i.e. 1,2,4,8,16,32,64,128,127

With the bags of these above denominations, we can obtain any value from 1 to 382.

pls explain in detail where i am wrong.. help me (and others who will browse this thread) get my (our) basics right.

:)

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

mohit_353 Says

I did not get ur point....pls explain......

What i meant was, that, only 9 bags are required to get any combination of 382

i.e. 1,2,4,8,16,32,64,128,127

With the bags of these above denominations, we can obtain any value from 1 to 382.

pls explain in detail where i am wrong.. help me (and others who will browse this thread) get my (our) basics right.

:)

Hey Guys Can Ny1 Tell Me Which Buks Should Be Referred For Preparing Maths, Di, Lr.
As I M Not Able To Join Ny1 Institute...

Plzz Guide Me....for Cat07

i beg to differ....

with these 9 bags (1,2,4,8,16,32,64,128,127) you can get any value. i.e. from 1 to 382.

Even after all the ramayan and mahabharat, no one has posted the correct option. :)

I did not get ur point....pls explain......

256 = 127 + 128 + 1


Mohit seems correct.

See, you can have any amount from 1 to 255 for sure.
Now, for any amount from 256 to 382, you will require bag carrying 127.
****I got this much
If you just exclude 127 from the range (256 to 382), you get (129 to 255), which can be achieved by employing the other 8 bags from 1 to 128.
**** I didn't understand this

i beg to differ....

with these 9 bags (1,2,4,8,16,32,64,128,127) you can get any value. i.e. from 1 to 382.

Even after all the ramayan and mahabharat, no one has posted the correct option.

karthikeyantcs Says

Yeah buddy you are right. Thanks a lot for the explanation.@Mohit,I am so sorry for my wrong explanation. You are correct. The answer is indeed 9 bags. Could you please check the solution from where you got this problem. Thanks a lot for the problem and I hope you will come with more problems like this.

hey ...sory 4 wht..we r here to learn from each other..i must be thankful to u guys for taking interest ........

some more stuff......

1. A watch ticks 90 times in 95 seconds and another watch ticks 315 times in 323 seconds. If both the watches are started together, how many times will they tick in the first hour?

A] 100
B] 101
C] 102
D] 99

2.How many 5 digit numbers can be made using 1, 2, 3, 4, 5 such that the digit in the unit

place is greater than the digit in the hundred's place ?

(1) 40 (2) 120 (3) 60 (4) 720


3. 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


4. A number when divided by 3 and 5 gives remainders 1 and 3. What will be the

remainder when the same number is divided by 15?

5. Two Trains A and B start from stations X and Y towards Y and X respectively. After

passing each other, they take 6 hours and 48 minutes and 3 hours and 20 minutes to

reach Y and X respectively. If train A is moving at 49 kms/hr, the speed of train B is..

Happy solving.........

256 = 127 + 128 + 1


Mohit seems correct.

See, you can have any amount from 1 to 255 for sure.
Now, for any amount from 256 to 382, you will require bag carrying 127. If you just exclude 127 from the range (256 to 382), you get (129 to 255), which can be achieved by employing the other 8 bags from 1 to 128.

ya ...horrible is right...

256 = 127 + 128 + 1


Mohit seems correct.

See, you can have any amount from 1 to 255 for sure.
Now, for any amount from 256 to 382, you will require bag carrying 127. If you just exclude 127 from the range (256 to 382), you get (129 to 255), which can be achieved by employing the other 8 bags from 1 to 128.

Yeah buddy you are right. Thanks a lot for the explanation.@Mohit,I am so sorry for my wrong explanation. You are correct. The answer is indeed 9 bags. Could you please check the solution from where you got this problem. Thanks a lot for the problem and I hope you will come with more problems like this.

karthikeyantcs Says

Still you did not understand the concept. Ok jus tell me how will you give me Rs.256?

256 = 127 + 128 + 1


Mohit seems correct.

See, you can have any amount from 1 to 255 for sure.
Now, for any amount from 256 to 382, you will require bag carrying 127. If you just exclude 127 from the range (256 to 382), you get (129 to 255), which can be achieved by employing the other 8 bags from 1 to 128.

why can't we put 127 in one bag......i think in this way we can get all combinations...

& at last we need 9 bags...

wht to say......

Still you did not understand the concept. Ok jus tell me how will you give me Rs.256?

karthikeyantcs Says

Hi buddy,In you solution tell me how are you going to give me say, Rs.350. You need to have 8 bags for Rs. 255. The remaining 127 should again be started with 1,2,4,8,16,32,64 bags ( 7 bags)we need to have 8 + 7 = 15 bags minimum. I hope it is clear.

why can't we put 127 in one bag......i think in this way we can get all combinations...

& at last we need 9 bags...

wht to say......

well guys ...i can't wait any more ...i am posting the first problem on number system.....
hope u guys will respond....

Ramu is given Rs 382 in one-rupee coins. He has been asked to
allocate them into a number of bags such that any amount required between Re
1 and Rs 382 can be given by just handling out a certain number of bags
without opening any of them. What is the minimum number of bags possible?
*****
***** a) 15
***** b) 13
***** c) 11
***** d) 17

i tried to come to the solution....

i tried with powers of 2 ( since we cannot open a bag )
the various bags he can have are, 1, 2, 4, 8, 16, 32, 64, 128 ( since 382 = 255 + 127 )
that is 8 bags

but it does not match any of the options...

tell me where i am wrong....

Hi buddy,In you solution tell me how are you going to give me say, Rs.350. You need to have 8 bags for Rs. 255. The remaining 127 should again be started with 1,2,4,8,16,32,64 bags ( 7 bags)we need to have 8 + 7 = 15 bags minimum. I hope it is clear.

well guys ...i can't wait any more ...i am posting the first problem on number system.....
hope u guys will respond....

Ramu is given Rs 382 in one-rupee coins. He has been asked to
allocate them into a number of bags such that any amount required between Re
1 and Rs 382 can be given by just handling out a certain number of bags
without opening any of them. What is the minimum number of bags possible?
*****
***** a) 15
***** b) 13
***** c) 11
***** d) 17

i tried to come to the solution....

i tried with powers of 2 ( since we cannot open a bag )
the various bags he can have are, 1, 2, 4, 8, 16, 32, 64, 128 ( since 382 = 255 + 127 )
that is 8 bags

but it does not match any of the options...

tell me where i am wrong....

Hi,

Can neone tell me where i can find this book:

Abhijit Guha - Quantitative Apptitude for MBA.

I tried at fountain area and dadar, but no luck.

Hi to all,

I have tried searching all of pagalguy for a thread on Quant for 2007, but have failed and whatever threads are there are either closed or non quantitative. So i wish to start this thread, and hope it to be the official thread for CAT 2007 for quant section. Mods, please make this a sticky thread so that more threads are not opened.

I would like all quant gurus warrior, vineet and the likes to please contribute to this thread.

Lets start with the basics and fundas and get set to CRACK CAT 07.

Common PUYS!!!!!!!!!

hi , first of all thanku very much for starting this thread......i was looking for quant threads...but all of them r closed ...so my wish has come true with this thread.....wellll now coming to our business i am another registered pagal for CAT.....i gave cat in 2004 in last yr of engg......screwed it big times....now willing to go for cat 2007....now the question is how will we start here....see i was going through last year of papers...& i also discussed wih many other CAT aspirants....acc. to all ...number system is the most important chapter for CAT...CAT has around 12 -13 q's directly or indirectly from number system......so let move ahead with number system.....we can pick up the questions from the prepatory books ....& discuss them on this comman platform.....we will give a perticular time for each section......i have always believed that i am quite good in mathematics...but when i was going through old quant threads...it scared me.....i don't think CAT comes with these type of tough problems.....i think it checks on the concept....wht u guys advise .......suggestions r welcome.....

This is a nice one.... but where to start... if we want to make full use of the forum we need to be more professional in approach.... we need to take a specific topic... not the generalised one like geometry, number system.... instead of that we will consider in geometry itself we will go with 2 D then move on to higher sections... giving basic idea... how to approach the problem.. short cuts.... and more over what i ve seen in these threads are we get carried away from the main idea... we should not let this happen... we should be more focussed...
an takers???
yu can count my name in this thread...
some one take the intitative and start a good topic since i am new i dont know much....

Krack_CAT Says

common dudes, post some more questions....lets get da fundas clear.......

Hey !! Dont worry !!!

If the response to the thread is luke warm right now...

The freshers are probably not fully into yet..

and the old timers are probably just taking a break after a gruelling round of exams..or prepping for GDPI sessions..

So give it some time and I m sure it will pick up..

Cheers!!

Regards
Tanveer

common dudes, post some more questions....lets get da fundas clear.......

Q) Find the remainder when 2000^1000 is divided by 13.

(2000^1000) mod 13 = ({154*13-2}^1000) mod 13
= ({-2}^1000) mod 13 (...coz all other terms would
involve a power of 13, which on dividin by 13 will give
rem as 0)
= ({-2}^{12*83} * {-2}^{4}) mod 13...(This is where
Euler's thm Comes in..coz 12 is the euler no for 13)
= 1*({-2}^{4}) mod 13
= 16 mod 13
= 3

-Maulin

i guess the remainder when 2000^1000 is divided by 13 is: 3
here z how i did it:
2000/13 gives a remainder of (-2)
so we need to find the remainder when (-2)^1000/13

(-2)^4/13 gives a remainder of 3
so we need 2 find out the remainder of (3)^250/13

now 3^3/13 gives a remainder 1
so we break 3^250= 3^249 * 3
the 1st part wil give a remainder 1.

hence the final remainder is (1*3)=3

Krack_CAT Says

Can some one please throw some light on this Euler concept.

Euler's thm (a generalization of Fermat's little thm) states tht
if n is a +ve integer
a is co-prime to n
then a^{(n)} = 1 (mod n)
where (n) is Euler's totient function of n
This means tht a^{(n)} when divided by n gives the remainder as 1.

Finding (n) :
Express n in terms of its prime factors..
Say n = (x^p)(y^q)(z^r)..where x,y,z are prime nos..
then (n) = n * (1-{1/x}) * (1-{1/y}) * (1-{1/z})

NOTE : if n is prime, then (n) will always be (n-1)..After all,all the nos less than a Prime No are Co-Prime to it..

- Maulin

Q) Find the remainder when 2000^1000 is divided by 13.

This involves some Euler number thingy which makes this problem very simple. Can some one please throw some light on this Euler concept.

OK... Consider this...

Mentally Divide any 2, 3 or 4Digit Number by any 1-digit Number with Decimal Precision

When asked to do division in your head, it's very impressive to be able to carry the answer out to several decimal places. When dividing by a 1-digit number, it's not that difficult, either.
First, you need to remember all the possible decimal equivalents for each single digit divisor. Most people will already know the first few decimal equivalents:

1/2=.5

1/3=.333...
2/3=.666...

1/4=.25
2/4=1/2=.5
3/4=.75

What are the rest? 5ths are easy, as you simply double the dividend, and place the decimal in front of that number:

1/5=.2
2/5=.4
3/5=.6
4/5=.8

With 6th, you already know 3 of the 5 decimals from above:

2/6=1/3=.333...
3/6=1/2=.5
4/6=2/3=.666...

You simply need to learn just two more 6ths:

1/6=.1666...
5/6=.8333...

7ths have a very unique pattern! Let's start with 1/7:

1/7=.142857142857142857...

All you have to do for 7ths is to remember the sequence 142857 (which repeats over and over again). Each 7th will always contain this same sequence, and only the starting point will change!

To find the appropriate starting point, take the dividend and multiply it by 14 (you should be able to do up to 14*6 in your head fairly quickly of course). Find the place in the 142857 sequence closest to this number, and you'll have the appropriate starting place!

Starting again with 1/7, we simply think (1*14=14, so 1/7 starts at the 14, and is thus equal to .142857142857...)

Here's how you figure the rest of the 7ths, with the thought process in parentheses:

2/7 (2*14=2=.2857142857142857...
3/7 (3*14=42)=.42857142857142857...
4/7 (4*14=56)=.57142857142857...
5/7 (5*14=70)=.7142857142857...
6/7 (6*14=84)=.857142857142857...

Once you see the pattern and practice it, 7ths are very simple.

8ths are also very simple, as they are half-steps in-between the 4ths. Simply multiply the dividend by 125, and place the decimal in front of it:

1/8=.125
2/8=1/4=.25
3/8=.375
4/8=1/2=.5
5/8=.625
6/8=3/4=.75
7/8=.875

9ths seem like they should be hard, but all you have to do is repeat the dividend over and over:

1/9=.111...
2/9=.222...
3/9=.333...
4/9=.444...
5/9=.555...
6/9=.666...
7/9=.777...
8/9=.888...

Just for reference, 10ths and 11ths aren't hard, and can be done in your head easily, as well.

For 10ths, simply place a decimal in front of the dividend:

1/10=.1
2/10=.2
3/10=.3
4/10=.4
5/10=.5
6/10=.6
7/10=.7
8/10=.8
9/10=.9

For 11ths, you need to know your 9 times table up to 10:

1/11=.090909...
2/11=.181818...
3/11=.272727...
4/11=.363636...
5/11=.454545...
6/11=.545454...
7/11=.636363...
8/11=.727272...
9/11=.818181...
10/11=.909090...

With a little practice, these decimal equivalents will come to mind quickly.

Now for the full feat. To keep things simple, start working with 2 digit numbers.

Have someone choose any 2-digit number and any 1-digit number, and you can announce the result of dividing the larger number by the smaller one. For example, let's say they choose 59 divided by 6.

You should quickly realize that the closest multiple of 6 to 59, without going over, is 54 (6*9). So, the answer is 9 and 5/6ths. Instead of saying it that way, however, you remember 5/6 = 0.833, and so the answer is 9.833.

People see decimals as very complex, so this is very impressive, yet not hard to do.

For 3- and 4-digit numbers, you need to practice working through the division problem from left to right in your head.

Starting with a 3-digit example, let's try 698 divided by 7. Beginning with the leftmost digit, we quickly see that 7 won't go into 6, so we move to the next digit. 7 will go into 69 nine times, so our answer is 90-something. Taking away 63 (7*9), that leaves us with 68 to work with. 7 can go into 68 nine times, as well, so that gives us 99, with 5 as a remainder, or 99 and 5/7ths. Remember the decimal equivalent of 5/7ths? This means you can give the answer as 99.7142857 in short order.

4-digits work the same way, with one extra step, of course. 4732 divided by 6? Let's try it:

4/6=won't work
47/6=7, carrying the 5 (47-42=5), so it's 700 something
53/6=8, carrying the 5 (53-48=5), so it's 780 something
52/6=8, carrying the 4 (52-48=4), so it's 788 and 4/6, or 788 2/3

Translated into decimal form, you say "788.666".

Even if you never get comfortable with 4-digit numbers, dividing 3-digit numbers by 1-digit numbers is still impressive, especially when you can carry it out to many decimal places.