Thanks!... mucool & sausi for your reply. Really, I have got benefited from your answers.
@ Sausi: Just before 3 months, I have started up with GMAT preparation. But currently I'm studying using IMS material. Once I have done with that, will move to arun sharma.
@ mucool & Sausi: Which material is good for concept building?.. I will finish my Quants preparation by end of this month. So, I am thinking to take quantitative tests alone as many as possible before I start with Verbal? Suggest some websites which contain challenging questions?...
Mohan Kumar T SaysThere are 6 boxes numbered 1, 2,....6. Each box is to be filled up either with a red or a green ball in such a way that at least 1 box contains a green ball and the boxes containing green balls are consecutively numbered. The total number of ways in which this can be done is?....
answer 21 ???
This is a standard CAT question.
What needs to be calculated is the sum of: ( here[] means greatest integer function )
=21
=10
=5
=2
=1
Thus max n = 21+10+5+2+1 = 39
Can you tell us where 21 is coming from or 10, 5, 2...basically the denominator.
I am not saying it will not give the correct answer. It will. And I totally agree with the trick. But again its not logical, because it does not follow straight line of thinking. And thats why he did not understand it the first time. People here ask for explanations, not solutions.
Anyway everyone is entitled to their own opinions. Ultimately what matters is that the guy who asked it, did he understand it.
I have a 770 on GMAT and a engg degree from IIT, but I can still slow my pace to write everything down, so that it makes sense.
My 2 cents.
GMATing: Realizing your B-School dream!!How I got a 770!!
Thanks!... mucool & sausi for your reply. Really, I have got benefited from your answers.
@ Sausi: Just before 3 months, I have started up with GMAT preparation. But currently I'm studying using IMS material. Once I have done with that, will move to arun sharma.
@ mucool & Sausi: Which material is good for concept building?.. I will finish my Quants preparation by end of this month. So, I am thinking to take quantitative tests alone as many as possible before I start with Verbal? Suggest some websites which contain challenging questions?...
I virtually did not prepare for quants on GMAT becuase I had prepped for CAT during my 4th at IIT. I screwed up DI but had a 99%le in quants. Basically I hav studied from Career Forum's package which was very good. i believe all CAT Quant packages are good. But they are way above GMAT requirement.
I would suggest you go to my website and download the tough quant problem docs and the formulae sheet.
Kaplan GMAT 800 is also good. There are a couple of tricy DS questions. I recommend doing them.
GMATing: Realizing your B-School dream!!How I got a 770!!
achintbvp Saysanswer 21 ???
I think it's 21 too.
If this is the answer, then this explanation might help.
Consider that only one has a green ball and calculate the number of ways.
Then for two, three, four, so on.
The way to do that is:
case 1: If there is one green ball, all else will be red balls.
so G R R R R R --> this has to be arranged = 6!/5! 6! is for 6 places and 5! is for 5 repetitions in the form of red balls = 6
case 2: GGRRRR or RGGRRR or RRGGRR so on = 5!/4!
where 5! is because two green balls will alsways be together and so clubbed as one and rest 4 reds...total 5 units. and divide by repetitions of 4 reds. which gives 5
and so on for the rest....
which will be 6+5+4+3+2+1
21
Hope that helps
GMATing: Realizing your B-School dream!!How I got a 770!!
Maverick buys 20 candies for a dollar. How many should he sell for three dollars to earn a 20% profit?
Please explain me the logic behind it?
@ Mucool: I have seen your Mixtures & Allegations Videos on gmating.com. I have watched it twice. Nicely done with explanation. Thanks a lot. Is there any other videos posted by you?....
Maverick buys 20 candies for a dollar. How many should he sell for three dollars to earn a 20% profit?
Please explain me the logic behind it?
Price per candy= 5 cents
20% profit means an SP of 6 cents (5 X 120%)
in 3 dollars he shud sell=300/6=50 candies.
now dun be confused by the question and fall into thinking that he bought only 20 candies...he bought at the rate of 20 for 1.
I hope that clarifies it.
M
GMATing: How I made it to an Ivy League B-School!!How I got a 770!!
Mohan Kumar T Says@ Mucool: I have seen your Mixtures & Allegations Videos on gmating.com. I have watched it twice. Nicely done with explanation. Thanks a lot. Is there any other videos posted by you?....
Tell me what topics you want covered. I will make it happen :)
GMATing: How I made it to an Ivy League B-School!!How I got a 770!!
Let S be the set of all two digit natural numbers with distinct digits. In how many ways can the ordered pair (P, Q) be selected such that P & Q belongs to S and have atleast one digit in common?
Thanks in advance.
Let S be the set of all two digit natural numbers with distinct digits. In how many ways can the ordered pair (P, Q) be selected such that P & Q belongs to S and have atleast one digit in common?
Thanks in advance.
I am not 100% sure, but this approach will be able to guide you:
Total number of integers is the set S
since, tens digit can be 1-9 (9 ways) and units digit can be 0-9, remove the digit used for units, so there will be 9 ways.
thus total S= 9*9 = 81
So total available ways to choose P and Q from this set = 81*81 = 6561
From this value we will need to substract the no of combinations of P and Q that dont have any digit in common.
To calculate this,
Say P = p1p2 (p1 is tens digit, p2 is units)
similarly Q = q1q2
p1 can be anything between 1-9 (9 choices)
p2 can be anything between 0-9, but excluding p1 (9 choices)
q1 can be anything between 1-9, but excluding p1 and p2 (7 choices)
q2 can be anything between 0-9, but excluding p1, p2, q1 (7 choices)
total combinations of P and Q = 9*9*7*7 = 3969
Thus combinations of P, Q that have at least one digit in common = 6561-3969 = 2592
Please confirm if this answer is correct or not.
@sausi007: Thanks a lot.....I have got your idea. I feel yours is 100% correct.
@mucool: Please give me your approach and answer for my previous question.
Given that the two digit natural number should contain two distinct digits. So, Say N=ab,
'a' can be from 1 to 9 and so there are 9 chances.
'b' can be 0 to 9, but except a and hence it also has 9 chances.
This actually negotiates two digit natural numbers contain same digits like 11, 22, 33......
so, now N = ab = 9*9 = 81.
Now (P, Q) = can contain 81*81 = 6561 pairs. But as per given statement it should not contain pairs which are not having any digits in common.
Let say P= p1*p2 & Q= q1*q2 as given by sausi007.
Let us take two examples:
Ex:1
P= 90 & Q = 85. In this p1 = 9, p2= 0, q1 = 8 & q2 = 5
From this,
p1 has 9 chances from 1 to 9
Let say, p2 has 9 chances, but we chose only 0. Then,
q1 has 8 chances from 1 to 9, except p1 & p2.
q2 has 7 chances from 0 to 9, except p1, q1 & p2 (or ''0'')
Ex:2
P= 85 & Q = 90. In this p1 = 8, p2= 5, q1 = 9 & q2 = 0
From this,
p1 has 9 chances from 1 to 9
p2 has 9 chances from 0 to 9
q1 has 7 chances from 1 to 9, except p1, p2 & 0
q2 has 7 chances from 0 to 9, except p1, p2 & q1
From these two examples,
If P ends with 0 (say 10, 20, 30, 40, 50, 60, 70, 80 & 90) then it gives Q = q1*q2 = 8*7 = 56 chances and P can be any two digit natural number ends with zero and so P has 9 chances in this. Hence, 9* (8*7) = 9*56 = 504 pairs are there without even a single digit as common.
If it not ends with zero, then P (= 9*9 - 9 = 72 chances), then Q = (7*7 = 49 chances). Hence 72*49 = 3528 pairs are there without even a single digit as common.
So, totally = 3528 + 504 = 4032 two digit natural number ordered pairs are there without a single digit as common.
Hence, 6561 - 4032 = 2529 two digit natural number ordered pairs are there with at least a digit in them as common.
I think, I have tried my best in my explanation. If any mistake is there in my explanation, please revert me back....
@sausi007: Thanks a lot.....I have got your idea. I feel yours is 100% correct.
@mucool: Please give me your approach and answer for my previous question.
I think Sausi's expln is spot on. And your approach unnecessarily treats 0 differently. It doesnt matter if its zero or not. The only caveat is that tens digit cannot have 0, so that its a two digit number. Sausi's expln takes that into account correctly.
GMATing: How I made it to an Ivy League B-School!!How I got a 770!!
@mucool: Please re-think over it....Me and my friend sat for a long time and ended up with this. It's a tricky problem.....
@mucool : Please upload some videos on Permutations & combinations, probabilities, Inequalities & statistics.
Thanks in advance.
Thanks in advance.
Mohan Kumar T Says@mucool: Please re-think over it....Me and my friend sat for a long time and ended up with this. It's a tricky problem.....
Yes you are right. I am sorry. I re-did the problem. You are in fact correct.
Zero has to be treated differently. Thanks for correcting me.
Sorry for the confusion.
M
@mucool : Please upload some videos on Permutations & combinations, probabilities, Inequalities & statistics.
Thanks in advance.
I will. Tell me what do you feel is the best way? Shall I solve problems, or teach concepts or do both.
How is the current video on Alligations? Are they good enough, so that I follow the same line of instruction in future videos?
Please comment on youtube...I will be taking your feedback from there. This doesn't seem appropriate. Thank you so much for your time.
M
@mucool: Sure...I will give my feedback there....Your video on Alligations explains concept based on the problems which you have chosen....Thats really good...So, please do both i.e., Initially give all the important concepts on a topic and then proceed with the some tricky problems based on that, as like Allegations.
Thank you for your help mucool........
Given that the two digit natural number should contain two distinct digits. So, Say N=ab,
'a' can be from 1 to 9 and so there are 9 chances.
'b' can be 0 to 9, but except a and hence it also has 9 chances.
This actually negotiates two digit natural numbers contain same digits like 11, 22, 33......
so, now N = ab = 9*9 = 81.
Now (P, Q) = can contain 81*81 = 6561 pairs. But as per given statement it should not contain pairs which are not having any digits in common.
Let say P= p1*p2 & Q= q1*q2 as given by sausi007.
Let us take two examples:
Ex:1
P= 90 & Q = 85. In this p1 = 9, p2= 0, q1 = 8 & q2 = 5
From this,
p1 has 9 chances from 1 to 9
Let say, p2 has 9 chances, but we chose only 0. Then,
q1 has 8 chances from 1 to 9, except p1 & p2.
q2 has 7 chances from 0 to 9, except p1, q1 & p2 (or ''0'')
Ex:2
P= 85 & Q = 90. In this p1 = 8, p2= 5, q1 = 9 & q2 = 0
From this,
p1 has 9 chances from 1 to 9
p2 has 9 chances from 0 to 9
q1 has 7 chances from 1 to 9, except p1, p2 & 0
q2 has 7 chances from 0 to 9, except p1, p2 & q1
From these two examples,
If P ends with 0 (say 10, 20, 30, 40, 50, 60, 70, 80 & 90) then it gives Q = q1*q2 = 8*7 = 56 chances and P can be any two digit natural number ends with zero and so P has 9 chances in this. Hence, 9* (8*7) = 9*56 = 504 pairs are there without even a single digit as common.
If it not ends with zero, then P (= 9*9 - 9 = 72 chances), then Q = (7*7 = 49 chances). Hence 72*49 = 3528 pairs are there without even a single digit as common.
So, totally = 3528 + 504 = 4032 two digit natural number ordered pairs are there without a single digit as common.
Hence, 6561 - 4032 = 2529 two digit natural number ordered pairs are there with at least a digit in them as common.
I think, I have tried my best in my explanation. If any mistake is there in my explanation, please revert me back....
Ok...got it....this was really tricky.
Now my question is, how good is this question from GMAT perspective,
do they really ask such lengthy problems. I dont think I can solve this kind of a problem in the exam scenario.