GMAT Problem Solving Discussions

MBA
4306
The following is an DS Question:-

If a, b, c, d & e are non- negative integers and p = 2^a*3^b and q = 2^c*3^d*5^e, Is p/q a terminating decimal????

1) a > b
2) c > d

I am confused with this problem....Please explain your answers....
Thanks in advance.
4307
The following is an DS Question:-

If a, b, c, d & e are non- negative integers and p = 2^a*3^b and q = 2^c*3^d*5^e, Is p/q a terminating decimal????

1) a > b
2) c > d

I am confused with this problem....Please explain your answers....
Thanks in advance.


I think i have seen a similar problem being discussed in the DS Section.
Anyways:

The only problem we face in deciding the nature of p/q is because of the "3". Since the 2 and 5, wherever they stay (numerator or denominator), they dont change the behaviour of the fraction, it will always remain terminating.

If we figure out what happens with the 3, we can conclude on the nature of the fraction

statement 1: a>b does not help in deciding the exponent of 3 so insufficient

statement 2: b>d , in the fraction, the exponent of 3 is essentially (b-d), so we know b>d, then we get that 3^(b-d) will always be more than 1 and stay in the numerator.
Thus with this information we can safely say that the fraction will be terminating. Sufficient.

I hope this is correct and i have not goofed up the explanation.
4308
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


I got confused again with this problem, can u guys elaborate why the zero in the end has to be treated differently
4309
@ sausi007: After i have posted this, I struck with the same idea...only 3 & 9 can lead this p/q to non-terminating decimals if one of them were in the denominator.

Thanks your answer.....
4310
DS Question:

Mr. Tolstoy bought 100 CDs at $x and sold them at $y. Did Mr. Tolstoy profit from the deal?

1) 40% of x > 30% of y
2) 30% of x > 40% of y
4311
@ sausi007: After i have posted this, I struck with the same idea...only 3 & 9 can lead this p/q to non-terminating decimals if one of them were in denominator.

Thanks your answer.....


3 in the denominator, we know will cause a non-terminating fraction,
what are the other primes which can do the same (7,13,17....)?
4312
@mucool: I also watched Allegations video in youtube... It is really simple and good. Please post maximum of quants videos like Probability......
4313
There are 100 tokens numbered from 1 to 100. In how many ways can 2 tokens be drawn simultaneously so that their sum is more than 100?
a) 4950
b) 5050
c) 2500
d) 2550
e) 5000

Thanking in advance.......
4314

i m having lots of problem with remainders pls help me:oops:

4315
sausi007 Says
I got confused again with this problem, can u guys elaborate why the zero in the end has to be treated differently


Let's say the digits are p1 p2 q1 q2

Now if p is ending with 0, p2 is zero...

Now earlier we were assuming this, and I am quoting you here...

q1 can be anything between 1-9, but excluding p1 and p2 (7 choices)

Now this is okay as long as q2 is not zero. Becuz if its zero, then you have 8 choices. Because p1 eats up one digit between 1-9. but p2 does not. becuz its zero. So q1 will have 8 options and not 7. but when p2 is not zero, then it will have 7. So thats why we have to calculate for both cases.

Hope that helps :)

GMATing: How I made it to an Ivy League B-School!!How I got a 770!!
4316
There are 100 tokens numbered from 1 to 100. In how many ways can 2 tokens be drawn simultaneously so that their sum is more than 100?
a) 4950
b) 5050
c) 2500
d) 2550
e) 5000

Thanking in advance.......


The answer should be 2500.

This is how.

let's say you chose 1...the next token can only be 100 = 1 way
chose 2...nect token can be 100, 99 = 2 ways
3...100, 99, 98 = 3 ways.
......

50 = 50 ways.

Now from 51 onwards things will be different.

51...you can chose 52, 53, 54,...100. = 49 ways. You do not consider 50, 49, 48....because they have been accounted for in the previous cases. you already had a combination of lets say 51 and 50 or 52 and 49...

so from here on...it will be:

51...49 ways
52 = 48 ways
....
99...1 way (only token no.100)

So total ways is=

1+2+3+....+50+49+48+47+....1
= 50X49 + 50
= 50X50
=2500

GMATing: How I made it to an Ivy League B-School!!How I got a 770!!
4317

Thanks a lot mucool.....

4318
raha21 Says
i m having lots of problem with remainders pls help me:oops:


Post a problem...and I will try to explain it, may be the concept too. Or even better make a tutorial video :)

GMATing: How I made it to an Ivy League B-School!!How I got a 770!!
4319
@mucool: Yes......I need it too. Post some tutorial videos on Permutations & combinations, Probability....Help me with these topics from small basics to tricky ones. I use to struggle at these topics....
4320
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....



Just came across this problem in this thread, thought of posting a diff approach.

S= (P,Q)
S= 99- 9(single digits) - 9( 11,22 ..99) = 81

For every P not ending in zero (which is 81-9 =72) Q can take (9+8 )*2 -1 = 33 values
Here how P = xy Q can take 9 digits starting with x and 8 digits ending with x similarly for y like for
P = 23, Q = ( 20, 21, 23, 24...29 and 12, 32, 42..92)

For every P ending in zero which is 9
P=x0 Q can take 17 values according to the above explanation.

Hence, The total sum would be 72*33+9*17 = 2529

Hope I didnt wrong anywhere!
4321

The probability of pulling a black ball out of a glass jar is 1/X. The probability of pulling a black jar and breaking the jar is 1/Y. What is the probability of breaking the jar?........

1) 1/(XY)
2) X/Y
3) Y/X
4) 1/(X+Y)
5) 1/(X-Y)

Please explain your answer.....

Thanking in advance.

4322
The probability of pulling a black ball out of a glass jar is 1/X. The probability of pulling a black jar and breaking the jar is 1/Y. What is the probability of breaking the jar?........

1) 1/(XY)
2) X/Y
3) Y/X
4) 1/(X+Y)
5) 1/(X-Y)

Please explain your answer.....

Thanking in advance.


1/X = pulling black ball from jar
1/Y = pulling black ball from jar AND breaking the jar

so since it is AND case,

Prob of pulling black ball from jar AND breaking the jar = (Prob of pulling black ball) * (Probability of breaking jar)

say, Probability of breaking jar = 1/Z

1/Y = (1/X)*(1/Z)
Therefore,
1/Z = X/Y

B is the answer.
4323
@mucool : Please upload some videos on Permutations & combinations, probabilities, Inequalities & statistics.

Thanks in advance.

one video uploaded.
cheers!
4324
@mucool: The video is really good. Please post more videos by covering more chapters. Your way of explaining the concept is really great. Your videos are making our fundamental stronger.
Thanks a lot mucool.

Cheers!.....
4325

How many 4 - digit numbers can be formed by using the digits 0 - 9, so that it contains exactly a distinct digits?.....

Please post your answers with approach....

Thanking in advance...