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

@ 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.....

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.

@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........

@mucool : Please upload some videos on Permutations & combinations, probabilities, Inequalities & statistics.

Thanks in advance.

@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.....

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.

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.

Every item in an apparel shop has a tag that is either blue, green or yellow in colour. The items with a blue tag are priced between $20 and $30. The items with a green tag are priced between $12 and $18 and the items with a yellow tag $5 and $10. Mary buys three items, each with a different color tag. Is the total discount on three items greater than 10%?......

1) The discount on the items with a blue tag is 20% and that on the items with a green tag is 25%.
2)The discount on the items with a yellow tag is 40%.


Please provide the answer with your explanation....Thanks in advance.