Ajay had N coins with him and he gave one more than half the number of coins with him to balu and then gave two more than one third of the remaining number of coins to chintu. If N is a two digit number, how many distinct values can N assume ?

Hi,After doing the operations, Ajay will have (2N-16)/6 coins left with him which must be a positive integer. First such 2 digit number will be 11. The next values will be 14, 17, 20 and so on. In this series, 2 digit numbers will be 30.Hope this answers your query.

can you once explain me the solution given for question 4 in probability topic test 1 dated 04.02.16

Hi Kaushik,Could you please ask the question or post the picture of the question? so that we can help you better.Thanks

How to do long problems

Hi Arpit,Can you please be a bit more specific with regards to the type of questions that you have a query about. There can be different strategies for long questions from different topics. So please post the topic in which you are facing problems.

Kulbushan takes every 3rd number counting from 1001 up to 4000 (1003, 1006, 1009, … , 3997, 4000) and writes them down in sequence. What will be the 251st digit (from the left) in this sequence?

How we came across with required digit is 8?

Hi Kumar The question asks for 251st digit, we know that all the numbers in the given range are 4 digit numbers 248/4 =62. So we know that the required digit will be 3rd digit of the 63rd term in the given sequence. The given sequence is an AP with common difference 3, so 63rd term will be1001 + 62*3 = 1001 +186 = 1187.So the required digit is 8.Hope that helps.

Masters in bussines Administration

find the value of I133-9I + [5-211] + I8-9I / I110-9I +[19-101] +8

|| implies modulus function. So anything inside it will always be positive. Hence. 124 - 206 + 1 -82 + 8 = 133 - 288 = -155

124-206+1/101-82+8125-206/109-8281/27=3

if x # y = 2(x+y/x-y), then find the value of (4#3)/7

Hi Sam,4 # 3 = 2(4+3)/(4-3) = 14So, 14/7 = 2

find the value of the expression 1+4 division 3*9 of (6-2)

Use BODMAS so we have 1+4/3*9 of 4=> 1+4/3*36=> 1+ 1.33*36=> 48+1 = 49

how to solve this find the value of 72540/5 +72540/25 +72540/125

Hi Sam,Well there's no shortcut to this and you have to do the calculations. However you can reduce the slightly by some smart work.The first term has to be divided so the result would be 14508. Now instead of dividing the second term by 25, you can rather take 14508 to be second term and divide it by 5.( We just took 5 common from numerator and denominator in second term). Yu will get 2901. Now instead of performing the division on 3rd term, divide 2901 by 5. You will get 580. Now add these values and you have your result. Note that I have assumed that all these terms are greatest integer less than equal to x. If that is not the case then it is not worth attempting this question in exam.

akhilesh sing sir is the ans of the question 7 to the power 13+6 to the power 12+ 1 to the power 11+4 to the power 7 to the power 13 713 is odd if we add 6 to it, it will still be odd. it if we add 6 with 7 am i right sir odd12 and odd 12 means 13 to the power 12 sir odd12 is odd, if we add 1 to it it will become even.and 1 with 13 it becomes 14 the even number Now (even)11(even)11 will be even it mens sir 14 to the power11 and adding 4 to it will give even only. and adding 4 with14 it becomes 18 sir (Even)7(Even)7 is even so the final result is even. and 18 to the power 7 is even and so the ans is even sir am i right

Hi Sam,I would request to post the questions in numerical forms using brackets. In this way, it is slightly difficult to comprehend the question.Anyway, I think you meant this ((((713+6)12+1)11+4)7)13((((713+6)12+1)11+4)7)13 If it something else please let me know.If the question is this then, you are right. The result would be even.Thanks