@sam05 said: Its coming to 4146....

@dead_alive said: from arun sharma , page 101 ... pre assessmen test .. "I am eight times as old as you were when I was as old as you are " . said a man to his son. Find out their present ages if the sum of their ages is 75 years

@chevy said: How many 5 digit numbers are there such that there is at least one digit "1", at least one digit "2", and at least one digit "3".

@chevy said: How many 5 digit numbers are there such that there is at least one digit "1", at least one digit "2", and at least one digit "3".

@peppie said: yaa i know the answer ..please explain in detail ..how is a hexagon obtained from a cube?

Don't know why I am getting (3*root(3))/4

@Foster87 said: For how many values of natural no A,where A<500,is the HCF of A and 2500 equal to 1?40020010050

@YouMadFellow said: (3*root(3))/ 4 ?

@YouMadFellow said: (3*root(3))/ 4 ?

@peppie said: A cube, one centimetre on a side, is sliced into two equal halves in such a way that the slice through the middle forms a regular hexagon. What's the area of the regular hexagon? plz tell me how is regular hexagon formed from a cube???? nt able to visualise ..

@YouMadFellow said: Total 5 digit numbers = 10^5 - 10^4 = 90000Now, 5 digit numbers which don't contain 1,2,3 are 6*10^4So, 90000-60000 = 30000 contain at least one 1, one 2, one 3

@chevy said:How many 5 digit numbers are there such that there is at least one digit "1", at least one digit "2", and at least one digit "3".

Bhai iska OA kya h

4500?

@YouMadFellow said: The minimum value of the expression given is 25, and it occurs at x = 7, so only one integral solution

@RoadKill said: @deepak_pgi @YouMadFellow 5 integral solutions: 3,4,5,6,7. Plot the numbers on a line and use the distance method.

@ziddi_armaan said: i dont know the right way but i am getting 2 3 4 5 6 7

@Enceladus said: 5 solutions? 3,4,5,6 and 7?

@RoadKill said: @deepak_pgi @YouMadFellow 5 integral solutions: 3,4,5,6,7. Plot the numbers on a line and use the distance method.

@deepak_pgi said: Find the number of integral solutions of |x - 1| + |x - 3| + |x - 7| + |x - 22| = 25.

@WildDreamer said: A milkman has six cans, each containing 10 litres of milk, of concentrations 20%, 30%, 40%, 50%, 70% and 80% respectively. If he mixed milk from exactly three of the cans to form one litre of milk of exactly 50% concentration, in how many ways could he have selected the three cans?(1) 14 (2) 15 (3) 16 (4) 17

The concentration of the milks are 20, 30, 40, 50, 70, 80.

Now to make 50% milk solution, you have to make sure that not all the three are less than or more than 50%.

a) One of the solutions is 20%, then the remaining two mixtures can be taken as, one of 30 or 40 => 2 ways and one of 70 or 80 => 2 ways so total 2*2 = 4ways.

Or, the 2nd mixture can be 50% and for 3rd one we can take from 70 or 80 => 2 ways.

Also we could take the remaining two mixtures as 70 and 80 both => 1 way.

=> Total of 4 + 2 + 1 = 7 ways.

b) Similarly with 30% mixture, one can be 40% and the third one can be either of 70 and 80 => 2 ways

Or, One can be 50 and the third one can be either of 70 and 80 => 2 ways.

Or, we could take the remaining two mixtures as 70 and 80 both => 1 way.

=> Total of 2 + 2 + 1 = 5 ways.

c) With 40% mixture => 2 + 1 = 3 ways

=> Total 7 + 5 + 3 = 15 ways., Option B)

A milkman has six cans, each containing 10 litres of milk, of concentrations 20%, 30%, 40%, 50%, 70% and 80% respectively. If he mixed milk from exactly three of the cans to form one litre of milk of exactly 50% concentration, in how many ways could he have selected the three cans?

(1) 14
(2) 15
(3) 16
(4) 17

@deepak_pgi said: Find the number of integral solutions of |x - 1| + |x - 3| + |x - 7| + |x - 22| = 25.

@chevy said: How many 5 digit numbers are there such that there is at least one digit "1", at least one digit "2", and at least one digit "3".

@Foster87 said: The number 2006! is written in base 22.How many zeroes are there at the end?? 500 450 200 199..Approach please!!!

@Foster87 said: The number 2006! is written in base 22.How many zeroes are there at the end?? 500450200199..Approach please!!!

@Foster87 said: The number 2006! is written in base 22.How many zeroes are there at the end?? 500450200199..Approach please!!!

@deepak_pgi said: Find the number of integral solutions of |x - 1| + |x - 3| + |x - 7| + |x - 22| = 25.

@YouMadFellow said: Yah there is a flaw. You are finding the remainder of 10^147 divided by 7 and cubing it to find the remainder of 10^147 divided 7^3.. this is not true. 342 is the right answer.

ya correct

thnk u

@Foster87 said:For how many values of natural no A,where A<500,is the HCF of A and 2500 equal to 1? 400 200 100 50

2500=(2^2) (5^4)

now less den 500

so all even numbrs shud b excluded,(as 2 is a commn fctr )

so 250 numbrs

now

odd multiples of 5(evn multiples we hav included in 2)

total odd=50

so hcf(10=250-50=200

@angel.mehra said: 7 is prime number,break,10^147 as (10^133)(10^7)(10^7)n 343 as 7*7*7, 10^7=10mod7=3mod 7 10^7)^19=3^19mod73^3=-1mod73^18=1mod73^19=3mod7 rem(3*3*3)mod7rem(27)mod7rem=6 if any mistak in approch pz post thnk u

@Enceladus said: Bhai coz this is actually 10^147*10^147. If we say 1. then we would go back to the original question,ie, 10^147 mod 343. Hence, 342.

@YouMadFellow said: How did you get it ? I don't have OA

@Angadbir said: You'll have to apply a different approach. (10^147 * 10^147) mod 343 = 1 10^147mod343 = +/-1 If 10^147mod343 = 1then 10^147 = 343k + 1That means 343k = 9999.... 147 timesHere, LHS is divisible by 7, but not RHSSeries of 147 9s cannot be divisible by 7 (Divisibility of 7 can be checked by subtracting alternate triplets. Here we have odd no. of triplets of '999's) So therefore 343k can never result in a seiries of 147 9s for a whole no. k.Thus, 10^147mod343 cannot be 1 Thus, 10^147mod343 = -1

Yes, now I am convinced. .... All I wanted was the logic to reject 1 as the answer..

@Foster87 said: Mee too got 200...ans not given though... MY APPROACH:multiples of 2 in the 1st 500 no's=250. multiples of 5 in the 1st 500 no's=100. multiples of 10 in the 1st 500 no's=50. hence 250+100-50=300 no's are divisible by either 2 0r 5..Thus 500-300=200 no's which are less than 500 are co-prime to 2500

@YouMadFellow said: Yes exactly my point.. But then can we find the remainder of this : 10^147 / 343 ? It will be either 1 or 342, but can we narrow down to one ?

@Foster87 said: For how many values of natural no A,where A<500,is the HCF of A and 2500 equal to 1?40020010050

@x2maverickc said: All of them do equal work. Thus each does 1/3rd of the work. So the time taken by Bhuvan = 6 days, Bheem = 8 days, Bahadur = 12 days, hence total days = 26 days. Now if Bheem does (1/y)th part of [Bhuvan + Bahadur] more, in addition to his quota of 1/3rd of the work then, Bheem's work = 1/3 + (1/y)(2/3) Bhuvan work= ( 1 - 1/y)(1/3) Bahadur work= ( 1 - 1/y)(1/3) Bhuvan takes [ (y-1)/y ] *18 days Bheem takes [ 1/3 + (1/y)(2/3) ]*24 days Bahadur takes [ (y-1)/y ] *36 days Hence, [{(y-1)/y}*18] + [(1/3 + 2/3y)*24] + [{(y-1)/y}*36]=25 we get y = 2. More work done by Bheem = [1/2]*[2/3] = 1/3

@YouMadFellow said: @Enceladus Yes exactly.. Same doubt here.. It can be 1 too.. That's why I asked if the opposite is true or not.. For example consider this : 6^50 mod 25 Now, E(25) = 20 so, 6^20 mod 5 = 1 ( good enough till now ) Now, let's take the power series of 6 to just see how the remainder varies with power 6^1 mod 25 = 66^2 mod 25 = 116^3 mod 25 = 166^4 mod 25 = 216^5 mod 25 = 1..So here we see that 6^5 in itself is also 1. So, obviously 6^20 is also 1. But if we go by the way the solution was proposed, then 6^10 should be -1, which is not true. 6^10 mod 25 is in fact 1.

@Angadbir said: Guys, be wary of this. We know that a coprime raised to power of euler of a number, leaves remainder 1, when divide by the number. But reverse is not necessarily true. E[n] is not necessarily the smallest power that would fetch a remainder 1 when a coprime to n is divided by n. Eg,3^40mod100 = 1, where E[100] = 40But 3^20 mod 100 = 1 as wellAnd so is 3^4mod100

@sowmyanarayanan said: set theory? how?

@just_nishu4u said:It takes Bhuvan,Bheem & Bahadur 18,24,36 days to do a piece of work.Each of them does the same amount of work & they complete the work in 26 days.Instead,if Bheem did extra work equal to the sum of yth parts of the work done by Bhuvan & Bahadur,they could have completed the work 1 day earlier.How much more work did Bheem do in the 2nd case as compared to that done by him in the 1st? A) 1/9 B) 1/6 C) 1/4 D) 1/3

@chevy said: A milkman has six cans, each containing 10 litres of milk, of concentrations 20%, 30%, 40%, 50%, 70% and 80% respectively. If he mixed milk from exactly three of the cans to form one litre of milk of exactly 50% concentration, in how many ways could he have selected the three cans?(1) 14 (2) 15 (3) 16 (4) 17 P.S.-- gs bhai aapne bulaaya aur hum chale aaye

@Enceladus said: 10^a mod 343 = 1. so a is of the form 294k, where k is a whole number. Yes, that is true always.

@ziddi_armaan said: @Enceladus Bhai main samajh ye cheez that -1*-1= 1 so remainder is 343-1= 342 but this is also possible that 1*1= 1 ?? plzz clear this doubt ,,, i mean 10^147 will give 1 ??is it possible /////??

@chevy said: A milkman has six cans, each containing 10 litres of milk, of concentrations 20%, 30%, 40%, 50%, 70% and 80% respectively. If he mixed milk from exactly three of the cans to form one litre of milk of exactly 50% concentration, in how many ways could he have selected the three cans? (1) 14 (2) 15 (3) 16 (4) 17 P.S.-- gs bhai aapne bulaaya aur hum chale aaye

@Enceladus said: 200 is it? Can be done by Set theory.

@ziddi_armaan said: Isme kya ... Mixing ka sequence bhi matter karega >??

@Foster87 said: For how many values of natural no A,where A<500,is the HCF of A and 2500 equal to 1?40020010050

@chevy said: A milkman has six cans, each containing 10 litres of milk, of concentrations 20%, 30%, 40%, 50%, 70% and 80% respectively. If he mixed milk from exactly three of the cans to form one litre of milk of exactly 50% concentration, in how many ways could he have selected the three cans? (1) 14 (2) 15 (3) 16 (4) 17 P.S.-- gs bhai aapne bulaaya aur hum chale aaye

@Foster87 said: For how many values of natural no A,where A<500,is the HCF of A and 2500 equal to 1?40020010050

@YouMadFellow said: You mean to say that Rem(10^147) = Xand X^2 = 1 so, X = -1 ( you rejected 1 because E(343) = 294 and hence for all powers less than 294.. there would be any power will would give the remainder as 1 ? ) (We know that 10^E(343) mod 343 will give 1, but does the opposite hold true too i.e. if 10 ^ a mod 343 = 1 => a = 294k where k is an integer ? )