@chevy said: mujhe toh ye samajh nhi aaya,,,aap hi bata do,,

@sam05 said: mene ese b try kia tha but kuch lafra kia kai pe.....nai a raha tha....then reverted to the above method...Inclusion Exclusion jai ho

@chandra12 said: guys can somebody help me out in this problem..

@chevy said: 90000 - 3*8*9*9*9*9 + 3*7*8*8*8*8 - 6*7*7*7*7 = 4146 OA:4146

@Faruq said: a boat rows30 km upstream and 44 km downstream in 10 hours.......it also rows 40km upstream and 55 km downstream in 13 hours.....find the speed of boat in still water.......

@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".

@chandra12 said: guys can somebody help me out in this problem..

@pagalaadmi said: Two cars A and B start simulatneously from points P and Q respectively towards each other. At a certain point R, the speed of car A decreases by 1/3. It then meets B at point S. Where SQ=2PR. If the speed of S had become1/3 less at the mid point of RS, the cars would have met at T where ST = PR/4. Find the ratio of RS/PR? options A)6:1 B)8:1 C)5:1 D)10:1

@Foster87 said: @MishraG Gt it thanks.......jus one thing...had the base been 121 instead of 22,the ans would be 199/2??

@chevy said: it is 4146,,, Approach batao sam bhai!!!

@peppie said: its still not clear hexagon nai ban rha

How can I make a 3D drawing here.. .. In basic terms, Rotate the cube about a diagonal and cut.

@chandra12 said: How many of first 1200 natural nos are either prime to 6 or to 15 1)3202)4803)4004)640

@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

@YouMadFellow said: You need to cut the cube in an incline. Imagine a cube ABCDPQRS (ABCD being the top face, and PQRS the bottom face, P is below A and so on). Now, imagine a line joining the midpoint of AP and CR. (It will pass through the body of the cube) Initially, the cube was perpendicular to the surface. Now, tilt the cube about this line by some degree (say 45). Then cut the cube with a perpendicular line passing through the mid-point of A-B and B-C. You will get a hexagon. Don't know why I am getting (3*root(3))/4

its still not clear hexagon nai ban rha :(

@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