There is a 200 miles long tunnel. one train enters the tunnel at a speed of 200mph while the other trains enter the tunnel in the opposite direction at a speed of 1000 mph. A bee travels at a speed of 1500 mph enters the tunnel goes to and back until it reaches the train. What is the distance covered by the bee when the two train collides (the bee survives)
Thanks
Distance between the trains = 200 miles Time taken by the trains to meet = 200/(200 + 1000) = 200/1200 = 1/6 hr = 10 mins
=> Bee will travel for 10 minutes => Distance travelled = 1500/6 = 250 miles
I came across this question of probability in Arun Sharma.
There are 8 orators A,B,C,D,E,F,G and H in how many ways the arrangements be made so that A always comes before B and B always comes before C
The options are as follows :
1.) 8!/3! 2.) 8!/6! 3.) 5!.3! 4.) 8!/(5!.3!)
The book says the correct answer is option 1 :shocked:, but I think it is option 3 which is correct.
My approach is since A,B,C are always going to be in a particular order only we can consider them as 1 letter so in all we have 6 i.e. D,E,F,G, H and ABC together as 1. The number of possible arrangements for 6 letters is
6! which is equal to 5!.3! (option c)
I doubt Arun Sharma's answer sheet as many times the answers metioned in them have been found wrong.
The problem in your approach is that, you are considering ABC as 1 entity and considering only 6 positions as a result. Where as A, B, C can appear any where out of the available 8 spaces. Its just that they have to appear in a certain order. SO thats why we first select 5 places out of 8 for D, E, F, G and H and then A, B, C are arranged according to the condition in the remaining 5 spaces.
there are four machines and it is known that exactly two of them are faulty.they are tested one by one in a random order till both the faulty machines are identified.then the probability that exactly three tests will be required to identify the faulty machines will be
1)1/2 2)1 3)1/3 4)2/3
As there are 2 faulty machines so sample space will be 4c2... and determining the faulty machines in exactly 3 attempts is possible if F NF F NF F F
The problem in your approach is that, you are considering ABC as 1 entity and considering only 6 positions as a result. Where as A, B, C can appear any where out of the available 8 spaces. Its just that they have to appear in a certain order. SO thats why we first select 5 places out of 8 for D, E, F, G and H and then A, B, C are arranged according to the condition in the remaining 5 spaces.
The problem in your approach is that, you are considering ABC as 1 entity and considering only 6 positions as a result. Where as A, B, C can appear any where out of the available 8 spaces. Its just that they have to appear in a certain order. SO thats why we first select 5 places out of 8 for D, E, F, G and H and then A, B, C are arranged according to the condition in the remaining 5 spaces.
There is a 200 miles long tunnel. One train enters the tunnel at a speed of 200mph while the other trains enter the tunnel in the opposite direction at a speed of 1000 mph. A bee travels at a speed of 1500 mph enters the tunnel goes to and back until it reaches the train. What is the distance covered by the bee when the two train collides (the bee survives)
Firstly the question says "one or two may get zero objects" and not "must get"
So, if that be the case, you need to consider so many different distributions as well...like 1,1,5 ; 1,2,4 ; 1,3,3 ...and so on.
So, the best way is to check it the other way round i.e. which object goes to whom method. Clearly, each object can go to any one out of the 3 people. So, that makes it 3*3*...7 times
Firstly the question says "one or two may get zero objects" and not "must get"
So, if that be the case, you need to consider so many different distributions as well...like 1,1,5 ; 1,2,4 ; 1,3,3 ...and so on.
So, the best way is to check it the other way round i.e. which object goes to whom method. Clearly, each object can go to any one out of the 3 people. So, that makes it 3*3*...7 times
Need help with this one from Progressions LOD3: If x, y, z are in GP and ax, by, cz are equal then a, b, c are in 1. AP 2. GP 3. HP 4. None of these 5. Cannot be determined
Need help with this one from Progressions LOD3: If x, y, z are in GP and ax, by, cz are equal then a, b, c are in 1. AP 2. GP 3. HP 4. None of these 5. Cannot be determined
My take:- Since ax,by & cz are equal Therefore (by)^2 = (ax)(cz) Now y^2 = xz (they're in g.p.) Therefore b^2 = ac & a,b,c are in g.p. (by canceling y^2 & xz from both sides)
Hey guys, I am so sorry...the post was wrong...The actual problem is If x, y, z are in GP and a^x, b^y, c^z are equal then a, b, c are in 1. AP 2. GP 3. HP 4. None of these 5. Cannot be determined
The answer is 3 - HP...I want the procedure.....Apologies once again....
Ok being very blunt, i am kinda in a fix with number theory..... especially the ones following remainder theorem :
Doubt 1> best way to find answers to problems of type:"Find the remainder for: (83^261)/17
doubt 2>how exactly does the table to find 'units digits' in terms of power work???
M pretty sure i'l be hitting this thread a hulluva lot of times....quant is currently not my strongest side.....:p so please excuseth the amateurish doubts.....
This is a post with a different example , In that post N is the divisor , in your case it's 17.
Doubt - 2 It's by virtue of cyclic nature of powers. If a^b ends with x , i.e, x is the unit digit and x*x also give the unit digit as x then we can say that (a^b)^n will always end with x. e.g, x*x=...x now x can be 1,5,6,0 as 1*1=1 , 5*5=25 , 6*6=36,0*0=0 (Say they are the recurrent digits as they will recur for every power ) Now for any power of abc (abc^n) to find the unit digit we should only consider the unit digit c^n So only 10 digits need to be considered (0,1,.....9) 0->0*0=0 So a0^n will always end with 0. 1->1*1=1 so a1^n will always end with 0 2->2*2=4,2*2*2=8,2*2*2*2=6 (it's 16 but we can only consider the unit digit ) Now 6*6 =36 (6 repeats ) So for 4 2s i.e, 2^4 we get those recurrent digits . So for any digit(n) among 0...9 there is number (x) for which n^x will end with one of those recurrent digits . For ,n=3 ->x=4 n=4->x=4 n=5->x=1 n=6->x=1 n=7,x->4 ........
Now for unit digit ->
Example for a number n=7 7^4n will always end with a recurrent number which is 7*7*7*7=1 7^(4n+1) will end with 1*7=7 7^(4n+2) will end with 7*7=9 7^(4n+3) will end with 9*7=3 7^(4n+4) will end with 3*7=1
You can do that for n=2,3,8,9 as well.
Thanks a lot subha(if i may call you that).....but it will be really helpful if you could please: 1>work out 83^261/17 using the numbers given 2>solve for unit digit:((173^45)*(152^77)*(777^999))
Then i can really relate to it and i am pretty sure i can take it smoothly from there....Also, i am a little confused about the 4n,(4n+1)....part for the unit digit thing...i understood 'x' which when used as power for a number returns the recurrent digits in units place... but m still confused about 4n,(4n+1)....can you please elucidate
Thanks a lot subha(if i may call you that).....but it will be really helpful if you could please: 1>work out 83^261/17 using the numbers given 2>solve for unit digit:((173^45)*(152^77)*(777^999))
Then i can really relate to it and i am pretty sure i can take it smoothly from there....Also, i am a little confused about the 4n,(4n+1)....part for the unit digit thing...i understood 'x' which when used as power for a number returns the recurrent digits in units place... but m still confused about 4n,(4n+1)....can you please elucidate
7^1 has a unit digit of 7. 7^2 has a unit digit of 9. (7^2 = 49) 7^3 has a unit digit of 3 (7^3 = 343) Actually you dont need to find the exact value of the power of 7 when you are to deal with the unit digit only. Just multiply 7 with the previous unit digit that you have obtained and you will get the unit digit of the next power of 7. So going by this process from now on, 7^4 has a unit digit of 1.(Taking the unit digit of the previous result ie. 3 and multiplying it by 7 which is 21)
Now 7^5 will give a unit digit of 7. 7^6 --> 9 7^7 --> 3 and it continues..... ........ So you can see the pattern followed wrt to the unit digits- 7,9,3,1,7,9,3,1......
Hence the result mentioned by Subhakimi : Example for a number n=7 7^4n will always end with a recurrent number which is 7*7*7*7=1 7^(4n+1) will end with 1*7=7 (7 having the power as 1,5,9.... will have 7 as its unit digit) 7^(4n+2) will end with 7*7=9 (7 having the power as 2,6,10.... will have 9 as its unit digit) 7^(4n+3) will end with 9*7=3 (7 having the power as 3,7,11.... will have 3 as its unit digit) 7^(4n+4) will end with 3*7=1 (7 having the power as 4,8,12.... will have 1 as its unit digit)
7^1 has a unit digit of 7. 7^2 has a unit digit of 9. (7^2 = 49) 7^3 has a unit digit of 3 (7^3 = 343) Actually you dont need to find the exact value of the power of 7 when you are to deal with the unit digit only. Just multiply 7 with the previous unit digit that you have obtained and you will get the unit digit of the next power of 7. So going by this process from now on, 7^4 has a unit digit of 1.(Taking the unit digit of the previous result ie. 3 and multiplying it by 7 which is 21)
Now 7^5 will give a unit digit of 7. 7^6 --> 9 7^7 --> 3 and it continues..... ........ So you can see the pattern followed wrt to the unit digits- 7,9,3,1,7,9,3,1......
Hence the result mentioned by Subhakimi : Example for a number n=7 7^4n will always end with a recurrent number which is 7*7*7*7=1 7^(4n+1) will end with 1*7=7 (7 having the power as 1,5,9.... will have 7 as its unit digit) 7^(4n+2) will end with 7*7=9 (7 having the power as 2,6,10.... will have 9 as its unit digit) 7^(4n+3) will end with 9*7=3 (7 having the power as 3,7,11.... will have 3 as its unit digit) 7^(4n+4) will end with 3*7=1 (7 having the power as 4,8,12.... will have 1 as its unit digit)
Hope I havent jumbled it up :D
Nota at all warrior....i got the cyclic thing....what i wanna know is how does the power take the form of : '4n'?as you said right,lets take n=7....maybe it'l be great if you can take the example of say,3 and show so that i can see a pattern. Also how do i apply this to stuff like>> solve for unit digit:((173^45)*(152^77)*(777^999))
Hey one more problem: find remainder for :(54^124)/17
