@bs0409 said:Ten tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. prob that largest no. appearing on the selected ticket is 7 is ??
7^5/ 10^6 ??
@Budokai001 said:The tickets should not have 8,9,10Total =10^6Favourable cases= (All cases less thanor equal to 7) - (cases less than 7 when 7 is not selected)7^6-6^6 ( 1,2,3,4,5,6,7 ->7 cases1,2,3,4,5,6 -> 6 cases)So (7^6 - 6^6)/10^6 ?
@ravi.theja said:7^5/ 10^6 ??
bros, which of them is the correct answer and what is the correct method of approaching this question? I often get confused between the two and end up leaving the question.
@Budokai001 said:The probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it .The least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is????/(a)7 bombs (b) 3 bombs(c) 8 bombs (d) 6 bombs (e) 9 bombsQ2) what is the probability of the destruction of the bridge if only 5 bombs are dropped ?(a) 62.32(b) 81.25(c) 45.23(d) 25.23(e) 31.32Do post approach as well . Am not getting the answer
1) Let there be 'n' bombs dropped.
Then prob of destruction is:
nC2+nC3+..........nCn/2^n>0.9
(2^n-n-1)/2^n>0.9
n=7
2)0.8125 is the prob
@nikemen said:bros, which of them is the correct answer and what is the correct method of approaching this question? I often get confused between the two and end up leaving the question.
Correct answer is (7^6-6^6)/10^6.....
If F(x) = x^4 -360x^2 +400, x is Integer, if F(x) is a prime number, then what is the sum of all possible F(x)??
@Budokai001 said:The probability of a bomb hitting a bridge is 1/2 and two direct hits are needed to destroy it .The least number of bombs required so that the probability of the bridge being destroyed is greater than 0.9 is????/(a)7 bombs (b) 3 bombs(c) 8 bombs (d) 6 bombs (e) 9 bombs
prob of hitting >= 0.9 ===> prob of not hitting
nc0+nc1 (1/2 )^n n= 7
@Budokai001 said:Q2) what is the probability of the destruction of the bridge if only 5 bombs are dropped ?(a) 62.32(b) 81.25(c) 45.23(d) 25.23(e) 31.32Do post approach as well . Am not getting the answer
5c2+5c3 +....5c5 / 2^5 =26/32 = .8125
@bs0409 said:Correct answer is (7^6-6^6)/10^6.....If F(x) = x^4 -360x^2 +400, x is Integer, if F(x) is a prime number, then what is the sum of all possible F(x)??
0. The integers satisfying the condition are 1 and -1. answer in both cases is 41. even if there are other numbers that satisfy this equation they'll cancel out to give the answer 0.
@ravi.theja said:F(x) will always be even since F(x) = x^2 ( x^2 -360) + 400 ..that is Even + even = even ==> sum = 2??
even kaise hoga ? :O
@Zedai said:
Here you go.........
SUPERB PROBLEM!!(I have explained in detail hence the solution looks long)
I have tried to put below the normal multiplication process (the way we multiply two 3 digit numbers)
..........................A.................. B.................A
..........................2...................3.................1
_______________________________________
..........................A.................. B.................A
...........3A..........3B..................3A................x
.2A......2B..........2A..................x..................x
_______________________________________
2A....(3A+2B).....3(A+B)...........(B+3A)..........A
________________________________________
now we compare this with the result given i.e BA4AA
(Clearly A and B are not zeros. as they are first digits of the numbers involved in the question)
I have tried to put below the normal multiplication process (the way we multiply two 3 digit numbers)
..........................A.................. B.................A
..........................2...................3.................1
_______________________________________
..........................A.................. B.................A
...........3A..........3B..................3A................x
.2A......2B..........2A..................x..................x
_______________________________________
2A....(3A+2B).....3(A+B)...........(B+3A)..........A
________________________________________
now we compare this with the result given i.e BA4AA
(Clearly A and B are not zeros. as they are first digits of the numbers involved in the question)
Step1
now the second last digit comes from (B+3A) which is A as given in this question.
also we observe that (B+3A) = (B+2A)+A. This means that (B+2A) must be equal to the base of the number system and give a carry to the third place.
fox example in decimal system if (3*A+B) gives ending digits as A then (2A+B) must be equal to 10. as we can see in decimal system (3*3+4)= (2*3+4)+3 =10+3
so we can write (B+2A) = b (base of the system)................(i)
now the second last digit comes from (B+3A) which is A as given in this question.
also we observe that (B+3A) = (B+2A)+A. This means that (B+2A) must be equal to the base of the number system and give a carry to the third place.
fox example in decimal system if (3*A+B) gives ending digits as A then (2A+B) must be equal to 10. as we can see in decimal system (3*3+4)= (2*3+4)+3 =10+3
so we can write (B+2A) = b (base of the system)................(i)
Step2
Now we look at the third last digit.
From step 1 we know there is a carry over that comes after addition of second last digits.
so 3(A+B)+1 ends in 4
=> 3(A+B) ends in 3
=> (2A+B) +(A+2B) ends in 3
=> b+(A+2B) ends in 3 (replacing 2A+B =b)
=> (A+2B) ends in 3
(we remove the base because base+digit always ends in the digit... example 10+2=12)
now if (A+2B) does not give any carry then A=1 and B=1 is the only possible solution, which is not possible. Hence we are forced to conclude that A+2B must give some carry
=>A+2B=b+3......................................(ii)
Now we look at the third last digit.
From step 1 we know there is a carry over that comes after addition of second last digits.
so 3(A+B)+1 ends in 4
=> 3(A+B) ends in 3
=> (2A+B) +(A+2B) ends in 3
=> b+(A+2B) ends in 3 (replacing 2A+B =b)
=> (A+2B) ends in 3
(we remove the base because base+digit always ends in the digit... example 10+2=12)
now if (A+2B) does not give any carry then A=1 and B=1 is the only possible solution, which is not possible. Hence we are forced to conclude that A+2B must give some carry
=>A+2B=b+3......................................(ii)
Step3
now add for (i) and (ii)
to get 3(A+B) =2b+3
=>As LHS is multiple of 3 then b has to be multiple of 3 as well. So possible values of b=3,6,9
we reject b=3 because we have digit 4 in the final answer.
if we put b=6 then we get B=4 and A=1.
now add for (i) and (ii)
to get 3(A+B) =2b+3
=>As LHS is multiple of 3 then b has to be multiple of 3 as well. So possible values of b=3,6,9
we reject b=3 because we have digit 4 in the final answer.
if we put b=6 then we get B=4 and A=1.
ATDH.
@desiboy2389
@desiboy2389 said:@getupsid there is a short cut for no of zeros of any factorial of a number say( n!) all you need to do isdivide the no with 5 , 5^2,5^3..untill the value after division is less than 1(considering the integer part only)...36!...>36/5+36/5^2 (considering only integer part) = 7 + 1 =8 ..since the next value isthis short will help me in other problems also...
@bs0409 said:Ten tickets are numbered 1,2,3.........10. six tickets are selected at random one at a time with replacement. prob that largest no. appearing on the selected ticket is 7 is ??
6/10c6 ?
How many triangular no's less then 1000 have the property that they are the difference of squares of two consecutive natural no's ?
no OA ! plz share approach
no OA ! plz share approach
ravi...is it 751 ?
@ravi.theja said:How many triangular no's less then 1000 have the property that they are the difference of squares of two consecutive natural no's ?no OA ! plz share approach
499 hoga kya ?>
@gs4890 @joyjitpal dnt have OA bhai..pls share ur approach 😃 if its natural numbers then all odd numbers..but its mentioned triangular numbers..i.e., series lik 1,2,3,6,10,15,21.....
@gs4890 @joyjitpal i hav dis solution :- N = (n+1)^2- n^2 = 2n+1 ..so it has to be a odd number
Number which can be representdd as sum of consecutive naturl no.
or nC2 numbr - TRIANGULAR number...
now nc2 n can hav max of 43 ==> 21 odd values of n
ans : 21
Number which can be representdd as sum of consecutive naturl no.
or nC2 numbr - TRIANGULAR number...
now nc2 n can hav max of 43 ==> 21 odd values of n
ans : 21
@ravi.theja said:How many triangular no's less then 1000 have the property that they are the difference of squares of two consecutive natural no's ?
no OA ! plz share approach
Triangular numbers are numbers which can be expressed as the sum of first n natural numbers.
So we need Fn=(k+1)^2-k^2
or, n(n+1)/2=2k+1
So we need odd triangular numbers.
so n must be of the form 4a+1 or 4a+2, i.e. 21 numbers for Fn
@ravi.theja said:How many triangular no's less then 1000 have the property that they are the difference of squares of two consecutive natural no's ?
no OA ! plz share approach
21??
actually this problem can be solved by observation...
we know that all odd numbers can be expressed as a diff of sq of cons. numbers...
now see, triangular numbers are -> 1,3,6,10,15,21,28,36,45...990
this follows a pattern of 2 odd 2 even...
so total = 44/2 - 1 = 21 such numbers..
@ravi.theja said:@gs4890 @joyjitpal i hav dis solution :- N = (n+1)^2- n^2 = 2n+1 ..so it has to be a odd number
Number which can be representdd as sum of consecutive naturl no.
or nC2 numbr - TRIANGULAR number...
now nc2 n can hav max of 43 ==> 21 odd values of n
ans : 21
i guess u hv made a small mistake....
largest n can be 44 (44*45/2 = 990)
44/2 = 22
but we need to minus 1 because 1^2-0^2 is not a valid case here (as diff of sq of natural numbers is being asked for)..