You want to find someone whose birthday matches yours. What is the least number of strangers whose birthdays you need to ask about to have a 50-50 chance?
@Cat.Aspirant123 said:The average age of Ram and Rita at the time of their marriage was 25 years. A son was born to them two years after their marriage. The present average age of all the three of them is 24 years. How many years is it since the couple got married?57810
Can do very simply without writing equations. Without writing anything in fact.
Starting average age 25.
After 2 years 27, but kid is added => average become 2/3rd of 27 i.e. 18 at that point.
New average of all 3 = 24 => 6 more years have passed.
So total 8 years.
Edit : And I see that ATDH has already given an elegant solution sans paper :)
regards
scrabbler
@ScareCrow28 said:You want to find someone whose birthday matches yours. What is the least number of strangers whose birthdays you need to ask about to have a 50-50 chance?
p(stranger not having same b'day) = 364/365
say we need n strangers
1-(364/365)^n=1/2
n~253
say we need n strangers
1-(364/365)^n=1/2
n~253
Find the remainder when 51^203 is divided by 7?
a) 4 b) 2 c) 1 d) 6
Pls. Note:- Providing steps of the solution will be helpful
a) 4 b) 2 c) 1 d) 6
Pls. Note:- Providing steps of the solution will be helpful
@anilapex said:Find the remainder when 51^203 is divided by 7?a) 4 b) 2 c) 1 d) 6Pls. Note:- Providing steps of the solution will be helpful
51^203 mod 7 = (49+2)^203 mod 7 = 2^203 mod 7 = 8^ 67 mod 7 * 4 mod 7 = 4 mod 7
Hence ans is 4 ..
@anilapex said:Find the remainder when 51^203 is divided by 7?a) 4 b) 2 c) 1 d) 6Pls. Note:- Providing steps of the solution will be helpful
a) 4
E(7) = 6
203 mod 6 = 5
51^5 mod 7 = 2^5 mod 7 = 32 mod 7 = 4
E(7) = 6
203 mod 6 = 5
51^5 mod 7 = 2^5 mod 7 = 32 mod 7 = 4
@anilapex said:Find the remainder when 51^203 is divided by 7?a) 4 b) 2 c) 1 d) 6Pls. Note:- Providing steps of the solution will be helpful
51^203mod7
=2^203mod7
=(2^6)^33*2^5mod7
=64^33*32mod7
=4
=2^203mod7
=(2^6)^33*2^5mod7
=64^33*32mod7
=4
@anilapex said:Find the remainder when 51^203 is divided by 7?a) 4 b) 2 c) 1 d) 6Pls. Note:- Providing steps of the solution will be helpful
4?
@anilapex said:Find the remainder when 51^203 is divided by 7?a) 4 b) 2 c) 1 d) 6Pls. Note:- Providing steps of the solution will be helpful
4 .
@anilapex said:Find the remainder when 51^203 is divided by 7?a) 4 b) 2 c) 1 d) 6Pls. Note:- Providing steps of the solution will be helpful
7 is a prime number.. so euler number for 7 = 6
so (51^203/7)R = (51^5/7)R = (2^5/7)R = (32/7)R =4
@soham2208 said:Weights are 3^0, 3^1, 3^2 Every weight upto 13gm (1 + 3 + 9) -> 13 ?
how about this one ?
13
every weight has three options 2 for either side of pan and 1 when it had not been chosen
=> 3*3*3 - 1(when there is no weight in any of pan) = 26
but here both pans are identical thats why we need divide 26 by 2
so , required answer = 26/2 = 13
every weight has three options 2 for either side of pan and 1 when it had not been chosen
=> 3*3*3 - 1(when there is no weight in any of pan) = 26
but here both pans are identical thats why we need divide 26 by 2
so , required answer = 26/2 = 13
next term ?
3, 8, 13, 24, 41, ...
@krum said:next term ?3, 8, 13, 24, 41, ...
3, 8, (8+3+2), (13+8+3), (13+24+4), (24+41+5)
Hence, 70?
@krum said:next term ?3, 8, 13, 24, 41, ...
70?
(3 + 8) + 2 = 13
(8 + 13) + 3 = 24
(13 + 24) + 4 = 41
(24 + 41) + 5 = 70
(3 + 8) + 2 = 13
(8 + 13) + 3 = 24
(13 + 24) + 4 = 41
(24 + 41) + 5 = 70
@jain4444 said:how about this one ? 13every weight has three options 2 for either side of pan and 1 when it had not been chosen=> 3*3*3 - 1(when there is no weight in any of pan) = 26 but here both pans are identical thats why we need divide 26 by 2so , required answer = 26/2 = 13
I did it this way:
Case 1: I use only one weight => 3 ways (1,3,9)
Case 2: I use two weights => Select the weights in 3C2 ways and then it can be sum or difference => 6 ways
Case 3: I use all three weights, but group them differently => (1,2) and (3,0) weights on either side => (1,2) -> 3 ways + (3,0) -> 1 way = 4 ways
Total = 13 ways
Your approach is nice :thumbsup:
(AA)^B = ABBA
find A+B