1) Tina
2) Mina
3) Gina
4) Lina
5) Bina
Tina, Mina, Gina, Lina and Bina are 5 sisters, aged in that order, with Tina being the eldest. Each of them had to carry a bucket of water from a well to their house. Their buckets €Ÿ capacities were proportional to their ages. While returning, equal amount of water got splashed out of their buckets. Who lost maximum amount of water as a percentage of the bucket capacity?
@ScareCrow28 said:Tina, Mina, Gina, Lina and Bina are 5 sisters, agedin that order, with Tina being the eldest. Each ofthem had to carry a bucket of water from a wellto their house. Their buckets €Ÿ capacities wereproportional to their ages. While returning, equalamount of water got splashed out of theirbuckets. Who lost maximum amount of water as apercentage of the bucket capacity?1) Tina2) Mina3) Gina4) Lina5) Bina
Bina..??
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 probability of the bridge being destroyed is greater than 0.9 is.
a)7 bombs
b)3 bombs
c)8bombs
d)6 bombs
e)9 bombs
a)7 bombs
b)3 bombs
c)8bombs
d)6 bombs
e)9 bombs
@ScareCrow28 said:Tina, Mina, Gina, Lina and Bina are 5 sisters, agedin that order, with Tina being the eldest. Each ofthem had to carry a bucket of water from a wellto their house. Their buckets €Ÿ capacities wereproportional to their ages. While returning, equalamount of water got splashed out of theirbuckets. Who lost maximum amount of water as apercentage of the bucket capacity?1) Tina2) Mina3) Gina4) Lina5) Bina
bina.
@ScareCrow28 said:Tina, Mina, Gina, Lina and Bina are 5 sisters, agedin that order, with Tina being the eldest. Each ofthem had to carry a bucket of water from a wellto their house. Their buckets €Ÿ capacities wereproportional to their ages. While returning, equalamount of water got splashed out of theirbuckets. Who lost maximum amount of water as apercentage of the bucket capacity?1) Tina2) Mina3) Gina4) Lina5) Bina
Let ages be 5,4,3,2,1[T,M,G,L,B]
Bucket capacities be 5,4,3,2,1
Let they loose 1 unit water
1/5 Bina
Bucket capacities be 5,4,3,2,1
Let they loose 1 unit water
1/5 Bina
@nole 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 probability of the bridge being destroyed is greater than 0.9 is.a)7 bombsb)3 bombsc)8bombsd)6 bombse)9 bombs
7 hoga....P(at least 2 hits in n bombs) is 1 - (nC0+nC1) /2^n = 1 - (1+n)/2^n so we need (1+n) to be less than 1/10 of 2^n which happens for 7.
regards
scrabbler
F(x) is a fourth order polynomial with integer coeff and no common factor.The roots of F(x) are -2,-1,1,2 .If p is a prime no. greater than 97 ,then the largest integer that divides f(p) for all values of p ???
72,120,240,360 plzz explain the approach
@Ashmukh said:F(x) is a fourth order polynomial with integer coeff and no common factor.The roots of F(x) are -2,-1,1,2 .If p is a prime no. greater than 97 ,then the largest integer that divides f(p) for all values of p ???72,120,240,360 plzz explain the approach
(p - 2)(p - 1)(p + 1)(p + 2)
(p - 1) and (p + 1) both are consecutive even (so one of them has to be a multiple of 4)
So, the expression will always be a product of 8
(p - 2)(p - 1)p(p + 1)(p + 2) is divisible by 5, but p is prime greater than 97, so (p - 2)(p - 1)(p + 1)(p + 2) will also be divisible by 5
Since p is prime, then it will of form 6k + 1 or 6k - 1.
If it is of form 6k + 1, then (p - 1) and (p + 2) both are multiples of 3
If it is of form 6k - 1, then (p - 2) and (p + 1) both are multiples of 3
So, the expression is multiple of 9
Hence, f(p) will always be divisible by 8*5*9 = 360
@nole said:@scrabbler can u derive the formula pls ?
This is the logic I used:
Binomial probability : Probability of exactly r successes in n trials (each with equal probability of success p) is nCr p^r (1-p)^(n-r). (Fairly standard formula and easy to derive logically if comfortable with PnC and Prob.)
So for example if we toss a die 5 times, probability of exactly 2 fours coming up is 5C2 (1/6)^2 (5/6)^3
Now here p = (1-p) = 1/2 so the formula simplifies to nCr /2^n (check it for yourself)
For the bridge to not be destroyed we want 0 hits or 1 hit - giving us nC0/2^n + nC1/2^n. So for the bridge to be destroyed, the remaining cases will occur i.e. probability will be 1 - (the two cases above) = 1 - [(nC0 +nC1)/2^n]. In case of 7 this will be 1 - 8/128 (>0.9) while for 6 it will be 1 - 7/64 (
What's the OA btw? :)
regards
scrabbler
For how many integer values of n, n/(1450 - n) is a perfect square?
@chillfactor said:For how many integer values of n, n/(1450 - n) is a perfect square?
6?
k^2 = n/(1450-n)
n = 1450 - 1450/(k^2+1)
n = 1450 - 2* 5^2 * 29/(k^2+1)
Values of k satisfying are : k= 1, 2, 3, 7 , 12 and 17
Hence 6 values ?? Sirjee __/\__
@chillfactor said:For how many integer values of n, n/(1450 - n) is a perfect square?
1??
let n/(1450-n)=k^2
so by rearrging we get :
n=1450*k^2/(k^2+1)
1450=2*25*29
so only for k=+1 or -1 we get a soln
@chillfactor said:For how many integer values of n, n/(1450 - n) is a perfect square?
Ummm, 5?
Edit: 6 values. Missed out 17 waala case
regards
scrabbler
Edit: 6 values. Missed out 17 waala case
regards
scrabbler
How many different numbers which are smaller than 2.10^8 can be formed by using digits 1 and 2 only.?
I am getting 256 because in the first place 1 is fixed then for the rest 8 places 2^8 ( we can have both 1 and 2 for each place) .
But in arun sharma the answer is 766.Where did i go wrong ?
@nole said:How many different numbers which are smaller than 2.10^8 can be formed by using digits 1 and 2 only.?I am getting 256 because in the first place 1 is fixed then for the rest 8 places 2^8 ( we can have both 1 and 2 for each place) .But in arun sharma the answer is 766.Where did i go wrong ?
It will be 2^8 + 2^8 + 2^7 + 2^6 + 2^5 + 2^4 + 2^3 + 2^2 + 2^1 = 766
9 digit numbers + 8 digit + 7 digit + 6 digit + 5 digit + 4 digit + ...+1 digit
@nole said:How many different numbers which are smaller than 2.10^8 can be formed by using digits 1 and 2 only.?I am getting 256 because in the first place 1 is fixed then for the rest 8 places 2^8 ( we can have both 1 and 2 for each place) .But in arun sharma the answer is 766.Where did i go wrong ?
you are considering only 9 digit numbers (what about 1, 2, 3, .., 8 digit numbers)
2 + 4 + 8 + .. + 256 + 256 = 510 + 256 = 766 such numbers
@chillfactor said:For how many integer values of n, n/(1450 - n) is a perfect square?
k=1 , n=725
k=2 , n=1160
k=3 , n=1305
k=7, n=1421
k=12, n=1440
k=17, n=1445
Ans. six values of n
Re-edited
k=2 , n=1160
k=3 , n=1305
k=7, n=1421
k=12, n=1440
k=17, n=1445
Ans. six values of n
Re-edited