A cistern has a capacity of 40,000 L. Initially, it is half full. Everyday, a certain quantity of water is used but at the end of the day, 4/3 of the used quantity is replaced into the cistern. If the amount of water used is the same everyday, then what is the total quantity of water used for refilling till the cistern is full?
a.60,000 L
b.20,000 L
c.80,000 L
d.40,000 L
e.50,000 L
A fair coin is tossed 10 times. find the prob two heads do not occur consecutively?
@heylady said:For how many integer values of 'x' is x^2-3x-19 divisible by 289?
x^2 - 3x - 19 = 289k (let)
so discriminant should be a perfect square
discriminant = 9 -4(-19 -289k) = 9 + 76 + 289*4k = 17( 5 + 68k) the term inside the bracket would never be divisible by 17, hence discriminant is not a perfect square, hence no integer value possible
@heylady
For how many integer values of 'x' is x^2-3x-19 divisible by 289?
x^2 - 3x - 19
= (x + 7)(x - 10) + 51
51 is a multiple of 17, so for the expression to be a multiple of 17 one of (x + 7) and (x - 10) has to be a multiple of 17
But if one of (x + 7) and (x - 10) is a multiple of 17, then other one is also a multiple of 17
=> If the expression is a multiple of 17, then it will be of form 289k + 51 and hence never a multiple of 289
So, no such integral x exists
@shattereddream said:A cistern has a capacity of 40,000 L. Initially, it is half full. Everyday, a certain quantity of water is used but at the end of the day, 4/3 of the used quantity is replaced into the cistern. If the amount of water used is the same everyday, then what is the total quantity of water used for refilling till the cistern is full?a.60,000 L b.20,000 L c.80,000 L d.40,000 L e.50,000 L
80 right? Extra added = 20 = 1/4th...
regards
scrabbler
regards
scrabbler
@Buck.up said:A fair coin is tossed 10 times. find the prob two heads do not occur consecutively?
144/1024 = 0.140625
@Buck.up said:A fair coin is tossed 10 times. find the prob two heads do not occur consecutively?
144/1024?
edited for addition error :splat:
regards
scrabbler
edited for addition error :splat:
regards
scrabbler
@Buck.up said:A fair coin is tossed 10 times. find the prob two heads do not occur consecutively?
total no. of outcomes = 2^10
we have to place heads in the spaces b/w tails
for 10 tails we have 11C0 space
so for 9 tails , we have 10C1 spaces
for 8 tails we have 9C2 spaces
till 5 tails we have 6C5 spaces
so no. of cases = 1 + 10 + 36 + 56 + 35 + 6 = 144
so 144/1024 = 9/64
@shattereddream said:A cistern has a capacity of 40,000 L. Initially, it is half full. Everyday, a certain quantity of water is used but at the end of the day, 4/3 of the used quantity is replaced into the cistern. If the amount of water used is the same everyday, then what is the total quantity of water used for refilling till the cistern is full?a.60,000 L b.20,000 L c.80,000 L d.40,000 L e.50,000 L
strt=20000L
usd/day=xL......filled=4/3xL...fr n days
increase =1/3xL...totl increase= 1/3xnL=20000L
=>nxL=60000L.....totl increase
used to fill=4/3xnL=4/3*60000=80000L
HAPPY HOLI TO ALL PUYZ...
:drinking::drinking::drinking:drinking:
@Buck.up said:A fair coin is tossed 10 times. find the prob two heads do not occur consecutively?
You can use recursion also.
Suppose a(n) is the number of such cases when a coin is tossed 'n' times
-> If in first toss tail appears, then for remaining (n - 1) tosses we will have a(n - 1) ways
-> If in first toss head appears, then on next toss there should be a tail. For remaining (n - 2) tosses we will have a(n - 2) ways
=> a(n) = a(n - 1) + a(n - 2)
a(1) = 2 and a(2) = 3
Now, use the formula to get a(10) which comes out to be 144
@Buck.up said:A fair coin is tossed 10 times. find the prob two heads do not occur consecutively?
[1+10+36+56+35+6]/2^10=[52+92]/1024=144/1024..
The solution is option 1:
for every x in the denominator, the numerator doubles so the initial divisor factor is 100000/1040.... for the given series it is, 99999............999996 with a common remainder of 160
for every x in the denominator, the numerator doubles so the initial divisor factor is 100000/1040.... for the given series it is, 99999............999996 with a common remainder of 160
Q 19 n 20
@Buck.up said:A fair coin is tossed 10 times. find the prob two heads do not occur consecutively?
2haeds not together=1+10+36+56+35+6=144
total cases = 2^10=1024
144/1024