[warrior]

Originally Posted by prade

A milkman had two cows. The daily collection of milk from each cow varies randomly and the combined collection of milk from them varies between 20 litres and 50 litres. On Sunday the milkman had a total milk collection of 30 litres. Harry bets Rs 100 on the possibility that the total milk collection on Monday will not differ from that on Sunday by more than 5 litres, while the milkman thinks otherwise. What is the maximum amount the milkman can bet if he can not afford to make a loss?
(a) Rs 200 (b) Rs 250 (c) Rs 300 (d) none of the foregoing

answer is option b

ans (a) 200

harry's range 25-35 =10

cow can give 20-50 =30

so chances of harry win =1/3
let x be the bet so -1/3 * x+ 2/3 * 100

give x =200

[singh_rudresh]

Originally Posted by Aarav

Friday evenings are not meant to solve QQAD :-)

Consider all points A' such that FA' = 3. They lie on a circle center F. The area of A'PQ is PQ/2 times the distance of A' from PQ. That distance is maximal for A'F perpendicular to PQ (because the distance is the distance of F from PQ + FA' sin(x), where x is the angle between A'F and PQ). Hence FP must be perpendicular to PQ. Similarly QF must be perpendicular to PR, so F must be the orthocenter.

Have a great weekend, and do well in mocks :-)
We will meet on Monday morning.

such a cute soln

[singh_rudresh]

If he cant afford to loose ... he cant bet so d) ..........
seriously , how can he bet if he cant afford to loose ....therez a non-zero probability of loss....
how does it matter how small that is.... ????

or he can bet any damn amount for 25 to 35
and same amount against 25 to 35
whatever the outcome....no profit no loss
so again d)

or i havent understood the ques at all....

[jalan]

Probability that he wins = 2/3
Expected Gain = 2/3*100

Proabbility that he loses =1/3
Let x be the maximum money he can bet
So the expected loss = 1/3*x

Now 2/3*100 - 1/3*x = 0
so x=200

Hence answer is a) 200...
wat say guys

[jalan]

oh sorry didnt see that warrior had already posted the same approach...

[raj_bhatia]

He cannot afford to lose?

even if he bets for 200 and he looses the bet then he will loose....only case where he will not loose is that his cow will give him extra milk of 200 Rs as minimum he expects it to be 20 litres a day and he can loose it if cow gave somwhere between 25 to 30 litres..which means an extra of 5 to 15 litres..so min he can bet for price of 5 litres of milk so that he will not loose anything even after losing the bet...but we dont have price of milk/litre so my answer is (d).

somewhere i still have feeling that i have missed something and it can be solved but my mind keep stating as above even after seeing above mentioned solutions...

[CATBUSTERZ]

Originally Posted by warrior

Okay junta solve this as a sat sun QQAD

If f(x) is a polynomial of degree 2005 that satisfies f(k) = 1 + 1/k for k = 1, 2, . . . , 2006, find f(2007).
(A)1/2007
(B)4028050/2007
(C)1/4022030
(D)1/4026042
(E)1/4030056

Any solutions for this?????

[nagi_nov3]

Originally Posted by warrior

ans (a) 200

harry's range 25-35 =10

cow can give 20-50 =30

so chances of harry win =1/3
let x be the bet so -1/3 * x+ 2/3 * 100

give x =200

harry range is 25 to 35 ( 11 values)
total range is 20 to 50 ( 31 values)

harry bets 100 for (11/31 probaility) for 1/31 value is 100/11
so probability that milk man can win is 20.100/11 = 181.8 rs

so i will go with d).

[TheCat]

Answer to #158

If the range is from 20 to 25 and 35 to 50 then milk man win ..if the range if from 25 to 50 harry wins..

EXcpected value of win for milkman = (6/31+16/31)*100 = 22/31*100

Excepted value of win for harry = 11/31 * p if p is the value milkman bet

Both shud be equal....so P= 200 ans option (a)

[siddhesh_j]

Quant Question # 159
------------------------------------------------------------
Question:
Quadrilateral ABCD is inscribed in a circle with diameter AD = 4. If sides AB=BC = 1 , then CD equals
(a) 3 (b) 7/2 (c) 4 (d) none of the foregoing


ans D)

ans comes to around 11.4/8 = 1.42

used theorem on any quadrilateral .... sum of squares of diagonals waala !