difficulty: moderateq4: The professor plays a game of "pass the baton" with his class of N students. He randomly selects one student, say A, and that student again randomly selects another student (say B)and passes him the baton, then B selects another student and passes him the baton and the game ends. What is the probability that the first student who got the baton from the professor ends up with the baton in the end.1. 1/(n-1)2. 1/2(n-1)3. 1/n4. none of these
Ratio of incomes of a and b is 3:4,ratio of expenditure is 4:5.The ratio of savings can be. 1)4:5 2)6:7 3)8:13 4)7:9
I did it like this ratio of savings =(3x-4y)/(4x-5y). then i equated the ratio with each of the options to get the answer. the correct answer is 8/13. But is there any shorter approach?
q2) If a-b varies directly with a+b,then a^2 - b^2 will vary directly with 1) a^2 + b^2 2)ab 3)a^2 + b^2 + 3ab 4) more than one of the above
I did it like this:
a-b=k(a+b) multiplying both sides by a+b a^2-b^2=k(a+b)^2
What to do next to reach the answer ? Answer is 4) more than one of the above.
-999 to 1000 ? :/@jain4444Bhai woh 3*3 wale ka OA ?
its 49/512
solution copied
label the squares 1-9. now, after rotation, the new squares are 7,4,1,8,5,2,9,6,3. i.e. square 7 occupies square 1's old place, 4 occupies 2's old place...and so on.
now count up the favourable cases for each possibility(remember that square 5(centre square) retains its original position after rotation. Thus, it must be black. So we consider only the 8 peripheral squares in our favourable cases):
all black squares -> 1 possibility
1 white, rest black -> 8 possibilities (as long as the centre square(5) is not white, any other square being white will become black after rotation )
2 white, rest black -> 8C2 - 8 = 20 possibilities (of the 8 peripheral squares, we can choose 2 squares in 8C2 ways. but if you choose 7 to be white, then 1 can't be white. similarly, if 4 is chosen white then 2 shouldn't be white....8 such pairs which won't result in a favorable selection. thus 8C2 - 8 )
similarly, studying the case of 3 white, rest black, we get -> 16 possibilities
4 white, rest black-> 4 possibilities
>4 white -> 0 possibilities, obviously
That's a total of 1+8+20+16+4 = 49 favourable out of a total of 2^9=512 possibilities
Let S be a set of positive integers, not necessarily distinct, in which the number 68 appears. The arithmetic mean of the numbers in S is 56. However, if 68 is removed, the arithmetic mean of the remaining numbers is 55. Determine the largest number that can appear in S.
Let S be a set of positive integers, not necessarily distinct, in which the number 68 appears. The arithmetic mean of the numbers in S is 56. However, if 68 is removed, the arithmetic mean of the remaining numbers is 55. Determine the largest number that can appear in S.
Let S be a set of positive integers, not necessarily distinct, in which the number 68 appears. The arithmetic mean of the numbers in S is 56. However, if 68 is removed, the arithmetic mean of the remaining numbers is 55. Determine the largest number that can appear in S.
Let numbers in set be N 56N-68=55(N-1) N=13 for largest number rest 11 should be minimum, positive int so 1 1,1,1...11times, ,68, L will be numbers 11+L=55*12 L=649?
Let S be a set of positive integers, not necessarily distinct, in which the number 68 appears. The arithmetic mean of the numbers in S is 56. However, if 68 is removed, the arithmetic mean of the remaining numbers is 55. Determine the largest number that can appear in S.
Let S be a set of positive integers, not necessarily distinct, in which the number 68 appears. The arithmetic mean of the numbers in S is 56. However, if 68 is removed, the arithmetic mean of the remaining numbers is 55. Determine the largest number that can appear in S.
649
56n - 68 = 55n - 55 n = 13 So total = 55(n-1) = 55*12 = 660 660 = L + 11S Smallest value that S can take is 1. So L = 660 - 11 = 649
Q1.A= (-3,0) B= (3,0) and L(OA) 0+ L(OB) =10 What is the probability that O lies inside the circle x^2+y^2= 25? I solved this using equations and got the answer, but that was tedious and time consuming...please suggest a better more logical approach!
Q2: There are 5 envelopes, 5 letter boxes and 5 cities.Each letter has only one correct letter box corresponding to it, and each of the letter boxes correspond to only one city. The letters are first posted in the letterboxes and then the letterboxes deliver mail to the cities. What is the probability of all the letters, as well as letterboxes reaching the wrong letterbox and city respectively?
Q1.A= (-3,0) B= (3,0) and L(OA) 0+ L(OB) =10 What is the probability that O lies inside the circle x^2+y^2= 25?I solved this using equations and got the answer, but that was tedious and time consuming...please suggest a better more logical approach
Well, L(OA) + L(OB) = 10 suggests that it is an ellipse, with the focal points at (-3,0) and (3,0). It intersects the x axis at (5,0) and (-5,0) and y axis at (0,4) and (0,-4).
Is the answer 1 ? I drew the ellipse, and noticed that it is contained inside the circle (x^2 + y^2 = 25 ) ..
