Quant Boosters  Swetabh Kumar  Set 2

Three biased coins A, B and C are such that the probability of getting a head when A is tossed is 1/4, the probability of getting a head when B is tossed is 3/4 and the probability of getting a head when C is tossed is 2/3. The three coins are tossed. If it is found that there are two heads and one tail, what is the probability that coin A shows a head?
(1/4 * 3/4 * 1/3 + 1/4 * 1/4 * 2/3) = 5/48
(3/4 * 2/3 * 3/4 + 3/4 * 1/4 * 1/3 + 1/4 * 2/3 * 1/4)
= 23/48 so 5/23In how many ways can seven balls of different colours be put into 4 identical boxes such that none of the box is empty?
(4,1,1,1): 7C4 * 3 * 2 * 4 = 840
(3,1,1,2): 7C3 * 4C2 * 2C1 * 4!/2! = 5040
(2,2,2,1): 7C2 * 5C2 * 3C2 * 4 = 2520
so(840+5040+2520)/4! = 350A team is planning to participate in a Dahihandi competition. On average, they succeed in breaking the handi on 2 out of every 11 attempts. How many attempts will they have to make to have at least a 50% chance of succeeding in breaking the handi?
2/11 + 9/11 * 2/11 + 9/11 * 9/11 * 2/11..n terms.
r=9/11
{ 1(9/11)^n} > = 0.5
(9/11)^n < = 0.5
so n=4In a 1000 m race, when Sumit is given a head start of 20 seconds and 200 metres, he runs for 2 minutes and beats Amit by 400 metres. Who wins the race and by how much, if Amit gets a head start of 200 metres?
a) Sumit, 91 m
b) Amit, 91 m
c) Sumit, 111 m
d) Amit, 111 m.S(120) = 800
A(100) = 600 so A=6 S=20/3
so A/S = 800/d = 18/20
so d=889. so Amit wins by 1000889 = 111mIn how many ways can 1,000,000 be written as sum of a square number and prime number?
p+q^2 = 10^6
p = (10^3+q)(10^3q)
since p is prime, 10^3q=1 so q=(10^31)= 999
so p = 1999
hence only 1 way, 1999+(999)^24 different digits are chosen and all possible positive 4 digit numbers are constructed out of them. The sum of the 4digit numbers is 186,648. How many different sets of such 4 digits can be chosen?
abcd so 24 arrangements.
a (6000+600+60+6) +.... = 186648
(a+b+c+d) = 186648/6666 = 28
so (9,8,6,5), (9,8,4,7)The ratio of the factorial of a number x to the square of the factorial of another number, which when increased by 50% gives the required number , is 1.25. Find the number x.
a) 6
b) 5
c) 9
d) Nonex=6
x and 2x/3
so (x!) / (2x/3)^2 = 5/4
x=6 pe 6!/(4!)^2 = 720/576 = 5/4 so 6A person has just sufficient money to buy either 30 guavas, 50 plums or 70 peaches. He spends 20% of the money on travelling, and buys 14 peaches, 'x' guavas and 'y' plums using rest of the money. If x, y > 0, what is the minimum value of the sum of x and y?
30a =50b = 70c =5T
4T = 14c + xa + yb
4T = 14 * 5T/70 + x * 5T/30 + y * 5T/50
4 = 1+ x/6 + y/10
x/6+y/10 = 3
5x+3y = 90
so min. sum when x=15, y=5 so 20S = {1, 2, ..., 5n}, where n is a natural number. If R is a subset of S such that the average of no two elements of R is n or an integral multiple of n, then the maximum number of elements in R is:
Options:
a) 4n1
b) 2n
c) 3n
d) 3n13n1
n=1 so (1,2,3,4,5)
so avg should not be integral multiple of 1, so avg should not be integral.
so R=(1,2) no other number can come. so 2 elements= 3*11=2After travelling for 30mins , a train met with an accident and was stopped there for 45 mins. Due to accident it's speed reduces to 2/3 of its former speed and the train reached the station 1.5 hours late. Had the accident occurred 60km after the point it occurred earlier , the train would have reached reached 30mins earlier. What is the length of the journey?
speed=60, distance=120
a/v+3/4 + 3(da)/2v = d/v +3/2
(a+60)/v + 3/4 + 3(da60)/2v = d/v+1
subtract.
60/v  90/v = 1/2
30/v = 1/2 so v=60.
so 1/2 +3d/2v = 2d/2v+3/2
d/2v = 1 so d=2v = 2 * 60=120