Quant Boosters  Hemant Malhotra  Set 10

Number of Questions  30
Topic  Quant Mixed Bag
Solved ?  Yes
Source  Elite's Grid Prep Forum

Q1) What is the approximate probability that two integers a and b chosen at random from 0 to 100 are such that a^2 − 15a + 56 > 0 and a− b > 60?
a) 0.15
b) 0.20
c) 0.23
d) 0.25
e) 0.29

Total cases = 101 * 101
a^2 − 15a + 56 > 0
(a− 7)(a − 8) > 0
so a< 7 or a > 8
ab > 60
(ab) > 60 or (ab) < −60Case 1: (ab) > 60
a=61,b = 0
a= 62, b =0, 1
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a = 100, b= 0, 1, 2, …, 39Case 2: (ab) < −60
b= 61,a = 0
b = 62, a= 0, 1
For b = 63, a = 0, 1, 2
For b = 64, a = 0, 1, 2, 3
For b = 65, a = 0, 1, 2, 3, 4
For b = 66, a = 0, 1, 2, 3, 4, 5
For b = 67, a = 0, 1, 2, 3, 4, 5, 6
For b = 68, a = 0, 1, 2, 3, 4, 5, 6 (a cannot be 7)
For b = 69, a = 0, 1, 2, 3, 4, 5, 6 (a cannot be 7 and 8)
For b= 70, a = 0, 1, 2, 3, 4, 5, 6, 9
For b = 71, a= 0, 1, 2, 3, 4, 5, 6, 9, 10
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For b = 100, a = 0 to 6 and 9 to 39so Total number of favourable cases = (1 + 2 + 3 + … + 40) + (1 + 2 + 3 + 4 + 5 + 6 + 7 + 7 + 7 + 8 + … + 38) = 1575
so probability = 1575/10201 = 0.15

Q2) Two integers, x and y are chosen with replacement out of the set {0,1,2,3,....10). What is the probability that x  y > 5

x  y > 5 or x  y < 5 so y  x > 5
now check
x  y > 5 means x > 5 + y
so when y = 0 then x > 5 so x = 6 to 10 so 5 cases
when y = 1 then 4 cases
when y = 2 then 3 cases
when y = 3 then 2 cases
when y = 4 then 1 case
so 1+ 2 + 3 + 4 + 5 = 5 * 6/2 = 15 cases
and same cases for y  x > 5 so total 30 cases
and total =11^2
so required probability = 30/11^2

Q3) Four numbers are multiplied together. The probability that the product is divisible by 5 or 10 is

When dividing a number by 5, the remainders can be 0,1,2,3,4 > means 5 cases
so when one number is div by 5 and rest are not then 4c1 * 4^3
now all div by 5 : 1 case
3 numbers div by 5 : 4c3 * 4
2 numbers div by : 4c2 * 4^2
Now (1 + 4c3 * 4 + 4c2 * 16 + 4c1 * 64)/5^4
= (1 + 16 + 96 + 256)/625
=369/625

Q4) What is the probability that there are exactly 5 cars between 2 particular cars if 10 cars are parked in a row?

A could be placed at 1st, 2nd, 3rd, 4th place only
So A has 4 options and A and B can shuffle in 2! Ways and rest 8 can shuffle in 8! Ways
so prob = 4 * 2! * 8!/10! = 4/45

Q5) The probabilities that a student passes in Mathematics, Physics and Chemistry are m,p and c respectively. Of these subjects a student has a 75% chance of passing in atleast one, a 50% chance of passing in atleast two and a 40% chance of passing in exactly two.Which of the following relations hold true
a) p + m + c = 27/20
b) p + m + c = 13/20
c) pmc = 1/10
d) both a and c

P(A U B U C) = 3/4
P(A) + P(B) + P(C)  P(A ∩ B)  P(A ∩ C)  P(B ∩ C) + P(A ∩ B ∩ C) = 3/4
P(A ∩ B) + P(B ∩ C) + P(A ∩ C)  2 * P(A ∩ B ∩ C) = 1/2
P(A ∩ B) + P (B ∩ C) + P( C ∩ A)  3 * P(A ∩ B ∩ C) = 2/5
Solve 2nd and 3rd
SO P (A ∩ B ∩ C) =1/2  2/5 = 1/10
NOW P(A) * P(B) * P(C) = 1/10
SO P(A) + P(B) + P(C)  (1/2 + 2/10) + 1/10 = 3/4
SO P(A) + P(B) + P(C) = 27/20

Q6) When three dice are thrown simultaneously , what is the probability that the sum of the three numbers that turn up is equal to 12 ?
a) 7/54
b) 25/216
c) 28/216
d) None of the above

Min value = 1
so a + b + c = 9
where a < = 5
9 + 3  1c3  1 = 11c2 = 55
now when a > 5
so 6 + a' + b + c = 9
so a' + b + c = 3
so 5c2 = 10
so 3 * 10 = 30
so 55  30 = 25
so prob = 25/216

Q7) A fair coin is tossed 10times. Let x/y be the probability that no two consecutive tosses show tails. where x & y are natural numbers such that HCF(x, y) = 1 find x+y.
a) 53
b) 83
c) 73
d) none of the above

10 things out of which we want to select r things in which no two are consecutive
10  r + 1 C r = 11  r C r
let
11c0 + 10c1 + 9c2 + 8c3 + 7c4 + 6c5
1 + 10 + 36 + 56 + 35 + 6 = 144
so 144/2^10=9/64

Q8) The minimum number of times a fair coin needs to be tossed, so that the probability of getting at least two heads is at least 0.96, is
(a) 6
(b) 7
(c) 8
(d) 9

atleast 2 = total  (no head + one head))
probability of gettin no head = 1/2
and in n times = (1/2)^n
now prob of getting one head in n turns
nc1 * (1/2)^n
so 1(1/2)^n  n * (1/2)^n > = 0.96
1((n+1)) * (1/2)^n > = 0.96
(n+1) * (1/2)^n < = 0.04
(n+1) * (1/2)^n < = 1/25
so n = 8

Q9) If 3 consecutive letters are selected at random from the English alphabet, then the probability that all three are consonants is:
a) 1/2
b) 11/24
c) 5/12
d) 7/12

A , E , I , O , U
now A will be in (1 consecutive)
_ _ E or _ E _ or E _ _ ( 3consecutive)
same in case of I ,O ,U so total consecutives which include vowels = 3 * 4 + 1 = 13
so not include vowels wll be 24  13 = 11
so OA=11/24

Q10) A is a set consisting 20 elements . Find probability of selecting two subsets P and Q of set A such that Q = (not P)