# Quant Boosters - Swetabh Kumar - Set 1

• a+b+c= f(1) and c=f(0).
Since there is no real root so f(0) and f(1) will have same sign.
So product is +ve always.

• Q4) Three dice are rolled simultaneously. What is the probability of getting a sum which is more than 15?

• (6-a)+(6-b)+(6-c) >15
a+b+c < 3
a+b+c=0 : 1 case
a+b+c=1 : 3 cases
a+b+c=2: 4C2=6 cases
so 1+3+6=10 cases. so 10/216 = 5/108

After a+b+c < 3, you can also use dummy variable to get rid of inequality.
a+b+c+d=3 But min. d should be 1
so a+b+c+(d+1) =3 so a+b+c+d=2 so 5C3 = 10 cases. so 10/216 = 5/108.

• Q5) If the natural numbers starting from 1 are written one after the other to form a 121-digit number, then what will be the last digit of the resultant number?

• 1-9: 9 digits.
112 more digits.
so (112/2) = 56 numbers.
so 10 to 65. so ends in 5

• 2 values. (2.5 and 3)

0-2
[x]>1 so [x]=2, 3.
when [x]=2 5=4+2{x} so {x}=0.5 so 2.5
when [x]=3 {x}=0 so 3

• Q7) Find the number of zeroes at the end of 11^( (5!)! ) - 1

• 11^(120!) -1
(10+1)^(120!)-1
last term of binomial = 120! * 10
number of zeroes = 28 + 1 = 29

• Q8) A nine digit number N is formed using 0 to 9 at-most once.what's the probability that the no is divisible by 45

• div. by 45 means div. by 5 and by 9. so unit digit is 0 or 5.
case 1: unit digit 5: so rest 8 digits can be 1 to 8 so 8! (to make overall digit sum=9k) or they can be 0 to 8 so 8!-7! (1st digit cannot be 0 so subtracting the 7! cases)
so 2 * 8!-7! = 15 * 7! here
case 2: unit digit 0 : so rest 8 digits can only be 1 to 8 to give an overall 9k digit sum...so 8! here or 8 * 7!
so total favourable= 15 * 7! +8 * 7! = 23 * 7!
total cases = 9 * 9 * ... 2 * 1 = 9 * 9! = 9 * 9 * 8 * 7!=648 * 7!
so probability = 23/648

• Q9) Number of ways of distributing 20 chocolate among 4 boys so that each boy received atleast one chocolate and the number of chocolate with each boy is different

• a+b+c+d = 20 so 19C3 = 969 total
ONLY 2 equal:
if a=b then 2a+c+d=20
a=1 so c+d=18 so (2,16)...(16,2) so 14
a=2 c+d=16 so (1,15)., (3,13)...(13,3),(15,1)..so 12
a=3 c+d=14 so (1,13), (2,12)....(13,1) so 10
a=4 c+d=12 so (1,11)...(11,1) so 8
a=5 c+d=10 so (1,9)..(9,1) so 8
a=6 c+d=8 so (1,7),..(8,8,) : 5
a=7 c+d=6 so (1,5), (2,4),(4,2),(5,1) so 4
a=8 c+d = 4 so (1,3),(3,1) so 2
a=9 c+d= 2 0 cases here.
so cases here: 63
hence 66 * 4C2
ONLY 3 equal:
3a+d = 20 so (1,17)...(6,2) so 5 cases (excluding 555) so 4C3 * 5
all 4 equal: (5555) so 1 case so 1 * 4C4=1
so (969-66 * 4C2 - 4C3 * 5-1) ~ 552

• Q10) What is the sum of all three digit palindromes which are multiples of 13 ?

• aba
101a+10b
10a+10b (since 101=10 mod 13)
10(a+b) so a+b =13
so 494, 585, 676, 767, 858, 949 so sum=4329

• Q11) 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 5/23

• Q12) In 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! = 350

• Q13) A team is planning to participate in a Dahi-handi 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?

61

61

61

45

61

63

61

61