# Quant Boosters - Swetabh Kumar - Set 1

• Number of Questions - 30
Topic - Quant Mixed Bag
Solved ? - Yes
Source - Compilation of my posts from various CAT prep forums.

• Q1) How many even integers between 100 and 200 (both included) are divisible neither by 7 nor by 9

• A is set of even nos divisible by 7
B is set of even nos divisible by 9
so required = A' ∩ B' = (A U B ) ' (De morgan law)
A U B= 7 + 6 - 1 = 12 ( 1 even no. div by 63)
so (AUB)' = total evens- 12 = 51-12=39

• Q2) Find minimum value of |x - 2| + |2x - 3| + |3x - 4|

• minima of |x-a|+|x-b| + |x-c|...and so on occurs at x= median of a,b,c...
since median is the middle value when a,b,c..etc are arranged in ascending/descending order. So on both sides of this middle value,function goes up. Hence, this is minima

|x-2| + |x-3/2|+|x-3/2| + |x-4/3|+|x-4/3|+|x-4/3|
median of 2, 3/2,3/2, 4/3,4/3,4/3 = (3/2+4/3)/2 = 17/12
so x=17/12 => |2-17/12| + |3-17/6|+ |4-17/4| = 7/12 + 2/12 + 3/12 =1

• Q3) If the equation a^2 + bx + c = 0 has no real roots then c ( a + b + c ) = ? (given a,b,c belongs to R and a is not 0)
a) +ve
b) -ve
c) non -ve
d) not

• 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 ?

61

64

42

61

61

63

34

61