Topic - Quant Mixed Bag

Solved ? - Yes

Source - Compilation of my posts from various CAT prep forums. ]]>

Topic - Quant Mixed Bag

Solved ? - Yes

Source - Compilation of my posts from various CAT prep forums. ]]>

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 ]]>

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

a) +ve

b) -ve

c) non -ve

d) not ]]>

Since there is no real root so f(0) and f(1) will have same sign.

So product is +ve always. ]]>

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.

112 more digits.

so (112/2) = 56 numbers.

so 10 to 65. so ends in 5 ]]>

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 ]]>

(10+1)^(120!)-1

last term of binomial = 120! * 10

number of zeroes = 28 + 1 = 29 ]]>

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 ]]>

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 ]]>

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 ]]>

(3/4 * 2/3 * 3/4 + 3/4 * 1/4 * 1/3 + 1/4 * 2/3 * 1/4)

= 23/48 so 5/23 ]]>

(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 ]]>