1) Ans B
The 3 digits that will be '6' can be selected in 5c3 = 10 ways
The remaining two digits can be any digit from 0 to 9 other than 6 i.e 9 digits.
so favorable outcomes = 5c3 *9*9*1*1*1 = 810
total outcome = 10*10*10*10 *10 = 10^5
P(exactly three 6) = 810/10^5
2) Ans E
y = (3^5-3^2)^2/ (5^7 -5^4)^-2
Numerator (3^2(3^3-1))^2 = (3^2(26) )^2 = 3^4*13^2*2^2
Denominator (5^4(5^3-1))^-2 = 1/(5^4(124))^2 = 1/5^8*4^2*31^2 = 1/5^8*2^4*31^2
Numerator/denominator = 3^4 *13^2 *2^2 *5^8 *2^4*31^2 = 3^4*13^2*31^2*2^6
From answer choices we see that 52^4 is not divisble by y
3) Ans B
Old sum/k = 55
old sum = 55 k
old sum +100 /k+1 = 60
old sum+100 = 60k+60
55k+100 = 60k+60
5k = 40 k= 8
4) Ans D
126 * sqrt k = a^2 where a is a +ve integer
7*2*3^2 *sqrt k = a^2
So we need one two and one 7 to make 'a' a perfect square
if k = 196 , 7*2*3^2*sqrt 196 = 2^2*7^2*3^2
5) For selecting 8 marbles from 12 marbles, a maximim of 4 marble can remain in
the jar . The scenarios for atleast 1 red marble and 1 blue marble to remain are:
1r1b 7c1*5c1 = 35
1r2b 7c1*5c2 = 70
2r1b 7c2*5c1 = 105
2r2b 7c2*5c2 = 210
3r1b 7c3*5c1 = 175
1r3b 7c1*5c3 = 70
Total 665 . Not sure if the logic is right..
6) Ans E
Work done by pump a,b,c in 1 hr = 1/x , 1/y, 1/z respectively
Working simultaneously, work done by them in 1 hr is:
1/x+1/y-1/z
yz +xz -xy/xyz job done in 1 hr
So whole job will be completed in xyz/yz+xz-xy hrs
Since pump A work for x hrs, fraction of job completed by pump A is :
x/(xyz/yz+xz-xy) = yz+xz-xy/yz
8) Ans) D
let the 7 consecutive odd integers be:
x, x+2, x+4, x+6, x+8, x+10, x+12 (in ascending order)
Sum of the 5 largest integers is x+12 +x+10 +x+8 +x+6 +x+4 = 5x+40
So 5x +40 = -185
5x = -225
x = -45
Sum of the 5 smallest integers = x+x+2+x+4+x+6+x+8 = 5x+20
5x+20 = (5*-45) +20 = -225+20 = -205
9) Ans c
sqrt x^2 = |x| = -x (since X<0)
sqrt (-y*|y|) = sqrt(-y*-y) sqrt y^2 = y
sqrt x^2/x - sqrt(-y*|y|) = -x/x - y = -1-y
10) Ans c
x^2<81 -9<x<9 y^2<25 -5<y<5
For x -2y to be a maximim prime number, x should be maximim prime number
and y minimum prime number
So x-2y = 7 - (2*-3) =13
11) Ans B
nth term of GP sereis is an = a1*r^n-1 where r =3 common ratio , a1 -=1
a13 = 1*3^12
a15 = 1^3^14
a13+a15 = 3^12 +3^14 = 3^12(1+3^2) = 3^12*10
Similarly a12 = 3^11 a14 = 3^13
a12+14 = 3^11+3^13 = 3^11*10
(a13+a15) - (a12+a14) = 3^12*10 - 3^11*10
3^11*10(3-1) = 3^11 *20
12)Ans B
x = yq1+3 y>3 since divisor should be greater than remainder
y = zq1+8 z>8 since divisor should be greater than remainder
so least value of z is 9 and least value of y 8 since for 8/9 remainder is 8
least value of x is 3 since for 3/8 remainder is 3
Least value of x+y+z = 3+8+9 = 20
13) Ans D
x = (8!)^10 - (8!)^6 / (8!)^5 -(8!)^3
Numerator, taking 8!^6 common ,(8!)^6 ((8!)^4-1) = (8!)^6 [(8!)^2+1)(8!)^2-1)]
Denominator, taking 8!^3 common (8!)^3 [(8!)^2-1]
x = (8!)^6 [(8!)^2+1)(8!)^2-1)] / (8!)^3 [(8!)^2-1]
x = (8!)^3[(8!)^2+1)
x = (8!)^5 +(8!)^3
x/(8!)^3 -39 = [(8!)^5 +(8!)^3/(8!)^3] -39
= [(8!)^2 +1]-39
=(8!)^2-38
8!^2 is a big number and will have 2 trailing zeros..
So some big number ending in 00 - 38 will have
digits 2,6 in units and tens palce respectively
So product is 2*6 =12