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

You figure it out by knowing that there is a 5 as a factor in the final number.

example 8! = 8*7*6*5...*1

Now if u know how many fives are there you will have that number of zeros in the end because at least that many number of 2 will there for sure ...

Now IF u want to figure out how may fives are there in (which is equivalent to finding how many zeros in the end of the number )
then simply keep dividing the factorial by 5 and adding it to the result until 5^x exceeds the factorial, i know i have used very confusing language but a example will simplify things for sure

how many zeros at the end of 312!
answer :- 312!/5 + 312!/5^2 + 312!/5^3 = 62 + 12 + 2 (now note i stopped at 5^3, because 5^4 =25*25=625 which is more that 312 ) hence :- 62 + 12 + 2 =76 number of 5s in the factorial and same number of zeros in the end because these many number of 2s will obviously be there in factorial and will form 10s .

IF you are still confused I would suggest you to download Bunuel's math's notes and even if you are not confused i would still suggest you to do the same ,
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[quote2]9. If x and y are negative numbers, what is the value of \frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y

Correction/suggestion/advice/confirmation --- needed
Answer to 9th question If x and y are negative numbers, what is the value of \frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}?[/b]
Sqrtx^2/x-sqrt(-y*|y|) = |x|/x-|y|= -1-y
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-y = +y and |y| no matter what, comes out as positive hence Sqrt (-y*|y| ) comes out as possitive y and sqrtx^2 =|x|= positive x , of course from your tone i get that i'm going wrong some where but
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What I meant is that if y<0 then |y|=-y and -1-|y|=-1-(-y)=-1+y.
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Quote:

13. If x=\frac{(8!)^{10}-(8!)^6}{(8!)^{5}-(8!)^3}, what is the product of the tens and the units digits of \frac{x}{(8!)^3}-39?
A. 0
B. 6
C. 7
D. 12
E. 14

x=((8!)^6*((8!)^4-1))/((8!)^3*((8!)^2-1))=(8!)^3*((8!)^2+1)


((8!)^3*((8!)^2+1) /(8!)^3)) -39=(8!)^2+1)-39
note, that 8! has 5 and 2. so 8! ends with 0. then (8!)^2 will end by two 0s.
***00-38=62
6*2=12
answ is

[Reveal] Spoiler:

D

4. What is the smallest positive integer k such that 126*\sqrt{k} is the square of a positive integer?
A. 14
B. 36
C. 144
D. 196
E. 441

126*\sqrt{k}=3*7*3*2*sqrt{k}=3^2*7*2
so we need one more 7*2 (14)
sqrootk=14
k=196
answ is

[Reveal] Spoiler:

D

what an irony. all the 3 questions are

[Reveal] Spoiler:

D

heh


9. If x and y are negative numbers, what is the value of \frac{\sqrt{x^2}}{x}-\sqrt{-y*|y|}?
A. 1+y
B. 1-y
C. -1-y
D. y-1
E. x-y
-x/x-|y|=-1-(-y)=y-1 since y<0 ; x<0

[Reveal] Spoiler:

D

is the answ
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10. If x^2<81 and y^2<25, what is the largest prime number that can be equal to x-2y?
A. 7
B. 11
C. 13
D. 17
E. 19

-9<x<9
-5<y<5

-9<x<9
-10<2y<10
x-2y<9-(-10)
x-2y<19

the answer is 17. D again the mystical letter for today heh


11. In an infinite sequence 1, 3, 9, 27, ... each term after the first is three times the previous term. What is the difference between the sum of 13th and 15th terms and the sum of 12th and 14th terms of the sequence?
A. 10*3^11
B. 20*3^11
C. 10*3^12
D. 40*3^11
E. 20*3^12

1, 3, 9, 27 ok please pay attention to the fact that the term of sequence has 3 one less than the number of the term,i.e.
the 3d term is 9, but the number has two 3s (one less than the term). the 4th term is 3*3*3 ,which has three 3s.

ok, then we have the following-
the q. asks us to find out -
(13th term+15th term) -(12th term +14th term)

(3^12+3^14)-(3^11-3^13)=
3^11(3-1)+3^13(3-1)=
2*3^11(1+3^2)=
20*3^11
the answer is B

p.s. also note, that answer choices B and D are leaders of the day heh
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1) B
2) E
3) B
4) A
5) E
6) B
7) D
8) D
9) D
10) D
11) B
12)
13) D

sqroot of (negative number ^2) = (negative number ^2)^1/2=negative number

for example sqroot (-5*|5|)=sqroot (-5^2)=((-5)^2)^1/2=-5[/quote]

agreed but here isn't the concept little different, as it is already mentioned that y is negative hence -y= positive y
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