Latest TCS Aptitude Question SOLUTION: 1)in how many ways can the letters of english alphabet be arranged so that there are 7 letters between A & B n no repeated. 2)Find no.of zeroes in 15*32*25*22*40*75*98*112*125 3)wh

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(#M40115500) TCS Category question Keep an EYE

1)in how many ways can the letters of english alphabet be arranged so that there are 7 letters between A & B n no repeated.
2)Find no.of zeroes in 15*32*25*22*40*75*98*112*125
3)what must be added to 5678 to get remainder 35 when divided by 460?
4) f(x)=1+x+x^2+......x^6.
Find remainder when f(x^7) is divided by f(x).
5)How many integers satisfy (x^2-x-1)^(x+2)=1?
6)there are 3 single digit distinct numbers A,B,C in geometric progression.
how many possible different values are for abs(A+B-C).

Asked In TCS Tripti Agarwal (9 years ago)

Unsolved Read Solution (8)

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Read Solution (8) :

davish nigam  (9 years ago)

Sol 1 : A and B are fixed and can be arranged by 2!
remaining letters=24
Arrange 7 letters out of 24= 24P7
Now consider A +7 letters+B as 1 letter, now remaining letters=26-9=17
Then 1letter and 17 letters can be arranged by 18!
i.e ans : 24P7*2!*18!

Sol 2 : (3*5)*(2^5)*(5^2)*(11*2)*(5*2^3)*(3*5^2)*(49*2)*(7*2^4)*(5^3)
so no of 5 = 9
no of 2=14
5*2=10
hence no of zero=9

sol :3 firstly take 5678 and divide it with 460
reminder = 158
now, we have to add a no. in 158 that can make vale 35 more then 460 i.e. 495
hence simply we have to add 337 in 158 to get reminder 35
hence
ans : 337

sol : 4 put x=0
so f(x)=1
f(x^7)=1
1/1=0
so 0 is answer

sol: 5 at x = 0 => (-1)^2 = 1
x = -1 => (1 + 1 -1)^1 = 1
x = -2 => (4 + 2 - 1)^0 = 1
x = 2 => (4 - 2 - 1)^4 = 1
hence only 4 can satisfy
ans : 4

sol : 6 A,B,C may be
(1,2,4) & (4,2,1)
(1,3,9) & (9,3,1)
(2,4,8) & (8,4,2)
(4,6,9) & (9,6,4)

find abs(A+B-c) for these 8 GPs
1, 5, 5, 11 , 2, 10, 1 , 11

we get 5 different values (1,2,5,10,11)

ans : 5

Is this solution Helpfull? Yes (5) | No (1)  

Emanuel Abhishek Murmu  (9 years ago)

solution to question 4 will be 7.

f(x)= 1+ x+ x^2+ x^3+......+x^n for x>1

now when f(x^(n+1)) is divided by f(x) remainder is simply (n+1)

so, in this case remainder is (6+1)= 7

Is this solution Helpfull? Yes (5) | No (3)  

davish nigam  (9 years ago)

@ vibk
approach is wright but logic is slightly mistaken.....
u said that u have taken 7 letters , a & b ans shuffled them in 3! ways.......
but see the condition given in the question... i.e. a 7 letters arranged in between a and b
hence 7 digits shuffled in 24P7 ways....
now.... A _ _ _ _ _ _ _ B hence A & B only shuffled in two ways i.e.
A _ _ _ _ _ _ _ B itself
ans B _ _ _ _ _ _ _ A
i.e. 2! ways
and once this operation completed consider all of the digits as a single digit to be shuffled in rest 17 letters....
........got my point

Is this solution Helpfull? Yes (2) | No   

Tripti Agarwal  (9 years ago)

A -B or B - A --------------------> 2 WAYS
from rest 24 letters, 7 can be arranged in 24P7 = 24! ways
also position of A-------B can be 1-------9, 2--------10, ..., 18-------26 ( 18 ways )
total no. of ways = 24P7 * 2 * 18! ways.

Is this solution Helpfull? Yes (2) | No (1)  

Tripti Agarwal  (9 years ago)

davish thanku bt cn u tell in sol 1 y are treating A+7letters+B as one letter

Is this solution Helpfull? Yes (1) | No (3)  

davish nigam  (9 years ago)

simply in solution 1 you have taken 24P7*2! for the three of these i.e. 24P7 for all 7 letters and 2! for a & b hence these three are treated as a single letter for our convenience... got

Is this solution Helpfull? Yes (1) | No (1)  

vibk  (9 years ago)

@davish i have dont regarding first solution...i think it shud be 24P7*3!*7!
consider 7 letters as 1 nd we get three arrangements to make i.e. A+1letter+B i.e. 3! nd rest 7! for 7 letters..plz correct me if ma approach iz wrong

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Roshan singh  (9 years ago)

how you r selecting digits of the number for qs 6

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