(Paper) : TCS Placement Paper Pattern (Only Aptitude)

Paper : TCS Placement Paper Pattern (Only Aptitude)

1) If log 0.317=0.3332 and log 0.318=0.3364 then find log 0.319 ?

Sol) log 0.317=0.3332 and log 0.318=0.3364, then
log 0.319=log0.318+(log0.318-log0.317) = 0.3396

2) A box of 150 packets consists of 1kg packets and 2kg packets. Total weight of box is 264kg. How many 2kg packets are there ?

Sol) x= 2 kg Packs
y= 1 kg packs
x + y = 150     .......... Eqn 1
2x + y = 264   .......... Eqn 2
Solve the Simultaneous equation; x = 114
so, y = 36

ANS :  Number of 2 kg Packs = 114.

3) My flight takes of at 2am from a place at 18N 10E and landed 10 Hrs later at a place with coordinates 36N70W. What is the local time when my plane landed?

a) 6:00 am
b) 6:40am
c) 7:40
d) 7:00
e) 8:00

Sol) The destination place is 80 degree west to the starting place. Hence the time difference between these two places is 5 hour 20 min. (=24hr*80/360).
When the flight landed, the time at the starting place is 12 noon (2 AM + 10 hours).
Hence, the time at the destination place is 12 noon - 5:20 hours = 6: 40 AM

4) A plane moves from 9?N40?E to 9?N40?W. If the plane starts at 10 am and takes 8 hours to reach the destination, find the local arrival time ?

Sol) Since it is moving from east to west longitide we need to add both
ie,40+40=80
multiply the ans by 4
=>80*4=320min
convert this min to hours ie, 5hrs 33min
It takes 8hrs totally . So 8-5hr 30 min=2hr 30min
So the ans is 10am+2hr 30 min
=>ans is 12:30 it will reach

5) The size of the bucket is N kb. The bucket fills at the rate of 0.1 kb per millisecond. A programmer sends a program to receiver. There it waits for 10 milliseconds. And response will be back to programmer in 20 milliseconds. How much time the program takes to get a response back to the programmer, after it is sent? Please tell me the answer with explanation.

Sol) see it doesn't matter that wat the time is being taken to fill the bucket.after reaching program it waits there for 10ms and back to the programmer in 20 ms.then total time to get the response is 20ms +10 ms=30ms...it's so simple....

6) A file is transferred from one location to another in 'buckets'. The size of the bucket is 10 kilobytes. Each bucket gets filled at the rate of 0.0001 kilobytes per millisecond. The transmission time from sender to receiver is 10 milliseconds per bucket. After the receipt of the bucket the receiver sends an acknowledgement that reaches sender in 100 milliseconds. Assuming no error during transmission, write a formula to calculate the time taken in seconds to successfully complete the transfer of a file of size N kilobytes.

(n/1000)*(n/10)*10+(n/100)....as i hv calculated...~~!not 100% sure

7) A fisherman's day is rated as good if he catches 9 fishes, fair if 7 fishes and bad if 5 fishes. He catches 53 fishes in a week n had all good, fair n bad days in the week. So how many good, fair n bad days did the fisher man had in the week

Ans:4 good, 1 fair n 2 bad days

Sol) Go to river catch fish
4*9=36
7*1=7
2*5=10
36+7+10=53...
take what is given 53
good days means --- 9 fishes so 53/9=4(remainder=17)  if u assume 5 then there is no chance for bad days.
fair days means ----- 7 fishes so remaining 17 --- 17/7=1(remainder=10) if u assume 2 then there is no chance for bad days.
bad days means -------5 fishes so remaining 10---10/5=2days.
Ans: 4 good, 1 fair, 2bad. ==== total 7 days.

x+y+z=7--------- eq1
9*x+7*y+5*z=53 -------eq2
multiply eq 1 by 9,
9*x+9*y+9*z=35 -------------eq3
from eq2 and eq3
2*y+4*z=10-----eq4
since all x,y and z are integer i sud put a integer value of y such that z sud be integer in eq 4 .....and ther will be two value y=1 or 3 then z = 2 or 1 from eq 4

for first  y=1,z=2 then from eq1 x= 4
so 9*4+1*7+2*5=53.... satisfied
now for second y=3 z=1 then from eq1 x=3
so 9*3+3*7+1*5=53 ......satisfied
so finally there are two solution of this question
(x,y,z)=(4,1,2) and (3,3,1)...

8) Y catches 5 times more fishes than X. If total number of fishes caught by X and Y is 42, then number of fishes caught by X?

Sol) Let no. of fish x catches=p
no. caught by y =r
r=5p.
r+p=42
then p=7,r=35

9) Three companies are working independently and receiving the savings 20%, 30%, 40%. If the companies work combinely, what will be their net savings?

suppose total income is 100
http:/
so amount x is getting is 80
y is 70
z =60
total=210

but total money is 300
300-210=90
so they are getting 90 rs less
90 is 30% of 300 so they r getting 30% discount

10) The ratio of incomes of C and D is 3:4.the ratio of their expenditures is 4:5. Find the ratio of their savings if the savings of C is one fourths of his income?

Sol) incomes:3:4
expenditures:4:5
3x-4y=1/4(3x)
12x-16y=3x
9x=16y
y=9x/16
(3x-4(9x/16))/((4x-5(9x/16)))
ans:12/19

11) If G(0) = -1 G(1)= 1 and G(N)=G(N-1) - G(N-2) then what is the value of G(6)?

Ans: -1

bcoz g(2)=g(1)-g(0)=1+1=2
g(3)=1
g(4)=-1
g(5)=-2
g(6)=-1

12) If A can copy 50 pages in 10 hours and A and B together can copy 70 pages in 10 hours, how much time does B takes to copy 26 pages?

Sol) A can copy 50 pages in 10 hrs.
A can copy 5 pages in 1hr.(50/10)
now A & B can copy 70 pages in 10hrs.
thus, B can copy 90 pages in 10 hrs.[eqn. is (50+x)/2=70, where x--> no. of pages B can copy in 10 hrs.]
so, B can copy 9 pages in 1hr.
therefore, to copy 26 pages B will need almost 3hrs.
since in 3hrs B can copy 27 pages.

13) what's the answer for that : A, B and C are 8 bit no's. They are as follows:

A -> 1 1 0 0 0 1 0 1
B -> 0 0 1 1 0 0 1 1
C -> 0 0 1 1 1 0 1 0 ( - =minus, u=union)

Find ((A - C) u B) =?

To find A-C, We will find 2's compliment of C and them add it with A,
That will give us (A-C)
2's compliment of C=1's compliment of C+1
=11000101+1=11000110
A-C=11000101+11000110
=10001001

Now (A-C) U B is .OR. logic operation on (A-C) and B
10001001 .OR . 00110011
The answer is = 10111011,
Whose decimal equivalent is 187.

14) One circular array is given(means memory allocation tales place in circular fashion) diamension(9X7) and sarting add. is 3000, What is the address of (2,3)........

Sol) it's a 9x7 int array so it reqiure a 126 bytes for storing.b'ze integer value need 2 byes of memory allocation. and starting add is 3000
so starting add of 2x3 will be 3012.

15) In a two-dimensional array, X (9, 7), with each element occupying 4 bytes of memory, with the address of the first element X (1, 1) is 3000, find the address of X (8, 5).

Sol) initial x (1,1) = 3000 u hav to find from x(8,1)so u have x(1,1),x(1,2) ... x(7,7) = so u have totally 7 * 7 = 49 elementsu need to find for x(8,5) ? here we have 5 elements each element have 4 bytes : (49 + 5 -1) * 4 = 212 -----( -1 is to deduct the 1 element ) 3000 + 212 = 3212

16) Which of the following is power of 3

a) 2345
b) 9875
c) 6504
d) 9833

17) The size of a program is N. And the memory occupied by the program is given by M = square root of 100N. If the size of the program is increased by 1% then how much memory now occupied ?

Sol) M=sqrt(100N)
N is increased by 1%
therefore new value of N=N + (N/100)
                =101N/100
M=sqrt(100 * (101N/100) )
Hence, we get M=sqrt(101 *  N)

18)
1)SCOOTER --------- AUTOMOBILE--- A. PART OF

2.OXYGEN----------- WATER ------- B. A Type of

3.SHOP STAFF------- FITTERS------ C. NOT A TYPE OF

4. BUG -------------REPTILE------ D. A SUPERSET O
F

1)B   
2)A
3)D    
4)C

19) A bus started from bustand at 8.00a m and after 30 min staying at destination, it returned back to the bustand. the destination is 27 miles from the bustand. the speed of the bus 50 percent fast speed. at what time it returns to the bustand

this is the step by step solution:
a bus cover 27 mile with 18 mph in =27/18= 1 hour 30 min. and it wait at stand =30 min.
after this speed of return increase by 50% so 50%of 18 mph=9mph
Total speed of returnig=18+9=27
Then in return it take 27/27=1 hour
then total time in joureny=1+1:30+00:30 =3 hour
so it will  come at 8+3 hour=11 a.m.
So Ans==11 a.m

20) In two dimensional array X(7,9) each element occupies 2 bytes of memory.If the address of first element X(1,1)is 1258 then what will be the address of the element X(5,8) ?

Sol) Here, the address of first element x[1][1] is 1258 and also 2 byte of memory is given. now, we have to solve the address of element x[5][8], therefore, 1258+ 5*8*2 = 1258+80 = 1338 so the answer is 1338.

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