1) 1(1!)
+ 2(2!)... 2012!
Ans : 2013-1!
2). If X^Y denotes X raised to the power of Y, find out last two
So, time difference=80X4=320mins=5 hrs 20 mins
Since the place is located towards west
Therefore local time of that place during take off=8:40 p.m.
So, local time during landing=8:40p.m.+10hrs=6:40a.m
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Ans : 2013-1!
2). If X^Y denotes X raised to the power of Y, find out last two
digits of (2957^3661)+(3081^3643)
Ans; 98 is the ans
7^1 is 7
and
5*1=5
so last to digit of (2957^3661) is 57
same for
1^3 is 1
8*3=24 unit digit of 24 is 4
last 2 digit of(3081^3643) is 41
41+57=98....
7^1 is 7
and
5*1=5
so last to digit of (2957^3661) is 57
same for
1^3 is 1
8*3=24 unit digit of 24 is 4
last 2 digit of(3081^3643) is 41
41+57=98....
3). if f(x) is a function for all real numbers x holds good
for maximum of 2x+4 and 12+3x..then wat is the value of x so that f(x) is equal
to 2x+4..Ans is -9
Ans: ans is -9
maximum (2x+4 ,12+3x)
and ques says maximum is 2x+4
if x= -9 2x+4 = -14 and 12+3x= -15 maximum is -14
so ans is -9
maximum (2x+4 ,12+3x)
and ques says maximum is 2x+4
if x= -9 2x+4 = -14 and 12+3x= -15 maximum is -14
so ans is -9
4). 15 women or 10 men complete in 55 days..how many days 4
women and 5 men vil take
1 Women= 10/15 Men
4 women= 4*10/158=8/3
total= 5+8/3 men= 23/3
chain rule
55*10 = (23/3)*D
D= 550*3/23 =71.73 days
4 women= 4*10/158=8/3
total= 5+8/3 men= 23/3
chain rule
55*10 = (23/3)*D
D= 550*3/23 =71.73 days
5). My flight takes off 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) 6:40 am c) 7:40 am d)
7:00 am e) 8:00 am
Ans;
Latitudianal difference=(10+70)=80coordinates 36N70W.What is the local time when my plane landed? a) 6:00 am b) 6:40 am c) 7:40 am d)
7:00 am e) 8:00 am
So, time difference=80X4=320mins=5 hrs 20 mins
Since the place is located towards west
Therefore local time of that place during take off=8:40 p.m.
So, local time during landing=8:40p.m.+10hrs=6:40a.m
6)
. In a class there are 60% of girls of which 25% poor.
What is the probability that a poor girl is selected is leader?
a)15% b)50% c)80% d)30%
a)15% b)50% c)80% d)30%
Ans: nswer will be 15%
girls=60%
poor gilrs=25%60 ie 15%
so probability that a girl will be poor is 15%
girls=60%
poor gilrs=25%60 ie 15%
so probability that a girl will be poor is 15%
7)(sin nx/n)=?
a)90 b)0 c)1 d)2
a)90 b)0 c)1 d)2
Ans; here n is a natural no. let us put n=1 u vl get x=sinx
wich means x is very small or tending to 0 so
answer is 0
answer is 0
8).302th of 1st day is saturday.what is the 15th day of 5th
year?
Ans; WEDNESDAY IS D ANSWER
302TH YR HAS 1 ODD DAY
SO 1ST DAY IS FRIDAY
SO 1ST DAY OF FIFTH YR HAS 4 ODD DAYS
+ 15 DAYS HAS 14 ODD DAYS
SO IN ALL 14 + 4 = 18 ODD DAYS = 4 ODD DAYS
THEREFORE WEDNESDAY
302TH YR HAS 1 ODD DAY
SO 1ST DAY IS FRIDAY
SO 1ST DAY OF FIFTH YR HAS 4 ODD DAYS
+ 15 DAYS HAS 14 ODD DAYS
SO IN ALL 14 + 4 = 18 ODD DAYS = 4 ODD DAYS
THEREFORE WEDNESDAY
9).Raj invested in Indigo, HUL and SBI shares at Rs. 300,
Rs. 200 and Rs. 5 per share, 100 shares for Rs. 10000. The number of Indigo and HUL shares he bought are
respectively
1.15, 25
2.23, 17
3.17, 23
4.17, 60
Rs. 200 and Rs. 5 per share, 100 shares for Rs. 10000. The number of Indigo and HUL shares he bought are
respectively
1.15, 25
2.23, 17
3.17, 23
4.17, 60
Ans; ans will be 17,23 coz..
left sbi share will be 100-(17+23)=60
then indigo=300*17=5100
and HUL=200*23=4600
sbi=5*60=300
............
total=5100+4600+300=10000
left sbi share will be 100-(17+23)=60
then indigo=300*17=5100
and HUL=200*23=4600
sbi=5*60=300
............
total=5100+4600+300=10000
10)A series of book was published at 7 year
intervals.When the 7th book was issued the sum of publication year is
13524.When was the 1st book published ?
,ans:1911. .
,ans:1911. .
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TCS 2013 placement new pattern questions 12
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