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  TCS PLACEMENT PAPERS

1. jake runs faster than paul.jake and paul each walk 24kms.the sum of the speeds is 7km/hr and the sum of the time taken by them is 14 hrs.then jake's speed is?

Asked In TCS 

ANS:

question is complete..!!
sol: 24/7-j + 24/j =14

24j + 24(7-j)/ (7-j)j 
here j is jack speed n 7-j is paul
on solving we get a eqn
14j^2-98j+168=0
solving we get j=4 n 3
but jack is faster hence ans is 4 km/hr..!!

2. A cow & Horse are bought for Rs.200000. The cow is sold at a profit of 20% and the Horse is sold at a loss of 10%. The overall gain is Rs.4000. The cost price of the cow is :
a) 70000
b) 80000
c) 120000
d) 130000

Asked In IBPS

ANS:

let c.p of cow=x, therefore c.p of horse= 200000-x
acc to problem S.P of cow = 6x/5 nd S.P of horse=9(200000-x)/10
ass overall gain is 4000, therefore total S.P-total C.P=4000
{6x/5+9(200000-x)/10}-200000=4000 solving equation,
we get x=80000, therefore C.P of Cow= 80000(ans)

3. If a lemon and an apple together cost Rs. 12.00, a tomato and a lemon cost Rs. 4.00 and an apple cost Rs.8.00 more than a tomato or a lemon then which of the following can be

Asked In TCS

ANS:

lemon+apple=12rs
tomato+lemon=4
then apple=8+t or 8+l
by solving we get
l+(8+l)=12
2l=4
l=2
so apple cost=8+2=10;

4. George does 3/5 of a piece of work in 9 days. He then calls paul and they finish the work in 4 days. How long would paul take to do the work by himself?
1. 28
2. 35
3. 30
4. 32

Asked In TCS

ANS:

30 days
gorge: does 3/5 th in 9 days  (3/5 = 9   è3/45 =1/15)
in 1 day he can do 1/15
so in 4 days he can do =4/15
:. total work done by gorge =3/5+4/15=13/15

work left= 1-13/15=2/15

which is done by paul in 4 days so paul can do 2/15*4 in 1 day=1/30
:. paul will take 30 days to do the job himself

5. Which of the following numbers must be added to 5678 to give a reminder 35 when divided by 460?

Asked In TCS

ANS:

ans is 797,

If you need of remainder 35, while dividing a number by 460. The Unit place of that number must be 5. So 5678+797= 6475
6475/460 gives reminder 35

6. Two identical circles intersect so that their centers , and their point of which they intersect , form a square of 1cm. Find the area of the portion which is common to the two circles.

Asked In TCS

ANS:

So, the side of the square is 1, which means the radius of each circle is also 1.
Area of squares = 1
Area of two circles not included in the square = 3(pi)/2
Area of oval overlap region of circles = [(pi)/2] - 1
Area of entire diagram = 3(pi)/2 + 1

suppose center of two circle is o and p.and two circle intersect points are a and b.so area of square oapb is=1 cm^2.Now area of triangle oab=triangle bpa=1/2*oapb.So area of triangle oab=1/2 cm^2.And area of arc oab=1/4*area of circle with center o.Area of circle with center o is=pi*1*1=pi.so area of arc oab=pi/4.so area between arc and triangle is=(pi/4)-(1/2).So another circle calculation is same.so total intersected area=2*(pi/4-1/2)=(pi/2)-1

7. What is the distance in cm between two parallel cords of length 32cm and 24cm in a circle of radius 20 cm?
1. 4 or 28
2. 2 or 14
3. 1 or 7
4. 3 or 21

Asked In TCS

ANS:

1. 4 or 28
let the perpendicular distance of the chords from the center be x & y respectvly

x^2 + 16^2 = 20^2 => x = 12
y^2 + 12^2 = 20^2 => y = 16

as, chords are parallel, they can same or opposite sides from centre
distance = 16+12= 28cm or 16-12=4cm

8. A semicircle is drawn with AB as its diameter. From C, a point on AB, a line perpendicular to AB is drawn, meeting the circumference of the semicircle at D. Given that AC=2cm and CD=6cm, the area of the semicircle in square cm will be ?

Asked In TCS

ANS

let O be center of circle and radius be x.
C can be anywhere on AB.Let us suppose that C is somewhere between A and O.
AC=2cm....so CO=(x-2)
DO=x,CD=6cm
CDO form a right angled triangle with right angle at C.
Using Pythagoras theorem,
x^2=6^2+(x-2)^2
4x=40
x=10cm
area=[pi(x)^2]/2
area=50pi

9. Two full tanks one shaped like the cylinder and the other like a cone contain liquid fuel the cylindrical tank held 500 lts more then the conolical tank After 200 lts of fuel is pumped out from each tank the cylindrical tank now contains twice the amount of fuel in the canonical tank How many lts of fuel did the cylindrical tank have when it was full?

Asked In TCS

ANS;

consider the colonical tank contains x lts
cylindrical tank then contains x+500 lts
after pumping of 200 lts from each tank..... now the tanks cntain x-200 lts & x+300 lts
x+300=2(x-200)
x=700 lts
originally cylindrical tank contains x+500 lts
therefore ans is 700+500= 1200 lts

10. What is probability of 53 Sundays in a leap year?

ANS:

2/7 or 0.2857 is the probability of 53 Sundays in a leap year.

Leap year consist of 366 days.
On that we have, 52 weeks and 2 days.

That 2 days may be

Sun,Mon
Mon,Tue
Tue,Wed
Wed,Thu
Thu,Fri
Fri,Sat
Sat,Sun

We have 7 options.
Among that we have to choose only 2 options which has Sundays. (i.e, [Sun Mon] or [Sat Sun]

So that, the answer is 2/7 for having 53 Sundays.

11. the value of a scooter depreciates in such a way that at the end of each year, is ¾ of its value at the beginning of same year. If the initial value of the scooter is rs40,000. What is the value at the end of 3yrs?
a)rs23125 b)rs19000 c)13435 d)16875

Asked In TCS

ANS:

As, it is given that , the cost becomes 3/4 at the end of year.
so, after 3 years ,the price of Scooter=40000*3/4*3/4*3/4= Rs.16875

(OR)
Initial amount=rs.40000
at 1st yr=40000*3/4=30000
2nd yr=30000*3/4=22500
end of 3rd yr=22500*3/4=16875.
Ans=16875

12. raj earns 25% on an investment but loses 10% on another investment.if the ratio of the two investment is 3:5.what is the gain or loss on two investments taken together?

Asked In Interview

ANS:

d qstn shd b 25% and 10%

taking the 2 investments to be 3x and 5x
total income of raj = (3x)*1.25+(5x)*0.9 = 8.25
therefore gain%=0.25/8*100=3.125%

(OR)

gain=25% and loss=10%
two investments are 3:5
total incom=3*125/100+5*90/100=8.25
gain%=0.25/8.25=0.3030*100=3.03
3.03%

13. A man goes upstream for a distance of 10 kms in 5 hours. Another man goes downstream for the same distance of 10kms in 3 houes. what is the difference of the speeds of the two men if the river speed is 0.5 Kms. (Round the answer to two decimals).)

Options:
A. 2.45
B. 1.56
C. 2.3
D. 1.33

Asked In TCS

ANS:

speed for 1st person=2 m/s ,speed for 2nd person=3.33 m/s.for upstream=2+0.5=2.5
for downstream=3.33+0.5=3.83.then diff=3.83-2.5=1.33
so ans=1.33

14. raj drives slowly along the perimeter of a rectangular park at 24kmph and completes one full round in 4mins. The ratio of length is to breadth is 3:2. What are its dimensions?

Asked In TCS

ANS:

length=3x ; breadth=2x

total distance = perimeter =2(3x+2x)
distance=10x ; distance= speed * time = 24 * 4/60(since time is in minute)
distance=10x=1.6;
x=.16
length=3x=.48;
breadth=2x=.32

480m*320m

15. a and b starts from their house at 10am. they travek from their house on MG road at 20kmph and 40kmph. they meet at T junction at 12:00 pm B reaches the T junction earlier and turns right .Both of them continue travelling till 2:00pm what is distance between a& b at 2:00pm?

Asked In TCS

ANS:

a reaches the junction at 12 after travelling 40 km
but b reaches junction at 11 after travelling 40 km.
at 2 am. a will travel 40 km from junction
at 2 am. b will travel for 3 hrs that is 120 km from junction so it travel 10 km.
distance between them is 120 +40=160 km



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