TCS Latest Placement Paper Questions - 2014 (6)


1. A cow and horse are bought for Rs.2,00,000. The cow is sold at a profit of 20% and the horse is sold a t a loss of 10%.  The overall gain is Rs.4000, the Cost price of cow?
a) 130000
b) 80000
c) 70000
d) 120000
Ans: Overall profit = $\displaystyle\frac{{4000}}{{200000}} \times 100 = 2\% $
By applying alligation rule, we get

So cost price of the cow = 2/5 x 200000 = 80,000

2. A circle has 29 points arranged in a clock wise manner from o to 28.  A bug moves clockwise manner from 0 to 28.  A bug moves clockwise on the circle according to following rule.  If it is at a point i on the circle, it moves clockwise in 1 sec by (1 + r) places, where r is the remainder (possibly 0) when i is divided by 11.  If it starts in 23rd position, at what position will it be after 2012 sec. 
Ans: After 1st second, it moves 1 + (23/11)r = 1 + 1 = 2, So 25th position
After 2nd second, it moves 1 + 25/11 = 1 + 3 = 4, So 29th position = 0
After 3rd second, it moves 1 + 0/11 = 1 + 0 = 1, So 1st position
After 4th second, it moves 1 + 1 = 3rd position
after 5th, 1 + 3/11 = 4 So 7th
After 6th, 1 + 7/11 = 8 so 15th
After 7th, 1 + 15/11 = 5 so 20th
After 8th, 1 + 20/11 = 10th, So 30th = 1st
So it is on 1st after every 3 + 5n seconds.  So it is on 1st position after 2008 seconds (3 + 5 x 401) So on 20th after 2012 position.

3. In a city 100% votes are registered, in which 60% vote for congress and 40% vote for BJP.  There is a person A, who gets 75% of congress votes and 8% of BJP votes.  How many votes got by A?
Assume total votes are 100. So A got
75% of 60 = 45
8% of 40 = 3.2
A total of 48.2 %

4. Mean of 3 numbers is 10 more than the least of the numbers and 15 less than greatest of the 3.  If the median of 3 numbers is 5, Find the sum of the 3 numbers?
Ans: Median is when the given numbers are arranged in ascending order, the middle one. Let the numbers are x, 5, y where x is the least and y is greatest.
Given that $\displaystyle\frac{{x + 5 + y}}{3} = x + 10$
and $\displaystyle\frac{{x + 5 + y}}{3} = y - 15$
Solving we get x = 0 and y = 25.
So sum of the numbers = 0 + 5 + 25 = 30

5. A and B start from house at 10am. They travel fro their house on the MG road at 20kmph and 40 kmph.  there is a Junction T on their path.  A turns left at T junction at 12:00 noon, B reaches T earlier, and turns right.  Both of them continue to travel till 2pm. What is the distance between A and B at 2 pm.
Distnace between House and T junction = 20 x 2 = 40.
ie., B reached T at 11 am.
B continued to right after 11 am and travelled upto 2. So distance covered by him = 3 x 40 = 120
A reached T at 12 noon and travelled upto 2 So distanced travelled by him = 2 x 20 = 40
So total distance between them = 120 + 40 = 160 km

6. In a particular year, the month of january had exactly 4 thursdays, and 4 sundays.  On which day of the week did january 1st occur in the year. 
a) monday
b) tuesday
c) wednesday
d) thursday
Ans: If a month has 31 days, and it starts with sunday, Then Sundays, Mondays, tuesdays are 5 for that month. If this month starts with monday, then mondays, tuesdays, and wednesdays are 5 and remaining days are 4 each. so this month start with Monday.

7. A, E, F, and G ran a race. 
A said "I did not finish 1st /4th
E said "I did not finish 4th"
F said "I finished 1st"
G said "I finished 4th"
If there were no ties and exactly 3 children told the truth, when who finishes 4th?
a) A
b) E
c) F
d) G
Ans: Option D

8. A child was looking for his father. He went 90 m in the east before turning to his right.  he went 20 m before turning to his right afain to lok for his father at his uncles place 30 m from this point.  His father was not there.  From there he went 100m north before meeting hiss father in a street.  How far did the son meet his father from the starting point. 
a) 90
b) 30
c) 80
d) 100

From the diagram, AB = 90 - 30 = 60 and BD = 100 - 20 = 80
$AD = \sqrt {A{B^2} + B{D^2}}  = \sqrt {{{60}^2} + {{80}^2}}  = 100$

9. In an office, at various times during the day the boss gives the secretary a letter to type, each time putting the letter on top of the pile in the secretary's inbox. Secretary takes the top letter and types it.  Boss delivers in the order 1, 2, 3, 4, 5 which cannot be the order in which secretary types?
a) 2, 4, 3, 5, 1
b) 4, 5, 2, 3, 1
c) 3, 2, 4, 1, 5
d) 1, 2, 3, 4, 5
Ans: Option B

10. At 12.00 hours, J starts to walk from his house at 6 kmph. At 13.30, P follows him from J's house on his bicycle at 8 kmph. When will J be 3 km behind P?
By the time P starts J is 1.5 hr x 6 = 9 km away from his house.
J is 3 km behind when P is 3 km ahead of him. ie., P has to cover 12 km.  So he takes 12 / (8 - 6) = 6 hrs after 13.30. So the required time is 19.30Hrs

11. J is faster than P.  J and P each walk 24 km.  Sum of the speeds of J and P is 7 kmph.  Sum of time taken by them is 14 hours.  Then J speed is equal to
a) 7 kmph
b) 3 kmph
c) 5 kmph
d) 4 kmph
Given J > P
J + P = 7, only options are (6, 1), (5, 2), (4, 3)
From the given options, If J = 4 the P = 3. Times taken by them = $\displaystyle\frac{{24}}{4} + \frac{{24}}{3} = 14$

12. In a G6 summit held at london. A french, a german, an italian, a british, a spanish, a polish diplomat represent their respective countries. 
(i) Polish sits immediately next to british
(ii) German sits immediately next to italian, British or both
(iii) French does not sit immediately next to italian
(iv) If spanish sits immediately next to polish, spanish does not sit immediately next to Italian
Which of the following does not violate the stated conditions?
a) FPBISG
b) FGIPBS
c) FGISPB
d) FSPBGI
e) FBGSIP
Ans: Option D

13. Raj drives slowly along the perimeter of a rectangular park at 24 kmph and completes one full round in 4 min.  If the ratio of length to bredth of the park is 3 : 2, what are the dimansions?
a) 450 m x 300 m
b) 150 m x 100 m
c) 480 m x 320 m
d) 100 m x 100 m
24 kmph = $\frac{{24 \times 1000}}{{60}} = 400$ m / min
In 4 minutes he covered 4 x 400 = 1600 m
This is equal to the perimeter 2 ( l + b) = 1600
But l : b = 3:2
Let l = 3k, b = 2k
Substituting, we get 2 ( 3k + 2k ) = 1600 => k = 180
So dimensions are 480 x 320

14. M is 30% of Q, Q is 20% of P and N is 50% of P. What is M / N
ans: Take P = 100, then N = 50, Q = 20, M = 6. So M/N = 3/25

15. At what time between 6 and 7 are the hands of the clock coincide?
Ans.  Total = ${360^0}$
For hour = 360/12 = ${30^0}$/hr
For Minute = full rotation = ${360^0}$/hr
Let the line is 't' , for 6 = 6*30=${180^0}$
then
30 t + 180=360 t
330t = 180
t = 180/330
t = 6/11 hr 6/11*60=360/11=32$\frac{6}{{11}}$
Ans. is 6:32

16. Series 1, 4, 2, 8, 6, 24, 22, 88 ?
Sol :  The given series is in the format: x 4, -2, x4, -2, x4, -2, x4....
        1x4 = 4
4-2=2
8-2=6
6x4=24
24-2=22
22x4=88
88-2=86
Ans: 86

17. 4 Women & 6 men have to be seated in a row given that no two women can sit together. How many different arrangements are there.
Sol : Let us first sit all the 6 men in 6 positions in 6! ways.  Now there are 7 gaps between them in which 4 women can sit in ${}^7{P_4}$ ways.
So total ways are 6! x ${}^7{P_4}$

18.  ${x^y} + {y^x} = 46$ Find x & y values ?
Sol: ${1^{45}} + {45^1} = 46$
      Hence x = 1, y = 45

19.  In 10 years, A will be twice as old as B was 10 years ago.  If A is  now 9 years older than B the present age of B is 
Soln: A +10=2(B-10) ........(1)    
      A =B + 9 ......... (2)
      from equations. 1 & 2
      we get B = 39 A will be 39+9=48 years old.

20. A student can select one of 6 different math book, one of 3 different chemistry book & one of 4 different science book.In how many different ways students can select book of math, chemistry & science.
Sol:  ${}^6{C_1} \times {}^3{C_1} \times {}^4{C_1}$ = 6x3x4=72 ways

21. Sum of two number is 50 & sum of three reciprocal is 1/12 so find these two numbers
Sol :  x+y = 50 .....(1)  x=50-y ....(2)
$\displaystyle\frac{1}{x} + \frac{1}{y} = \frac{1}{{12}}$ $ \Rightarrow \displaystyle\frac{{y + x}}{{xy}} = \frac{1}{{12}} \Rightarrow 12(y + x) = xy$ ...(3)
put (2)  in  (4)
$ \Rightarrow $ 12(y+50-y)=(50-y)y
$ \Rightarrow $ 12y+600-12y=50y-${y^2}$
$ \Rightarrow $ ${y^2}$-50y+600=0
$ \Rightarrow $ ${y^2}$-30y-20y+600=0
$ \Rightarrow $ y(y-30)-20(y-30)=0
$ \Rightarrow $ (y-20) (y-30)=0
y=20 or y=30
if y=20 then x = 30
or y=30 then x = 20
two numbers are 30 & 20