TCS-Questions- 14

 TCS – Questions – 14

1.       A and B start from house at 10am. They travel from 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.
Ans: 160 km.

Distance between House and T junction = 20 x 2 = 40km since A travels at 20 kmph and reaches the junction at 12.00 noon.
B reached T at 11 am and continued walking till 2.00 pm ie another 3 hours in which time he covered further 20 km and thus in four hours he has travelled 160 km.   

A reached the T junction after travelling 2 hours covering a distance of 40 km and travelled another two hours in the opposite direction to that of B and  covered further 40 km. Thus B in total travelled 80km.

Thus the distance between them now is 120 + 40 = 160km. (Note that at the T junction they travel ibn opposite directions) (This question is repeated with change in variables and the directions)

2.       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, then who finishes 4th?
a) A        b) E        c) F         d) G
Ans: (d)

3.       4 Women & 6 men have to be seated in a row on condition that no two women can sit together. In how many ways these arrangements can be made?

Ans: 6! X 7P4 ways.

The six men can be seated in 6! Ways. There are now seven places available for the women to sit. (One before the first male and another after the last male, and in between there are five places for the women to sit) The women can be seated in these seven places in 7P4 ways.

4.       Find the total number of combinations of 5 letters a,b,a,b,b taking some or all at a time?
Ans: 11 ways.

As these five letters consists of only a and b, the

First letter can be chosen in 2 ways either a or b
2 letters can be chosen in 3 ways:  aa, ab, bb
3 letters can be chosen in 3 ways:  bbb, aab, bba
4 letters can be chosen in 2 ways: . aabb, bbba
5 letters can be chosen in 1 way (all the letters)
Thus, the total ways of combinations are 11

5.       Find the 55th word of SHUVANK in dictionary.

Ans: AHSNKUV
The letters of SHUVANK if arranged in alphabetical order we get:  A H K N S U V
The first letter will be A and the remaining 6 letters can be arranged in 6! Ways. This is far ahead of our requirement of 55^th word.

If we take the first two alphabets A and H, the remaining five letters can be arranged in 5! Ways ie 120 ways. This again is head of our requirement of 55^th word.

If we take the first three alphabets A H K then the remaining four letters can be arranged in

4! Ways ie 24 ways.

The next arrangement can be A H N. Again the remaining four letters can be arranged in 4! Ways ie 24 ways.

Next arrangement of letters will be A H S K and the remaining three letters can be arranged in 3! Ways.

We now have, alphabets starting from A H K can be arranged in 24 ways

Alphabets starting from A H N can be arranged in 24 ways

Alphabets starting from A H S K can be arranged in 6 ways. Totalling 54 ways

Hence the 55^th arrangement of word will be AHSNKUV  

6.       Car A leaves city C at 5pm and is driven at a speed of 40kmph.  2 hours later another car B leaves city C at 60 kmph and is driven in the same direction as car A. After how much time the car B will be 9 km ahead of car A.  

Ans: 4 hours 27 minutes (267 minutes)

Car A would have covered 80 kms in the two hour time. If car B were to lead car A by 9 km then it should cover totally 89km while both the vehicles continue running. The relative speed of the two vehicles comes into play.
Relative speed = 60 - 40 = 20 kmph

Hence to bridge the gap of 89 kms distance Car B will be taking 89/20 ie 4 9/20hours

Thus, the answer is 4 hours 27 minutes.

7.       n is a natural number and n^3 has 16 factors.  Then how many factors can n^4 have?
Ans: either 21 or 25

We know the total factors of a number N: (a^p)(b^q)(c^r) is equal to (p+1)(q+1)(r+1)

Since n^3 has 16 factors n^3 could be in either of the following formats namely

A^15 or a^3.b^3  In either case the number of factors are 16.

Hence the number of factors  

If n3 =a15 then n = a5 and number of factors of n4 = 21
n3 = a3.b3 then n = ab and number of factors n4 = 25

8.       W, X, Y, Z are integers.  The expression X - Y - Z is even and the expression Y - Z - W is odd.  If X is even what must be true?
a) W must be odd                            b) Y - Z must be odd
c) Z must be even                            d) Z must be odd
Ans: (a)

Since X is even, both Y, Z are either even or odd.
Y - Z in both cases is however even.  So (Y - Z ) - W = odd is possible only when W is odd

9.       1(1!)+2(2!)+3(3!)....2012(2012!) = ?

Ans: 2013!-1
1(1!)=1  ⇒ 2!-1
1(1!)+2(2!)=1+4=5 ⇒ 3!-1
1(1!)+2(2!)+3(3!)=1+4+18=23 ⇒ 4!-1
On the above rationale
1(1!)+2(2!)+3(3!)+........+2012(2012!)=2013!-1

10.   In a G6 summit held at London, a French, a German, an Italian, a British, a Spanish,and a Polish diplomats represented  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: (b)