Tcs new pattern 2013 set 3

Tcs written exam :: 7th September , New Delhi

1.There is a circle with 2 triangles inscribed in it. in opposite direction making a star. The triangle is equilateral of side 12. u have to tell the area of the remaining portion of the circle.    

Q2. MOTHER +DAUGHTER+INFANT AGE IS 74. MOTHER AGE IS 46 MORE THEN DAUGHTER AND INFANT. AND INFANT AGE IS 0.4 OF DAUGHTER. FIND DAUGHTERS AGE.     

Q3.  SOME CARDS QUESTION. PROBABILITY OF 1 ACE WHICH IS RED AND ONE SPADE. SOMETHING LIKE THIS. DO GO THRU THE CARDS QUES. THE DECKS AND EVERYTHING.

Q4.THERE ARE EXACTLY 4 THURSDAYS AND 4 SUNDAYS IN A MONTH OF 31. FIND THE FIRST DAY. MY ANS IS MONDAY.

Q5. THEN THERE WERE 2-3 QUESTIONS BASED ON SPEED AND DISTANCE. EASY ENOUGH.

ONE WAS LIKE 2 CARS START FROM A POINT X. CAR1 START  AT 10 AM AND CAR2 AT SOME TIME. BOTH THE CARS GO TILL THEY REACH A T POINT. ONE TAKES LEFT OTHER RIGHT. IT WAS TO BE FOUND OUT HOW  MUCH WAS THE DISTANCE BETWEEN THE 2 CARS AT 2 PM.

Q6. ANOTHER TIME DISTANCE, A PERSON RUNS AROUND A RECTANGULAR FIELD. TIME GIVEN SPEED GIVEN. FIND THE DISTANCE. APPLY PERIMETER FORMULA 2(Length+breadth). And length and breadth ratio given. Find area.

Q7.THEN THERE WAS A SERIES COMPLETION QUESTION.

7^0,7^1 , 7^1+7^0,7^2… kind of a series.

Q8 THREE GROUPS :: X1,X2,X3

A,B COMMON TO ALL THREE. THEN ATLEAST 1 ELEMNT COMMON TO 2 IE X1,X2,X2,X3,X3,X1. FINDMINIMUM ELEMENTS IN EACH GROUP. ANSWER IS 4.    

Q9. U PEOPLE PROBABLY MUST HAVE DONE THE OPENSEESAME SAMPLE PAPER. THERE WAS 1 FLY

QUESTION THERE. IT JUMPS AT EVEY TICK 4 STEPS. THERE WAS A SIMILAR QUESTION AND THE INSECT WAS JUMPING IN A CIRCULAR QUESTIONS.    

Q10. Now A NEW ELEMENT.

THERE WERE2-3 VISUALIZING QUESTIONS ALSO.

A CUBE WAS GIVEN  IN 1D. WITH DOTTED LINES. FACE WERE MARKED. WITH 1 FACE AS X. U HAD TO TELL THAT WHAT WILL BE FACE OPPOSITE X WHEN IT IS FOLDED AS A CUBE.

Q11:

DI QUES

THERE WERE AT LEAST 6-7  QUESTIONS

LINES WERE GIVEN TO INTERPRET. WHOLE LOT OF LINES THEN WE HAD TO ANSWER 1 QUESTION BASED ON THE CONCLUSION

SOMETHING LIKE THIS(THIS QUES IS JUST A SAMPLE)

Directions for Q. 1 to Q. 5: Refer the data:

J, K, L, M and N collected stamps. They collected a total of 100 stamps. None of them collected less than 10. No two among them collected the same number.

(i) 3    collected the same number as K and M together.

(ii) L collected 3 more than the cube of an integer

(iii) The no. collected by J was the square of an integer.

(iv) Total no. collected by K was either the square or cube of an integer

1. The no. collected by J was:

(1) 27 (2) 49 (3) 36 (4) 64       

2. The no. collected by K was:

(1) 16 (2) 27 (3) 25 (4) 36      

3. The difference of numbers collected by L & M was:

(1) 3 (2) 2 (3) 5 (4) 9        

Solution

TCS New Aptitude Test New Delhi –Solution

Answer

in which at the center there common part of two triangle is hexagon so from he center of the circle draw 6 triangle in hexagon.
so total there is equilateral 12 triangle
after that their find altitude of one of the main triangle that is
(sqrt(3)/4)*(a*a)=1/2 * 12(base) * altitude

so altitude is=6sqrt(3)

now the small triangle below that altitude having the altitude is
(sqrt(3)/4)*(a*a) where a=4 because all 12 triangle is equilateral = 1/2 * 4(base) * altitude

sp altitude=2sqrt(3)

so diameter of circle=6sqrt(3) + 2sqrt(3) = 8sqrt(3)
so radius is = 4sqrt(3)

remaining area (ans) = pi * r * r – 12* (sqrt(3)/4) * a * a where a=4

so ans is=67.58

2.   m+d+I = 74

M=46+i+d

I=0.4*D

46+d+d0.4d+d+0.4d=74

D=10

6. APPLY PERIMETER FORMULA 2(Length+breadth)

8.  A, B are common to all Three. And another is common

1              2              3

A B            A B           A B

Now one is common in 1 and 2

One is common in  2 and 3

One is common in 3 and 1

So total no of common is 4(a+b+1+1)

9.

Calculate after 5 sec the fly will come to no 1.

So as 5 is a factor of 60 so fly will be in 1 after 60 min.

Directions for Q. 1 to Q. 5: Refer the data:

L=12,N=11,J=27,M=14,K=36

Given,

K+M+J+L+N=100 ———(i)

K,M,J,L,N>10 ———(ii)

K≠M≠N≠J≠L ———(iii)

K+M=J+L+N ———(iv)

L=3 + X^3 , where X is an integer ———(v)

J= Y^3, where Y is an integer (Modify the question here it should be like that) ———(vi)

K=Z^2 or Z^3, where Z is an integer ———(vii)

Now from the ques value of J can be 27 or 64

but if we take 64 then eqn iv would not satisfy so J=27

Again considering eqn ii, iv & ii we can say that L+N = 23 & K+M=50

Now from eqn v the value of L must be 11, thus L=11

That conclude N = 12

Now from the question part 3,

The difference of numbers collected by L & M was:

(1) 3 (2) 2 (3) 5 (4) 9

Considering the fact M-L=any of those values

Now M=L+any of those values

M=11+any of those values

Now considering each values, M= 14, value =3; K=50-14 =36 which is a square of integer, thus satisfies

again M=13, value = 2; K=37 which is neither square or cube of any integer

M=16, value = 5; K=34 which is neither square or cube of any integer

M=20, value = 9; K=30 which is neither square or cube of any integer

Thus M = 14 & K = 36.