GS 2010 :: TIFR exam questions

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Re: GS 2010 :: TIFR exam questions

Postby somyagupta » Thu Dec 17, 2009 9:43 pm

prabhakar97 wrote:3. Consider the following relations in a Database where the relations indicate certain Customers purchasing clothes in certain shops.
C(CustName)
S(ShopName)
B(CustName, ShopName)

What does the following Relational Algebra equation return ?

Pi(CustName)(CxS-B)

a. Customers who have brought clothes from at least 1 Shop.
b. Customers who have not brought clothes from any Shop.
c. Customers who have brought clothes from all the Shops.
d. Customers who have brought clothes from only one Shop.
e. Customers who have brought clothes from more than Shop.

I marked A but probably I got this one wrong in a hurry. I believe it should be B


CxS gives cartesian product of C and S, and pi(CxS-B) will result in those cust names which DID NOT buy from all the shops, which is correct acc. to this question.
However, the question asked in the Exam was: C- pi custname(CxS -B) , therefore we get: Customers who bought clothes from all the shops. : 'c'

prabhakar97 wrote:8. The minute and hour hand of a clock meet at 12 noon and midnight. In between these two times they meet N times. N is:
a. 6
b. 11
c. 12
d. 13
e. None of the above

I marked E. The answer is 10
The answer is None of the above, alright... but isn't the reason because the answer is Actually "0"?
The hour hand moves 1 unit in the clock every 5 minutes, while the minute hand moves a unit of clock every minute. They Do Not overlap at anytime except for at 12noon and midnight.


prabhakar97 wrote:11. What will be the coefficient of x^3 in the polynomial [(1+x)^3][(2+x^2)^10]?
a. 2^14
b. (3C3) + (10C3)
c. (3C3)*(10C3)*2^10
d. 32
e. (3C3)*(2^9)

I marked E. Use binomial expansion, and its pretty easy.

(1+x)^3 = 3C0 x^0 + 3C1 x^1 + 3C2 x^2 + 3C3 x^3
(2+x^2)^10 = 10C0 x^0 *2^10 + 10C1 x^2 * 2^9 + 10C2 x^4 *2^8+ ......
multipying the above terms for coeffient of x^3 we get:
(3C1)*(10C1)*(2^9) + (3C3)*(10C0)*(2^10)
= 3*10*2^9 + 2^10
= 2^9*(30+2)
=2^9 * 32
= 2^9 * 2^5
= 2^14
so answer should be A

prabhakar97 wrote:13. An unsorted array of n elements is sorted by determining the Maximum value from the array, and eliminating it. This is followed by evaluating the maximum value from the rest of the array elements and eliminating it... and so on. What is the maximum number of comparisons that will be performed in the worst case:
a. linear order of n
b. O(nlogn)
c. O(n^2)
d. O(n^1.5) but not more
e. same as heap sort

I marked A

err... isnt the algorithm they are telling the same as Heap sort in a crude sense? ... in the Worst case... wont the number of comparisons be same as Heap Sort? thats what I think.... btw... what logic did you use?
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Re: GS 2010 :: TIFR exam questions

Postby prabhakar97 » Fri Dec 18, 2009 2:30 pm

11. What will be the coefficient of x^3 in the polynomial [(1+x)^3][(2+x^2)^10]?
a. 2^14
b. (3C3) + (10C3)
c. (3C3)*(10C3)*2^10
d. 32
e. (3C3)*(2^9)

Oh.... I got it wrong too

8. The minute and hour hand of a clock meet at 12 noon and midnight. In between these two times they meet N times. N is:
a. 6
b. 11
c. 12
d. 13
e. None of the above

Man, this will be definitely 10. They will meet at 1:05, 2:11, 3:17, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49, 10:54 and finally at (11:60 thats 12:00 which is excluded) . Count them. You will get 10
Luck favours the lucky!!!
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Re: GS 2010 :: TIFR exam questions

Postby x94510man » Wed Dec 23, 2009 8:06 pm

this question was at the 1st page, bottom most question of the page .. i dont remember it exactly , neither the options, pls sum1 complete it

Q>> IF A MAN, DOES SUMTHING ..... HE WILL GO BANKKRUPT. BUT IF HE TAKES LOAN FROM BANK, HE WONT GO BANKKRUPT.HE TAKES LOAN FROM THE BANK .
OPTIONS WERE :
1> he goes bankkrupt.
2>he didnt go bankkrupt.
3>none

(pls some1 complete it)
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Re: GS 2010 :: TIFR exam questions

Postby x94510man » Wed Dec 23, 2009 8:08 pm

how many questions did u all attempt ? how many were correct ? what;s ur xpected marks ?
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Re: GS 2010 :: TIFR exam questions

Postby prabhakar97 » Fri Jan 15, 2010 1:05 pm

Result announced. Ankit Shah cracked it http://univ.tifr.res.in/gs2010/WT_Results/Interview%20Shortlist%20-%20CSS.htm

Total 32 candidates from CS have been selected.
Luck favours the lucky!!!
Practice makes a man(and women too) perfect!!!
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Re: GS 2010 :: TIFR exam questions

Postby subhadeepmaji » Fri Jan 22, 2010 5:03 pm

I also cleared the test..can anyone tell me how many guys they select finally!
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Re: GS 2010 :: TIFR exam questions

Postby sameer2009 » Sat Jan 23, 2010 11:10 am

prabhakar97 wrote:8. The minute and hour hand of a clock meet at 12 noon and midnight. In between these two times they meet N times. N is:
a. 6
b. 11
c. 12
d. 13
e. None of the above
Man, this will be definitely 10. They will meet at 1:05, 2:11, 3:17, 4:22, 5:27, 6:33, 7:38, 8:44, 9:49, 10:54 and finally at (11:60 thats 12:00 which is excluded) . Count them. You will get 10


yes 10 is correct...
they will meet
[1]while hour hand moves from 1 to 2
[2]while hour hand moves from 2 to 3
..
..
[9]while hour hand moves from 9 to 10
[10]while hour hand moves from 10 to 11
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