Here are some of the solutions of GATE 2010 CS paper
1-C 2-D 3-C 4-A 5-B 6-A 7-B 8-A 9-B 10-A
11-A 12-C 13-B 14-C 16-D 17-B 18-C 19-C
20-B 21-A 22-B 24-A 25-D 26-A 27-A 28-D 29-D
Disclaimer :We do not guarantee the correctness of answers
GATE 2010 more Answers and solutions coming soon
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question paper was easy .could u plz let me answer?
it was an easy paper for cs people…….
cheers guyzz
where r the answersss … post them asap…
There are errors in this solutions….
I think for Qn 11 , it would be (d) 0 2
# cat ptr.c
#include
int f(int *p, int *q)
{
p = q;
*p = 2;
}
int i = 0, j = 1;
int main()
{
f(&i, &j);
printf ("%d %dn", i, j);
return;
}
# cc -go ptr ptr.c
# ./ptr
0 2
#
I think ans 4 Q (62) is (B) SGHI
here is right key made by a group of cusat(myself Rajeev) and is verified.
1-C,2-d,3-c,5-b,6-a,8-a,9-b,10-0,11-d,19-c,24-a,
26-b,29-d,33-b,35-c,41-c,44-d,51-b,52-c,59-d,
61-a,64-d
nt f(int *p, int *q)
{
p = q;
*p = 2;
}
int i = 0, j = 1;
int main()
{
f(&i, &j);
printf ("%d %dn", i, j);
return;
}
In this program i=0 and j=1 is passed to the functions so *p is 0 and * q is 1
The statement p=q will make p to point to q address . so p and q are pointing to same variable 1 . The statement p=2 will make both p and q point to 2 , so the output will be 2,2
Yeah Abhishek's right…
boss 11 ans is 0 1 bcoz function doesnt return anything more over it doesnt effect i, j
the answer for Q.11 is (0,2)…
i have chkd it on TC…
ha ha ha
@ above The question is based on call by reference not call by value , so its not necessary that it should return .All calculations are done on address . So 2,2 is the correct one
Man ans. is (0,2), because p=q, deferences the pointer p of i's memory. so p points to j not i,
so changing value of referenced p will change j only !!
Moreover GNU C++ Compiler also gave 0,2
i ran it on boraland tc
it is (0,2)
and btw abhishek, i suppose gate implicitly assumes that any code in the paper is in C and uses all the features stated by it unless stated other wise.
so think the actual ans would be the actual output of the program after executing the code fragment in a computer.
correct ans : 2-d,9-b,11-d,13-a,14-c,15-d,16-d,17-b,19-c,20-b,22-b,26-b,29-d,30-b,31-a,32-d,35-c,38-c,39-b,40-b,43-d,47-b,54-b,55-a
Q61 : c)1623
Q 61) A. 534
u add 1 to sum of digits@ units place,
add 2 to sum of digits@ tens place,
add 1 to sum of digits@ hundereds place.
where is the answer for instrumentation
int f(int *p, int *q)
{
p = q;
*p = 2;
}
int i = 0, j = 1;
int main()
{
f(&i, &j);
printf ("%d %dn", i, j);
return;
}
this will give ans "0 2"
after function call i value remains i=0;
but when p=q p point to j&q point to j.*p=2 means j=2.but not i va;ue changed
so the out put is i=0 j=2
i compiled in turbo c compiler
this gives the i=0 j=2 result
CAN SUM1 POST CS QUESTION PAPER (PHOTOS) PLS
Q 61) c
137+276 =435(octal)
therfore,
731+672=1623(octal)
hai every one…
i executed the 11th qn… it is option 'D'- 0 2
can any explain me solution for 54 and 55…
i got ans as option c,b
and 51 is option A..
the path is 1->0->3->4->2
total weight is 7..
check it out
I think option for 33 is C..
since an instruction cannot perform OF operation when another inst is performing PO or WO operation on same register…
eta ekta bal er moto answer shet
the answer for 51 is (A)
ya i got 41 marks according to the above key considering 51 as A and 11 as D
see
yea 51 is 'A' good job dude!!
what about 50
Q 51) B
1->0->4->2
option A would have been correct if the condition was NOT mentioned ie.AT MOST THREE EDGES….
any idea wat might be the cut-off??
answer of q 30 is c
Somebody please Post the Question paper also
Please somebody post the question paper
All questions papers are here
http://www.techvyom.com/2010/04/gate-cs-it-20-years-questions-papers.html
I want to crack the GATE Exam. Thanks for the help.