CAT Dates and Other Exam Dates
Important Dates for CAT, XAT, MAT, IIFT, SNAP and JMET Management Exams:
Common Admission Test (CAT 2009) Dates
CAT Advertisement : Sunday, 30 August 2009
Sale of CAT Vouchers : Wednesday, 9 September 2009 – Thursday, 1 October 2009
Registration for CAT 2009 : Online CAT form Wednesday, 9 September 2009 – Thursday, 1 October 2009
CAT Test Dates : Saturday, 28 November 2009 – Monday, 7 December 2009
CAT Result : Friday, 22 January 2010 onwards
Rescheduling Window : Wednesday, 9 September 2009 – Thursday, 1 October 2009
XAT Dates
XAT Test Dates: 3rd January, 2010
XAT 2010 Prospectus: 1st September, 2009 to 30th November, 2009
Online Registration for XAT and XLRI: September 1, 2009
MAT Exam Dates
Date of MAT(Paper Based test): 6th September, 2009
Date of MAT(Online test): 7th September, 2009 onwords
Last date of availability of forms: 10th August, 2009
Last Date for receipt of complete application form: 13th August, 2009
Availability of Admit Cards: Seven days before the date of test
Online Filling of Management Institute Codes (MI Codes) of five chosen institutes where your MAT score to be sent: Up to the date of test.
SNAP Dates
Last Date for Registration: November 24, 2009 Tuesday
Payment Closes on: November 27, 2009 Friday
SNAP 2009 Test Date: 20th December 2009
SNAP 2009 Result: January 12, 2010 Monday
IIFT Dates
Last date for sale of prospectus by post: 20th August 2009
Last date for sale of prospectus across the IIFT counter: 3rd September 2009
Last date for receipt of completed application: 3rd September
Written test: 22nd November 2009 (10 am – 12 noon)
JMET Dates
Commencement of submission of data for ONLINE Application: Monday, September 7, 2009
Last date for submission of data for Online Application (website closure): Friday, 9th October (6.00Â p.m.)
Last date for Receipt of completed Offline and Online Application Form at IISc Bangalore Office: Friday, October 16, 2009
JMET Exam Date: Sunday, December 13, 2009
JMET 2010 Results: Monday, January 4, 2010
Related posts:
September 16th, 2009 at 12:14 am
1508 inches
September 16th, 2009 at 12:55 am
@Somit
how did u say that?
September 16th, 2009 at 12:59 am
Is it 48*5=240 inch?
September 16th, 2009 at 1:07 am
@Smita
Can you explain?
@All
Try to solve by drawing a picture.It will be easy for you.
September 16th, 2009 at 1:43 am
no its [( 48^2+ 5^2)^0.5]*5=241 accurately but as the verticle height is negligible in front of circumference , we can consider last case too
September 16th, 2009 at 1:46 am
if we’ll open the trajectory of creeper as a single lin then the verticle height( pitch ) will be the height , circumference would be base so the length covered by creeper will be given by pythagorus theorem , for one full twis … later it’ll be multiplied by 5 as the total height is 5 times the height covered in one twist
September 16th, 2009 at 2:19 am
@Smita
in “( 48^2+ 5^2)^0.5″ from where did 5^2 come ?Can u tell us?
September 16th, 2009 at 2:50 am
the answer should be 510 inches..
September 16th, 2009 at 2:58 am
@Ankit
Can you justify your answer with some explanation?
September 16th, 2009 at 3:48 am
ard 325!!! nt sure thg
September 16th, 2009 at 3:52 am
sorry kindly delete d prev post!!
d answr is 510 inches
x^2 = 90^2 + 48^2
x=102
102*5 = 510 inches!!!
September 16th, 2009 at 3:59 am
is it right admin??
September 16th, 2009 at 4:06 am
@Ginu
I will give answer tomorrow.Now I can’t tell you if it is right or not.Let other people also take part.
September 16th, 2009 at 10:10 am
964.2857 inches is the answer……………
since a twist of 90 inches on circumference of 48 inches will give it height 42 cms…………….
upto height 420 it will take 10 twists so height is 900 inches ….
now for remaining 30 inches, in 21 inches it will grow 45 inches
for remaining 9 inches for every 7 inches it grows 15 inches.
for remaining 2 inches it grows 90/21 = 4.2857 inches
so total height is 900+45+15+4.2857=964.2857 inches
September 16th, 2009 at 10:15 am
964.2857
for 420 inches creep will have 10 twist each of height(90-48=42) inches,so it will grow 900 inches long.
for 30 inches height it will grow 30×90/42=64.2857
so total length of creep is 964.2857 inches
September 16th, 2009 at 10:43 am
I think the answer is 240 inch
September 16th, 2009 at 11:39 pm
The right answer is 510 inches.
Here is the solution:
It may be noted that:
Width of the rectangle = Circumference of the cylinder = 48 inches.
Height of the rectangle = Vertical distance on the cylinder = 90 inches (in one twist).
Using the Pythagorean Theorem for a right-angled triangle,
Length of the hypotenuse = (48^2 + 90^2)^1/2 = 102 inches.
Now, the number of twists the creeper makes around the tree trunk is 5 (= 450 / 90).
If the length of the creeper (as given by the hypotenuse) is 102 inches in one twist, then the total length of the creeper in 5 twists is 510 inches.
September 18th, 2009 at 8:59 pm
Nice content indeed! i will visit as often as i can.
cheers
October 8th, 2009 at 11:42 am
hai.. can u tel me, whether MAT exams will be conducted in march 2010 or in which month it will be conducted before may in 2010? .. leave me a reply…
October 8th, 2009 at 11:37 pm
@Arunbal: Date of MAT February 2010 is 7th February 2010 (Sunday).