Well m planning for abroad. Is US a better option then UK ..?? can anyone plz let me know about some colleges over there ..??? how is canada for an option ..?? Are there any reputed colleges in CAnada..?
hello,
In canada, you can consider McGill Univ., HEC Montreal, York Univ. etc...
I'll suggest you to go for US rather UK, if you can afford.. US is bit costly than UK. In US, MBA is for 2 years than UK's MBA of 1 year.
US business school selection depends on your academic, GMAT score and work exp. 
Refer US news ranking for MBA program....
Thnaks
thanks for the ans
Please solve the below question.....please explain.....
In a 4 person race, medals are awarded to the fastest 3 runners. The first-place runner receives a gold medal, the second-place runner receives a silver medal, and the third-place runner receives a bronze medal. In the event of a tie, the tied runners receive the same color medal. (For example, if there is a two-way tie for first-place, the top two runners receive gold medals, the next-fastest runner receives a silver medal, and no bronze medal is awarded). Assuming that exactly three medals are awarded, and that the three medal winners stand together with their medals to form a victory circle, how many different victory circles are possible?
A) 24 B) 52 C) 96 D) 144 E) 648
hello,
In canada, you can consider McGill Univ., HEC Montreal, York Univ. etc..............
swagatamgupta SaysWell m planning for abroad...............
hello@swagatamgupta
are you looking for MBA from abroad or India?.....................
swagatamgupta Saysplz let me inform about some colleges where i can apply after my GMAT. plz guys waiting for ur replys
Guys. This is a GMAT "math" problem thread. Not a personal problems thread. Please use the correct threads to get better answers. This thread http://www.pagalguy.com/discussions/gmat-query-center-2010-13-25055904 probably already has similar questions and more people who could help you
Please solve the below question.....please explain.....
In a 4 person race, medals are awarded to the fastest 3 runners. The first-place runner receives a gold medal, the second-place runner receives a silver medal, and the third-place runner receives a bronze medal. In the event of a tie, the tied runners receive the same color medal. (For example, if there is a two-way tie for first-place, the top two runners receive gold medals, the next-fastest runner receives a silver medal, and no bronze medal is awarded). Assuming that exactly three medals are awarded, and that the three medal winners stand together with their medals to form a victory circle, how many different victory circles are possible?
A) 24 B) 52 C) 96 D) 144 E) 648
Is the answer C?
3 people (to win medals) have to be picked out of 4 people. So 4C3 = 4 ways
3 people can be arranged in a circle in (n-1)! ways = 2 ways
These are independent events so 4x2 = 8
Medals won can be
#1 G-S-B - 3 medals given to 3 people in 3! ways. = 6
#2 G-G-S - 3 medals given to 3 people with 2 ways being identical = 3! / 2 = 3
#3 G-S-S - 3 medals given to 3 people with 2 ways being identical = 3! / 2 = 3
Thus, total number of ways = (6+3+3) * 8 = 96
P.S: Personally I think there should be another case G-G-G. But I couldn't find 104 in the options. So went with C. Curious to know the official answer though.
Yeah official answer is c......even i calculated the answer 104...so was bit confused :-)......thanks for the explanation.....
sidhjenu SaysYeah official answer is c......even i calculated the answer 104...so was bit confused :-)......thanks for the explanation.....
I put in the last case only for the sake of completion and prejudice due to thinking the way CAT sets its papers.
On the GMAT you are expected to solve the question within two minutes. There was already enough of complexity in the problem.
a) Selecting people and selecting medals
b) Arranging them in a circular fashion
c) Realizing that G-S-S will still have two variations in a circle because of the two people winning silver being unique.
In such a case GMAT is unlikely to test you on a situation that was not mentioned (A 3 way tie) unless you were in the range of a 50 score in quant. Also, remember that process of elimination is one of the best methods of solving problems in any multiple choice exam.
All the best.
Gave a 2nd thought......i think 3 Gold is not possible because then the 4th one will be 2nd...and according to the problem exactly 3 persons can b awarded. So the answer must be 96. Let me know what you think.:-)
1)A number 15 is divisible into three parts which are in AP and the sum of their squares is 83.find the smallest number..the given answer is 3
2)The sum of three numbers in GP is 14 and the sum of of their squares is 84.find the largest number..the given answer is 8..
pls explain in details with Steps
2)The sum of three numbers in GP is 14 and the sum of of their squares is 84.find the largest number..the given answer is 8..
pls explain in details with Steps
1)A number 15 is divisible into three parts which are in AP and the sum of their squares is 83.find the smallest number..the given answer is 3
2)The sum of three numbers in GP is 14 and the sum of of their squares is 84.find the largest number..the given answer is 8..
pls explain in details with Steps
1)Let the nos. be a-d,a,a+d.
Sum is 15
=>(a-d)+a+(a+d) = 15
=>a = 5.
Sum of squares = (a-d)^2+a^2+(a+d)^2 = 83
=>3a^2+2d^2 = 83
=>2d^2 = 8
=>d = +-2.
So, the nos. are 3,5,7(when u choose d=2), or 7,5,3(when u choose d=-2).
In either case, smallest no. is 3.
2)Let the nos. be a,ar,ar2 (I'm writing ar^2 as ar2 for this problem).
a+ar+ar2 = 14...(i)
Sum of squares = a2+a2r2+a2r4 = 84....(ii)
Take square of (i)....
a2+a2r2+a2r4+2a2r+2a2r2+2a2r3 = 196
=>2ar(a+ar+ar2) = 112 (substitute (ii))
=>2ar.14 = 112
=>ar = 4
=>a = 4/r
Put this value in (i)...
4/r+4+4r = 14
=>4/r+4r = 10
=>4r2-10r+4 = 0
=>(r-1/2)(r-2) = 0
=>r=1/2 or r=2
So, the nos. are 8,4,2(when u choose r=1/2) or 2,4,8(when u choose r = 2).
In either case, the largest number is 8
Hello mates, I want to order one of the Manhattan guides to get the 6 free tests.I am split between RC and CR guide.Which one would you suggest?I already have powerscore Cr and Aristotle SC guide.Will the Manhattan guide be useful? What about the RC guide?
The problem is I only want to order for the sake of getting those free tests.
Hello mates, I want to order one of the Manhattan guides to get the 6 free tests.I am split between RC and CR guide.Which one would you suggest?I already have powerscore Cr and Aristotle SC guide.Will the Manhattan guide be useful? What about the RC guide?
The problem is I only want to order for the sake of getting those free tests.
go for the RC !
Hello mates, I want to order one of the Manhattan guides to get the 6 free tests.I am split between RC and CR guide.Which one would you suggest?I already have powerscore Cr and Aristotle SC guide.Will the Manhattan guide be useful? What about the RC guide?
The problem is I only want to order for the sake of getting those free tests.
This thread is meant for discussing math problems on the GMAT. Please use
http://www.pagalguy.com/discussions/gmat-query-center-2010-13-25055904 thread for other questions. Thanks.
How many factors does 36 raised to the power '2' have?
a. 2
b. 8
c. 24
d. 25
e. 26
Someone please explain the answer pls. I find it difficult to imagine/understand the explanations in the OA.
How many factors does 36 raised to the power '2' have?
a. 2
b. 8
c. 24
d. 25
e. 26
Someone please explain the answer pls. I find it difficult to imagine/understand the explanations in the OA.
One easy way to find the total no of factors of a number --
1. break the number into prime factors: 36^2 = 6^4 = 3^4 * 2^4
2. Add 1 to the length(exponent value) of all prime factors:
2.a length of 3 is 4.. adding 1 gives 5,
2.b length of 2 is also 4.. adding 1 gives 5
3. multiply the outcoming lengths to get the total factors : 5 * 5 = 25
This happens because to be a factor, a number must have the same
primes, and raised to the same or lower powers. Consider 36 = 2^2 * 3^2
total factors = (2+1)*(2+1) = 9 , which are 1, 2, 3, 4, 6, 9, 12, 18, 36
2 as a factor can be present in the form -2 raised to power 0,1 or 2.
Same is the case for 3 here.
We need to add 1 to consider the factor where other number(2 or 3 raised to power 0) is not involved.
1 = 2^0 * 3^0 //neither 2 not 3
2 = 2^1 * 3^0 // 3 is not involved
3 = 2^0 * 3^1 // 2 is not invloved
4 = 2^2 * 3^0 //3 is not involved
6= 2^1 * 3^1
9 = 2^0 * 3^2 // 2 is not invloved
..
..
36 = 2^2 * 3^2
So just remember the golden rule..
if n = a^x * b^y , where a and b are prime numbers,
then total factors of n = (x+1)(y+1)
if prime factos increases say n =a^x * b^y * c^z , likewise u can add the multiplication factor, total factors = (x+1)(y+1)(z+1)
Hi guys does anybody have OG Quant Review 2nd Edition !1 need help with resource.Can u please send it to
Kfactor21 SaysHi guys does anybody have OG Quant Review 2nd Edition !1 need help with resource.Can u please send it to
Kindly edit your post and remove ur personal id..it against the rules
an ore contains 25% of an alloy that has 90% iron.other than this i the remaining 75% of the pre there is no iron.how many kgs of the ore are needed to obtian 60 kg of pure iron?..
pls explain the proceedings..thanks in advance
an ore contains 25% of an alloy that has 90% iron.other than this i the remaining 75% of the pre there is no iron.how many kgs of the ore are needed to obtian 60 kg of pure iron?..
pls explain the proceedings..thanks in advance
1 Kg Ore has 0.25*1 Alloy , and 0.25*1*0.9 Iron
Inversely 60 Kg Iron have 60/(0.25*1*0.9) = 266.67 Kg Ore
..........
Cheers
Is x an even integer?
(1) is divisible by 4.
(2) is divisible by 16.
A. Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B. Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C. BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D. EACH statement ALONE is sufficient.
E. Statements (1) and (2) TOGETHER are NOT sufficient to answer the question asked, and additional data are needed.