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Re: N is a natural number having the sum of its digits as 3. [#permalink]
15 Jan 2013, 02:58
chiccufrazer1 wrote:
akashganga wrote:
For the sum of digits to be equal to 3 only three combinations are possible
1. 1,1,1 and 10 0s (zeroes) 2. 1,2 and 11 (zeroes) 3. 3 and 12 (zeroes)
So the answer shall be arranging a thirteen digit number with cases 1,2 and 3
1. For this case total arrangements = (3(for first digit) * 12! (for rest of the digits))/3!*10!(because 3 1s and 10 zeroes are similar) this equals 66
2. Similarly for 2nd case arrangements = (2(again for first digit)*12!(for rest of the digits))/11!(for similar 11 zeroes) this equals to 24
3. For this case only one arrangement is possible, hence 1
total is 66+24+1 = 91 So answer is C
DJ
hey how did you come to a conclusion that it is a combination problem?
Posted from my mobile device
I think the more you practice, the easier it gets to tell what kind of a problem it is.
But also the question says how many values are possible. And when we look at options, if they would have been less than 10 I would have done each and every case manually, but since they are large in number, there is no other way of doing it in short period of time. So it is a Permutation and Combination problem.
In the following figure find angle ACB. Given C is the poin [#permalink]
15 Jan 2013, 03:23
In the following figure find angle ACB. Given C is the point of contact of two circles and A, B are the points of contact of the common tangent PP1 to the two circles. (Find figure in the attched file) a. 60 b. 90 c. 120 d. 89 e. Cannot be determined
Re: N is a natural number having the sum of its digits as 3. [#permalink]
15 Jan 2013, 03:41
wow that now makes sense..we are really not supposed to solve for large numbers manually but with a simple trick like combination..i was really confused..thanks
Re: In the following figure find angle ACB. Given C is the poin [#permalink]
15 Jan 2013, 10:29
Expert's post
pariearth wrote:
In the following figure find angle ACB. Given C is the point of contact of two circles and A, B are the points of contact of the common tangent PP1 to the two circles. (Find figure in the attched file) (A) 60 (B) 90 (C) 120 (D) 89 (E) Cannot be determined
Dear Pariearth
This is a very hard Geometry problem, much harder than you will see on the GMAT. I'm pretty good with Geometry, and it took me several minutes to puzzle out this particular problem.
Attachment:
tangent circles tangent to line.JPG [ 26.62 KiB | Viewed 1193 times ]
In the diagram above, D and E are centers of the circle. Of course, AD = CD, so ACD is an isosceles triangle, and similarly, CE = BE, so BCE is an isosceles triangle as well. We will need Mr. Euclid's remarkable theorem, the Isosceles Triangle Theorem.
Let angle DAC = p, and let the complement of p be q ----> p + q = 90, and 2p + 2q = 180.
In triangle ACD, by the Isosceles Triangle Theorem, angle DAC = angle ACD = p, and because the three angles of triangle ACD must sum to 180, we know angle D = 180 - 2p = 2q. The angles of triangle ACD are p & p & (2q).
Now, look at angle E. Segments AD and BE are parallel (they are both perpendicular to line AB, and perpendicular to the same thing means they're parallel). Because AD // BE, angles D & E are same side interior angles, which must be supplementary. Since angle D = (2q), angle E = (2p).
We know angle (BCE) = angle (CBE), again by the ITT. Because the three angles of triangle BCE must sum to 180, we can deduce that angle (BCE) = angle (CBE) = q. The angles of triangle BCE are q & q & (2p).
Now, look at the three angles at point C. The three angles form a straight line, so their sum must be 180.
Re: In the following figure find angle ACB. Given C is the poin [#permalink]
16 Jan 2013, 02:50
Dear Mike
May be this problem is not the "GMAT" type but the following solution is less time consuming-
We draw QO which is a common tangent to both the circles (as in attached figure). Now, AQ=QC & BQ= QC (tangents from an external point of a circle are equal. Let angle CAB = x & CBA=y therefore, angle ACQ = x & BCQ = y now, ACB+CAB+CBA=180 this implies, (x+y)+ x + y = 180 i.e., x+y = 90 or angle ACB= 90
In the figure below, AB, BC, BD & AC are chords of a circle. [#permalink]
16 Jan 2013, 22:12
In the figure below, AB, BC, BD & AC are chords of a circle.
The angle bisectors of \(\angle\)CBD & BCA intersect at E. EF is perpendicular to EC. If \(\angle CAB = 20^{\circ}\) and \(\angle BEF = \angle ABD\), then find\(\angle ACD\).
a. \(30^{\circ}\) b. \(35^{\circ}\) c. \(40^{\circ}\) d. \(25^{\circ}\) e. None of these
A group consists of m equally efficient workers. A piece of [#permalink]
20 Jan 2013, 22:14
A group consists of m equally efficient workers. A piece of work was started by one of them. The \(p^th\) worker (where \(2\leq\)p \(\leq m\)) joined the work (p - 1)x days after the \((p-1)^th\) worker joined. The work was completed in mx days after the \(m^th\) worker joined. The \(1^st\) worker received a share of $5500 out of the total $38500 paid to the group for completing the work. Find m.
A person invested a sum of $91.30. In how many years will th [#permalink]
20 Jan 2013, 22:26
A person invested a sum of $91.30. In how many years will the sum become $5000, if it is compounded every moment at the rate of 100% per annum? (given e = 2.72)
Re: A shopkeeper mixes three varieties of rice costing Rs.10, Rs [#permalink]
20 Jan 2013, 23:50
Expert's post
pariearth wrote:
A shopkeeper mixes three varieties of rice costing Rs.10, Rs.12 and Rs.17 per Kilogram. Which of the following represents the ratio in which the three varieties are mixed, if the shopkeeper makes a profit of 20% by selling the miture at Rs.15.60 per kilogram?
a. 9:14:36 b. 11:14:25 c. 14:36:43 d. 2:6:3 e. None of the above
Source - T.I.M.E CAT'11 Series
At first, we calculate the CP of mixture. CP = [SP/(100+% profit)]*100 = 15.60*100/120 = 13/-.
Now, look at the cost prices of 3 varieties of rices. These CPs are 10, 12, and 17. The mixture CP=13 is closest to II variety of rice (12). [Concept of weighted average] We can infer that among all 3 varieties, II variety ratio would be highest. Clearly, options A, B, & C are eliminated, because in all 3 options III variety ratio is highest. So the answer would be either D or E. Let's try with option D.
As per option D. CP of mixture would be.. =(2*10+6*12*3*17)/(2+6+3) = 13. It means option D is the correct answer.
The figure below shows the network connecting cities A, B, C [#permalink]
21 Jan 2013, 03:13
The figure below shows the network connecting cities A, B, C, D, E & F. The arrows indicate permissible direction of travel. What is the number of distinct paths from A to F?
a. 9 b. 10 c. 11 d. 12 e. 8
Source - T.I.M.E. CAT'11 Series
Attachments
Network.jpg [ 85.71 KiB | Viewed 1237 times ]
Last edited by pariearth on 22 Jan 2013, 03:26, edited 1 time in total.
Re: The figure below shows the network connecting cities A, B, C [#permalink]
21 Jan 2013, 19:47
Expert's post
pariearth wrote:
The figure below shows the network connecting cities A, B, C, D, E & F. The arrows indicate permissible direction of travel. What is the number of distinct paths from A to F?
a. 9 b. 10 c. 11 d. 12 e. 8
Source - T.I.M.E. CAT'11 Series
What about the lines which have no arrows? Are both directions permissible there? _________________
In a three digit number, the difference between its hundreds digit and its tens digit is equal to the difference between its tens digit and its units digit. Also the sum of the digits is 9. How many numbers are possible which satisfy the given condition?
Re: In the following figure find angle ACB. Given C is the poin [#permalink]
22 Jan 2013, 04:54
Expert's post
pariearth wrote:
In the following figure find angle ACB. Given C is the point of contact of two circles and A, B are the points of contact of the common tangent PP1 to the two circles. (Find figure in the attched file) a. 60 b. 90 c. 120 d. 89 e. Cannot be determined
Source: T.I.M.E- CAT'11 Series
As you have mentioned, this question is CAT relevant. The study material of CAT and that of GMAT are quite different. Although basic concepts are the same, it is not advisable to use CAT material for GMAT because it complicates things for no reason. There are a ton of things you need to 'know' for CAT which you don't need for GMAT. Your starting point in a CAT question is much ahead of your starting point in a GMAT question. Hence, this question isn't very relevant as far as GMAT is concerned.
Nevertheless, if you do get something like this, remember that you can solve by making the diagram according to your needs in PS questions. Make the two circles equal so that you can easily see the required angle. Mind you, GMAT doesn't have 'cannot be determined' as an option.
Attachment:
Ques3.jpg [ 8.87 KiB | Viewed 1218 times ]
You can see that it is an isosceles triangle with the ratio of the sides as \(\sqrt{2}R:\sqrt{2}R:2R i.e. 1:1:\sqrt{2}\). Hence it must be a right triangle and the required angle must be 90 degrees.
If the required angle is 90 in this case, it will be 90 no matter how you draw the diagram. _________________
A father starts from home at 3:00p.m. to pick his son from s [#permalink]
23 Jan 2013, 03:33
A father starts from home at 3:00p.m. to pick his son from school at 4:00p.m. One day the school got over early, at 3:00p.m. The son started walking towards home. He met his father on the way and both returned home 15 minutes earlier than the usual time. If the speed of the father is 35 km per hour, find the speed of the son (in km/hr).
Re: In the following figure find angle ACB. Given C is the poin [#permalink]
23 Jan 2013, 04:47
[quote="VeritasPrepKarishma As you have mentioned, this question is CAT relevant. The study material of CAT and that of GMAT are quite different. Although basic concepts are the same, it is not advisable to use CAT material for GMAT because it complicates things for no reason. There are a ton of things you need to 'know' for CAT which you don't need for GMAT. Your starting point in a CAT question is much ahead of your starting point in a GMAT question. Hence, this question isn't very relevant as far as GMAT is concerned.[/quote]
Dear Karishma, I do agree with you on the above fact and I would be very thankful to you if you could give a comparision between the Verbal section of GMAT & CAT as well as the IR section of GMAT with DI section of CAT. I am particularly tensned about the IR section. Since I have given CAT once, I know its reasoning section is highly tricky and tough. So do I need to put in the same level of effort for IR section?? Thanks in advance... -Pari
Re: In the following figure find angle ACB. Given C is the poin [#permalink]
23 Jan 2013, 21:40
VeritasPrepKarishma wrote:
The two tests are not very comparable but I can give you my thoughts.
Verbal Comparison - RC is similar in both the tests though I feel GMAT RCs are easier on average. GMAT has CR which is not there in CAT. It's very logical and easy to get through using reasoning skills. The SC section is tougher in GMAT than any section that tests grammar in CAT.
The IR section of GMAT is very new and hence untested. The questions will not be very tough. Once you get the hang of what to do in them, they should be fairly easy - certainly much easier than some DI questions that CAT can throw at you. Try some IR sample questions here: http://www.veritasprep.com/gmat/integra ... questions/
The point is, GMAT needs logic and reasoning. If your concepts are clear, there is no reason you wouldn't be able to figure the questions. CAT is a much harder test since you need to 'know' a lot of things.
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