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5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16

Clearly there are more than 8 vouchers as each of four can get at least 2. So,

Quote:

basically 120 ways vouchers can the distributed are the ways to distribute x-8 vouchers

, so that each can get from zero to x-8 as at "least 2", or 2*4=8, we already booked. Let x-8 be k.

Answer: C (15).

P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r-1C_{r-1}\).

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\).

Hope it helps.

Bunuel, I have a question on direct formula as well as on the question itself-Can we generalize the formula in case if out of n items, if r people has to share say more than 1 items (at least >=k items, where k >=2) .

Also could not get really why 120 should be the way of distributing x-8 vouchers
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10. How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3?[/b] (A) 72 (B) 76 (C) 78 (D) 80 (E) 84

It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...

From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.

We'll have 8 sets of collinear points of three: 3 horizontal {(1,1),(2,1),(3,1)} {(1,2)(2,2)(3,2)}... 3 vertical 2 diagonal {(1,1)(2,2)(3,3)}{(1,3)(2,2)(3,1)}

So the final answer would be; 9C3-8=84-8=76

Answer: B.

Hope it's clear.

Bunuel,

1- Collinear point issue will arise in case of overlapping values of x, y ? (as in here we have all the overlapping range for x & y). Also since range here is small for both x, y (ie.=3) we can manually calculate the collinear points but in case of large range how do we go about it ? should it be = # overlapping points on X + # overlapping points on Y + # diagonal points (which will essentially be min(# overlapping points on X , Y) -1 )-- Not so sure on this though ...

2- I see a similar Question in OG12 PS Q.229- The method explained here in the above example does not seems to fit too well there. basically in the question we have -4 <= X <=5, 6<= Y <=16. Can you please throw some light in the context of OG question....

1- Collinear point issue will arise in case of overlapping values of x, y ? (as in here we have all the overlapping range for x & y). Also since range here is small for both x, y (ie.=3) we can manually calculate the collinear points but in case of large range how do we go about it ? should it be = # overlapping points on X + # overlapping points on Y + # diagonal points (which will essentially be min(# overlapping points on X , Y) -1 )-- Not so sure on this though ...

2- I see a similar Question in OG12 PS Q.229- The method explained here in the above example does not seems to fit too well there. basically in the question we have -4 <= X <=5, 6<= Y <=16. Can you please throw some light in the context of OG question....

1- Collinear point issue will arise in case of overlapping values of x, y ? (as in here we have all the overlapping range for x & y). Also since range here is small for both x, y (ie.=3) we can manually calculate the collinear points but in case of large range how do we go about it ? should it be = # overlapping points on X + # overlapping points on Y + # diagonal points (which will essentially be min(# overlapping points on X , Y) -1 )-- Not so sure on this though ...

2- I see a similar Question in OG12 PS Q.229- The method explained here in the above example does not seems to fit too well there. basically in the question we have -4 <= X <=5, 6<= Y <=16. Can you please throw some light in the context of OG question....

Yeah thanks for this (however i had found this through customized search) however I am still struggling with my original question- consider this if I have range say 1<=x<=5 and -9<=y<=3 then (instead of 1<x<=3, 1<=y<=3 as stated in the above question) then how do I about (the question explained there is for right angle triangle- slightly different case than this one)....

I hope i am not complicating too much here.
_________________

Thanks a bunch Bunuel for all this nice set of questions (really helped me to do more and more!). However Just one note to all those who are trying this set - Please solve these sets once you have gained some confidence !
_________________

8. How many positive integers less than 10,000 are such that the product of their digits is 210? (A) 24 (B) 30 (C) 48 (D) 54 (E) 72

210=1*2*3*5*7=1*6*5*7. (Only 2*3 makes the single digit 6).

So, four digit numbers with combinations of the digits {1,6,5,7} and {2,3,5,7} and three digit numbers with combinations of digits {6,5,7} will have the product of their digits equal to 210.

{1,6,5,7} # of combinations 4!=24 {2,3,5,7} # of combinations 4!=24 {6,5,7} # of combinations 3!=6

24+24+6=54.

Answer: D.

10. How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤3 and 1≤y≤3? (A) 72 (B) 76 (C) 78 (D) 80 (E) 84

It would be better if you draw it while reading this explanation. With the restriction given (1≤x≤3 and 1≤y≤3) we get 9 points, from which we can form the triangle: (1,1), (1,2), (1,3), (2,1)...

From this 9 points any three (9C3) will form the triangle BUT THE SETS of three points which are collinear.

We'll have 8 sets of collinear points of three: 3 horizontal {(1,1),(2,1),(3,1)} {(1,2)(2,2)(3,2)}... 3 vertical 2 diagonal {(1,1)(2,2)(3,3)}{(1,3)(2,2)(3,1)}

So the final answer would be; 9C3-8=84-8=76

Answer: B.

Hope it's clear.

i just want to thank you bunuel but i still have some question to make it clear:

lets say that i was given 5 points for y and the same 5 for x: so it will be choosing 25c3- 5 vertical-5 horizontal and 2 diagonals

and to make it even more difficult: lets say that there where 6 points for x and 3 for y: so it will be 18c3-6 horizontal and 3 vertical - 2 diagonals or that is a bit surprise in here?

let's say that the Q was: How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6? how will it be solved: will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

let's say that the Q was: How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6? how will it be solved: will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance

While I seriously doubt whether one could encounter such long range question (esp. because calculating # diagonals is going to be little tricky here) unless you are shooting for 51 in quant.

That said let me try my hands-

Think about when it will be horizontal collinear- all the y values are same for a given set of X values. so we have 6 values where Y can be same (it has to be integer coordinate)- so total # horizontal collinear points- 6 You can have similar argument for vertical (constant X and vary Y) set of collinear points- 6

For # diagonals (please refer tot the attachment, I sketched only one side of the diagonals ) - you should be able to count the numbers now. For one side it comes out that we will have 16 such pairs (of 3 points) so by symmetry you need to multiply by 2. SO a total # diagonals will be 32.

let's say that the Q was: How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6? how will it be solved: will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance

I think your question is quite similar to yogesh1984's question above. I missed answering his question (thought of doing it later due to the diagram involved but it skipped my mind). Anyway, let me show you how I would solve such a question. Both the questions can be easily answered using this method.

How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6?

5. Mrs. Smith has been given film vouchers. Each voucher allows the holder to see a film without charge. She decides to distribute them among her four nephews so that each nephew gets at least two vouchers. How many vouchers has Mrs. Smith been given if there are 120 ways that she could distribute the vouchers? (A) 13 (B) 14 (C) 15 (D) 16 (E) more than 16

Answer: C.

Clearly there are more than 8 vouchers as each of four can get at least 2. So, basically 120 ways vouchers can the distributed are the ways to distribute \(x-8\) vouchers, so that each can get from zero to \(x-8\) as at "least 2", or 2*4=8, we already booked. Let \(x-8\) be \(k\).

In how many ways we can distribute \(k\) identical things among 4 persons? Well there is a formula for this but it's better to understand the concept.

Let \(k=5\). And imagine we want to distribute 5 vouchers among 4 persons and each can get from zero to 5, (no restrictions).

Consider:

\(ttttt|||\) We have 5 tickets (t) and 3 separators between them, to indicate who will get the tickets:

\(ttttt|||\) Means that first nephew will get all the tickets,

\(|t|ttt|t\) Means that first got 0, second 1, third 3, and fourth 1

And so on.

How many permutations (arrangements) of these symbols are possible? Total of 8 symbols (5+3=8), out of which 5 \(t\)'s and 3 \(|\)'s are identical, so \(\frac{8!}{5!3!}=56\). Basically it's the number of ways we can pick 3 separators out of 5+3=8: \(8C3\).

So, # of ways to distribute 5 tickets among 4 people is \((5+4-1)C(4-1)=8C3\).

For \(k\) it will be the same: # of ways to distribute \(k\) tickets among 4 persons (so that each can get from zero to \(k\)) would be \((K+4-1)C(4-1)=(k+3)C3=\frac{(k+3)!}{k!3!}=120\).

\((k+1)(k+2)(k+3)=3!*120=720\). --> \(k=7\). Plus the 8 tickets we booked earlier: \(x=k+8=7+8=15\).

Answer: C (15).

P.S. Direct formula:

The total number of ways of dividing n identical items among r persons, each one of whom, can receive 0,1,2 or more items is \(n+r-1C_{r-1}\).

The total number of ways of dividing n identical items among r persons, each one of whom receives at least one item is \(n-1C_{r-1}\).

Hope it helps.

Awesome...hats off...this is totally new to me...widens my realm..and strengthens my reasoning...thanks a lot
_________________

The night is at its darkest just before the dawn... never, ever give up!

yangsta, i liked your solution for 4. I didnt know we can use the definition of linear equation to solve such problems.

I used the guessing method. we have two relationships...6--30 and 24---60. This means when R is increased 4 times, S increases 2 times, so if R is increased 2 times S will increase 1 time. Now, 30*3 ~ 100, so 3 times increase in S will have atleast a 6 times increase in R, i.e. R should be something greater than 36..closest is 48

Another method (let me call it intuition method) :

6 on scale R corresponds to 30 on scale S and 24 on scale R corresponds to 60 on scale S. If we notice the relationship, we will see that for every 6 points on scale R, 10 points move on scale S. So, 90 points on scale S corresponds to 42 points on Scale R and another 6 points of scale S for another 10 points on scale R. Hence 100 on scale S corresponds to 42+6 = 48 on scale R.

I hope I am making sense
_________________

Ifmypostdida dancein your mind, send methe stepsthrough kudos :)

My MBA journey at http://mbadilemma.wordpress.com/

let's say that the Q was: How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6? how will it be solved: will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance

I think your question is quite similar to yogesh1984's question above. I missed answering his question (thought of doing it later due to the diagram involved but it skipped my mind). Anyway, let me show you how I would solve such a question. Both the questions can be easily answered using this method.

How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6?

Ok, so we have a total of 36 co-ordinates (as shown below by the red and black dots). We need to make triangles so we need to select a triplet of co-ordinates out of these 36 which can be done in 36C3 ways. Out of these, we need to get rid of those triplets where the points are collinear. How many such triplets are there? Look at the diagram:

Attachment:

Ques2.jpg

The Black dots are the outermost points. Red dots are the inside points. Now each of these red dots is the center point for 4 sets of collinear points (as shown by the red arrows). Hence the 4*4 = 16 red dots will make 16*4 = 64 triplets of collinear points. These 64 triplets account for all collinear triplets except those lying on the edges. Each of the 4 edges will account for 4 triplets of collinear points shown by the black arrows. Hence, there will be another 4*4 = 16 triplets of collinear points. Total triplets of collinear points = 64 + 16 = 80 Therefore, total number of triangles you can make = 36C3 - 80

Similarly you can work with 1<=x<=5 and -9<=y<=3. The number of red dots in this case = 11*3 = 33 So number of collinear triplets represented by red arrows will be = 33*4 = 132 Number of black arrows will be 3 + 11 + 3 + 11 = 28 Total triplets of collinear points = 132 + 28 = 160 Total triangles in this case = 65C3 - 160

Ma'am, It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw. my way: no: of collinear points=? horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs. hence horz and vert. lines give a total of 2*6*6C3. next 2 diagonals give same no: of such possibs. consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3.

along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3).

total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3).

It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw. my way: no: of collinear points=? horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs. hence horz and vert. lines give a total of 2*6*6C3. next 2 diagonals give same no: of such possibs. consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3.

along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3).

total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3).

Yes, I did miss out on the non-adjacent collinear points! And on the face of it, your calculation looks correct. I will put some more time on this variation tomorrow (since today is Sunday!) and get back if needed.
_________________

let's say that the Q was: How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6? how will it be solved: will 3C36 minus 6 vertical and 6 horizontal minus 2 diagonals will be the answer or will the answer be different.

thank's in advance

I think your question is quite similar to yogesh1984's question above. I missed answering his question (thought of doing it later due to the diagram involved but it skipped my mind). Anyway, let me show you how I would solve such a question. Both the questions can be easily answered using this method.

How many triangles with positive area can be drawn on the coordinate plane such that the vertices have integer coordinates (x,y) satisfying 1≤x≤6 and 1≤y≤6?

Ok, so we have a total of 36 co-ordinates (as shown below by the red and black dots). We need to make triangles so we need to select a triplet of co-ordinates out of these 36 which can be done in 36C3 ways. Out of these, we need to get rid of those triplets where the points are collinear. How many such triplets are there? Look at the diagram:

Attachment:

Ques2.jpg

The Black dots are the outermost points. Red dots are the inside points. Now each of these red dots is the center point for 4 sets of collinear points (as shown by the red arrows). Hence the 4*4 = 16 red dots will make 16*4 = 64 triplets of collinear points. These 64 triplets account for all collinear triplets except those lying on the edges. Each of the 4 edges will account for 4 triplets of collinear points shown by the black arrows. Hence, there will be another 4*4 = 16 triplets of collinear points. Total triplets of collinear points = 64 + 16 = 80 Therefore, total number of triangles you can make = 36C3 - 80

Similarly you can work with 1<=x<=5 and -9<=y<=3. The number of red dots in this case = 11*3 = 33 So number of collinear triplets represented by red arrows will be = 33*4 = 132 Number of black arrows will be 3 + 11 + 3 + 11 = 28 Total triplets of collinear points = 132 + 28 = 160 Total triangles in this case = 65C3 - 160

Ma'am, It would like to point out tht the resoning given is wrong. the triplets need not necessarily be adjacent. tht's the flaw. my way: no: of collinear points=? horizontal and vertical lines both give the same no: and each line of 6 points gives 6C3 possibs. hence horz and vert. lines give a total of 2*6*6C3. next 2 diagonals give same no: of such possibs. consider any diagonal direction. it gives 3,4,5,6,5,4,3 collinear points along 6 parallel lines corresponding to any diagonalic direction and each of the points gives us their corresponding triples-3C3+4C3+5C3+6C3+5C3+4C3+3C3.

along 2 such dirs. this adds up to 2*(2*(3C3+4C3+5C3)+6C3).

total no: of line forming selections= 2*6*6C3+ 2*(2*(3C3+4C3+5C3)+6C3).

Can you please elaborate on the bolded part in details...
_________________

This is a 6x6 square. For each diagonal of this square, you have 8 parallel lines, you can draw within the square by joining the vertices that lies on the edges of the square. eg: Join (1,2) & (2,1); (1,3) & (3,1); (1,4) & (4,1); (1,5) & (5,1); to get 4 parallel lines along the diagonal (1,6)-(6,1) Similarly you can get 4 lines on the other side of the diagonal.

Of these, (line joining (1,2) to (2,1) is of no use to us since it contains only 2 points within the square) the line joining point (1,3) & (3,1) contains total of 3 integer co-ordinates, the line joining point (1,4) & (4,1) contains total of 4 integer co-ordinates, and so on.....

Any 3 points that you select from these lines will be collinear and not form a traingle. Thus, you have 3,4,5,6,5,4,3 points collinear along the lines parallel to the diagonal. Rest as akhilesh has mentioned.

You may draw a figure by plotting these points.

My 1st post on this forum, so Apologies for the weird explanation.

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