Somebody please explain me this Either the question and its explaination are wrong or may be I am not able to understand..............
Q>If you draw two cards from a deck, what is the probability that you will get the Ace of Diamonds and a black card? A>There are two ways you can satisfy this condition: (a) You can get the Ace of Diamonds first and then a black card or (b) you can get a black card first and then the ace of diamonds. Let's calculate Case A. The probability that the first card is the Ace of Diamonds is 1/52. The probability that the second card is black given that the first card is the Ace of Diamonds is 26/51 because 26 of the remaining 51 cards are black. The probability is therefore 1/52 x 26/51 = 1/102. Now for Case B: the probability that the first card is black is 26/52 = 1/2. The probability that the second card is the Ace of Diamonds given that the first card is black is 1/51. The probability of Case 2 is therefore 1/2 x 1/51 = 1/102, the same as the probability of Case 1. Recall that the probability of A or B is P(A) + P(B) - P(A and B). In this problem, P(A and B) = 0 since a card cannot be the Ace of Diamonds and be a black card. Therefore, the probability of Case A or Case B is 1/102 + 1/102 = 2/102 = 1/51. So, 1/51 is the probability that you will get the Ace of Diamonds and a black card when drawing two cards from a deck.
cant see nething wrong wid this explanation.. with replacement is not mentioned in da question.. so this is da only way..
there r only two cases.. case 1: (1st card(ace of diamond) AND(x) 2nd card(black card)) OR(+) case 2: (1st card(black card) AND(x) 2nd card(ace of diamond))
i think this is da shortest possible method 2 solve this question..
and as phemanth has mentioned if the question requires ne ordering of draw of the cards or drawing of cards wid replacement, then it'll b very clearly mentioned in da question..
yeah...i spent a while trying to explain this. So I think it would be better if we can categorize a problem that we post into: a) "hey this is a great problem, I have solved it and want to share with you" b) "plz help me out with this"
That way others know how to respond. What say? Don't mean any offence but just a suggestion from a newbie pagalguy.
this will b difficult 2 do.. i dont know how i shd rate this problem.. the solution juz needed multiplication of 3 numbers.. so if i mention dat this is a tough problem then most probably i might b lying.. :)
for X: x-cordinates will have 10 values, y-cordinates 11 values = 110 total values..
for Y: since XY is parallel to x-axis, y-cordinates can have only 1 value for each value of y-cordinates of X, and x-cordinates have 9 values(if X n Y dont overlap) = 9x1 = 9 total values.
for Z: since the triangle is right-angled at X, XZ will b parallel(or concident) to y-axis.. same reasons as for Y, cordinates of Z can have 10x1 ways = 10 ways..
so total ways = 110 x 9 x 10 = 9900 (a simple multiplication)..
Guys please help me solve this problem...i am pretty weak in p&c;:
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420 B. 2520 C. 168 D. 90 E. 105
I approached it as follows: First select 4: 8C4 Then put rest 4 with the 4 selected initially = 4x3x2x1 = 4! So answer = 8C4 x 4! But this is not the answer i guess. plz help 😞
Guys please help me solve this problem...i am pretty weak in p&c;:
A group of 8 friends want to play doubles tennis. How many different ways can the group be divided into 4 teams of 2 people?
A. 420 B. 2520 C. 168 D. 90 E. 105
I approached it as follows: First select 4: 8C4 Then put rest 4 with the 4 selected initially = 4x3x2x1 = 4! So answer = 8C4 x 4! But this is not the answer i guess. plz help :(
either subtract the number of cases u've counted more than once in ur answer or use the direct formula ((4x2)!)/(2!)^4 x 4!
Can you let me know the formula to usefor such type of questions?
Thanks, Anurag...
Anurag, I wouldn't advice remembering formulas for permutations. You should be able to get to the GMAT level problems by practicing a few sets.
the question can be solved like this: 8 people - A B C D W X Y Z
One of the 8 can be paired with other in 7 ways. Remaining 6 people. In this one of them can be paired with the other in 5 ways. Remaining 4 people. Can be paired with the remaining in 3 ways. Remaining 2 people have to be paired. So total = 7 * 5 * 3 * 1 = 105.
I felt this is the best approach. I didn't solve it this way but I found this on the net and thought it will be useful to share. (no credit for me please)
Anurag, I wouldn't advice remembering formulas for permutations. You should be able to get to the GMAT level problems by practicing a few sets.
the question can be solved like this: 8 people - A B C D W X Y Z
One of the 8 can be paired with other in 7 ways. Remaining 6 people. In this one of them can be paired with the other in 5 ways. Remaining 4 people. Can be paired with the remaining in 3 ways. Remaining 2 people have to be paired. So total = 7 * 5 * 3 * 1 = 105.
I felt this is the best approach. I didn't solve it this way but I found this on the net and thought it will be useful to share. (no credit for me please)
Thanks. Hemanth
i dont agree with this solution.. its looking ridiculous.. One of the 8 can be paired with other in 7 ways. how?? even for choosing one of the 8, there will b 8 ways of doing so.. now where has this 8 gone in da calculation.. n wat abt those 6, 4 n 2 ppl..
but if u had known da answer b4 solving, then u can plug-in ne values n use ne reasoing as the reasoning used above.. :)
@anurag xy items can be divided into x groups, each containing y items(order of group not important) = ((xy)!)/((y!)^x)*(x!) [* denotes multiplication)
i dont agree with this solution.. its looking ridiculous.. One of the 8 can be paired with other in 7 ways. how?? even for choosing one of the 8, there will b 8 ways of doing so.. now where has this 8 gone in da calculation.. n wat abt those 6, 4 n 2 ppl..
but if u had known da answer b4 solving, then u can plug-in ne values n use ne reasoing as the reasoning used above.. :)
@anurag xy items can be divided into x groups, each containing y items(order of group not important) = ((xy)!)/((y!)^x)*(x!) [* denotes multiplication)
:) Buddy Dare2.... You can form 7 groups using 1 person. Makes sense? Lets say I pick A for this job. I have AB or AC or AD or AE or AW or AX or AY or AZ = 7 right?
Lets say you choose any of these groups. Lets say you form AB. Now from the remaining 6 guys how many groups of 2 can you form? Say you pick C for this job. So CD, CW, CX, CY, CZ. = 5 right?
Now lets say you make CZ from the above group. So you have formed AB, CZ. Remaining 4 guys. You can form 3 groups similarly.
This logic sounds ok to me. If you can objectively point out what the mistake is I am more than happy to correct myself.
:) Buddy Dare2.... You can form 7 groups using 1 person. Makes sense? Lets say I pick A for this job. I have AB or AC or AD or AE or AW or AX or AY or AZ = 7 right? this 7 ways is possible only if u make 1 of them fixed n choose 1 from the remaining 7..
Lets say you choose any of these groups. Lets say you form AB. Now from the remaining 6 guys how many groups of 2 can you form?2 can be choosen from 6 in 6C2 ways ie 15 ways Say you pick C for this job. So CD, CW, CX, CY, CZ. = 5 right? here again this is possible only if u r making C fixed..
Now lets say you make CZ from the above group. So you have formed AB, CZ. Remaining 4 guys. You can form 3 groups similarly.
This logic sounds ok to me. If you can objectively point out what the mistake is I am more than happy to correct myself. if this logic is ok.. then justify it in case if out of 15 players 5 teams are to be made with 3 players in each team..
Thank you. Hemanth
hemanth.. from this logic a better n standard method will b.. select 2 members 4 1st team = 8C2 select 2 members 4 2nd team = 6C2 (since 2 are already selected only 6 remains) .......................... 3rd team = 4C2 .......................... 4th team = 2C2
total ways = (8C2 x 6C2 x 4C2 x 2C2)/4! (since each team can be selected in each of the 4 groups) = 105 this logic will b a generic one..
direct formula always helps save time.. (i hope all software engineers will agree to this specially when calling functions n classes which r already there, instead of writing fresh code 4 them)
:) Buddy Dare2.... You can form 7 groups using 1 person. Makes sense? Lets say I pick A for this job. I have AB or AC or AD or AE or AW or AX or AY or AZ = 7 right? this 7 ways is possible only if u make 1 of them fixed n choose 1 from the remaining 7..
Lets say you choose any of these groups. Lets say you form AB. Now from the remaining 6 guys how many groups of 2 can you form?2 can be choosen from 6 in 6C2 ways ie 15 ways Say you pick C for this job. So CD, CW, CX, CY, CZ. = 5 right? here again this is possible only if u r making C fixed..
Now lets say you make CZ from the above group. So you have formed AB, CZ. Remaining 4 guys. You can form 3 groups similarly.
This logic sounds ok to me. If you can objectively point out what the mistake is I am more than happy to correct myself. if this logic is ok.. then justify it in case if out of 15 players 5 teams are to be made with 3 players in each team..
Thank you. Hemanth
hemanth.. from this logic a better n standard method will b.. select 2 members 4 1st team = 8C2 select 2 members 4 2nd team = 6C2 (since 2 are already selected only 6 remains) .......................... 3rd team = 4C2 .......................... 4th team = 2C2
total ways = (8C2 x 6C2 x 4C2 x 2C2)/4! (since each team can be selected in each of the 4 groups) = 105 this logic will b a generic one..
direct formula always helps save time.. (i hope all software engineers will agree to this specially when calling functions n classes which r already there, instead of writing fresh code 4 them)
You got it buddy !!!
This approach can be explained using a smaller number:- 4 friends : A,B,C,D. All the teams of 2 that can be formed are : ------------------------------------------ AB AC AD BA BC BD CA CB CD DA DB DC --------------------------------------------------- Totaling 12. But as you see AB=BA and AD=DA and so on. This is happening because a team can be selected in the other groups so we divide the result by 2 giving us 6 which is actually the number of unique teams if you count. --------------------------------------------------- Using P&C; we can solve it as:- ------------------------------- Choose 2 out of 4 : 4C2 Choose 2 out of remaining 2 : 2C2 total ways are 4*3*2/2 = 12. since the team can be chosen in other groups hence we divide the result by 2! giving us 6. -------------------------------------------------
This approach can be explained using a smaller number:- 4 friends : A,B,C,D. All the teams of 2 that can be formed are : ------------------------------------------ AB AC AD BA BC BD CA CB CD DA DB DC --------------------------------------------------- Totaling 12. But as you see AB=BA and AD=DA and so on. This is happening because a team can be selected in the other groups so we divide the result by 2 giving us 6 which is actually the number of unique teams if you count. --------------------------------------------------- Using P&C; we can solve it as:- ------------------------------- Choose 2 out of 4 : 4C2 Choose 2 out of remaining 2 : 2C2 total ways are 4*3*2/2 = 12. since the team can be chosen in other groups hence we divide the result by 2! giving us 6. -------------------------------------------------
Peace out everybody. :)
Cool ..... we solved the problem in many ways... thats the point. Well done Dare2 and Mr.Gail!
1) If a triangle has sides 8, 10, and z, which of the following cannot be the area of the triangle? (A)3 (B) 6 (C) 12 (D) 24 (E) 46 2) A store sells wrapping paper only in square sheets of integer length. If Robert has to wrap a cylinder of radius 3 centimeters and height 5 centimeters, what is the length, in centimeters, of the smallest sheet he can buy? (A)13 (B) 12 (C) 10 (D) 8 (E) 7 3) The ratio of men to women employed at Company X is m : w and the ratio of employees with a bachelors degree to employees with a masters degree is b : n. There are 75 men and 85 employees with masters degrees among the 150 employees at Company X. If all employees at Company X have a degree, which of the following could be the number of women with bachelors degrees? (A)85 (B)75 (C)66 (D)51 (E)It cannot be determined from information given. 4) Four identical cylinders are to be packed standing upright in the same direction into a rectangular shipping box with dimensions 3 x 12 x 4. What is the maximum possible volume of one of the cylinders? (A)48 (B)27 (C)12 (D)9 (E)6.75 5) How many different positive even integers are factors of 294? (A)4 (B) 6 (C) 7 (D) 9 (E) 11
1) If a triangle has sides 8, 10, and z, which of the following cannot be the area of the triangle? (A)3 (B) 6 (C) 12 (D) 24 (E) 46 2) A store sells wrapping paper only in square sheets of integer length. If Robert has to wrap a cylinder of radius 3 centimeters and height 5 centimeters, what is the length, in centimeters, of the smallest sheet he can buy? (A)13 (B) 12 (C) 10 (D) 8 (E) 7 3) The ratio of men to women employed at Company X is m : w and the ratio of employees with a bachelors degree to employees with a masters degree is b : n. There are 75 men and 85 employees with masters degrees among the 150 employees at Company X. If all employees at Company X have a degree, which of the following could be the number of women with bachelors degrees? (A)85 (B)75 (C)66 (D)51 (E)It cannot be determined from information given. 4) Four identical cylinders are to be packed standing upright in the same direction into a rectangular shipping box with dimensions 3 x 12 x 4. What is the maximum possible volume of one of the cylinders? (A)48 (B)27 (C)12 (D)9 (E)6.75 5) How many different positive even integers are factors of 294? (A)4 (B) 6 (C) 7 (D) 9 (E) 11
1) A. area of 3 is not possible. The smallest area works out when the z is really small, Z cannot be less than 2, and the triangle is a obtuse triangle. Assume Z is 2.1. then using (area) = (s*(s-10)(s-8 )(s-2.1))^1/2 we get around 4. Even if you take Z infinitesimally close to 2, area wouldn't change significantly from the value 4.
2) C. If you imagine to unfold the cylinder it forms a rectangle of side 5 and another side 2*pi*3(assume it is ~19). So a sheet of 85sqcm is good enough. If he buys a 10*10 sheet he will have sufficient paper. I am assuming you dont have to wrap the ends too.
3) D. If you work out using a table to put the values you see that 51 works out fine. The question which of these "could be" a value. 51 could definitely be one. A,B,C can never be those values. The value of women with bachelors degree can be 65 or less. There are basically 75 women and men. and 85 with masters and 65 with bachelors.
4)D. Assuming the box is L*B*H. I get the L*B = 3*12. This is a rectangular cross section with somewhat longish breadth. The best way you can fit the 4 cylinders is in a straight line (along the breadth) and each of them will have a diameter of 3. there the volume is closest to 9*pi.
5)B. 294 = 2*3*7*7. Form as many odd factors and multiply with 2. You get 2,6,14,42,98,294.
Please let me know if I have missed any corner case in these problems. I liked these problems and enjoyed doing them.
1) If a triangle has sides 8, 10, and z, which of the following cannot be the area of the triangle? (A)3 (B) 6 (C) 12 (D) 24 (E) 46 The best way of doing this problem is using the area of the triangle formula = .5*ab*Sin(included angle) This will give area as 40*Sin(included angle) Since Sin value can never be greater than 1 the answer would be 46. 2) A store sells wrapping paper only in square sheets of integer length. If Robert has to wrap a cylinder of radius 3 centimeters and height 5 centimeters, what is the length, in centimeters, of the smallest sheet he can buy? (A)13 (B) 12 (C) 10 (D) 8 (E) 7 2*pi*5*3=3.14*30 ~ 96 so answer should be 10 3) The ratio of men to women employed at Company X is m : w and the ratio of employees with a bachelors degree to employees with a masters degree is b : n. There are 75 men and 85 employees with masters degrees among the 150 employees at Company X. If all employees at Company X have a degree, which of the following could be the number of women with bachelors degrees? (A)85 (B)75 (C)66 (D)51 (E)It cannot be determined from information given.
Drawing a table for men/women and bachelors/graduates will give us the solution. It has to be less than 65 and question asks for possible solutions which gives us 51. 4) Four identical cylinders are to be packed standing upright in the same direction into a rectangular shipping box with dimensions 3 x 12 x 4. What is the maximum possible volume of one of the cylinders? (A)48 (B)27 (C)12 (D)9 (E)6.75 Diameter needs to be maximum and that can be achieved by keeping it at 3 which gives us the option D. 5) How many different positive even integers are factors of 294? (A)4 (B) 6 (C) 7 (D) 9 (E) 11 take the 2 out of the factors of 294 which gives us 7^2 and 3. So no. of factors is (2+1)*(1+1)=6
1) x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT x=w x>w x/y is an integer w/z is an integer x/z is an integer 2) For any integer k > 1, the term length of an integer refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 2 2 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y x and the length of y? 5 6 15 16 18
1) A. area of 3 is not possible. The smallest area works out when the z is really small, Z cannot be less than 2, and the triangle is a obtuse triangle. Assume Z is 2.1. then using (area) = (s*(s-10)(s-8 )(s-2.1))^1/2 we get around 4. Even if you take Z infinitesimally close to 2, area wouldn't change significantly from the value 4.
2) C. If you imagine to unfold the cylinder it forms a rectangle of side 5 and another side 2*pi*3(assume it is ~19). So a sheet of 85sqcm is good enough. If he buys a 10*10 sheet he will have sufficient paper. I am assuming you dont have to wrap the ends too.
3) D. If you work out using a table to put the values you see that 51 works out fine. The question which of these "could be" a value. 51 could definitely be one. A,B,C can never be those values. The value of women with bachelors degree can be 65 or less. There are basically 75 women and men. and 85 with masters and 65 with bachelors.
4)D. Assuming the box is L*B*H. I get the L*B = 3*12. This is a rectangular cross section with somewhat longish breadth. The best way you can fit the 4 cylinders is in a straight line (along the breadth) and each of them will have a diameter of 3. there the volume is closest to 9*pi.
5)B. 294 = 2*3*7*7. Form as many odd factors and multiply with 2. You get 2,6,14,42,98,294.
Please let me know if I have missed any corner case in these problems. I liked these problems and enjoyed doing them.
Thank you for a nice post. Hemanth
Sorry for the late reply The OAs are 1) E 2) A 3) D 4) D 5) B
1) x is the sum of y consecutive integers. w is the sum of z consecutive integers. If y = 2z, and y and z are both positive integers, then each of the following could be true EXCEPT
x=w x>w x/y is an integer w/z is an integer x/z is an integer 2) For any integer k > 1, the term "length of an integer" refers to the number of positive prime factors, not necessarily distinct, whose product is equal to k. For example, if k = 24, the length of k is equal to 4, since 24 = 2 2 2 3. If x and y are positive integers such that x > 1, y > 1, and x + 3y x and the length of y? 5 6 15 16 18
1) If a triangle has sides 8, 10, and z, which of the following cannot be the area of the triangle? (A)3 (B) 6 (C) 12 (D) 24 (E) 46 The best way of doing this problem is using the area of the triangle formula = .5*ab*Sin(included angle) This will give area as 40*Sin(included angle) Since Sin value can never be greater than 1 the answer would be 46. 2) A store sells wrapping paper only in square sheets of integer length. If Robert has to wrap a cylinder of radius 3 centimeters and height 5 centimeters, what is the length, in centimeters, of the smallest sheet he can buy? (A)13 (B) 12 (C) 10 (D) 8 (E) 7 2*pi*5*3=3.14*30 ~ 96 so answer should be 10 3) The ratio of men to women employed at Company X is m : w and the ratio of employees with a bachelors degree to employees with a masters degree is b : n. There are 75 men and 85 employees with masters degrees among the 150 employees at Company X. If all employees at Company X have a degree, which of the following could be the number of women with bachelors degrees? (A)85 (B)75 (C)66 (D)51 (E)It cannot be determined from information given. Drawing a table for men/women and bachelors/graduates will give us the solution. It has to be less than 65 and question asks for possible solutions which gives us 51. 4) Four identical cylinders are to be packed standing upright in the same direction into a rectangular shipping box with dimensions 3 x 12 x 4. What is the maximum possible volume of one of the cylinders? (A)48 (B)27 (C)12 (D)9 (E)6.75 Diameter needs to be maximum and that can be achieved by keeping it at 3 which gives us the option D. 5) How many different positive even integers are factors of 294? (A)4 (B) 6 (C) 7 (D) 9 (E) 11 take the 2 out of the factors of 294 which gives us 7^2 and 3. So no. of factors is (2+1)*(1+1)=6
Sorry for the late reply The OAs are 1) E 2) A 3) D 4) D 5) B
VPITC: Agree with 1. I made a calc error. What about 2? Can you post the steps?
The cylinder needs to be wrapped up completely which gives 2*pi*r*h + 2*pi*r^2. by substituting the values ~(96+57)=153 the nearest available square is 13
The cylinder needs to be wrapped up completely which gives 2*pi*r*h + 2*pi*r^2. by substituting the values ~(96+57)=153 the nearest available square is 13