In a group of 68 students, each student is registered for at least one of three classes History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes? a. 19 b. 10 c. 9 d. 8 e. 7
Typical Venn Diagram problem. Total students = 68 H - 25 M - 25 E - 34 H & M & E = 3 Let H & M = x; M & E = y and H & E = z
So, we have: Only H = 25 - x - z - 3 Only M = 25 - x - y - 3 Only E = 34 - y - z - 3
Jeff Smith cannot completely remember his four-digit ATM pin number.He only remembers the first two digits, and he knows that each of the last two digits is less than 3. The ATM machine will allow him 3 tries before Ace is blocked for further access to his account. If he randomly guesses the last two digits each time, what is the probability that he will get access to his account?
Dont have answer options.
I am skeptical on my answer, but here goes my approach. Please YELL if it is wrong!
Since the last two digits are less than 3 each one of them could be {0,1,2} So total possibilities are: 00,01,02 10,11,12 20,21,22 - overall 9 Out of these one combination is the correct one.
Jeff's attempts: Probability of Jeff getting it right the first time: 1/9 Probability of Jeff failing first time followed by success second time: 8/9 * 1/9 Probability of Jeff failing first and second times and successful on 3rd: 8/9 * 8/9 * 1/9
Hey Vikram, Interesting approach ..scratchy on this one ... But just a few logical reservations from my side ...
In ur evaluation of 2nd try , after first failure, logically he knows 1 is obviously incorrect..hence chances of getting it right the 2nd time increases to 1/8 ( Since the correct code has to be 1 of the rem 8 codes)
Hence, prob of getting right 2nd time is 8/9 * 1/8 = 1/9 ...i dont think prob for any attempt shd change ...
Similarly , for 3rd try = 8/9 * 7/8 * 1/7 = 1/9
So, prob = 1/9 + 1/9 + 1/9 = 1/3 Either am thinking too straight or not thinking at all ...
Other 2 ways i thought off : total 9 ...hence chances of getting access in 3 attempts is 3/9 = 1/3 too easy to believe it is true .
Another approach : prob (success) = 1 - prob( failure on every occasion) = 1 - (8/9*7/8*6/7) = 1 - 2/3 = 1/3
If given 9 chances, unless the person is retarded, he shd be able to crack it ...so if we proceed ur approach to find success to 9 attempts, i dont think it gets to 1 ...
Hi, Can anyone please explain me this problem? Just don't seem to get it.
Thanks.
In a group of 68 students, each student is registered for at least one of three classes History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes? a. 19 b. 10 c. 9 d. 8 e. 7
Originally Posted by amsey1382 which of the following lists the number of point at which a circle can intersect a triangle A?) 1 B) 1,2 C) 2, 4, 6 D) 1,2,3,4,5 & 6
Hi, Can anyone please explain me this problem? Just don't seem to get it.
Thanks.
@jamifahad The ques means - > at how many different points can a triangle and circle touch/intersect (assuming ofcourse that they are touching.)
@BizzareNym: the answer is right, but the calculations are wrong.. I wonder how you nailed this one.. copy mara huh!! .. :biggrin:
IMO B 67/256
Using combinatorics: Total possible outcomes are 4*4*4*4= 256 Total possible outcomes of getting one 4 = (1*3*3*3)*4 = 108 Total possible outcomes of getting NO 4 = 3*3*3*3 = 81
Total possible outcomes of getting one 4 OR NO 4 = 108 + 81 = 189
Therefore total possible outcomes of getting atleast two times 4 = 256 - 189=67
Probablity = 67/256
QUOTE=pj02;1632030]A four-sided die is rolled 4 times. What is the probability of getting a 4 at least twice.
Jeff Smith cannot completely remember his four-digit ATM pin number.He only remembers the first two digits, and he knows that each of the last two digits is less than 3. The ATM machine will allow him 3 tries before Ace is blocked for further access to his account. If he randomly guesses the last two digits each time, what is the probability that he will get access to his account?
Dont have answer options.
I am skeptical on my answer, but here goes my approach. Please YELL if it is wrong!
Since the last two digits are less than 3 each one of them could be {0,1,2} So total possibilities are: 00,01,02 10,11,12 20,21,22 - overall 9 Out of these one combination is the correct one.
Jeff's attempts: Probability of Jeff getting it right the first time: 1/9 Probability of Jeff failing first time followed by success second time: 8/9 * 1/9 Probability of Jeff failing first and second times and successful on 3rd: 8/9 * 8/9 * 1/9
Hey Vikram, Interesting approach ..scratchy on this one ... But just a few logical reservations from my side ...
In ur evaluation of 2nd try , after first failure, logically he knows 1 is obviously incorrect..hence chances of getting it right the 2nd time increases to 1/8 ( Since the correct code has to be 1 of the rem 8 codes)
Hence, prob of getting right 2nd time is 8/9 * 1/8 = 1/9 ...i dont think prob for any attempt shd change ...
Similarly , for 3rd try = 8/9 * 7/8 * 1/7 = 1/9
So, prob = 1/9 + 1/9 + 1/9 = 1/3 Either am thinking too straight or not thinking at all ...
Other 2 ways i thought off : total 9 ...hence chances of getting access in 3 attempts is 3/9 = 1/3 too easy to believe it is true .
Another approach : prob (success) = 1 - prob( failure on every occasion) = 1 - (8/9*7/8*6/7) = 1 - 2/3 = 1/3
If given 9 chances, unless the person is retarded, he shd be able to crack it ...so if we proceed ur approach to find success to 9 attempts, i dont think it gets to 1 ...
pj02, OA pls ..
OA is : 217/729
Bhavin, I also solved the problem like this, i was totally wrong in my approach : each of the last two digits is less than 3 --> 0,1,2, which equals 3 possibilities for the 3. digit and same goes for the last digit. Thus, 3*3 possibilities in total.
1. Heights of women in a city follows a normal distribution with mean 160 cm and standard deviation of 6 cm. In a normal distribution, only 0.0063% of the population is not within 4 standard devistions of the mean. If 5 women are more than 184 cm tall, approximately how many women stay in the city? a>16000 b>40000 c>80000 d> 100,000 e> 160,000
is the ans for this c i-e approx 80000 women stay in the city ....
the value 5 is given for 4SD away from mean which is approximately (1 in 15787)=> 5*15787 =78935
In a group of 68 students, each student is registered for at least one of three classes History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes? a. 19 b. 10 c. 9 d. 8 e. 7
Nice catch buddy! I think I wrongly typed the values from my note book and was concerned to check only the end result Even number of mistakes often nullify one another.
I wouldnt mind as long as I am consistent in doing this.
In a group of 68 students, each student is registered for at least one of three classes History, Math and English. Twenty-five students are registered for History, twenty-five students are registered for Math, and thirty-four students are registered for English. If only three students are registered for all three classes, how many students are registered for exactly two classes? a. 19 b. 10 c. 9 d. 8 e. 7
@BizzareNym: the answer is right, but the calculations are wrong.. I wonder how you nailed this one.. copy mara huh!! .. :biggrin:
Hey Vikram, Interesting approach ..scratchy on this one ... But just a few logical reservations from my side ...
In ur evaluation of 2nd try , after first failure, logically he knows 1 is obviously incorrect..hence chances of getting it right the 2nd time increases to 1/8 ( Since the correct code has to be 1 of the rem 8 codes)
Hence, prob of getting right 2nd time is 8/9 * 1/8 = 1/9 ...i dont think prob for any attempt shd change ...
Similarly , for 3rd try = 8/9 * 7/8 * 1/7 = 1/9
So, prob = 1/9 + 1/9 + 1/9 = 1/3 Either am thinking too straight or not thinking at all ...
Other 2 ways i thought off : total 9 ...hence chances of getting access in 3 attempts is 3/9 = 1/3 too easy to believe it is true .
Another approach : prob (success) = 1 - prob( failure on every occasion) = 1 - (8/9*7/8*6/7) = 1 - 2/3 = 1/3
If given 9 chances, unless the person is retarded, he shd be able to crack it ...so if we proceed ur approach to find success to 9 attempts, i dont think it gets to 1 ...
pj02, OA pls ..
OA is : 217/729
Bhavin, I also solved the problem like this, i was totally wrong in my approach : each of the last two digits is less than 3 --> 0,1,2, which equals 3 possibilities for the 3. digit and same goes for the last digit. Thus, 3*3 possibilities in total.
3 tries --> 3/9 ==> 1/3
Vikram got it right... Good Going Vikram !!!
Bhavin, The reason why I did not chose 8/9 * 1/8 and 8/9 * 7/8 * 1/7 for the second and third attempts is due to the following sentence in the question stem: ............... If he randomly guesses the last two digits each time..........
If Jeff is randomly guessing....I understood that he is NOT learning from his "wrong attempts". So, the wrong choice is still included in the number of outcomes.
If he did not randomly guess each time, instead, just tried (knowledgeably) then it makes perfect sense that the outcomes reduce to 8 and 7 analogously.
I don't know if I thought too much about it Your Jeff is smarter than mine :)
In a funfair,each mother had exactly 2 children.At end of the fair,36 mothers have lost either one or both the children and 62 children have lost their mother.How many mother's lost single children and huv many lost both the children?
32. A set of numbers has the property that for any number t in the set, t + 2 is in the set. If -1 is in the set, which of the following must also be in the set?
I. -3 II. 1 III. 5
A. I only B. II only C. I and II only D. II and III only E. I, II, and III
let total num of mothers who lost one child = x total num of mothers who lost two children = y
x + y =36 --I x + 2y = 62 -- II this is total chidlren lost
solving I & II we get x=10, Y =26
Solve this
In a funfair,each mother had exactly 2 children.At end of the fair,36 mothers have lost either one or both the children and 62 children have lost their mother.How many mother's lost single children and huv many lost both the children?
Problem says that for every t , we must have t + 2 but it doesnt say that t can there only when we have t-2. Therefore I dont think -3 must be there in the set. The set may as well start from -1 only.
32. A set of numbers has the property that for any number t in the set, t + 2 is in the set. If 1 is in the set, which of the following must also be in the set?
I. 3 II. 1 III. 5
A. I only B. II only C. I and II only D. II and III only E. I, II, and III
32. A set of numbers has the property that for any number t in the set, t + 2 is in the set. If 1 is in the set, which of the following must also be in the set?
I. 3 II. 1 III. 5
A. I only B. II only C. I and II only D. II and III only E. I, II, and III
This set is: {.....-5, -3, -1, 1, 3, 5, 7....} All of these numbers comply to the condition of t and t+2 to be in the set.
I found the following batch of PS questions just posted on Scribd (here). I am having trouble with # 5. Any suggestions?
Wharton89, Just for the sake of one question, you would not need to go through the hassle of uploading a document and provide a Scribd URL....unless you want all of us to go there and there by increase your visitor count or hit ratio
j/k
The Question is as follows:
z is a positive integer and a multiple of 2; p = 4^z, What is the reminder when p is divided by 10?
A) 10 B) 6 C) 4 D) 0 E) Cannot be Determined.
Ans: IMO: B z=2,4,6,.... 2z=4,8,12,... p=16, 256, 4096...etc p/10 always will have a reminder of 6 as p has 6 as the last digit in all of these.
z is a positive integer and a multiple of 2; p = 4^z, What is the reminder when p is divided by 10?
A) 10 B) 6 C) 4 D) 0 E) Cannot be Determined.
Ans B .. 4 has a cyclicity of 2 All odd powers of 4 end with 4 and all all even powers of 4 end with 6 ... If z is even, 4^z always has a unit digit 6 and hence remainder when divided by 10 is 6 ...
32. A set of numbers has the property that for any number t in the set, t + 2 is in the set. If 1 is in the set, which of the following must also be in the set?
I. 3 II. 1 III. 5
A. I only B. II only C. I and II only D. II and III only E. I, II, and III
I think the answer is E and the set will have an infinte no. of values...
...
