@sandeepp007
can u explain how u arrived at 16
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
@sandeepp007
can u explain how u arrived at 16
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
It is given that x + 3y
starting from 2(since K>1) the maximum value of x
we are left with 487 for 3y( since 999-512). the neares possible value for this is 3*2^7=484--------(2)
The value for sum of length of x and y is sum of 9( from 2^9) and 7(2^7) which gives the answer as 16.
(anything more than 2 will only make the sum less)
i hope it answers...
Please let me know your method tosolve this question:
How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,339
Thanks,
Anurag...
How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,339
if this comes in GMAT the best way to do early is (4586-3247)=1339, since 13 is fixed and only the rest of the digits between those to numbers which can be changed are 1339. what is the OA and pls explain if there is any other way.
How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,339
if this comes in GMAT the best way to do early is (4586-3247)=1339, since 13 is fixed and only the rest of the digits between those to numbers which can be changed are 1339. what is the OA and pls explain if there is any other way.
I don't know any way to solve this problem other than counting manually.
Answer is correct.
Thanks,
Anurag...
There will be only one number in hundred numbers which has 1 in tens digit and 3 in unit digit. eg, 13, 113, 213, 313, 413,......
no of hundres b/w 324,700 and 458,600 is (458,600-324,700)/100=1339.
for each 1339 hundreds, there will be one number each. So, answer is 1339.Hope this clarifies!
1) There are 68 children in the cafeteria of a school and all of the children have something for lunch. Thirty-four of the children brought lunches from home, 23 of the children bought a drink from the cafeteria beverage machine, and 32 of the children bought fruit in the cafeteria. If 18 children did at least 2 of these things, how many children did exactly two of these things?
3
6
9
13
15
How many integers between 324,700 and 458,600 have tens digit 1 and units digit 3?
(A) 10,300
(B) 10,030
(C) 1,353
(D) 1,352
(E) 1,339
if this comes in GMAT the best way to do early is (4586-3247)=1339, since 13 is fixed and only the rest of the digits between those to numbers which can be changed are 1339. what is the OA and pls explain if there is any other way.
I don't know any way to solve this problem other than counting manually.
Answer is correct.
Thanks,
Anurag...
Using P&C;:
Number = _ _ _ , _ 1 3 .
Now we take cases :
3 2 4 , _ 1 3 : No of choices = 3. Since the blank can take only 7,8,9.
3 2 _ , _ 1 3 : No of choices = 5*10, since 1st blank takes 5...9=5 and 2nd 0...9=10
3 _ _ , _ 1 3 : 1st blank = 3...9=7, 2nd=0..9=10, 3rd=0..9=10; total = 700
for 1st digit 4: we go the reverse way:
4 5 8 , _ 1 3 : blank can take 0....5 = 6.
4 5 _ , _ 1 3 : 1st blank can take 0....7 = 8 and 2nd takes = 0...9=10; total =80.
4 _ _ , _ 1 3 : 1st blank takes 0...4=5, 2nd = 0..9=10, 3rd = 0...9=10; total = 500.
Hence total numbers = 3+50+700+6+80+500=1339
1) There are 68 children in the cafeteria of a school and all of the children have something for lunch. Thirty-four of the children brought lunches from home, 23 of the children bought a drink from the cafeteria beverage machine, and 32 of the children bought fruit in the cafeteria. If 18 children did at least 2 of these things, how many children did exactly two of these things?
3
6
9
13
15
posting after a long long time !!
since there are no takers for this question, i'll attempt d same ..
IMO Ans E : 15
By set theory,
If region I represents exactly one grp, region II represents exaclty 2 grps, region III represents exactly 3 grps, then ..
I + II + III = 68 (i.e total ppl involved, since neither is zero)
Also,
I + 2II + 3III = 89 (34+23+32)
Solving both equations, we get II + 2III = 21
And we know that, II + III = 18 ( atleast 2 is given as 1
Hence, III = 3 and therefore II = 15
Hence, 15 students did exactly 2 things (Ans) ..
Let me know OA and feel free to clarify ..
Cheers !
1) There are 68 children in the cafeteria of a school and all of the children have something for lunch. Thirty-four of the children brought lunches from home, 23 of the children bought a drink from the cafeteria beverage machine, and 32 of the children bought fruit in the cafeteria. If 18 children did at least 2 of these things, how many children did exactly two of these things?
3
6
9
13
15
Drawing a venn diagram,we can solve this q. Given that no.of ppl dng atleast 2 things is 18.So no.of ppl dng only one thing will be 50.When u add all the regions we get no.of ppl dng 3 things is 3.So,no.of ppl dng exactly 2 things will be 18-3 = 15.
I will go with option E)15.
-Deepak.
1) Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
20%
30%
40%
50%
60%
1) Anthony and Michael sit on the six-member board of directors for company X. If the board is to be split up into 2 three-person subcommittees, what percent of all the possible subcommittees that include Michael also include Anthony?
20%
30%
40%
50%
60%
No. of groups = 6c3/2 (as forming a group is same as forming the corresponding next group) = 10.
No. of groups where A and M are together = 4. A & M can be together with any one of the 4 remaining guys. So they can be part of 4 such teams.
Hence, 4/10*100 = 40%.
1) If there are 85 students in a statistics class and we assume that there are 365 days in a year, what is the probability that at least two students in the class have the same birthday (assuming birthdays are distributed independently)?
(Can't have choices given here, as they are all JPEG's and I am too lazy to do that, so to make it more challenging, no choices. Please do not resolve the fractions and just present them as it is in the factorial forms).
2)If x is a positive integer and z is a non-negative integer such that (2,066)^z is a divisor of 3,176,793, what is the value of z^x - x^z?
A)-81
B)-1
C)0
D)1
E)It Cannot Be Determined
3)A box contains either blue or red flags. The total number of flags in the box is an even number. A group of children are asked to pick up two flags each. If all the flags are used up in the process such that 60% of the children have blue flags, and 55% have red flags, what percentage of children have flags of both the colors?
A)5%
B)10%
C)15%
D)20%
E)It can not be determined
4)There are six different models that are to appear in a fashion show. Two are from Europe, two are from South America, and two are from North America. If all the models from the same continent are to stand next to each other, how many ways can the fashion show organizer arrange the models?
A)48
B)64
C)24
D)8
E)72
5)Set S consists of the following unique integers: -2, 17, 3, n, 2, 15, -3, and -27; which of the following could be the median of set S?
A)1
B)9
C)14
D)17
E)It Cannot Be Determined
6)At a technology consulting firm with x computers, all of which are desktops or laptops, 30% are laptops; if 80% of the total number of computers have more than 1GB of RAM and 10% of the computers with less than 1GB of RAM are laptops (and no computers have exactly 1GM of RAM), approximately what percent of the desktops have more than 1GB of RAM?
A)75%
B)60%
C)52%
D)40%
E)45%
posting after a long long time !!
since there are no takers for this question, i'll attempt d same ..
IMO Ans E : 15
By set theory,
If region I represents exactly one grp, region II represents exaclty 2 grps, region III represents exactly 3 grps, then ..
I + II + III = 68 (i.e total ppl involved, since neither is zero)
Also,
I + 2II + 3III = 89 (34+23+32)
Solving both equations, we get II + 2III = 21
And we know that, II + III = 18 ( atleast 2 is given as 1
Hence, III = 3 and therefore II = 15
Hence, 15 students did exactly 2 things (Ans) ..
Let me know OA and feel free to clarify ..
Cheers !
Grt to see you are back
The OA is indeed E
But i tried it little differently but got stuck in the middle
Assume
No. of people who had both lunch and drinks only is x
No. of people who had both lunch and fruits only is y
No. of people who had both fruits and drinks only is z
No. of people who had all three is p
Then,
34-(x+y+p)+23-(x+z+p)+32-(y+z+p)+x+y+z+p=68
But in this case LHS was not the same as RHS in the end
Can you tell me where i went wrong
1) if there are 85 students in a statistics class and we assume that there are 365 days in a year, what is the probability that at least two students in the class have the same birthday (assuming birthdays are distributed independently)?
(can't have choices given here, as they are all jpeg's and i am too lazy to do that, so to make it more challenging, no choices. Please do not resolve the fractions and just present them as it is in the factorial forms).
2)if x is a positive integer and z is a non-negative integer such that (2,066)^z is a divisor of 3,176,793, what is the value of z^x - x^z?
A) -81
b) -1
c) 0
d) 1
e) it cannot be determined
3)a box contains either blue or red flags. The total number of flags in the box is an even number. A group of children are asked to pick up two flags each. If all the flags are used up in the process such that 60% of the children have blue flags, and 55% have red flags, what percentage of children have flags of both the colors?
A) 5%
b) 10%
c) 15%
d) 20%
e) it can not be determined
4)there are six different models that are to appear in a fashion show. Two are from europe, two are from south america, and two are from north america. If all the models from the same continent are to stand next to each other, how many ways can the fashion show organizer arrange the models?
A) 48
b) 64
c) 24
d) 8
e) 72
5)set s consists of the following unique integers: -2, 17, 3, n, 2, 15, -3, and -27; which of the following could be the median of set s?
A) 1
b) 9
c) 14
d) 17
e) it cannot be determined
6)at a technology consulting firm with x computers, all of which are desktops or laptops, 30% are laptops; if 80% of the total number of computers have more than 1gb of ram and 10% of the computers with less than 1gb of ram are laptops (and no computers have exactly 1gm of ram), approximately what percent of the desktops have more than 1gb of ram?
A) 75%
b) 60%
c) 52%
d) 40%
e) 45%
imo
1) 1-365!/(280!*365^85)
2) b : -1
3) c : 15%
4) a : 48
5) a : 1
6) a : 75%
Let me know the OAs
imo
1) 1-365!/(280!*365^85)
2) b : -1
3) c : 15%
4) a : 48
5) a : 1
6) a : 75%
Let me know the OAs
All correct! Good job. VP.
Ankit, The first one is scary but it isn't. Try it.
thanks.
1) If there are 85 students in a statistics class and we assume that there are 365 days in a year, what is the probability that at least two students in the class have the same birthday (assuming birthdays are distributed independently)?
(Can't have choices given here, as they are all JPEG's and I am too lazy to do that, so to make it more challenging, no choices. Please do not resolve the fractions and just present them as it is in the factorial forms).
All correct! Good job. VP.
Ankit, The first one is scary but it isn't. Try it.
thanks.
Trying....
In atleast problems its always easier to do 1-Not(P) = P.
Not(P) in this case means all the birthday lie on unique days.
Number of ways of choosing unique B'days for all 85 are 365C85 = 365!/280!*85!.
Now we need to find the P of each student having a unique B'day and multiply them together.
Which is = (choose student*choose day)/(Total ways)
ways of choosing a student = 85C1, 84C1, 83C1...
Choosing day = 365C1,364C1,363C1....
Total ways = 85C85*365C85 = 365!/280!85!
Hence for student 1,2,3.....85th.
(85/365)*(280!85!/365!)..(1/281)*(280!85!/365!)...
I am lost beyond this point....
Does such a question appear on GMAT?
Trying....
In atleast problems its always easier to do 1-Not(P) = P.
Not(P) in this case means all the birthday lie on unique days.
Number of ways of choosing unique B'days for all 85 are 365C85 = 365!/280!*85!.
Now we need to find the P of each student having a unique B'day and multiply them together.
Which is = (choose student*choose day)/(Total ways)
ways of choosing a student = 85C1, 84C1, 83C1...
Choosing day = 365C1,364C1,363C1....
Total ways = 85C85*365C85 = 365!/280!85!
Hence for student 1,2,3.....85th.
(85/365)*(280!85!/365!)..(1/281)*(280!85!/365!)...
I am lost beyond this point....
Does such a question appear on GMAT?
Unlikely I would think. I am a bit sleepy to look into your solution but my approach is:
For the solution, this is like placing 85 students in 365 benches. right? Once a bench is taken its taken and so no 2 students will have the same bench. Call this probability as P(no-bench)
so our required answer = 1 - p(no-bench)
The probability that the first student has no one in his bench = 365/365.
The probability that the second student has no one in his bench = 364/365
...
..
The probability that the 84th student has no one in his bench = 282/365
The probability that the 85th student has no one in his bench = 281/365 (281 are the total remaining and he occupies any of the remaining 281 seats, once he takes it remaining are 280 benches for anyone else to come and occupy).
So the total probability = 365/365*364/365*.....281/365 = simplifying = 365!/365^85*1/280!
so our answer = 1 -
I wish these dont come in the exam.
thanks,
pH
hey i had a prob hope sm1 can help me out with it....
iv just graduated doing a bcom this year....im thoroughly confused whether i should do my mcom or i should go ahead and do a gmat or a cat or cet....what r the benfits & the drawbacks of either of them....pls reply asap....thanx
Unlikely I would think. I am a bit sleepy to look into your solution but my approach is:
For the solution, this is like placing 85 students in 365 benches. right? Once a bench is taken its taken and so no 2 students will have the same bench. Call this probability as P(no-bench)
so our required answer = 1 - p(no-bench)
The probability that the first student has no one in his bench = 365/365.
The probability that the second student has no one in his bench = 364/365
...
..
The probability that the 84th student has no one in his bench = 282/365
The probability that the 85th student has no one in his bench = 281/365 (281 are the total remaining and he occupies any of the remaining 281 seats, once he takes it remaining are 280 benches for anyone else to come and occupy).
So the total probability = 365/365*364/365*.....281/365 = simplifying = 365!/365^85*1/280!
so our answer = 1 -
I wish these dont come in the exam.
thanks,
pH
Bang-on. That is the correct solution indeed. Good work.