Probability question:
There are a total of 163,000 people in two groups
A: 31,200 people; B: 1,31,800; A and B are mutually exclusive
A total of 85,000 people are chosen following this procedure:
step 1: choose 20,000 people from A
then,
step 2: choose 65,000 people from the rest of the people (including unselected A and B people which is 11200+131800 = 1,43,000)
what is the probability of a person from Group A being selected at the end of step 2?
Hi,
Thanks so much for ur reply.But I somehow could not understand it at all:p
SO,that prompted me to find my own solution.
You can let me know if I am correct.Coz this is much simpler and I think I am correct.
OK here goes.We have 27numbers
the smallest if them being 111 and largest being 888
Now we know avg = sum/total
Hence sum = avg * total
avg of a seriesis avg of first and last(approximately)
Hence avg = (111+ 88/2
Givesme 499
Hence I get sum = 499*27 which is 13473....Quite closeto the real value..:smile:
Okay more more toughie .Can someone help me crack with their method?
If Dave works alone he will take 20 more hours to complete a task than if he worked with Diana to complete the task. If Diana works alone, she will take 5 more hours to complete the complete the task, then if she worked with Dave to complete the task? What is the ratio of the time taken by Dave to that taken by Diana if each of them worked alone to complete the task?
A. 4 : 1
B. 2 : 1
C. 10 : 1
D. 3 : 1
E. 1 : 2
And one more
In a class of 50 students, 10 did not opt for math. 15 did not opt of science and 2 did not opt for either. How many students of the class opted for both math and science?
I got the ans here as 23but it actually 27.I dont know how?Am I making a sillymistake?
Thanks in advance
hi suruchee...
1) yes ...as an approx u can rely on ur technique, but it is not full proof, since sum of n nos = n/2 * (T1 + Tn) only when the nos are in AP...n the above 27 nos r not in AP...
sum has already been explained earlier...actually with repetition allowed, 27 nos r poss and each of d digit appears 9 times in units, tens n hundreds place....
Hence, sum = 9*100*(1+5+
+ 9*10*(1+5+ + 9*1*(1+5+ 2) let them take x hours when they work together...
equate total work with sum of ind works for 1 hr...
HEnce, 1/(x+20) + 1/(x+5) = 1/x
solving, x = 10, dave= 30 hrs, diana = 15
ratio = 2:1
3) yes, the ans is indeed 27, maybe u must have done a calculation mistake...
opt for maths = 40
opt for science = 35
both = x
neither = 2
total = grp1 + grp2 - both + neither
50 = 40 + 35 -x + 2
x= 27 .........Ans
Probability question:
There are a total of 163,000 people in two groups
A: 31,200 people; B: 1,31,800; A and B are mutually exclusive
A total of 85,000 people are chosen following this procedure:
step 1: choose 20,000 people from A
then,
step 2: choose 65,000 people from the rest of the people (including unselected A and B people which is 11200+131800 = 1,43,000)
what is the probability of a person from Group A being selected at the end of step 2?
Very complicated question , i do not think GMAT will have such a tough questions , anyways can anybody explain the answer please . i shall study probability concepts now and try to post the answer ASAP
Probability question:
There are a total of 163,000 people in two groups
A: 31,200 people; B: 1,31,800; A and B are mutually exclusive
A total of 85,000 people are chosen following this procedure:
step 1: choose 20,000 people from A
then,
step 2: choose 65,000 people from the rest of the people (including unselected A and B people which is 11200+131800 = 1,43,000)
what is the probability of a person from Group A being selected at the end of step 2?
hey...can u pls give answer options or d OA answer...so maybe we can think of the approach...hv nt seen a similar sum on GMAT...
question is not very clear to me...are we wanting to know chance of 1 particular person from grp A being selected or the chance that atleast 1 person frm grp A gets selected in stage 2 ?
if its 1 particular person, what i could think the solution does nt look gud...
prob of a particular person getting selected = 1 - (p(not selected in stage 1) *p(not selected in stage 2)
= 1-
if its atleast 1 person from A in stage 2, then
p= 1 - (131800C65000 / 143000C65000 )...
can somebody pls clarify....am i missing a clue ?
i dont think this is the solution.....
hey...can u pls give answer options or d OA answer...so maybe we can think of the approach...hv nt seen a similar sum on GMAT...
question is not very clear to me...are we wanting to know chance of 1 particular person from grp A being selected or the chance that atleast 1 person frm grp A gets selected in stage 2 ?
if its 1 particular person, what i could think the solution does nt look gud...
prob of a particular person getting selected = 1 - (p(not selected in stage 1) *p(not selected in stage 2)
= 1-
if its atleast 1 person from A in stage 2, then
p= 1 - (131800C65000 / 143000C65000 )...
can somebody pls clarify....am i missing a clue ?
i dont think this is the solution.....
Thanks for trying Bhavin422, this question is framed by me. A real situation about H1B visas last year
I would like to know the chance of any one particular person from A.
I have arrived at this: 1- = 0.80
143000-65000 = 78000
But, I want to solve this the other way, probability of being selected rather than (1 - not selected).
Thanks for trying Bhavin422, this question is framed by me. A real situation about H1B visas last year
I would like to know the chance of any one particular person from A.
I have arrived at this: 1- = 0.80
143000-65000 = 78000
But, I want to solve this the other way, probability of being selected rather than (1 - not selected). :thumbsup:
thanx for clarification HK...question is clear to me now...
i cud think of 2 ways to solve this one...
a) P(particular person is selected) = 1-
=20000/31200 + [11200/31200 *65000/143000)
=0.64 + 0.36 *0.45
=0.64 + 0.16
=0.8....same ans ....so it convinces me that it is correct.....
hope this solves the query...
thanx for clarification HK...question is clear to me now...
i cud think of 2 ways to solve this one...
a) P(particular person is selected) = 1-
=20000/31200 + [11200/31200 *65000/143000)
=0.64 + 0.36 *0.45
=0.64 + 0.16
=0.8....same ans ....so it convinces me that it is correct.....
hope this solves the query...
Thats perfect, Bhavin
Hi,
Can someone help me understand this problem that I faced on Gmatprep?
A certain candy manufcaturer decreases the weight of candy M by 20% but leaves the price unchanged.What is the resulting increase in price per ounce of the candy M?
A.5%
B.10
C.15
D.20
D.25
OA is 25.Thanks in advance.
HI Puys,
I have some good geometry problems froim Gmatprep that I could not solve and am not able to upload.Is there any way that I can collect email ids from those of you interested in discussing those and we can solve them?
Please let me know.
I am saying this coz docs are still not getting uploaded even after messaging mods.I know it may be against rules,but is there another way?
Thanks
Hi,
Can someone help me understand this problem that I faced on Gmatprep?
A certain candy manufcaturer decreases the weight of candy M by 20% but leaves the price unchanged.What is the resulting increase in price per ounce of the candy M?
A.5%
B.10
C.15
D.20
D.25
OA is 25.Thanks in advance.
lets assume weight of the candyM is 100 ounce, price x
price per ounce is x/100
reduced weight 80
so price per ounce is x/80
=> (x/80-x/100)/x/100
which gives (20*100)/(80*100)= 0.25, hence 25%
Hope this clarifies.
[ HK ]
Hi,
Can someone help me understand this problem that I faced on Gmatprep?
A certain candy manufcaturer decreases the weight of candy M by 20% but leaves the price unchanged.What is the resulting increase in price per ounce of the candy M?
A.5%
B.10
C.15
D.20
D.25
OA is 25.Thanks in advance.
Suppose the weight of candy M is 100 ounce and the price is 100$. Weight of the candy after 20% reduction in weight = 80 ounce.
Initial price/ounce = 100/100 = 1$
Price/ounce after reduction in weight = 100/80 = 5/4$ =1.25$
Hence, the increase in price per ounce of the candy M = {(1.25-1)/1} x 100 = 25%
[quote= Originally Posted by suruchee View Post
Can anyone help me solve this problem with the method used?
What is the sum of all 3 digit positive integers that can be formed using the digits 1, 5, and 8, if the digits are allowed to repeat within a number?
A. 126
B. 1386
C. 3108
D. 308
E. 13986
Hi Suruchee,
Answer to this question can be made much simpler. You don't need to calculate the whole thing. You can apply the "elimination technique" and you should apply this only when you are running out of time.
Since you know that there are 27 such possible numbers (1,5,8 can be in 1s, 10s and 100s place) => (3x3x3 = 27) and the units place has to be (1+5+ x9 = 36 (you carry 3 of 36 to 10s place) = 6 in 1s place.
So you blindly eliminate answer options C and D.
Now you are left with options A,B and E.
Keeping the greatest no. ie. 8 in 100s place for 9 times, the total sum will obviously be greater than 811 * 9 > 7200.. So you can now eliminate options A and B. The only option that remains is E (which is the answer :))
Hope this helps
Best Regards,
Sidd
Hi,
This is a problem from 800Kaplan book.
At a speed of 50mph,a certain car uses 1 gallon of gas every 30 miles.If car starts with a full 12 gallon tank of gas and travels for 5 hours at 50mph,the amount of gas used will be what fraction of full tank?
Hi,
This is a problem from 800Kaplan book.
At a speed of 50mph,a certain car uses 1 gallon of gas every 30 miles.If car starts with a full 12 gallon tank of gas and travels for 5 hours at 50mph,the amount of gas used will be what fraction of full tank?
at 50mph, 1 gallon = 30 miles
5*50 = 250 miles
number of gallons used up = 250/30 = 25/3
so the fraction of gas used = 25/12*3 = 25/36
Hi,
A toughie here from GmatPrep.
If two of the four expressions x+y,x+5y,x-y and 5x-y are chosena t random,what is the probablity that their product will be of the form x^2-(by)^2,where b is an integer.
A.1/2
B.1/3
C.1/4
D.1/5
E.1/6
Please discs your method of solving.WIll disclose OA later.Thanks
two expressions out of 4 can be selected in 4C2 ways = 6
now for co-eff of X^2 is One, this can happen only if we 5x-y is excluded in multiplying. Hence total ways 3C2
Prob = 3C2/4C2 = 3/6 = 1/2
This does not seem to be the OA!!
Yes my mistake. I realized after posting. We can only choose (x+by) and (x-by) .. only x+y and x-y remain.
Is it 1/6 then ?
Yes!!Thanks so much.
A----------B---------C----------D
Is
CD > BC ?
(1) AD = 20
(2) AB = CD
A Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
B Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
C BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
D EACH Statement ALONE is sufficient.
E Statements (1) and (2) TOGETHER are NOT sufficient.
Answer is E as exxplined by most other pguys