Deepak, can u plz solve this one for me?
certain junior class has 1000 students and certain senior class has 800 students. among these students there are 60 sibling pairs consisting 1 junior and 1 senior. If one student is to be selected at random from each class what is the probability that the 2 students selected will be a sibling pair?
thanks.
@Mission:
Could tis be a possible solution ?:|
Stength of Class J : 1000
Strength of Class S : 800
Since there are 60 sibling pairs. There shall be be 60 in J class and 60 in S class.
Probability of one in that 60 (in J) getting selected is = 60/1000
Probability of one in that 60 (in S) getting selected is = 60/800
Probability of these two being a sibling pair = 1/60.
(60/1000) X (60/800) X (1/60) = 3/40,000 !!!! ;)
(Practically also, if u see, selecting 2 from a set of 1800 ppl, and expecting them to be a sibling pair is quite less probable) ;)
Numerically odd though!! I guess dis is the one.
Regards,
Ravi.
(Acclivity leading to infinity...)
Deepak, can u plz solve this one for me?
certain junior class has 1000 students and certain senior class has 800 students. among these students there are 60 sibling pairs consisting 1 junior and 1 senior. If one student is to be selected at random from each class what is the probability that the 2 students selected will be a sibling pair?
thanks.
@Mission:
Could tis be a possible solution ?:|
Stength of Class J : 1000
Strength of Class S : 800
Since there are 60 sibling pairs. There shall be be 60 in J class and 60 in S class.
Probability of one in that 60 (in J) getting selected is = 60/1000
Probability of one in that 60 (in S) getting selected is = 60/800
Probability of these two being a sibling pair = 1/60.
(60/1000) X (60/800) X (1/60) = 3/40,000 !!!! ;)
(Practically also, if u see, selecting 2 from a set of 1800 ppl, and expecting them to be a sibling pair is quite less probable) ;)
Numerically odd though!! I guess dis is the one.
Regards,
Ravi.
(Acclivity leading to infinity...)
thanks ravisekharan...
it is the correct answer. I forgot to multiply by probability of two being a sibling pair (1/60)... and picked 9/2000 as a wrong choice... GMAT tricked me here... thanks again...
Guys, one quick question. If a question mentions square root of a number, (like x = root of 25), do we automatically assume it as a positive square root? Like in above, we take x=5 and not x=-5.
Deepak, can u plz solve this one for me?
certain junior class has 1000 students and certain senior class has 800 students. among these students there are 60 sibling pairs consisting 1 junior and 1 senior. If one student is to be selected at random from each class what is the probability that the 2 students selected will be a sibling pair?
thanks.
no. of ways of selecting a student in junior class who has a sibling in senior class = 60/1000
any of these 60 students in da junior class will have only 1 sibling corresponding to him,
so no. of ways of selecting a senior with the given condition is 1/800.
so probability is (60/1000)*(1/800)
skyron SaysGuys, one quick question. If a question mentions square root of a number, (like x = root of 25), do we automatically assume it as a positive square root? Like in above, we take x=5 and not x=-5.
Hi Skyron,
Its better to consider both +5 & -5 and eventually eliminate one of the numbers based on the flow of the question.
-Deepak.
This is not a tough one ...but I am confused with multiple reasonings.
Please help
The participants in a race consisted of 3 teams with 3 runners on each team .A team was awarded 6-n points if one of the runners finished in nth Place, where 1
(1) No team was awarded more than a total of 6 points
(2) No pair of teammates finished in consecutive places among the top five palces.
This is a DS prob. I see conlifctory answers.
Please reply with explanation. Thanks.
This is not a tough one ...but I am confused with multiple reasonings.
Please help
The participants in a race consisted of 3 teams with 3 runners on each team .A team was awarded 6-n points if one of the runners finished in nth Place, where 1
(1) No team was awarded more than a total of 6 points
(2) No pair of teammates finished in consecutive places among the top five palces.
This is a DS prob. I see conlifctory answers.
Please reply with explanation. Thanks.
Im nt sure if I have got this prblm.There is a race of 9 participants.Points are given for teams if its members are placed in positions 1-5. So if a person comes in 6th place,he will get only 0(zero) points.So gng with this understanding and looking at the statments I have the following workout:
statement 1 : matrix will look smeting like this :
case 1
1 5 0
2 4 0
3 0 0
case 2
1 2 3
4 0 0
5 0 0
So statement 1 is suff
Statement 2 :
1 4 0
0 2 5
0 0 3
So statement 2 is suff
I will go with option D.Pls correct me if I'm wrong.
-Deepak
Im nt sure if I have got this prblm.There is a race of 9 participants.Points are given for teams if its members are placed in positions 1-5. So if a person comes in 6th place,he will get only 0(zero) points.So gng with this understanding and looking at the statments I have the following workout:
statement 1 : matrix will look smeting like this :
case 1
1 5 0
2 4 0
3 0 0
case 2
1 2 3
4 0 0
5 0 0
So statement 1 is suff
Statement 2 :
1 4 0
0 2 5
0 0 3
So statement 2 is suff
I will go with option D.Pls correct me if I'm wrong.
-Deepak
It looks like one more Matrix
5 3 1
2 4 0
0 0 0
Was not considered by you, as per the condition 2 , Hence the option D may not be correct. my opinion is with A.
Please correct me if I am wrong.
Rahul
Hi Skyron,
Its better to consider both +5 & -5 and eventually eliminate one of the numbers based on the flow of the question.
-Deepak.
I don't think we need to consider -5 in such situations. SQR root of any number will always be a positive number and not a negative one unless stated to be imaginary number (which GMAT does not believes in).
When we have equation (x)^2 = 25 then we need to consider x= +5 and -5.
But when x=sqr root of 25 then it needs to be (+5 X +5) and not (-5 X -5).
answer given is probably wrong.. so juz checking..
how many positive integers, from 2 to 100 inclusive, are not divisible by odd integers greater than 1?
a. 5
b. 6
c. 7
d. 46
e. 48
what is the largest integer n such that 43! is divisible by 2 raised to the power n
A. 31
B. 38
C. 39
D. 41
E. none of these
My guess, 2^n, i.e 6 , option (b), 2, 4, 8, 16, 32, 64
what is the largest integer n such that 43! is divisible by 2 raised to the power n
A. 31
B. 38
C. 39
D. 41
E. none of these
this is a direct formula question.
43/2 + 43/4 + 43/8 + 43/16 + 43/32
=21+10+5+2+1
=39 option C
this is a direct formula question.
43/2 + 43/4 + 43/8 + 43/16 + 43/32
=21+10+5+2+1
=39 option C
hi...thanks a lot...dare2 is right..i've found the formula as well...
sunnygoeldelhi Sayshi...thanks a lot...dare2 is right..i've found the formula as well...
Can someone please explain me the formula?
answer given is probably wrong.. so juz checking..
how many positive integers, from 2 to 100 inclusive, are not divisible by odd integers greater than 1?
a. 5
b. 6
c. 7
d. 46
e. 48
my answer
answer given is probably wrong.. so juz checking..
how many positive integers, from 2 to 100 inclusive, are not divisible by odd integers greater than 1?
a. 5
b. 6
c. 7
d. 46
e. 48
my answer
I think the above condition will be satisfied only by the nos those are powers of 2.
2,4,8,16,32,64 - only 6 nos. So I will go with option b.6
Hi,
Please let me know what formulae you guys used.. for the problem
what is the largest integer n such that 43! is divisible by 2 raised to the power n
Regards,
Asha
Hi,
Please let me know what formulae you guys used.. for the problem
what is the largest integer n such that 43! is divisible by 2 raised to the power n
Regards,
Asha
Divide 43 by 2 till u get a quotient less than 2 and add all the quotients
43/2 = 21
21/2 = 10
10/2 = 5
5/2 = 2
2/2 = 1
So, 21+10+5+2+1 = 39
Hope it helps
try this..
in how many ways can 1,2,3,4 and 5 be arranged so that the even numbers are not adjacent?
a. 108
b. 96
c. 72
d. 60
e. 48