At a certain university, if 50 percent of the people who inquire about admission policies actually submit applications for admission, what percent of those who submit applications for admission enroll in classes at the university?
(1) Fifteen percent of those who submit applications for admission are accepted at the university.
(2) Eighty percent of those who are accepted send a deposit to the university.
Is this an E?
Hey Bhawin, your answer is correct but argument is grossly flwed.
for st1: x= 10m, y = 10n, where m & n are integers is not sufficient to express the concept. m and n will have to be relatively coprime.
so x=10*2 and y=10*3 have gcd as 10 but xy is divisible by 8;
but x=10*3 and y=10*5, have gcd as 10 but xy is not divisible by 8;
So st 1 is inconclusive.
for st2:
lcm(x,y) is 100 =2^2*5^2 => x=2^a * 3^b, and y = 2^c * 3^d, where max(a,c)=2 and max(b,d)=2. For xy to be divisible by 8 we need a+c>=3.
For (a,c)=(2,0) xy is not divisible by 8
but for (a,c) = (2,1) it holds true. So st 2 alsone is not sufficient.
st1 and st2 taken together, xy = 100*10=1000, which gives a guarantee.
thanx for pointing d mistake bro..
yup...i have committed a grave mistake..
i also meant m and n have to be co-prime, but in a hurry, did not put it across effectively..
(1) the greatest common divisor of x and y is 10
Case 1: consider 20 and 30 , Not Divisible by 8
Case 2: Consider 40 and 30 , Divisible by 8
Condition 1 not sufficient when taken alone
(2) the lease common multiple of x and y is 100
Case 1 : Consider 25 and 4 , LCM 100 , divisible by 8
only case here can be that of co priomes coz if we have a number which is the factor of other number eg 50 and 2 , then lcm wd be 50
Cant think of ne such other pair right now so iguess We can find the answer with only option 2
ill go with that
hey ...Ans is not B..there are 2 co-prime pairs whose LCM is 100..
these pairs are..
1 and 100
4 and 25
so st 2 alone is not sufficient..we need both the statements combined to solve the problem
At a certain university, if 50 percent of the people who inquire about admission policies actually submit applications for admission, what percent of those who submit applications for admission enroll in classes at the university?
(1) Fifteen percent of those who submit applications for admission are accepted at the university.
(2) Eighty percent of those who are accepted send a deposit to the university.
Is this an E?
I would go with (C)....
At a certain university, if 50 percent of the people who inquire about admission policies actually submit applications for admission, what percent of those who submit applications for admission enroll in classes at the university?
(1) Fifteen percent of those who submit applications for admission are accepted at the university.
(2) Eighty percent of those who are accepted send a deposit to the university.
Is this an E?
this seems to be a more of a critical reasoning question than maths...
am not sure of this 1..even i feel ans is E..
St 1: Of those who are accepted by the university, everybody need not necessarily enrol for classes...not sufficient...
St 2 : Sending a deposit to university also does not necessarily mean you enrol for classes...u can always ask for a refund :p..
both statements combined also just tells us that applicant has been accepted by university and he/she has paid the deposit, again it does not necessarily mean u enrol for classes...nothing conclusive..
ANS E..
montaqes SaysI would go with (C)....
Yes C will be the answer.
and enrollment is 12% .
Yes C will be the answer.
and enrollment is 12% .
Jha,
Does paying a deposit imply enrolling at the university ? Is that a sufficient condn ?
What if the student decides to forego the amount for a greater good ?
Shouldnt it be E ?
Hey Bhawin, your answer is correct but argument is grossly flwed.
for st1: x= 10m, y = 10n, where m & n are integers is not sufficient to express the concept. m and n will have to be relatively coprime.
so x=10*2 and y=10*3 have gcd as 10 but xy is divisible by 8;
but x=10*3 and y=10*5, have gcd as 10 but xy is not divisible by 8;
So st 1 is inconclusive.
for st2:
lcm(x,y) is 100 =2^2*5^2 => x=2^a * 3^b, and y = 2^c * 3^d, where max(a,c)=2 and max(b,d)=2. For xy to be divisible by 8 we need a+c>=3.
For (a,c)=(2,0) xy is not divisible by 8
but for (a,c) = (2,1) it holds true. So st 2 alsone is not sufficient.
st1 and st2 taken together, xy = 100*10=1000, which gives a guarantee.
I believe that u must have meant that x=2^a * 5^b and not 2^a * 3^b.
If what u have written is what u meant, I have not understood the solution.
Could u pls clarify ?
This one has left me confused !
If Z1, Z2, ......., Zn is a series of consecutive positive integers, is the sum of all the integers in this series odd ?
a) (Z1+....Zn)/n = Odd Integer
b) n is odd
I believe that u must have meant that x=2^a * 5^b and not 2^a * 3^b.
If what u have written is what u meant, I have not understood the solution.
Could u pls clarify ?
Hey I hope u r intelligent enough, aint you?
That is a mistake, it sud be 5 instead.
Jha,
Does paying a deposit imply enrolling at the university ? Is that a sufficient condn ?
What if the student decides to forego the amount for a greater good ?
Shouldnt it be E ?
Well I have seen the OS, rest, u r entitled to to have ur own explanation.
This one has left me confused !
If Z1, Z2, ......., Zn is a series of consecutive positive integers, is the sum of all the integers in this series odd ?
a) (Z1+....Zn)/n = Odd Integer
b) n is odd
Hi Kingcat,
answer is (A)
because average of even number of consecutive positive integers will always be a decimal number and not integer. Try taking average of first 10 numbers:
sum of the integer (10*11) / 2 divided by number of terms (10) = 5.5
Aternatively, since it is an AP with common difference of 1. average of the series can be calculated by taking middle to integers (in this case 5 and 6) and finding out the average of these two integers. Therefore, an average of one odd integer and one even integer will always come as decimal number: 5+6/2 = 5.5
Coming back to question, it is given that average is odd integer, which means, there are odd number of consecutive integers. take first 11 positive integer, middle term i.e 6 will be the average of them.
Since, odd * odd will always be odd....statement 1 is sufficient to answer
This one has left me confused !
If Z1, Z2, ......., Zn is a series of consecutive positive integers, is the sum of all the integers in this series odd ?
a) (Z1+....Zn)/n = Odd Integer
b) n is odd
from a,
let z1= a, Z2=a+1, ....
sigmaZ(i)= an+ n*(n-1)/2 = M
so M/n is an integer only if n is odd.
Now for n = odd, say 2k+1, we will have k even numbers. and k+1 odd numbers. where k is any integer. But for k even or odd this(sum) can be either odd or even respectively.
So inconclusive.
from b, say n = 2p+1, so sum will have p even numbers and p+1 odd numbers
from same logic this is inconclusive too.
Taken both at a time: sum is odd. so answer is C.
Hi Kingcat,
answer is (A)
because average of even number of consecutive positive integers will always be a decimal number and not integer. Try taking average of first 10 numbers:
sum of the integer (10*11) / 2 divided by number of terms (10) = 5.5
Aternatively, since it is an AP with common difference of 1. average of the series can be calculated by taking middle to integers (in this case 5 and 6) and finding out the average of these two integers. Therefore, an average of one odd integer and one even integer will always come as decimal number: 5+6/2 = 5.5
Coming back to question, it is given that average is odd integer, which means, there are odd number of consecutive integers. take first 11 positive integer, middle term i.e 6 will be the average of them.
Since, odd * odd will always be odd....statement 1 is sufficient to answer
Let me analyse ur solution. :)
and first all let me prove that ur answer is wrong.
Sum of odd number of consecutive integers: may give odd or even based on from where it starts.
example: 3 to 7 ie. 3 4 5 6 7 sum is 25 an odd
then look at 2 to 6 ie. 2 3 4 5 6 that gives 20 an even.
Let me analyse ur solution. :)
and first all let me prove that ur answer is wrong.
Sum of odd number of consecutive integers: may give odd or even based on from where it starts.
example: 3 to 7 ie. 3 4 5 6 7 sum is 25 an odd
then look at 2 to 6 ie. 2 3 4 5 6 that gives 20 an even.
true..but AVERAGE of even number of consecutive numbers can't be an integer :wink:
now think abt it again...:-P
Let me analyse ur solution. :)
and first all let me prove that ur answer is wrong.
Sum of odd number of consecutive integers: may give odd or even based on from where it starts.
example: 3 to 7 ie. 3 4 5 6 7 sum is 25 an odd
then look at 2 to 6 ie. 2 3 4 5 6 that gives 20 an even.
jha buddy,
montaques said average of EVEN number of consecutive integers is always odd!
1 2 3 4
2 3 4 5
Well.. here is my attempt!
the sum of these numbers would be /2
now if a) is true, then the is odd.. but from here we can not say anything abt the sum of all numbers
now if b) is also true then n is odd..
only after combining a) and b) we can say that the sum is odd..
according to me, the answer shud be C..
This one has left me confused !
If Z1, Z2, ......., Zn is a series of consecutive positive integers, is the sum of all the integers in this series odd ?
a) (Z1+....Zn)/n = Odd Integer
b) n is odd
Hi Kingcat,
answer is (A)
because average of even number of consecutive positive integers will always be a decimal number and not integer. Try taking average of first 10 numbers:
sum of the integer (10*11) / 2 divided by number of terms (10) = 5.5
Aternatively, since it is an AP with common difference of 1. average of the series can be calculated by taking middle to integers (in this case 5 and 6) and finding out the average of these two integers. Therefore, an average of one odd integer and one even integer will always come as decimal number: 5+6/2 = 5.5
Coming back to question, it is given that average is odd integer, which means, there are odd number of consecutive integers. take first 11 positive integer, middle term i.e 6 will be the average of them.
Since, odd * odd will always be odd....statement 1 is sufficient to answer
hey montaque...ur ans is correct..
but sum of first 11 integers is not a good example, it does not validate statement 1...becoz its average is even and not odd..
let me have an alternate approach...
We know, in an AP, mean = median..
St 1 : average is odd => median is odd..
median is middle term of the series and median is odd implies there are odd no of terms or there are equal no of terms on either side of median. these nos on either side of median form pairs and sum of these pairs is always even. Finally even + median (odd) = odd ..
eg 1+2+3+4+5....avg = median =3
nos on each side : 2+4=6
1+5=6
6+6+3 = odd...
hence statement is sufficient...
St 2 :
n is odd...now first term is not necessarily 1..
if z1 is odd, then sum is odd when median is odd or n is of the form n = 4k+1...
eg 1+2+3=6 though no of terms is odd (becoz n is not in the form of 4k + 1)
and 1+2+3+4+5 = 15 (n is of the form n = 4k+1...)
if z1 is even,then sum is odd when median is odd or n is of the form 4k+3...
eg 2+3+4=9 (becoz n is not in the form of 4k + 3)
2+3+4+5+6=20 (becoz n is not in the form of 4k + 3)
so odd no of terms is not conclusive ...not sufficient..
Ans A..
true..but AVERAGE of even number of consecutive numbers can't be an integer :wink:
now think abt it again...:-P
Yeah that is fine ... I saw that.
so u have odd number of consecutive integers right?
but that can again be either odd or even. do u agree? So answer is not A.
Yeah that is fine ... I saw that.
so u have odd number of consecutive integers right?
but that can again be either odd or even. do u agree? So answer is not A.
yes jha...sum of odd no of conscutive integers can be even or odd...but given that average of consecutive integers is odd not only implies that no of terms are odd but also implies that sum of these odd no of consecutive integers is odd only..Ans A