CAT 2008: Quantitative Questions a Day 1 to 50 - The discussions

[prade]

Originally Posted by slam

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Originally Posted by slam

Quantitative Question # 050
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Each question is followed by two statements X and Y. Answer each question using the following instructions:

Choose 1 if the question can be answered by X only
Choose 2 if the question can be answered by Y only
Choose 3 if the question can be answered by either X or Y
Choose 4 if the question can be answered by both X and Y
Choose 5 if the question can not be answered by combining X and Y also



If 1 < a < 2 and k is an integer, then what is [ak/(2 - a)], where [x] denotes the greatest integer not larger than x.

(X) [a[k/(2 - a)] + a/2] = p
(Y) [a[k/(2 - a)] + (a+1)/2] = q and k is even

let 2-a = x => 0<x<1
=>[ak(2-a)] = [2k/x - k]

from (X)
[a[k/(2 - a)] + a/2]
= [(2-x)[k/x] + 1-x/2]
=[2[k/x] - x[k/x] + 1-x/2]
let 2[k/x]-x[k/x] = m+n where m is the integer part and n the fractional part
=>[m+n+1-x/2] = p
1/2<1-x/2<1 and 0<n<1
if 1/2<n+1-x/2<1 then m = p
if 1<=n+1-x/2<=2 then m = p-1

from (Y)

[a[k/(2 - a)] + (a+1)/2] = q and k is even
=>[(2-x)[k/x] + (3-x)/2] = q
=>[m+n+1.5-x/2] = q
m+1 = q

not able to figure out where from here... Aarav give some hints

[deep@k]

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Quantitative Question # 050
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Each question is followed by two statements X and Y. Answer each question using the following instructions:

Choose 1 if the question can be answered by X only
Choose 2 if the question can be answered by Y only
Choose 3 if the question can be answered by either X or Y
Choose 4 if the question can be answered by both X and Y
Choose 5 if the question can not be answered by combining X and Y also



If 1 < a < 2 and k is an integer, then what is [ak/(2 - a)], where [x] denotes the greatest integer not larger than x.

(X) [a[k/(2 - a)] + a/2] = p
(Y) [a[k/(2 - a)] + (a+1)/2] = q and k is even






My take.....
substitute the value of a and k

Let a = 1.4 and k = 2
then answer will come only by using (X)....
now, change the value of k and take any odd value...
then answer will come from (X) only........


so, option (1)......






Please tell me where I m wrong................................

[prade]

furthur to the discussion


let a=1.99999999
then 2-a->0
[k/(2-a)] is a huge value kx
now [kx*1.9999999+1.9999999/2 ] = [kx+1/2 + (kx+1/2)0.999999]
how do we decide p from here

[swati1606]

take condition X

[a[k/(2-a)]+a/2]=p
put [k/(2-a)]=B
thus
[aB+a/2]=p
=> p<=aB+a/2<p+1
=> (p-a/2)<aB<p

now a lies between 1 & 2, thus a/2 lies between 0&1
hence aB=a[k/(2-a)]=p-1

take condition Y
proceeding in a manner similar to the one shown above we reach,

q-(a+1/2)<aB<q-(a-1/2)
now this value would depend on the fact whether a lies between 1&1.5 or between 1.5&2. hence,in itself,condition Y is not sufficient

so,i'll go with option (a)---> only X is sufficient

[Aarav]

Originally Posted by swati1606

take condition X

[a[k/(2-a)]+a/2]=p
put [k/(2-a)]=B
thus
[aB+a/2]=p
=> p<=aB+a/2<p+1
=> (p-a/2)<aB<p

now a lies between 1 & 2, thus a/2 lies between 0&1
hence aB=a[k/(2-a)]=p-1

take condition Y
proceeding in a manner similar to the one shown above we reach,

q-(a+1/2)<aB<q-(a-1/2)
now this value would depend on the fact whether a lies between 1&1.5 or between 1.5&2. hence,in itself,condition Y is not sufficient

so,i'll go with option (a)---> only X is sufficient

This is very good stuff -> stuff of Champions and achievers. There is a small mistake in your solution but still deserves 4/4.

[swati1606]

Originally Posted by Aarav

This is very good stuff -> stuff of Champions and achievers. There is a small mistake in your solution but still deserves 4/4.

thanks a lot aarav, please help me with the mistake I'm making though.

[Varun Khullar]

here i go..

somehow get up too late.. these days



Each question is followed by two statements X and Y. Answer each question using the following instructions:

Choose 1 if the question can be answered by X only
Choose 2 if the question can be answered by Y only
Choose 3 if the question can be answered by either X or Y
Choose 4 if the question can be answered by both X and Y
Choose 5 if the question can not be answered by combining X and Y also



If 1 < a < 2 and k is an integer, then what is [ak/(2 - a)], where [x] denotes the greatest integer not larger than x.

(X) [a[k/(2 - a)] + a/2] = p
(Y) [a[k/(2 - a)] + (a+1)/2] = q and k is even


considering X

[t] lies between

p< [t]<p+1
so we get
p< a[k/(2-a)] +a/2 <p+1

we know a is between1 and 2
so we know that

a[k/(2-a)] = p-1
we know that from the equalities about that its has to p-1
so x can be used..

but we have to find [ak/(2 - a)] and commutativty is not
a[x] is not equal to [ax] or is it
stuck here..

(Y) [a[k/(2 - a)] + (a+1)/2] = q
so we know that a[k/(2 - a)] is q or q-1
we also know k is even !! does this help
i think we know here too its q-1

[ravishah17]

My take

From the given statements;
1<a<2
-1>-a>-2
0<2-a<1

thus 1/(2-a)>1
a/(2-a)>a>1
ak/(2-a)>=k

Let us take each statement
X==>


[a[k/(2-a)]+a/2]=p

a[k/(2-a)]+a/2>=p
[k/(2-a)]>=(2p-a)/2a

Let us consider [k/(2-a)]=k'(which is an integer)

k'>=(2p-a)/2a


Y==>

1<(a+1)/2<3/2


Still not able to move ahead.. would see if I can move ahead

[thebornattitude]

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Quantitative Question # 050
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Each question is followed by two statements X and Y. Answer each question using the following instructions:

Choose 1 if the question can be answered by X only
Choose 2 if the question can be answered by Y only
Choose 3 if the question can be answered by either X or Y
Choose 4 if the question can be answered by both X and Y
Choose 5 if the question can not be answered by combining X and Y also



If 1 < a < 2 and k is an integer, then what is [ak/(2 - a)], where [x] denotes the greatest integer not larger than x.

(X) [a[k/(2 - a)] + a/2] = p
(Y) [a[k/(2 - a)] + (a+1)/2] = q and k is even


Solution


Let [k/(2-a)] =m......(i)


so from (X), we get


p<=am+(a/2)<p+1-(a/2)


p-(a/2)<=am<p+1-(a/2)......(ii)


Now from (i)...we get
m<=k/(2-a)<m+1


so multiplying by a, we get,


am<=ak(2-1)<am+a


so putting values of a, p in different cases, we get that, [ak/(2-1)] , can take either p or (p-1 ) as values.....so cannot be confirmed here


Case (Y)...
It is nothing but an extension of statement (X)..


we get
q<=am+a/2 +1<q+1
so


q-1<=am+a/2<q
comparing it with (ii),, we can just infer that,


q=p+1....nothing much..




So answer- option (5),,,,cannot be answered by any of them.

[shivam_01]

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Quantitative Question # 050
------------------------------------------------------

Each question is followed by two statements X and Y. Answer each question using the following instructions:

Choose 1 if the question can be answered by X only
Choose 2 if the question can be answered by Y only
Choose 3 if the question can be answered by either X or Y
Choose 4 if the question can be answered by both X and Y
Choose 5 if the question can not be answered by combining X and Y also



If 1 < a < 2 and k is an integer, then what is [ak/(2 - a)], where [x] denotes the greatest integer not larger than x.

(X) [a[k/(2 - a)] + a/2] = p
(Y) [a[k/(2 - a)] + (a+1)/2] = q and k is even

Quite good question to end the series

ok let me try and give my solution
Since 1<a<2
now [ak/(2-a)] now since nothing is known about a or k hence
take the condition X
from this we get that
[a[k/(2 - a)] + a/2] = p.......................................1
if k = (2-a) then this will result in [k/(2 - a)] = 1 always
so our solution [a.1+a/2]=p which yields p = 1 or 2 hence the condition
[ak/(2 - a)] = 1

now when k > (2-a) then [k/(2 - a)] = t where t is any natural number from 1 to .....

From 1
[a.t+a/2]=p
since the relation is always equal to p in any case hence the
1<p<2
----->1<[a.t+a/2]<2 and 1<a<2 which gives t=0 which is not possible as t is a
natural number

now k < (2-a) the condition [k/(2 - a)] = -infinity to 0 let us take that equal to s
applying the above condition
[a.s+a/2]=p
1<p<2
1<[a.s+a/2]<2 and 1<a<2
which can give any value from -infinitu to 0 ..
hence X alone is not sufficient


From condition Y and the same condition the unique answer can not be obtained hence the answer is option 5 question can not be answered by combining X and Y