official quant thread for cat08

[madadi]

Originally Posted by selebratinglife

Answer can be any number..
50x=40y(x and y are respective run rates of australia and india)
Now even using the 2nd statement we get the same equation 30(4y/3) = 50x..No conclusion can be arrived at..

yep, definitely... from the question itself, we come to know, that australia played 50 overs, but India, equalled the score in 20% less overs, which means in 40 overs itself... and if India's run rate were 33.33% (or 1/3 rd) higher, the scored would've been equalled in 10 overs less... which means 30 overs itself...

move over to the options now... the first is 250... this means India made 250 runs in 40 overs... the run rate, hence, comes out to be 250/4..
if the run rate were 1/3rd higher, it'd have been, 25/4 + 25/12 = 100/12...
now, if we let the total overs played by India be x, then total runs scored by India= run rate* no of overs...
=> 100/12 * x = 250, which gives the value of x as 30...
and we've already been given that India equals the score in 30 overs...
hence, 250 is the answer....

But if we look at the other options, they suit well too... hence, no conclusion can be drawn...

[srikar2097]

Originally Posted by madadi

yep, definitely... from the question itself, we come to know, that australia played 50 overs, but India, equalled the score in 20% less overs, which means in 40 overs itself... and if India's run rate were 33.33% (or 1/3 rd) higher, the scored would've been equalled in 10 overs less... which means 30 overs itself...

move over to the options now... the first is 250... this means India made 250 runs in 40 overs... the run rate, hence, comes out to be 250/4..
if the run rate were 1/3rd higher, it'd have been, 25/4 + 25/12 = 100/12...
now, if we let the total overs played by India be x, then total runs scored by India= run rate* no of overs...
=> 100/12 * x = 250, which gives the value of x as 30...
and we've already been given that India equals the score in 30 overs...
hence, 250 is the answer....

But if we look at the other options, they suit well too... hence, no conclusion can be drawn...

It does not work that way here! It would be a mistake to do it like this...

Lets take option (b) i.e india scores 240 in 40 overs
=> run rate is 6 runs/over
it increases it by 4/3 and plays for 30 overs only.
=> 6*(4/3)*30 gives 240 !

For the other options too, this is true...

Hence (e) none of the above. So the 'crook' method might not always work

[madadi]

yes, srikar... true to the last word... solved through the options.. only to find out, the answer cud be calculated through the other options too... hance edited the post... (faux pas!!)... sorry....

[selebratinglife]

Originally Posted by srikar2097

It does not work that way here! It would be a mistake to do it like this...

Lets take option (b) i.e india scores 240 in 40 overs
=> run rate is 6 runs/over
it increases it by 4/3 and plays for 30 overs only.
=> 6*(4/3)*30 gives 240 !

For the other options too, this is true...

Hence (e) none of the above. So the 'crook' method might not always work

Answer definitely is not e)..since all he values satisfy ..So the question is best left unanswered

[shivam_01]

New Problem!!!!!

If one root of the equation x3 + 3x2 + 4x + D = 0, is the arithmetic mean of the other two roots, then the value of D is
-3 6 5 2 9

[Fuzon]

New Problem!!!!!

If one root of the equation x3 + 3x2 + 4x + D = 0, is the arithmetic mean of the other two roots, then the value of D is
-3 6 5 2 9

Is the answer -3?

[intello]

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[linksuresh]

Originally Posted by shivam_01

New Problem!!!!!

If one root of the equation x3 + 3x2 + 4x + D = 0, is the arithmetic mean of the other two roots, then the value of D is
-3 6 5 2 9

simple !
ans => D= 2

let roots be a,b,c .. a<b<c => b = (a+c)/2
a+b+c = -3
=> a+c = -2


ab+bc+ca = 4
=> b(a+c)+ca = 4
4/2+ca=4
=>ac = 2

D = -abc = -ac(a+c)/2
hence D=2

[srikar2097]

Originally Posted by shivam_01

New Problem!!!!!

If one root of the equation x3 + 3x2 + 4x + D = 0, is the arithmetic mean of the other two roots, then the value of D is
-3 6 5 2 9

Let roots be (a-d), a, (a+d)
sum of roots=-b/a=-3
=>a=-1 -->1

sum of the product taken 2 at a time
is 3-d^2 = 4
or (1-d^2) = 2 --->2

product of roots is -d/a = -D
=> (1-d^2) = D
from 2 we get D = 2. Hence choice(D)

[raghav507]

Originally Posted by shivam_01

New Problem!!!!!

If one root of the equation x3 + 3x2 + 4x + D = 0, is the arithmetic mean of the other two roots, then the value of D is
-3 6 5 2 9

hi,
i am getting D = 2...using the sum of roots, product of roots (all and taken 2 at a time)
is it the right answer...