AIMCAT 1203 (please do not open if you have not given the test yet)

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This thread is for AIMCAT 1203 starting from 02nd October 2011 till 09th October 2011. Lets post our scores and discuss the test. P.S: Please Don't discuss any question before the test window is over. All the best Puys !!!
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are nhi shillong wala dost was in gnral category
ek kozikhode me gya tha with 81 prctl he was SC ..reservation is bullshit


mera bhi ek friend shillong mei gya hai 93 percentile..
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@The bond : Great man!! THanks a lot...Totally forgot to incorporate the arrangements amongst the buyers...THanks man!!

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Shishir123 Says
Why have people stopped giving aimcats????? Only 16000 gave this when whhen at a time more than 28k were writing...?


I think most of the ppl are busy preparing hard and are avoiding mocks to not let down their morale. Or on the flip side it may be the case that after looking at an average performance in mocks they have lost hope and relinquished their dreams in their minds. But no doubt these 16000 are serious candidates. Better try to get at the top among these.
IMT GHAZIABAD
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I have a doubt regarding question 13th...

The way i solved it is as follows...What am i missing..

For exactly 2 buyers to have billing amount in the range 200 and 300

the probability is 0.1*0.1 *a(a is calculated as under)

For the remaining 2 buyers the billing amount can be

a) both between 0-100 ---- o.6*o.6
b) bith in 100-200 -----o.2*0.2
c) Both greater than 3oo ---- 0.1*0.1
similarly the other combinations aew

0.1*0.6,0.6*0.2 and 0.1*0.2

Therefore my answer came out to be

o.1*0.1(0.6*o.6 + 0.2*0.2 + 0.1*0.1 + 0.6*0.2+ 0.2*0.1+0.1*0.6)

But the answer that i am getting by this method is different!!

Please help Puys!!


Your method corrected:

For exactly 2 buyers to have billing amount in the range 200 and 300 the probability is 4C2*0.1*0.1 *a(a is calculated as under)(you need to select two among the 4 first.)

For the remaining 2 buyers the billing amount can be

a) both between 0-100 ---- o.6*o.6
b) bith in 100-200 -----o.2*0.2
c) Both greater than 3oo ---- 0.1*0.1
similarly the other combinations aew

2(0.1*0.6),2(0.6*0.2) and 2(0.1*0.2)
(the part considering different range for remaining two people (say, P3 and P4) would be multiplied by two: P3-range1, P4-range2, or P3-range2, P4-range1)

Therefore the answer is:
4C2*o.1*0.1(0.6*o.6 + 0.2*0.2 + 0.1*0.1 + 2(0.6*0.2+ 0.2*0.1+0.1*0.6))


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can anyone plz explain question no 26 the liar n truthteller one ... i am in a fix:banghead:

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bhar.anirban5 Says
But i have taken all cases...When the two remaining buyers belong to 2 different categories/....

when u have taken for diff categories multiply by 2 as two cases will arise. i did that and gt the correct ans.
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Why have people stopped giving aimcats????? Only 16000 gave this when whhen at a time more than 28k were writing...?

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As far as i remember there was nothing like other two shud have prices in same range. u have to consider cases when they are in different ranges.

hope it helps!!


But i have taken all cases...When the two remaining buyers belong to 2 different categories/....
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I have a doubt regarding question 13th...

The way i solved it is as follows...What am i missing..

For exactly 2 buyers to have billing amount in the range 200 and 300

the probability is 0.1*0.1 *a(a is calculated as under)

For the remaining 2 buyers the billing amount can be

a) both between 0-100 ---- o.6*o.6
b) bith in 100-200 -----o.2*0.2
c) Both greater than 3oo ---- 0.1*0.1
similarly the other combinations aew

0.1*0.6,0.6*0.2 and 0.1*0.2

Therefore my answer came out to be

o.1*0.1(0.6*o.6 + 0.2*0.2 + 0.1*0.1 + 0.6*0.2+ 0.2*0.1+0.1*0.6)

But the answer that i am getting by this method is different!!

Please help Puys!!


As far as i remember there was nothing like other two shud have prices in same range. u have to consider cases when they are in different ranges.

hope it helps!!
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I have a doubt regarding question 13th...

The way i solved it is as follows...What am i missing..

For exactly 2 buyers to have billing amount in the range 200 and 300

the probability is 0.1*0.1 *a(a is calculated as under)

For the remaining 2 buyers the billing amount can be

a) both between 0-100 ---- o.6*o.6
b) bith in 100-200 -----o.2*0.2
c) Both greater than 3oo ---- 0.1*0.1
similarly the other combinations aew

0.1*0.6,0.6*0.2 and 0.1*0.2

Therefore my answer came out to be

o.1*0.1(0.6*o.6 + 0.2*0.2 + 0.1*0.1 + 0.6*0.2+ 0.2*0.1+0.1*0.6)

But the answer that i am getting by this method is different!!

Please help Puys!!

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