@psk.becks said:question
18 ?
@next_big_thing said:This is a question from CAT 2010 paper. Can anyone plz solve it coz I am at a loss ??? -->A triangular garden is to be enclosed inside 3 straight fences. Raju started running at an uniform speed on the three sides of the proposed boundary of the garden & discovered that he was taking 48min, 102min, 90min to traverse the garden's three sides. However, in the latest colony meet, it was decided that the coverage area of the park should be reduced to accomodate a parking lot. Since the fence on the shortest boundary was already constructed, the length of the boundary covered by Raju in 90min was reduced by 60% & the longest was modified accordingly so that the park still remained triangular in shape.1) How much time will Raju take to run along the boundary of the garden thus formed? 120min, 148min, 128min, 160min, 144min2) If Raju's speed was 15metres/min, what is the approx area of the parking lot (in km^2)? 0.39, 0.28, 0.29, 0.45, 0.52Plz post the solution as well or else it wont be of any help.
1) 144 mins (pythagoras theorem) with old sides as 48s, 102s,90s and new sides being 48s, 36s, 60s while still maintaing a right angled triangle. 2) .29 km^2
@psk.becks said:question..
All letters except E can be arranged in 5!/3!
These 5 letters will have 6 gaps in which E can be placed in 6C3 ways.
Total ways=5!/3! *6C3
@Subhashdec2 said:bhai approach btana i got P to be equal to option AQ 1 hai kya?
P = nP2 (m+n-2)!
Q = (m+n-2)!/m!(n-2)!
P/Q = nP2 * mPm * (n-2)!
@Subhashdec2 said:18?abc as one unit=3!=6abd as one unit=3!=6ac and bd as two units=3!=6
Not able to understand well 
. Plz explain
. Plz explain @Buck.up said:Not able to understand well . Plz explain
Edited
Bhai
Possibilities are
-ABC
-ABD
-AC and BD
For first two cases if we take it as a unit (for ex- abc as unit and remaining d,e,f) then we can arrange in (n-1)! i.e. 3! . Since there are two such cases so 2 * 3!
Similarly for the last case we take (ac as one unit and bd as other unit and e,f) then we apply (n-1)! i.e. 3!
Total = 2*3! + 3! = 18
Bhai
Possibilities are
-ABC
-ABD
-AC and BD
For first two cases if we take it as a unit (for ex- abc as unit and remaining d,e,f) then we can arrange in (n-1)! i.e. 3! . Since there are two such cases so 2 * 3!
Similarly for the last case we take (ac as one unit and bd as other unit and e,f) then we apply (n-1)! i.e. 3!
Total = 2*3! + 3! = 18
@dushyantagarwal said:Bhai Possibilities are -ABC-ABD-ACDFor each case if we take it as a unit (abc as unit and remaining d,e,f) then we can arrange in (n-1)! i.e. 3! . Since there are three cases so 3 * 3!
Yaar yeh ACD ke case me "B must have C/D on its right" kaise fit hogi
Bodhi Vriksha has shifted its base to Facebook...
@Buck.up said:Yaar yeh ACD ke case me "B must have C/D on its right" kaise fit hogi
my bad .. I put that in other way. 
Check again that post .. I have edited
Check again that post .. I have edited
@dushyantagarwal said:my bad .. I put that in other way. Check again that post .. I have edited
Got it. Thanks 
Question..
A circle C is such that it touches two sides of an equilateral triangle as well as the circumcircle of the given triangle. what is the ratio of areas of incircle :circle C: circumcircle.
NO OA AVAILABLE.
@KhannaiiM said:
m=( 1051)1/3 -(151)1/3
logm=1/3log(1051/151)
logn=1/3log(1052/152)
logp=1/3log(1053/153)
since log is an increasing function
we need to compare
1051/151 1052/152 1053/153
a/b >a+x/b+x
so m>n>p
@Subhashdec2 said:m=( 1051)1/3 -(151)1/3logm=1/3log(1051/151)logn=1/3log(1052/152)logp=1/3log(1053/153)since log is an increasing functionwe need to compare1051/151 1052/152 1053/153a/b >a+x/b+xso m>n>p
m=x-y
then how can this be
log m= log x-log y ?????
@psk.becks said:Question..A circle C is such that it touches two sides of an equilateral triangle as well as the circumcircle of the given triangle. what is the ratio of areas of incircle :circle C: circumcircle.NO OA AVAILABLE.
Circle's radius would be: a(2-sqrt3)/2
Hence ratio would be: 1: (2-sqrt 3)^2: 4