Puyss..Im solving Quantum Cat by sarvesh kumar...Is there any dedicated thread for it??...If so pls provide the link..Help out puysss...
@dineshtce I" think no such dedicated thread is there for Sarvesh Kumar you may post the questions here only. And you shouldnt have asked this questions here .Keep it spam free
@rnishant231 said:Thanks..
check this out...u will get to know mass geometry....i was also not familiar with it...
@bodhi_vriksha said:11/24=.4583 which is a fraction greater than .45 .. so u may have got +3 .. but the motive of this question was to get the approach correct ..Act i am out ..so wont be able to upload the figure but i solve these kind of prob by mass geometry... as AQ is median .. ar(aqc)=1/2...now using mass geometry ,put weights of x at B and C .. hence weight at Q will be 2x .as APC=1:3 --> weight at c=x .. so weight at A will be 3x .. SO wight at R=4X..similarly, weights at A=3x and at Q=2x... so AOQ=2:3(inverse ratios of weights)..now ar(aor)/ar(aqc)= 2*1/5*4=1/10ar(aor)=1/10*1/2=1/20 =0.05 .. hence ar(ORCQ)=0.5 - 0.05= 0.45 Team BV--Pratik Gauri
Sir, can you explain the flaw in this?
regards
scrabbler
regards
scrabbler
@bodhi_vriksha said:no try again .. seems u r commiting a v small mistake .. close yet so far Team BV--Pratik Gauri
There are 50 successive discounts of 2%,4%,6%,8%....Find the net discount obtained?
@maroof10 said:There are 50 successive discounts of 2%,4%,6%,8%....Find the net discount obtained?
100%
:neutral:
@maroof10 said:There are 50 successive discounts of 2%,4%,6%,8%....Find the net discount obtained?
100% :mg:
regards
scrabbler
regards
scrabbler
@maroof10 said:There are 50 successive discounts of 2%,4%,6%,8%....Find the net discount obtained?
100%
Xavier and Haneef are competing with each other in a swimming competition in a pool of 50m length for 1000m Race . xavier crosses 50m length in 2 min's and haneef cross it in 3min 15secs . How many times they cross/meet each other ?
please share the detailed explanation .
Also had it been 5 min for crossing 50m in case of Haneef , what would have been the ans ?
Please explain in detail .
please share the detailed explanation .
Also had it been 5 min for crossing 50m in case of Haneef , what would have been the ans ?
Please explain in detail .
@Asfakul said:Xavier and Haneef are competing with each other in a swimming competition in a pool of 50m length for 1000m Race . xavier crosses 50m length in 2 min's and haneef cross it in 3min 15secs . How many times they cross/meet each other ?please share the detailed explanation .Also had it been 5 min for crossing 50m in case of Haneef , what would have been the ans ? Please explain in detail .
40 hai kya? if yes i will post my approach.. but i dont think 40 is thhe answer to the first part of the question. i am missing a case here
@Koushik98 said:Triangle ABC has unit area. P, Q, R divides the three sides in the ratios as shown in figure. Find the shaded area of triangle.
here u have to find ao/aq to arrive at the answer
ao/aq comes as 1/3
and area of apr comes as 1/8
area of quadilateral bprc is 7/8
so area of triangle boq=1/2/*2/3=1/3
area of triangle opb=1/2*1/3*1/2=1/12
area of the shaded portion =7/8 - (1/3+1/12)
= 7/8 - 5/12 = 11/24
so 11/24
@maroof10 said:@iLoveTorres@scrabbler approach plz
Last step discount of 100% means SP = 0. hence overall discount of 100%!
regards
scrabbler
regards
scrabbler
@Asfakul said:Xavier and Haneef are competing with each other in a swimming competition in a pool of 50m length for 1000m Race . xavier crosses 50m length in 2 min's and haneef cross it in 3min 15secs . How many times they cross/meet each other ?please share the detailed explanation .Also had it been 5 min for crossing 50m in case of Haneef , what would have been the ans ? Please explain in detail .
is it 19?
@bodhi_vriksha said:11/24=.4583 which is a fraction greater than .45 .. so u may have got +3 .. but the motive of this question was to get the approach correct ..Act i am out ..so wont be able to upload the figure but i solve these kind of prob by mass geometry... as AQ is median .. ar(aqc)=1/2...now using mass geometry ,put weights of x at B and C .. hence weight at Q will be 2x .as APC=1:3 --> weight at c=x .. so weight at A will be 3x .. SO wight at R=4X..similarly, weights at A=3x and at Q=2x... so AOQ=2:3(inverse ratios of weights)..now ar(aor)/ar(aqc)= 2*1/5*4=1/10ar(aor)=1/10*1/2=1/20 =0.05 .. hence ar(ORCQ)=0.5 - 0.05= 0.45 Team BV--Pratik Gauri
sir shouldnt we try to find out ao/aq as it will ease the pain of solving
i found out ao/aq as 1/3
and then area of shaded region as 11/24
whats wrong here??
@Asfakul said:Xavier and Haneef are competing with each other in a swimming competition in a pool of 50m length for 1000m Race . xavier crosses 50m length in 2 min's and haneef cross it in 3min 15secs . How many times they cross/meet each other ?please share the detailed explanation .Also had it been 5 min for crossing 50m in case of Haneef , what would have been the ans ? Please explain in detail .
As distance is constant ,
ratio of speeds are in inverse ratio of times taken ..
so , x/h= 195/120=13/8 here x and h denote their speeds
So x goes at 13v and h at 8v ... when X completes 26 rounds and H completes they would be at the same starting point again .. so they wont meet in 1st round, 27th round and so on ... 26k+1 :)
But in this ques 1000/50=20 rounds but they wont meet in first round ..
Hence they would meet 19 times :)
Team BV--Pratik Gauri
@bodhi_vriksha said:Nishant, I have drawn and attached the graph. Hope that helps.Team BV - Vineet
sir, yeah 3/5 and 5/3 ratio kon sa interval pe a raha hain??
@Asfakul said:Xavier and Haneef are competing with each other in a swimming competition in a pool of 50m length for 1000m Race . xavier crosses 50m length in 2 min's and haneef cross it in 3min 15secs . How many times they cross/meet each other ?please share the detailed explanation .Also had it been 5 min for crossing 50m in case of Haneef , what would have been the ans ? Please explain in detail .
when the two divers meet itravelling in opposite direction
t= 50(2k+1)/(5/12 +10/39)
t=74.28(2k+1) k= 0....... 14 satisfies
when the two divers meet travelling in same direction
t=50(2k+1)/(5/12 - 10/39)
=272.xx(2k+1)
k= 0 .....3 satisfies
upper limit of t= 20*2=40*60=2400 sec
15+4=19