Discussion Room : Quant
ABCD is a square with side length 2 cm. It is divided into five rectangles of equal areas, as shown in the figure. The perimeter of the rectangle BEFG is
Can anyone explain please.
Side length = 2 cm.
So area of square = 4 cm^2
Given square is divided into 5 rectangles of equal areas. So each rectangle has an area of 4/5 cm^2.
Proceed with this understanding (i.e, if we know one side of rectangle = 2 then other side should be 2/5 to get an area of 4/5) and we get EB = 4/3 cm and BG = 3/5 cm. Figure as below
Perimeter = 58/15 cm (does it match with the OA?)
Sir one more
sumit agarwal last edited by
One side comes out to be 4/3 and another one comes out to be 3/5
sumit agarwal last edited by zabeer
One way could be
You take two triplets:-
a=12, b=5, c=9
Area comes out to be 84...by heron's.
If u put these values in options...only third option would give 84 as answer.
zabeer last edited by
Tan(A + B) = (Tan A + Tan B)/(1 - TanA * TanB)
Tan70 = (Tan50 + Tan20)/(1 - Tan50 * Tan20)
Tan70 - Tan70 * Tan50 * Tan20 = Tan50 + Tan20
We know, Tan70 * Tan20 = 1
So, Tan70 - Tan50 = Tan50 + Tan20
Tan70 = Tan20 + 2Tan50.
A swimming pool is of length 50 m. A and B enter a 300 m race starting simultaneously at one end of the pool at speeds of 3 m/s and 5 m/s. How many times will they meet while travelling in opposite directions before B completes the race?
d) 5 times
In every meet their combined distance travelled is 100 m.
Let there be "a" meets.
=>[(100a) * 5/(5+3)]=300
Which gives a=4.8
Though it cant be decimal.
But Where am I proceeding wrong with this approach????could anyone help me?
I guess the problem lies when they might be travelling in same direction ..
P and Q are two diametrically opposite points on a circular track. A and B start from P at speed of 20 and 40 rounds per hr respectively. C starts from Q at the same time at 10 rounds per hour. If B and C move in clockwise direction and A in anticlockwise direction ,find the time after which they meet for the third time anywhere on the track?
Sol:--What i tried is as follows:-
A and C
They meet first in L/2(30)L hr
So 3rd time they meet in L/60L+L/30L+L/30L= 5/60 hr
A and B will meet for the third time = 3L/60L hour
B and C :
they meet first at L/2(40-10)=L/60L hour
Third time they meet in:
They meet together third time in LCM(5/60,3/60,5/60)
=15/60 hour=15 min
But options are: 5 ,7, 11and 13....
Assumption that they travel 100 m together between every meetings is not correct. Like if A is 10 m from one of the end while B crossed him, then the next meeting would happen once B comes back from that end. They won't cover 100 m together in this scenario right?
Yes sir, that's right.
But there is a similar question in gyan room-arithmetic shortcuts ,where the scenario is same.. the meetings are given...and one has to find speed....
The same logic of combined distance is applied there.
So,what else could one apply here, instead of tabulation as used in the original answer....
Question in Gyan room is not the same.
Will share some thoughts on the question in discussion here.
We know, A would take 50/3 = 16.66 seconds and B would take 10 seconds to complete one lap.Let X and Y be the two sides of the pool, X being the start point. So,
(0 - 10 seconds) : A : X --> Y & B : X --> Y (ignore this slot as both are traveling in same direction)
(11 - 20 seconds) : A : (X --> Y)(Y --> X) & B : (Y --> X) (there is one meeting point here as they are traveling in opposite directions and B completes one full lap in this interval)
(21 - 30 seconds) : A : (Y --> X) & B : (X --> Y) (second meeting here)
(31 - 40 seconds) : A : (Y --> X)(X --> Y) & B : (Y --> X) (third meeting here)
(41 - 50 seconds) : A : (X --> Y) & B : (X --> Y) (ignore as both are in same direction)
(51 - 60 seconds) : A : (Y --> X) & B : (Y --> X) (again, same direction. So ignore. Race ends here and B wins)
So together 3 meeting points.
Note : We took interval of 10 seconds as B (fastest) would complete a full lap during this time.
Sure, it took some time to write all this, but it won't take more than a minute to solve with a rough sketch.
sumit agarwal last edited by
Thank u. Got it :)
sir i got RS =6sqrt 3
dont know how to proceed further.......................................
Angle R =90
PR =12 (by similarity)
Apply pythagores in tri PRS
Hope it helps..
Rahul Jodavula last edited by
3 individuals john wright, greg chappell and gary kristen are in the race for the appointment of new
coach of team india. The probabilities of their appointment are 0.5, 0.3 and 0.2 respectively. If john
wright is appointed then probability of ganguly appointed as a captain will be 0.7 and corresponding
probability if greg chappell or gary kristen is appointed are 0.6 and 0.5 respectively. find the overall
probability that ganguly will appointed as a captain
If you have OR, ADD the probabilities.
If you have AND, multiply the probabilities.
Here, Ganguly can be the captain in the following manner
(John AND Ganguly) OR (Greg AND Ganguly) OR (Gary AND Ganguly)
(0.5 * 0.7) + (0.3 * 0.6) + (0.2 * 0.5) = 0.63 should be our answer.
What's the OA ?