Discussion Room : Quant

58/15.
One side comes out to be 4/3 and another one comes out to be 3/5

One way could be
You take two triplets:
9,12,15
5,12,13
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.

@vinaycat2017
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?
a) Twice
b) Thrice
c) Once
d) 5 timesIn every meet their combined distance travelled is 100 m.
Let there be "a" meets.
=>[(100a) * 5/(5+3)]=300
Which gives a=4.8Though 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 hrA and B will meet for the third time = 3L/60L hour
B and C :
they meet first at L/2(4010)=L/60L hour
Third time they meet in:
L/60+ 2L/30=5L/60LThey meet together third time in LCM(5/60,3/60,5/60)
=15/60 hour=15 minBut options are: 5 ,7, 11and 13....

@sumitagarwal
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?

@zabeer
Yes sir, that's right.
But there is a similar question in gyan roomarithmetic 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.

@zabeer
Thank u. Got it :)

Thank you [email protected] sir @sumitagarwal

sir i got RS =6sqrt 3
dont know how to proceed further.......................................

@vinaycat2017
RS=6Sqrt3
Angle R =90
PR =12 (by similarity)
Apply pythagores in tri PRS
PS=sqrt(144+108)=6sqrt7Hope it helps..

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

@rahuljodavula
In general,
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 ?

@zabeer zabeer sir ek profit loss ka doubt hai i cant calculate the initial and final amount plz help..
A reduction of 20% in the price of sugar enables a housewife to purchase 6 kg more for Rs. 240. What is the original price per kg of sugar?

@vinaycat2017
Let original price be x per kg
So initially, we could buy 240/x kg for 240 rupees
Now the new price is 0.8 x (20% reduction in x)
So now we can buy 240/0.8x kg for 240 rupees
The difference, 240/0.8x  240/x = 6 kg (given)
240/x (1.25  1) = 6
x = 240 * 0.25/6 = 40 * 0.25 = 10 rs / kg would be the original price.

Find number of whole number solutions of a + b + c + d = 20 where a, b, c and d ≤ 8
= 23c3  4 * 14c3 + 4c2 * 5c3
= 1771  1456 + 60
= 375why 60 is as added here, if we go by the common way for odthis instead of coefficient method:
(9+a')+(b+c+d) =20
a'+b+c+d=11so total ways= 23C3 4*14c3.=315
which 60 cases need to be added here?
Could anyone expalin the part?

Question : A does half as much work as B in 3/4th of the time. Together they take 18 days to complete the work. How much time shall B take to do it?
Did it as follows.
Assumed B takes 12 days to to do 12 work. (Rate 1w/day)
So, A does 6 work(half as much as B) in 9 days (3/4th of the time taken by B to do the same) so the rate is 6/9 = 2/3 work per day)
Given they take 18 days to complete the work. One day they do (1+2/3) = 5/3 work, in 18 days they do 30 work.
Hence B shall take 30 days to do 30 work since the rate of B is 1w/ day.But the answer given is 45 days.
Is there any logic fail in the method I used?

In how many ways 8 friends be arranged in a row if three of the friends do not want to sit at the
extremes?pls explain?

@aadirs 30 days is correct I guess.
1/a + 1/b = 1/18
we know, a = 3b/2 (half the work in 3/4th time means full work in 1.5 times the time taken by b)
2/3b + 1/b = 1/18
5/3b = 1/18
b = 18 * 5/3 = 30