Discussion Room : Quant

What is the greatest common divisor of the 2010 digit and 2005 digit numbers below?
333.........333(2010 times)
777........777(2005 times)Let A = 333..33 and B = 777..77, then A/3 and B/7 both contains the digit 1 only (2010 times and 2005
times respectively). As HCF(2010, 2005) = 5 that means HCF(A/3, B/7) = 11111.
Now A is also divisible by 7 as 2010 = 6 × 335 (Remember a number formed by repeating same digit 6 times is divisible by 7). So HCF(A, B) = 77777Here, As HCF(2010, 2005) = 5 that means HCF(A/3, B/7) = 11111. ???
Could anyone explain the solution?

In ∆ABC (not drawn to scale), the altitude from A, the angle bisector of ∠BAC, and the
median from A to the midpoint of BC divide ∠BAC into four equal angles. What is the measure in degrees of angle ∠BAC?Here the solution says apply sine rule
What I could get is AB sin3A/4 = AC sin A/4
But nt able to establish relation between AB and AC. How to find the angle BAC then?

@sumitagarwal
A = 3333 ...... (2010 times)
B = 7777 ...... (2005 times)A = 3 * 1111 ..... (2010 times)
B = 7 * 1111 ..... (2005 times)HCF(11111.... (m times), 1111 .... (n times)) = 1111.. (HCF(m,n) times)
A is divisible by 7. B is not divisible by 3
HCF(A, B) = 11111 * 7 = 77777

30,000 is divided into three equal installments.The remaining sum at rate 4 % is added to the intallment amounts.Find the value of each installment.

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
(a) 51/16
(b) 36/11
(c) 58/15
(d) 47/13

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 belowPerimeter = 58/15 cm (does it match with the OA?)


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@zabeer
Sir one more

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