Discussion Room : Quant

Suppose real numbers x and y satisfy x^2 + 9y^2  4x + 6y + 4 = 0.what is the maximum value of 4x  9y?
Geometric approach
We know that (x  2)^2 + (3y + 1)^2 = 1 and we need to maximise 4x  9y.
Now, let's replace 3y by Y, so that know equation turns into a circle
(x  2)^2 + (Y + 1)^2 = 1 with center at (2, 1) and radius 1.
And now we need to find the maximum value of 4x  3Y which will be equal to 4x  3Y = k (say) and will be achieved when this line is tangential to the circle.Doubt  what is the logic behind the point that " for the value of expression to be maximum , that line should be tangential to the circle " ..
Could anyone help me with understanding this concept please.

@sumitagarwal
We are trying to maximize the distance between the line and the centre of the circle. It is also known that the circle lies on the circle so it has to be a tangent.Sharing a solution from @kamal_lohia sir
x^2 + 9y^2  4x + 6y + 4 = 0
i.e. {x^2  2(x)^(2) + 2^2} + {(3y)^2 + 2(3y)(1) + 1} = 1
i.e. (x  2)^2 + (3y + 1)^2 = 1Now we want to maximise 4x  9y. Let 4x  9y = k, i.e. x = (k + 9y)/4
So eliminating x, above equation becomes: (k + 9y  8)^2 + 16(3y + 1)^2 = 16
i.e. 225y^2 + {18(k  8) + 96}y + (k  8)^2 = 0
i.e. 225y^2 + (18k  48)y + (k  8)^2 = 0Now as y is a real number, so discriminant of the above equation should be greater than or equal to zero.
So, (18k  48)^2  4(225)(k  8)^2 ≥ 0
i.e. (18k  48)^2  (30k  240)^2 ≥ 0
i.e. (3k  8)^2  (5k  40)^2 ≥ 0
i.e. (2k + 32)(8k  48) ≥ 0
i.e. (k  6)(k  16) ≤ 0
i.e. 6 ≤ k ≤ 16.Thus required maximum value of 'k' is 16.

A polygon has internal angles of measures 90° and 270° only. If it has18 angles of measure 270°, then what is the number of angles with measure 90°?
Sol:Let there be n angles of 90°, thn sum of all angles = (90n) + (18 × 270) = (n + 16) 180
How RHS= (n+16) 180?
Could anyone explain the part.?

Sum of the interior angles of a polygon is: (n  2) × 180° where n is the number of sides of the polygon.
now n  sided polygon has n interior angles. here we have 18 angles that measure to 270 and say y angles that measures to 90.
So n = number of sides = number of interior angles = 18 + y
sum of all angles = (n  2) × 180° = (18 + y  2) * 180° = (y + 16) * 180°Detailed solution :
Let there be x angles of 90°
Also let there be y angles of 90°
so we can say n = x + y
Sum of interior angles = (x + y  2) × 180°
we know, sum of interior angles = x * 90° + y * 270°
x * 90° + y * 270° = (x + y  2) × 180°
we know, y = 18
so x * 90° + 18 * 270° = (x + 16) × 180°
x * 90° = 18 * 270°  16 * 180°
x = (18 * 270°  16 * 180°)/90°
= 54  32 = 22
So there are 22 interior angles with measure 90°
what is the OA ?

Thanks a lot:)

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?)


@

@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....