Official Quant thread for CAT 2013

MBA CAT 2024
11861
@venomizer said:
a-2orange colored area is ur ans...area of yellow portion =area of bigger circle-area of squarewhich is (pi-2)-now area of orange portion =4*(area of semicircle of smaller circle-area of yellow portion/4)
hw did u find area of squre
my geometry is so week
11862
@venomizer said:
@hari_bang u understood it or not bro..well seegiven that a circle of radius of 2cm has a square inscribed in it that means the diagonal of the square is also the diameter of the circle from there we can find the side of the square which we vl get as _/2(root2 )area of square=2now the radius of the the smaller circle vl be _/2by 2i hope now its clear ..if not ask again
thnx bro:)
11863
@incognitos said:
a. 1LHS of the equation can be viewed as a number written in the number system with base 11.....you just need to find whether 151001 can be converted into a unique number in number system with base 11.....it turns out that we can do that.....we get v=10, w=3, x=4, y=10, z=4....
is dis only solution
11864
@hari_bang said:
is dis only solution
first of all tell me whether the answer is 'a' or not....
this is the simplest of the solutions.....there is always hit and trail method which is cumbersome.....
11865
@incognitos said:
first of all tell me whether the answer is 'a' or not....this is the simplest of the solutions.....there is always hit and trail method which is cumbersome.....
i dnt knw ans yr
11866
@hari_bang said:
how many 6 dig no.can be formed using dig 1 to 6,without repition such that number is divisible by the digit at unit places...402528648720
we only need to check when 4 is at units => 120*3/5 = 72

so total = 720-72=648

11867

@hari_bang said:
The circle o having a diameter 2 cm,has a square inscribed in it.Each side of the square is then taken as a diameter to form 4 smaller circles O".Find the total area of all four O" cicles which is outside the circleO :I dnt knw ans...Plz provide solutions..
answer is 2
11868
@krum said:
we only need to check when 4 is at units => 120*3/5 = 72so total = 720-72=648
??????????
11869
@venomizer a different and lengthier version :)

all numbers ending up with 2 will be divisible by 2 ( ends at 2 )
all numbers ending up with 3 will be divisible by 3 ( since digit sum is anyway 3 )
all numbers ending up with five will be divisible by 5 will be divisible by 5 ( ends at 5)
all numbers ending up with 6 will be divisible by 6 ( since ending at 6 makes it divisbile by 2 and digit sum of 3 makes it also divisible by 3 so in turn the number is divisible by 6 )

only numbers ending at 4 will be divisible are having 24 and 64 as their penultimate digits ( 14,34,54 ain't possible )

so 5*5! + 2*4! = 5*120 + 2*24 = 600+48 = 648 :)
11870
@venomizer said:
??????????
and in the previous solution, he has actually removed all combination of numbers with 14,34,54 out of all possible 14,24,34,54,64 as their penultimate combinations.

Such numbers can be written as 120*3/5 or 4!*(5-3) = 72
11871
how many integral values of a are there such that quadratic expression (x+a)(x+1991) + 1 can be factored as a product (x+b)(x+c) where b and c are integers

11872
x^2 + 7x - 14(q^2 +1) = 0,'q' is an integer. How many integer roots can the quadratic equation have?
11873
@JoyNick said:
x^2 + 7x - 14(q^2 +1) = 0,'q' is an integer. How many integer roots can the quadratic equation have?
Is the answer None??
11874

Which regular polygon has the ratio of its diagonals to its sides as 3:1?

11875
@pyashraj yes... please elaborate
11876
@rushikesh90

Should be a Nanogon(9 sided)

nc2 - n/n = 3/1, or, n=9..
11877
@rushikesh90 said:
Which regular polygon has the ratio of its diagonals to its sides as 3:1?
Nonagon---9 sided
11878
@JoyNick

Given, x^2 + 7x - 14(q^2 +1)= 0...(i)

Now, x= {-7 +/- [49 + 56(q^2 +1)]^1/2}/2..Let x be any integer k..

Thus, 2k = {-7 +/- [49 + 56(q^2+1)^1/2]

=> 2k+1 = +/-[49 + 56(q^2+1)^1/2]

=>(2k+1)^2 = 49 + 56(q^2+1)^1/2

=>4k^2 + 49 + 28 = 49 + 56(q^2 +1)

=>k(k+7)= 14(q^2 +1)

Now, Two case are possible..

Case(i):When k = 14..Then, k+7 = q^2 +1, or, 20=q^2..Thus, q is not an Integer..

Case(ii): When k+7=14, Then, k=q^2+1, or, q^2=6..Thus, q is not an Integer..

Thus, Conclusively we can say that when q is an Integer..the roots of the above Qudratic Eqn are not Integer..Hence None.. :)
11879
@JoyNick said:
how many integral values of a are there such that quadratic expression (x+a)(x+1991) + 1 can be factored as a product (x+b)(x+c) where b and c are integers
a can take two values.....1993, 1989 ?????
11880
@incognitos yes...detailed soln please