Official Quant thread for CAT 2013

@nole said:
@hatemonger yes m getting something else.please share the solution.
Reg hex in circle => radius of circle = side of hex (think about it! Childhood craft classes and all... 😃 )

So if area of circle = 144pi then radius = 12 = side of hex. Now area of reg hex = 3rt3/2 * side^2 = 3rt3/2 x 144 = 216rt3.

regards
scrabbler

@scrabbler thanks
@nole said:
@hatemonger yes m getting something else.please share the solution.
area of circle is 144pi so r= 12
and area of hexagon (sqrt(3))/4)*12^2*6 six equilateral triangle
@nole said:
1) The area of a circle curcumscribed about a regular hexagon is 144 pi. What is the area of the hexagon ?2) find the area of the triangle whose sides measuresqrt(x+y)sqrt(y+z)sqrt(z+x)


Cos A = y/_/(y+z)(x+y)

Calculate sin A ans use Area= 1/2 B*C * Sin A

Getting Area = (1/2 )*_/(xy+yz+zx)


Anybody having a short cut
@amresh_maverick said:
koi solve akro iskoRem 123123123.....123 ( 300 digits) when divided by 99
6???

@scrabbler If radius is equal to side of hex then i think we will have 6 equilateral traingle (full hexagon) then m getting altitude as sqrt(144-36) =sqrt(108)

now m doing 6 (6 equilateral traingle) * sqrt 3 /4 * 108= 162 sqrt (3) as answer .where did i go wrong ? can u tell ?
@amresh_maverick said:
koi solve akro iskoRem 123123123.....123 ( 300 digits) when divided by 99
Is it 33?

regards
scrabbler

@amresh_maverick said:
koi solve akro iskoRem 123123123.....123 ( 300 digits) when divided by 99
33?
66*50 mod 99 = 33 ..
@scrabbler said:
Is it 33?regardsscrabbler
Aaap aur galat , ho nahi sakta

Also plz solve @jain4444 sir's prob
@scrabbler m sorry i was taking altitude when it was side of the triangle . got it.
@scrabbler said:
Is it 33?regardsscrabbler
bhai 33 kaise ayega
??
@nole said:
@scrabbler If radius is equal to side of hex then i think we will have 6 equilateral traingle (full hexagon) then m getting altitude as sqrt(144-36) =sqrt(108)now m doing 6 (6 equilateral traingle) * sqrt 3 /4 * 108= 162 sqrt (3) as answer .where did i go wrong ? can u tell ?
It should be rt3/4 * side^2 not *ht^2

Are = 1/2 * base * ht
Here ht = rt3/2 * side
so area = 1/2 * side * rt3/2 * side = rt3 / 4 * side^2

regards
scrabbler

@scrabbler ya .got it thanks .
@amresh_maverick said:
Aaap aur galat , ho nahi saktaAlso plz solve @jain4444 sir's prob
Ans pata hai uska?? Pata hai to he btaunga m :P
@amresh_maverick said:
koi solve akro iskoRem 123123123.....123 ( 300 digits) when divided by 99
OK....maybe slightly non-conventional approach....Chinese-Remainder-type logic.

123123....(300 digits) is divisible by 11 (either using the fact that divisible by 1001, or by just applying divisibility rule of 11) so let me say it is 11K

Also when divided by 9, it leave a remainder of 6 (since sum of digits is 600 and 600/9 leaves rem 6). So it is of the form 9M+6.

Hence when divided by 99 which is 11 x 9, it should be a number of both these forms (and since the first form 11K occurs at steps of 11 and the second 9M+6 at steps of 9, we will get 1 such number 6, 15, 24, 33, 42, 51 and so on, 33 is the one which also satisfies 11K form. Hence went with 33....

regards
scrabbler

@amresh_maverick can u explain ur approach ?
@jain4444 said:
Find fifth digit from the end of 5^(5^(5^(5^5)))
Answer options? Slightly painful otherwise....maybe options se easy hoga.

regards
scrabbler

@scrabbler said:
Answer options? Slightly painful otherwise....maybe options se easy hoga.regardsscrabbler
Could it be done like this? :
R [ 5^5^5^5^5 ] / 10^5 = R [ 5^5k/10]^5 = 03125
Hence, 5th digit is = 0??
@amresh_maverick said:
koi solve akro iskoRem 123123123.....123 ( 300 digits) when divided by 99
33 ar raha hai bhai...
sai hai kya??
the number can be written as 11k or 9n+6
the ifrst number to satisfy this is 33...
so the number 123123...
will be part of an AP which has its first term as 33....

@scrabbler said:
Answer options? Slightly painful otherwise....maybe options se easy hoga.regardsscrabbler
0 , 1 , 5 , 4