Official Quant thread for CAT 2013

@subhakimi said:
1. 5 years ago the ratio of ages of father to that of his son was 4:1 and 15 years hence it will be
4x, x after 20 years = (4x+20)/(x+20) = 2:1, x= 10
Present age of son = 15
@subhakimi said:
2: 1. What is the present age of son ?Again try by without equation. (I know it's easy)2.Akbar says to Birbal "I am four times as old as you were when I was as old as you are". Birbal replies "10 years back I was 9 yours younger to you". Find the age of Birbal.
15 again
@subhakimi said:
3. How many zeros are there at the end of 626! - 625! ?

625!*625 = 5^4 * 5^(125+25+5+1) = 5^160
160 zeroes
@subhakimi said:
3. How many zeros are there at the end of 626! - 625! ?
4 + 125 + 25 + 5 + 1 = 160 ?
@subhakimi said:
What is the least area of a triangle whose vertices have integral co-ordinates and two of the sides are greater than 2013?
503 k aas paas hai kya...??

Three sides 2014 2014 and 1...??
@sid2222000 said:
503 k aas paas hai kya...??Three sides 2014 2014 and 1...??
Third vertex k integral co-ordinate nai honge na tab?
@ScareCrow28 said:
Third vertex k integral co-ordinate nai honge na tab?
@subhakimi said:
Sorry sid babu ... but dur dur bhi nahi hai Was posting few easy ones ... so many people answered them so quickly ... had to up the level a bit
pacch.....integral sides leke kar rha tha...
@subhakimi said:
What is the least area of a triangle whose vertices have integral co-ordinates and two of the sides are greater than 2013?
x^2 + y^2 = 2014^2

y = _/(2014^2 - x^2)

Yaha se koi x niklega such that y is integer :splat:

Is it the right method?..
@sid2222000 said:
pacch.....integral sides leke kar rha tha...
1012 hai kya answer...??

Coordinates:

(0,0), (0,1), (2014, 0)
@subhakimi said:
1.5 years ago the ratio of ages of father to that of his son was 4:1 and 15 years hence it will be 2:1. What is the present age of son ?Again try by without equation. (I know it's easy)2.Akbar says to Birbal "I am four times as old as you were when I was as old as you are". Birbal replies "10 years back I was 9 yours younger to you". Find the age of Birbal.3. How many zeros are there at the end of 626! - 625! ?
15
15
160..
A given line segment is divided at random into three parts. What is the probability that they form sides of a possible triangle ?
@sid2222000 said:
2013 hai kya answer...??Coordinates:(0,2013), (1,0), (-1,0)
Ye sai lagra hai :splat: ..

@subhakimi said:
What is the least area of a triangle whose vertices have integral co-ordinates and two of the sides are greater than 2013?
2013 ?

co-ordinates (1,0)(-1,0)(0,2013)..area = 1/2*2*2013 = 2013...
@subhakimi said:
What is the least area of a triangle whose vertices have integral co-ordinates and two of the sides are greater than 2013?
Seems its 2013...
@subhakimi said:
What is the least area of a triangle whose vertices have integral co-ordinates and two of the sides are greater than 2013?
1007 ..... y=1 line ko fix karo and form triangles.

coordinates will be 0,0 ... 2014,0 and 2014,1 ...


@subhakimi said:
Guys I said it is less that 503 (which was the answer of Sid)It has a very small value. Try to get the logic.BTW try the question posted by @albiesriram .
1/2
sorry misinterpreted the question
@subhakimi Dimaag ki dahi hogai sir, but 1007 se kam ni hora :banghead:
@Iamchaitu said:
Last Few years in CAT coaching centres did you find that ....

Sir, kindly do not spam here. I think it's against PG rules to advertise.
you may post this on shoutbox.
http://www.pagalguy.com/posts/4952946

#Mother of all Spams!

@subhakimi said:
Guys I said it is less that 503 (which was the answer of Sid)It has a very small value. Try to get the logic.BTW try the question posted by @albiesriram .
Answer is 1/2 na....:P

Coordinates:

(0,0), (-1,0) , (-2014,1)
@subhakimi said:
Is this the answer to my question or posted by albie ? If it's answer of my question then you know it's correct
am i? i thought i misinterpreted the question
@subhakimi said:
Answer is 1/2What we need to take is that the height cannot be less than 1 and so is the baseSo we can take height to be 1 and base to be 1 as well.Then we can draw an obtuse angled triangle and stretch it as far as possible. You need not bother about lengths as the height & base are fixed
Edit korechi.......typo hoye gechilo....:P
@subhakimi said:
Answer is 1/2What we need to take is that the height cannot be less than 1 and so is the baseSo we can take height to be 1 and base to be 1 as well.Then we can draw an obtuse angled triangle and stretch it as far as possible. You need not bother about lengths as the height & base are fixed
Couldn't understand sir, 2 of the lengths are > 2013 na..
You have fixed 2 of them as 1, so howcome they satisfy the question?

EDIT: Got it sir