Official Quant thread for CAT 2013

@bullseyes said:
A rope is cut into 3 pieces of unequal lengths. Find the length of shortest piece.1) Length of 2 longest pieces is 12 cm2) Length of 2 shorter pieces is 11 cm. a) can be answered by a aloneb) can be answered by b alonec) can be answered by using bothd) can be answered by eithere) cannot be answered
let the lengths be (6+x), (6-x) and (5+x)

using the conditions we get
(6+x)>(6-x) => x > 0
(6-x)>(5+x) => x

=> length of rope is (6+x)+(6-x)+(5+x) = 17+x

since we can any value of x between 0 and 0.5 hence option (e)

ATDH.
@bullseyes said:
A rope is cut into 3 pieces of unequal lengths. Find the length of shortest piece.1) Length of 2 longest pieces is 12 cm2) Length of 2 shorter pieces is 11 cm. a) can be answered by a aloneb) can be answered by b alonec) can be answered by using bothd) can be answered by eithere) cannot be answered
e) cannot be answered ?

Let the lengths of the 3 pieces be a,b,c. (a > b > c)
1) a + b = 12 (cannot be answered by 1 alone)
2) b + c = 11 (cannot be answered by 2 alone)

First set: a = 6.4, b = 5.6, c = 5.4
Second set: a = 6.3, b = 5.7, c = 5.3
Clearly, there are multiple sets of (a,b,c) which satisfy both the statements.

Hence, cannot be answered.
@ScareCrow28 said:
e) cannot be answered
@anytomdickandhary said:
let the lengths be (6+x), (6-x) and (5+x) using the conditions we get(6+x)>(6-x) => x > 0 (6-x)>(5+x) => x => length of rope is (6+x)+(6-x)+(5+x) = 17+x since we can any value of x between 0 and 0.5 hence option (e)ATDH.
yes it is
@sujamait said:
a right angled triangle with longest side 80,construct 3 paths from the corner point opposite to the longest side such that these three paths divide the longest side into four equal segments.determine the sum of the lengths of three paths.
I think there would be no constant answer to the problem....

if we think of extreme case when a right triangle with base almost equal to 0 then height will almost coincide with hypotenuse. Hence the paths drawn from the corner points will be almost vertical, and in this case sum of the lengths of paths would be 60+40+20 = 120.

Now take the other extreme case when other two sides are equal, i.e it is right angle isosceles triangle. In this case sum of the paths would be 40*{1+rt(5)}

Not sure if I missed anything in the problem!

ATDH.

same DS question.


IS x negative?

1) n^3(1-x^2)
2) x^2-1


a) can be answered by 1 alone
b) can be answered by 2 alone
c) can be answered by using both
d) can be answered by either
e) cannot be answered


@soumitrabengeri said:
Here, the length of the middle path will be half the longest side..i.e the hypotenuseHence, l(middle path) = 40mLength of other 2 paths = root (40^2+20^2)Sum of the lengths of the 3 paths = 40+2root(40^2+20^2)Please correct me if i am wrong
i had a doubt will middle side be perpendicular??? its not always true
@bullseyes said:
same DS question.IS x negative?1) n^3(1-x^2)2) x^2-1 a) can be answered by 1 aloneb) can be answered by 2 alonec) can be answered by using bothd) can be answered by eithere) cannot be answered
e) cannot be answered?
@Brooklyn said:
i had a doubt will middle side be perpendicular??? its not always true
Even i am not sure about it..that's why i wrote the last line..
Please correct me if i am wrong
@bullseyes said:
same DS question.IS x negative?1) n^3(1-x^2)2) x^2-1 a) can be answered by 1 aloneb) can be answered by 2 alonec) can be answered by using bothd) can be answered by eithere) cannot be answered
c) cannot be answered using both ?

1) x^2 > 1 and n^3 > 0
=> n^3(1 - x^2) Cannot be answered by 1 alone

2) x^2 => x^2 - 1 Cannot be answered by 2 alone

Put 2 in 1
1 - x^2 > 0
n^3 (1 - x^2) => n Cannot be answered by using both.
@anytomdickandhary said:
I think there would be no constant answer to the problem....if we think of extreme case when a right triangle with base almost equal to 0 then height will almost coincide with hypotenuse. Hence the paths drawn from the corner points will be almost vertical, and in this case sum of the lengths of paths would be 60+40+20 = 120.Now take the other extreme case when other two sides are equal, i.e it is right angle isosceles triangle. In this case sum of the paths would be 40*{1+rt(5)}Not sure if I missed anything in the problem!ATDH.
yeah actually sum of squares of path was asked..forgot to type that..in hurry.
5600 is correct.
@sujamait said:
yeah actually sum of squares of path was asked..forgot to type that..in hurry.5600 is correct.
in that case if "none of these" was not an option then I would directly write 60^2 + 40^2 + 20^2 without giving a thought.

Idea is that as there is no constraint on the other two sides, so I will chose one of the extreme cases where my calculations would be easiest.

If "none of these" was also an option then I would have rechecked by assuming it to be an isosceles triangle and checked the calculations once more.

So my take is that in worst case scenario this would not be more than 1 min question.

ATDH.
@jain4444 said:
you got the right place then just be regular on this thread Three pipes A, B and C can fill a tank in 6 hours. After working at it together for 2 hours, C is closed and A and B can fill the remaining part in 7 hours. The number of hours taken by C alone to fill the tank is:A. 10 B. 12C. 14 D. 16
14 hours!!
@bullseyes said:
same DS question.IS x negative?1) n^3(1-x^2)2) x^2-1 a) can be answered by 1 aloneb) can be answered by 2 alonec) can be answered by using bothd) can be answered by eithere) cannot be answered
Ans. Cannot be answered.
Question is asked whether 'x' is negative and in both the given statements the same even power of x i.e. x^2 is given as a part of information.
We can only comment on the absolute value of x if the only information we have is an even power of x.
@bullseyes said:
same DS question.IS x negative?1) n^3(1-x^2)2) x^2-1 a) can be answered by 1 aloneb) can be answered by 2 alonec) can be answered by using bothd) can be answered by eithere) cannot be answered
option 4

Q: with all the nos from 1 to 100 how many rt triangles are possible?

what is the smallest pythagorean triplet (i.e a+b+c is minimum) s.t all of a,b,c are perfect squares?



ATDH.
@anytomdickandhary said:
what is the smallest pythagorean triplet (i.e a+b+c is minimum) s.t all of a,b,c are perfect squares?ATDH.
a^2+b^2=c^2
(k1)^4+(k2)^4=(k3)^4

no solution if k1,k2,k3 are intezers..

are u asking abt non-integrals also?
@viewpt said:
Q: with all the nos from 1 to 100 how many rt triangles are possible?
52 triangles???
@wovfactorAPS said:
a^2+b^2=c^2(k1)^4+(k2)^4=(k3)^4no solution if k1,k2,k3 are intezers..are u asking abt non-integrals also?
thats right it will have no integral solutions.

ATDH.
@anytomdickandhary said:
what is the smallest pythagorean triplet (i.e a+b+c is minimum) s.t all of a,b,c are perfect squares?ATDH.
Let a=(x^2-y^2)=k1^2
Let b=2xy=k2^2
Then c=(x^2+y^2)=k3^2
x^2=k1^2+y^2 Another triplet{k1,y,x} or {y,k1,x}---1
x^2+y^2=k3^2 Another triplet{x,y,k3} or {y,x,k3}----2
No triplet satisfy 1 and 2 so no triplet exist in which all are perfect square
But any one of three can exist as a perfect square