Official Quant thread for CAT 2013

MBA CAT 2024
19026
@pyashraj said:
@The_LoserIs it 0?...Considering bth the numbers to be an integer..

O is not in option.
@Subhashdec2 said:
HCF=2^4*3^2*5*7LCM=2^10*3^5*5^2*7let the numbers beN1= 2^x1*3^x2*5^x3*7N2= 2^y1*3^y2*5^y3*7(x1,y1)=(4,10) or(10,4)=2similar for (x2,y2) (x3,y3) 2*2*2=8??
LCM also entails other prime numbers like 11 & 13.
Aren't we going to distribute those figure to the numbers?
Correct me if wrong.?
19027
@Estallar12 said:
This is the series for e^1.

e = 2.7182 and so smallest integer greater than S = 3.
bhai...approach ...
19028
@karan20 said:
bhai...approach ...
The expansion of
e^x = 1+x/1! +x^2/2! + x^3/3!+......
Put x=1 we get the series in the question
e^1 = 2.71
So S=3
19029

HCF = 7!

LCM = 13!
Number of Pairs.?
@Estallar12 plz help
19030
@Subhashdec2
32 ordered pairs, or 16 pairs existing :).
19031
@anantn said:
@Subhashdec232 ordered pairs, or 16 pairs existing .
for which one
19032
@The_Loser said:
HCF = 7!LCM = 13!Number of Pairs.?@Estallar12 plz help
HCF=2^4*3^2*5*7
LCM=2^10*3^5*5^2*7*11*13
let the numbers be
N1= 2^x1*3^x2*5^x3*7*11^x4*13^x5
N2= 2^y1*3^y2*5^y3*7*11^y4*13^y5
(x1,y1)=(4,10) or(10,4)=2
similar for (x2,y2) (x3,y3)
(x4,y4)=(0,1) or(1,0)
(x5,y5)=(0,1) or(1,0)
2*2*2*2*2=32

19033
@Subhashdec2 said:
HCF=2^4*3^2*5*7LCM=2^10*3^5*5^2*7*11*13let the numbers beN1= 2^x1*3^x2*5^x3*7*11^x4*13^x5N2= 2^y1*3^y2*5^y3*7*11^y4*13^y5(x1,y1)=(4,10) or(10,4)=2similar for (x2,y2) (x3,y3) (x4,y4)=(0,1) or(1,0)(x5,y5)=(0,1) or(1,0)2*2*2*2*2=32@The_Loser EDIT please check now

Shouldn't it be divided by 2 since 2*2*2*2*2 gives us repeated pairs....logically we should not require ordered pairs, even if the question doesn't specify anything.....so I feel 16 ho sakta hai.

regards
scrabbler
19034
@veertamizhan said:
In a right angle triangle, right angled at B, the angle subtended by AB at the in-center is 110*, If D is the midpoint of the hypotenuse, find the measure of angle BDC.
Is this 100 degrees? Trying to draw the figure....let me know OA please...

regards
scrabbler
19035
@veertamizhan said:
In a right angle triangle, right angled at B, the angle subtended by AB at the in-center is 110*, If D is the midpoint of the hypotenuse, find the measure of angle BDC.
Draw the figure , and use the property that exterior angle is equal to sum of remote interir angle.

I am getting the answer as 110'
19036

From a point inside an equilateral triangle of side 3cm, 3 perpendiculars are drawn to meet the opposite side. What is the sum of the length of these 3 perpendiculars?

19037
@catahead said:
From a point inside an equilateral triangle of side 3cm, 3 perpendiculars are drawn to meet the opposite side. What is the sum of the length of these 3 perpendiculars?
3rt3/2? If it has to be a unique answer, can take any point (including one of the vertices) and get the answer.

regards
scrabbler
19038
@scrabbler said:
3rt3/2? If it has to be a unique answer, can take any point (including one of the vertices) and get the answer.regardsscrabbler
Exactly, It has an unique answer and the approach is right
19039
@catahead
it forms 3 triangles, the sum of which add up to the area of the triangle.
so .5*3(p1+p2+p3)=root3/4*9
solve to get p1+p2+p3 as 3root3/2
19040
@Subhashdec2 said:
for which one
for the lcm one
19041
@Subhashdec2 said:
HCF=2^4*3^2*5*7LCM=2^10*3^5*5^2*7*11*13let the numbers beN1= 2^x1*3^x2*5^x3*7*11^x4*13^x5N2= 2^y1*3^y2*5^y3*7*11^y4*13^y5(x1,y1)=(4,10) or(10,4)=2similar for (x2,y2) (x3,y3) (x4,y4)=(0,1) or(1,0)(x5,y5)=(0,1) or(1,0)2*2*2*2*2=32
@scrabbler said:
Shouldn't it be divided by 2 since 2*2*2*2*2 gives us repeated pairs....logically we should not require ordered pairs, even if the question doesn't specify anything.....so I feel 16 ho sakta hai.regardsscrabbler
I guess Scrabbler is right. as options are : 5,12,16,20
so ordered might not be required. 16 ho sakta hai.
19042
Any one need trachtenberg speed Math ebook ,PM ur emails
19043
x is a number such that x^2 + 3x + 9 = 0. What is the value of x^3?
19044
@jain4444 said:
x is a number such that x^2 + 3x + 9 = 0. What is the value of x^3?
x^3 = (x^2)*(x) = (-3x-9)*(x) = (-3)*(x^2 + 3x) = (-3)*(-9) = 27 ?
19045
@YouMadFellow said:
x^3 = (x^2)*(x) = (-3x-9)*(x) = (-3)*(x^2 + 3x) = (-3)*(-9) = 27 ?
can you pls explain your method??

i would have done it like this..
x = -3+/-{sqrt(9-36)}/2 = -3+/-{sqrt(-25)}/2 = (-3+/-5i)/2 = (-3-5i)/2 or (-3+5i)/2
and then calculated x^3 by applying (a+/-b)^3 which is kind of arduous...your method seems less cumbersome...