Official Quant thread for CAT 2013

MBA CAT 2024
25748
@viewpt said:
Q: 107^109^101 div 111 R?
107^109^101 mod 111
E(111)=111(36/37)(2/3)=72
109^101 mod 72
E(72)=72(1/2)(2/3)=24
109^5 mod 72

72=8*9
109^5 mod 9=1
109^5 mod 8=5^5 mod 8=5
9a+1=8b+5
b=4
72n+37

107^37 mod 111

111=37*3
107^37 mod 37=107 mod 37=-4
107^37 mod 3=2
37a-4=3b+2
107

25749
@scrabbler sorry, the question is this "In triangle ABC, the angle subtended by side BC at the orthocenter is 110 degrees, find the angle subtended by BC at the incenter."Can I have a diagram too, please."

I AM EXTREMELY sorry sir.
25750
@bodhi_vriksha said:
LCM (phi(37),phi(3))=36 =>107^36= 1(mod 111). Also, 109^101=1(mod36)Hence, required remainder = 1Team BV - Vineet
I guess ans is 107
25751
@veertamizhan said:
@scrabbler sorry, the question is this "In triangle ABC, the angle subtended by side BC at the orthocenter is 110 degrees, find the angle subtended by BC at the incenter."Can I have a diagram too, please."I AM EXTREMELY sorry sir.
OK then will be 125 only....just make the relevant changes in the diagram I had attached!

regards
scrabbler
25752
@viewpt said:
I guess ans is 107
Yes. It will be 107 only. See below.

LCM (phi(37),phi(3))=36 =>107^36= 1(mod 111). Also, 109^101=1(mod36)
Hence, required remainder = 107^1 =107

Team BV - Vineet
25753
Q: arjun is long jump player. Once he was 1 m away frm the wall and with each jump he covers half of the remaining didtsnce. find out the no. of jumps req. to reach the wall.
25754
@viewpt said:
Q: arjun is long jump player. Once he was 1 m away frm the wall and with each jump he covers half of the remaining didtsnce. find out the no. of jumps req. to reach the wall.
Infinite? It will be a GP with common ratio 1/2
.
Team BV - Vineet
25755
@viewpt said:
Q: 107^109^101 div 111 R?
107
25756
@viewpt said:
Q: arjun is long jump player. Once he was 1 m away frm the wall and with each jump he covers half of the remaining didtsnce. find out the no. of jumps req. to reach the wall.
Infinite ??
25757
@mohitjain said:
hw did u get 33 as rem wid 37?
107^109^101 mod 37
E(37)=36
so, 109^101mod 36 gives 1
107^109^101 =107^(36k+1)
107^36k * 107 mod 37

107 mod 37 gives -4 or -4 + 37 = 33
25758
@albiesriram said:
Take xsinx=m and the entire expression as k.

(9m^2+4)/m = k = >9m^2-mk+4=0. Now D>=0 => k^2-144>=0=> K min = 12 :)

Team BV - Vineet
25759
@viewpt said:
Q: arjun is long jump player. Once he was 1 m away frm the wall and with each jump he covers half of the remaining didtsnce. find out the no. of jumps req. to reach the wall.
GP with 1/2 common ratio.
So, infinite number of steps
25760
@veertamizhan said:
@scrabbler no I don't have OA(?)the answer is 125 saar.
according to the property of orthocentre angle subtended at the orthocentre can be given as angle(A)=180-angle(BoC) where o is the orthocentre
Now for incentre you have the property angle(incentre)= 90+1/2 angle(any of the vertice A or B or C)

So applying these you get angle A = 180-110=70
angle (incentre) = 90+1/2(70)=125
25761
25762
@iLoveTorres thanks man

25763

a reduction in price of petrol by 10% enables a motorist to buy 5 gallons more for 180$. find the original price of petrol.

25764
@vbhvgupta said:
a reduction in price of petrol by 10% enables a motorist to buy 5 gallons more for 180$. find the original price of petrol.
$4.5?
25765
@vbhvgupta said:
a reduction in price of petrol by 10% enables a motorist to buy 5 gallons more for 180$. find the original price of petrol.
xy=180
0.9x(y+5)=180
x(y+5)=200
5x=20
x=4
25766
@iLoveTorres said:
$4.5?
@Subhashdec2 said:

x=4


OA $4

25767
@viewpt said:
Q: arjun is long jump player. Once he was 1 m away frm the wall and with each jump he covers half of the remaining didtsnce. find out the no. of jumps req. to reach the wall.
this would be indeterminable..