Official Quant thread for CAT 2013

MBA CAT 2024
21166

@amresh_maverick said:
There are 3 states and 3 students representing each state. In how many ways can 5 students be chosen such that at least one student is chosen from each state ?PS: sabha yahi samapt hoti hai
108??

3(3c1 *3c2*3c2) + 3(3c1*3c3*3c1)
81+27
=108
21167
@mohitjain said:
In an isosceles right angled triangle ABC, it is given that AB = BC. DE and GF are drawn parallel to BC and AB respectively such that DE as well as GF divide the triangle ABC into two equal parts. BO when extended meets AC at the point H. Find the ratio BO : OH.
@iLoveTorres ..here.
21168
@ChirpiBird said:
@scrabblerPlease see what i am doing wrong.too confused now to proceed. ..
here length of the side of the square is 1-rt2 = -.414
check your approach from there.. may be now you get the right answer
21169
@mohitjain said:
In an isosceles right angled triangle ABC, it is given that AB = BC. DE and GF are drawn parallel to BC and AB respectively such that DE as well as GF divide the triangle ABC into two equal parts. BO when extended meets AC at the point H. Find the ratio BO : OH.
@scrabbler bro wouldnt the point O be the centroid of the triangle? if its the centroid of the traingle then the ratio should be 2:1 right? kindly take some pain in clarifying my doubt if its not the centroid.
21170

1) Find the remainder when 12^13^14 is divided by 145.
2)Find the remainder when 3737^(77)(777!) is divided by 73.

21171
@nole said:
1) Find the remainder when 12^13^14 is divided by 145.
I'm applying Chinese Rem. here.

145 = 5*29.

Now, 12^13^14 mod 5 = 2^13^14 mod 5.
2^4 = 16 mod 5 = 1. And, 13^14 mod 4 = 1^14 = 1.
So, 2^1 mod 5 = 2.

Now, 12^13^14 mod 29.
12^2 mod 29 = - 1 mod 29. And 13^14 mod 2 = 1.
So, 12^13^14 mod 29 = 12^1 mod 29 = 12.

Thus, 5p + 2 = 29q + 12
For p = 2 and q = 0, it satisfies.

So, Remainder = 12.
21172
@nole said:

2)Find the remainder when 3737^(77)(777!) is divided by 73.
What is this ? :roll:

Is it 3737^77^777! or 3737^ { 77*777! } or something else. :splat:
21173
@Estallar12
hi can you explain this part
2^4 = 16 mod 5 = 1. And, 13^14 mod 4 = 1^14 = 1
thanks
21174
@justanshul said:
@Estallar12hi can you explain this part2^4 = 16 mod 5 = 1. And, 13^14 mod 4 = 1^14 = 1thanks
2^4 = 16 which gives 1 Remainder on dividing with 5.
So, 2^4 = 1 mod 5.

Now, here power of 2 is 13^14 so we need to express the power in the form of 4k + a.
So, Finding remainder when 13^14 is divided by 4.

13^14 mod 4 = 1^14 mod 4 (Because 13 gives Rem. as 1 on division with 4.)
So, 13^14 = 1 mod 4.
21175
If 12 persons are seated in a row .The number of ways to select 3 persons so that no two of them are sitting next to each other is?..Plz post proper answer with illustration
85
100
120
240
21176
@raopradeep said:
If 12 persons are seated in a row .The number of ways to select 3 persons so that no two of them are sitting next to each other is?..Plz post proper answer with illustration85100120240
_ P1 _ P2 _ P3 _

g = gap between any two person

g1 + g2 + g3 + g4 = 9 (where g2 and g3 should be greater than 0)

g2 = g2' + 1 and g3 = g3' + 1

g1 + g2' + g3' + g4 = 7
=> number of integral solution = 10C3 = 120
21177
@amresh_maverick said:
If a , b, and c are the roots of 2x^3+4x^2 -3x-1=0 find tha value of (1-a)(1-b)(1-c)

if you know the concepts related to theory of equations then this problem cab be solved in one line

(1-a)(1-b)(1-c) = -1*(constant term of 2*(x-1)^3 -4(x-1)^2 -3x+1) = 2

Related Concepts explained below

Concept1[changing the signs of coefficients alternately will change the sign of roots]
if a,b,c are the roots of the equation then -a,-b,-c must be the roots of

2x^3 -4x^2 -3x+1=0

Concept2 (to shift all the roots by 'k' units replace x by x-k in the orginal equation)

=>(-a+1), (-b+1) and (-c+1) will be the roots of eqn 2(x-1)^3 - 4(x-1)^2 - 3(x-1) + 1 = 0

Concept3 (co-efficients of equations)

we know -1*(constant term) of cubic equation gives the product of the roots

constant term of 2(x-1)^3 - 4(x-1)^2 - 3(x-1) + 1

= -1*[ 2*(-1)^3 -4*1^2 +3*1 + 1] = 2

=>(1-a)(1-b)(1-c) = 2


ATDH.

21178
@Koushik98 said:
rest letters +1.....yea 1 kaha se aia???pls elaborate
Can you please post the original source of the question?
I don't remember it now :neutral:
21179
@iLoveTorres said:
@scrabbler bro wouldnt the point O be the centroid of the triangle? if its the centroid of the traingle then the ratio should be 2:1 right? kindly take some pain in clarifying my doubt if its not the centroid.
No, a centroid is where the medians intersect. Each median divides the triangle into 2 equal parts; doesn't mean that any line dividing the triangle into two equal parts is a median.

In this case of an isosceles triangle, the height (taking unequal side as base) will be the median, centroid will be at 1/3 of this height and hence for example if we draw a line parallel to the base through the centroid it will give us an area ratio of (2 : 3)^2 = 4 : 9 and not the required 1/2.

Draw the figure and think about it!

regards
scrabbler
21180
@Estallar12 said:
I'm applying Chinese Rem. here.145 = 5*29.Now, 12^13^14 mod 5 = 2^13^14 mod 5.2^4 = 16 mod 5 = 1. And, 13^14 mod 4 = 1^14 = 1.So, 2^1 mod 5 = 2.Now, 12^13^14 mod 29.12^2 mod 29 = 1 mod 29. And 13^14 mod 2 = 1.

12^2 mod 29 = -1 no? Fortunately it still works as 12^4 mod 29 = 1 and 13^14 mod 4 = 1 as well.
@Estallar12 said:
So, 12^13^14 mod 29 = 12^1 mod 29 = 12.Thus, 5p + 2 = 29q + 12For p = 2 and q = 0, it satisfies.So, Remainder = 12.

However, we needn't do Chinese Remainder....

12^2 is 144 so seedha 12^2 mod 145 is -1, so 12^4 mod 145 is 1 so 12^13^14 = 12^(4k+1) is equiv to 12.

regards
scrabbler
21181
@scrabbler said:
12^2 mod 29 = -1 no? Fortunately it still works as 12^4 mod 29 = 1 and 13^14 mod 4 = 1 as well.However, we needn't do Chinese Remainder....12^2 is 144 so seedha 12^2 mod 145 is -1, so 12^4 mod 145 is 1 so 12^13^14 = 12^(4k+1) is equiv to 12.regardsscrabbler
Yeah I'd used the direct method previously and posted but Chinese Remainder was what to be used while solving so used that.

12^4 mod 145 = 1 went very easier! :P
21182
@nole said:

2)Find the remainder when 3737^(77)(777!) is divided by 73.
3737^77^777! mod 73

E(73) = 72
E(72) = 24

777! mod 24 = 0
77^0 mod 72 = 1

3737^1 mod 73 = 14
21183
@scrabbler said:
No, a centroid is where the medians intersect. Each median divides the triangle into 2 equal parts; doesn't mean that any line dividing the triangle into two equal parts is a median. In this case of an isosceles triangle, the height (taking unequal side as base) will be the median, centroid will be at 1/3 of this height and hence for example if we draw a line parallel to the base through the centroid it will give us an area ratio of (2 : 3)^2 = 4 : 9 and not the required 1/2.Draw the figure and think about it!regardsscrabbler
now i got it.. so if it was given that the line through the vertices divides the triangle into 2 equal halves only then it is to be treated as median, right?
and thanks bro..now this concept is clear to me
21184
@iLoveTorres said:

now i got it.. so if it was given that the line through the vertices divides the triangle into 2 equal halves only then it is to be treated as median, right?

Absolutely! Medians are good fun.

Try this:

Consider a triangle ABC such that AB = 11, AC = 12 and BC = 13. D, E and F are points on AC such that BD is an altitude, BE an angle bisector and BF a median. Find the lengths DF and EF.

(Also try generalising this to any triangle with sides k-1, k and k+1 with the three lines drawn to the line of length k. Something curious I noticed on a long journey when I ran out of better things to do 😛 )

regards
scrabbler
21185

plz HELP...

Balu has र216 in two rupee denomination. He wanted to allocate them into a number of bags in such a way that he could give any amount (even number of rupes) between र2 and र216 by handing out one or more bags without opening them. What is the minimum number of bags he would require