Official Quant thread for CAT 2013

gnehagarg · December 1, 2012, 1:48pm

@vijay_chandola said:
Let N be a 4 digit number in base 5. A converts it into decimal and finds that it is divisible by 7. M on the other hand assumes it to be a decimal number and observes that it is also divisible by 7. How many such N exist?

Let abcd is four digit number in base 5

So a,b,c,d are less than 5

d+5c+25b+125a=7k

5(c+5b+25a)=7k-d

c+5b+25a=(7k-d)/5

c+5b+25a=25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120

d value is greater than 5 for 8 cases

Total cases are 20-8=12 cases

Numbers are(1001),(1104),(1200),(1203),(2201),(2304),(2400),(3003),(3401),(4004),(4100),(4203)

sid_dumbo · December 1, 2012, 1:51pm

@Cat.Aspirant123 5 deer, 10peacock.... sitter!

kejriwalsonal1 · December 1, 2012, 1:58pm

@ankittripathi said:
d

using all three together!!

vijay_chandola · December 1, 2012, 2:03pm

@gnehagarg said:
Let abcd is four digit number in base 5 So a,b,c,d are less than 5d+5c+25b+125a=7k5(c+5b+25a)=7k-dc+5b+25a=(7k-d)/5c+5b+25a=25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120d value is greater than 5 for 6 casesTotal cases are 20-6=4 cases

Please follow the previous posts. Answer is 8 cases.

htomar · December 1, 2012, 2:09pm

@Cat.Aspirant123 said:
In a mixed collection of peacocks and deer, if legs are counted they are 40 but if heads arecounted, they are just 15. How many peacocks are there?

10 peacock

soham2208 · December 1, 2012, 2:11pm

If A+B+C=180 and tanA+tanB+tanC=x,
find the value of (tanA)(tanB)(tanC)

Without approach, I will just ignore the post

. Solve it using equations rather than putting values :), no fun in laziness 😃

rachit_28 · December 1, 2012, 2:22pm

@soham2208 said:
If A+B+C=180 and tanA+tanB+tanC=x,find the value of (tanA)(tanB)(tanC)Without approach, I will just ignore the post . Solve it using equations rather than putting values , no fun in laziness

is it x sir ?

tan(A+B) = -tanC

tanA+tanB/1-tanAtanB = -tanC

so x = tanA*tanB*tanC

vijay_chandola · December 1, 2012, 2:28pm

@soham2208 said:
If A+B+C=180 and tanA+tanB+tanC=x,find the value of (tanA)(tanB)(tanC)Without approach, I will just ignore the post . Solve it using equations rather than putting values , no fun in laziness

tan(A+B+C)=(tan A+tan B+tan C-tan A*tan B*tan C)/xxx

now, A+B+C=180=> tan 180=0

=> tan A+tan B+tan C-tan A*tan B*tan C=0

or, tan A+tan B+tan C=tan A*tan B*tan C

"Without approach, I will just ignore the post" ko apna signature bana lo. kahan baar 2 type krte rahoge :P :mg:

htomar · December 1, 2012, 2:29pm

@soham2208 said:
If A+B+C=180 and tanA+tanB+tanC=x,find the value of (tanA)(tanB)(tanC)Without approach, I will just ignore the post . Solve it using equations rather than putting values , no fun in laziness

x

tan(a+b) = tan(a) + tan(b) / 1 - tan(a)tan(b)

tan(180 -c) = tan(a) + tan(b)/ 1 - tan(a)tan(b)

-tanc = tan(a) + tan(b)/ 1 - tan(a)tan(b)

tan(a)tan(b)tan(c) = tan(a) + tan(b) + tan(c) = x

bullseyes · December 1, 2012, 2:35pm

The sequence N(1), N(2), N(3), ......... is defined by N(1) = 211, N(2) = 375, N(3) = 420, N(4) = 523, and N(n) = N(n–1) – N(n–2) + N(n–3) – N(n–4) for all N ≥ 5. What will be the value of N(531) + N(753) + N(975)???

vijay_chandola · December 1, 2012, 2:36pm

@gnehagarg said:
Let abcd is four digit number in base 5 So a,b,c,d are less than 5d+5c+25b+125a=7k5(c+5b+25a)=7k-dc+5b+25a=(7k-d)/5c+5b+25a=25,30,35,40,45,50,55,60,65,70,75,80,85,90,95,100,105,110,115,120d value is greater than 5 for 8 casesTotal cases are 20-8=12 casesNumbers are(1001),(1104),(1200),(1203),(2201),(2304),(2400),(3003),(3401),(4004),(4100),(4203)

Some cases need to be omitted. 1200 mod 7=!0

for reference follow the chillfactor sir's post here

http://www.pagalguy.com/forums/quantitative-ability-and-di/official-quant-thread-c-t-88456/p-3611604/r-4131922

OA is 8

vijay_chandola · December 1, 2012, 2:45pm

@bullseyes said:
The sequence N(1), N(2), N(3), ......... is defined by N(1) = 211, N(2) = 375, N(3) = 420, N(4) = 523, and N(n) = N(n–1) – N(n–2) + N(n–3) – N(n–4) for all N ≥ 5. What will be the value of N(531) + N(753) + N(975)???

First five terms are positive and next five terms are negative and so on..

So, N(531)+N(753) + N(975)=N(1)+N(3) + N(5)=898

kejriwalsonal1 · December 1, 2012, 2:54pm

@sujamait said:
Find all triples (a; b; c) of natural numbers such that lcm(a; b; c) = a + b + c.

(1,2,3)

htomar · December 1, 2012, 2:58pm

@bullseyes said:
The sequence N(1), N(2), N(3), ......... is defined by N(1) = 211, N(2) = 375, N(3) = 420, N(4) = 523, and N(n) = N(n–1) – N(n–2) + N(n–3) – N(n–4) for all N ≥ 5. What will be the value of N(531) + N(753) + N(975)???

Every nth term will be (n-10)th term for n > 10

In other words.. nth term will be equal to rth term where r is remainder when n is divided by 10.

so N(531) + N(753) + N(975) = N(1) + N(3) + N(5) = N(1) + N(3) + N(4) - N(3) + N(2) - N(1)

= N(2) + N(4) = 898

scrabbler · December 1, 2012, 3:04pm

@kejriwalsonal1 said:
(1,2,3)

What about (2, 4, 6)? and (3, 6, 9)? and so on...

regards
scrabbler

kejriwalsonal1 · December 1, 2012, 3:49pm

@scrabbler said:
What about (2, 4, 6)? and (3, 6, 9)? and so on...regardsscrabbler

oh yes...!!

mailtoankit · December 1, 2012, 4:00pm

@bullseyes said:
The sequence N(1), N(2), N(3), ......... is defined by N(1) = 211, N(2) = 375, N(3) = 420, N(4) = 523, and N(n) = N(n–1) – N(n–2) + N(n–3) – N(n–4) for all N ≥ 5. What will be the value of N(531) + N(753) + N(975)???

N(1)+N(3)+N(5)=N(2)+N(4)=523+375=898

tina_angel · December 1, 2012, 4:11pm

2 ques on probability..

1. From a group of 5 men and 4 women a committee is to be formed so that in any of these committees there are more women than men. How many different committees can be formed?

2. 15 ppl sit round a circular table. What are the odds against two particular ppl sitting together?

scrabbler · December 1, 2012, 4:26pm

@Tina_angel said:
2 ques on probability..1. From a group of 5 men and 4 women a committee is to be formed so that in any of these committees there are more women than men. How many different committees can be formed?

This isn't a question on probability...but koi nahin try karte hue phir bhi :P

1w 0m - 4C1 x 5C0 = 4 x 1 = 4 ways
2w 0/1 m - 4C2 x (5c0+5C1) = 6 x 6 = 36 ways
3w 0/1/2 m - 4C3 x (5c0+5C1+5C2) = 4 x 16 = 64 ways
4w 0/1/2/3 m - 4C4 x (5c0+5C1+5C2 +5C3) = 1 x 26 = 26 ways
So total = 130 ways?

@Tina_angel said:
2. 15 ppl sit round a circular table. What are the odds against two particular ppl sitting together?

Let the ppl be A and B. Now when A sits anywhere, there are 14 other places, 2 next to him and 12 not. Hence 12 : 2 or 6 : 1 are the odds against.

regards
scrabbler

torque024 · December 1, 2012, 4:33pm

@Tina_angel said:
2 ques on probability..1. From a group of 5 men and 4 women a committee is to be formed so that in any of these committees there are more women than men. How many different committees can be formed?2. 15 ppl sit round a circular table. What are the odds against two particular ppl sitting together?

(1W,0M), (2W,0M), (3W,0M), (4W,0M) (2W1M), (3W1M), (4W,1M), (3W2M), (4W,2M), (4W,3M)
(4C1+4C2+4C3+4C4)+(5C1)(4C2+4C3+4c4)+5C2(4C3+4C4)+5C3(4C4)
(4+6+4+1)+(5*11)+(10*5)+(10*1)
=15+55+50+10
130?
-----

When 2 always sit together WAYS = 2(n-2)!.

odd against=((n-1)!-2(n-2)!)/2(n-2)!=(n-3)/2 = 12/2=6