Official Quant thread for CAT 2013

@Abir1103 said:
Q . thr r t taps 1 , 2 ...t each of which can fill a cistern . the rate of filling of nth tap is such tht it is equal to twice tht of all the taps frm 1 to (n-1) put together . if the 18th tap can fill the empty cistern in 2 mins , then find the time in which the 15th tap alone can fill the empty cistern .need soln plz .
if first fills in x mins
2nd will fill in x/2 mins
3rd will fill in x/6 mins
4th will fill in x/18 mins
.
.
.
gp with r= 1/3 and a=x/2
17th term will be=x/2 * 1/3^16
x=4*3^16
fourteenth term=ar^13
4*3^3/2=54 minutes?
@Abir1103 said:
Q . thr r t taps 1 , 2 ...t each of which can fill a cistern . the rate of filling of nth tap is such tht it is equal to twice tht of all the taps frm 1 to (n-1) put together . if the 18th tap can fill the empty cistern in 2 mins , then find the time in which the 15th tap alone can fill the empty cistern .
need soln plz .
54
@ScareCrow28 If we have to make 7 boys sit alternately with 7 girls around a round table which is numbered then the no of ways in which this can be done is
why not 6! for boys??
@Subhashdec2 said:
if first fills in x mins2nd will fill in x/2 mins3rd will fill in x/6 mins4th will fill in x/18 mins...gp with r= 1/3 and a=x/217th term will be=x/2 * 1/3^16 x=3^16fourteenth term=ar^133^3/2=13.5 minutes?
Mistake hai idhar.. It should be x*1/3^16
And Then the ans should be 14th term = ar^13 = x * 1/3^13 = 2*3^16 * 1/3^13 = 54
@ScareCrow28 said:
Mistake hai idhar.. It should be x*1/3^16And Then the ans should be 14th term = ar^13 = x * 1/3^13 = 2*3^16 * 1/3^13 = 54
kar diya bhai edit
@ani6 said:
@ScareCrow28If we have to make 7 boys sit alternately with 7 girls around a round table which is numbered then the no of ways in which this can be done is why not 6! for boys??
Table is numbered. So seats are already distinct. Normally we divide n!/n because the seats are assumed to be effectively identical....

regards
scrabbler

@ani6 said:
@ScareCrow28If we have to make 7 boys sit alternately with 7 girls around a round table which is numbered then the no of ways in which this can be done is why not 6! for boys??
Because the seats are numbered.. So There are 7 "diff" positions for boys.. there is no symmetry left
Q: How many nos. (n) are there b/w 1 and 200 such that n/2, n/3 and (2n+1)/5 are all composite natural nos.?
@Abir1103 said:
Q . thr r t taps 1 , 2 ...t each of which can fill a cistern . the rate of filling of nth tap is such tht it is equal to twice tht of all the taps frm 1 to (n-1) put together . if the 18th tap can fill the empty cistern in 2 mins , then find the time in which the 15th tap alone can fill the empty cistern .need soln plz .
let say 1st tab = 1 ltr per minute
2nd = 2
3rd = 2*3
4th = 2*3*3
.
15th = 2*3^13
.
18th = 2*3^16

required answer = 2^2*3^16/2*3^13 = 54 minutes
@ani6 said:
If we have to make 7 boys sit alternately with 7 girls around a round table which is numbered then the no of ways in which this can be done is
7!x7!x2
@viewpt said:
Q: How many nos. (n) are there b/w 1 and 200 such that n/2, n/3 and (2n+1)/5 are all composite natural nos.?
no has to be of the form 6k
12
12k+1/5
12k should end in 4 or 9
9 is not possible
k should end in 2,7
200/6=33
k can be 2,7,12,17,22,27,32
12*2+1/5=5 not composite
12*7+1/5=17 not composite
12*12+1/5=29 not composit
e
12*17+1/5=41 not composite
12*22+1/5=53 not composite
12*27+1/5=65
12*32+1/5=77
2??


@viewpt EDIT
@jain4444 said:
correct hai bhai

approach batao

P.S - this is not a original question...ye self made hai
a^2+6b^2=x^2
6b^2=x^2-a^2
6b^2= (x+a)(x+a)
b^2= x+a
6= x-a
x = b^2+6/2
as x is an integer b has to be a multiple of 2
so all numbers from 0 to 998 will satisfy for b
but as a is a square number so is b then
if 1,2 satisfy then -1,-2 -1,2 1,-2 all will satisfy
but we need only a
so we have to take only 2 out of the above
so 2*449=998
but for 0 only -3,0 satisfies so
998 + 1 = 999
1) the area of the circle circumscribing three circles of unit radius touchnig each other is ?

2)find the ratio of diamter of the circles inscribed in and circumscribing an equilateral triangle to its height. ?
@Abir1103 said:
Q . thr r t taps 1 , 2 ...t each of which can fill a cistern . the rate of filling of nth tap is such tht it is equal to twice tht of all the taps frm 1 to (n-1) put together . if the 18th tap can fill the empty cistern in 2 mins , then find the time in which the 15th tap alone can fill the empty cistern .
need soln plz .
let the rate of first pipe be x
of second pipe 2x
of 3rd = 6x
of 18th pipe = 6*3^15*x
now volume of container = 2*6*3^15x
this volume = 6*3^12x*v
so equating the two v= 54 min
@shinoda said:
1) the area of the circle circumscribing three circles of unit radius touchnig each other is ?2)find the ratio of diamter of the circles inscribed in and circumscribing an equilateral triangle to its height. ?
1)pi/3 (4+2sqrt(3))??
@shinoda said:
1) the area of the circle circumscribing three circles of unit radius touchnig each other is ?2)find the ratio of diamter of the circles inscribed in and circumscribing an equilateral triangle to its height. ?
1)pi/3 (4+2sqrt(3))??
3) Circles are drawn with four vertices as the center and radius equal to the side of square. Ifthe square is formed by joining the mid points of another square of side 2rt6, find the area common to all the four circles ?
@shinoda said:
1) the area of the circle circumscribing three circles of unit radius touchnig each other is ?2)find the ratio of diamter of the circles inscribed in and circumscribing an equilateral triangle to its height. ?
1. pi/3*(2 + rt3)^2
2. 2 : 1 ?
@Subhashdec2 said:
1)pi/3 (4+2sqrt(3))??
nop..
@mailtoankit said:
1. pi/3*(2 + rt3)^22. 2 : 1 ?
1st is correct in 2nd we have to find the ratio of diamter of two circles and the height of the triangle.

plz share ur approach for the first one ...