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12662
@pavimai
@pavimai said:
There are 4 people Erick, Helena, Johny and David and they have 9 seats available to sit on. Now Johny and Helena don창€™t like each other and that창€™s why they always need to have one blank seat in between them and others can sit on any seat but no one can sit on the blank seat between Johny and Helena.First, the seats are arranged in a linear fashion and then the number of different linear arrangements possible are found to be M.Then the seats are arranged in circular fashion after which the number of different arrangements possible are found to be N. Then M + N is found to beOPTIONS1) 270 2) 450 3) 480 4) None of these pls explain the solution
270??? 210+60.. Linear ==> 7 ways u select for H_J and arrange them in 14 diff ways... 1. H_J 2. J_H... and reaming 6 postns 2 ppl ===> 6c2 =15 ways..so M= 14*15 =210

N => selectng a position for H_J in circular is 8/2 ways = 4 ways.. circular (N-1)/2 ways...nw selectng 2 out of 6 is 6c2 ways ==> 15*4=60 =n

M+N=270??
12663

Find the no.of factors of 1080 excluding the factors, which are perfect squares.
a) 28
b) 29
c) 30
d) 31

Do share the approach.

12664
@anilapex said:
Find the no.of factors of 1080 excluding the factors, which are perfect squares.a) 28b) 29c) 30 d) 31Do share the approach.

1080 = 2^3 *3^3 * 5
total factors = 4*4*2 = 32
subtract dose which are perfect squares that is 2^2 , 3^2, 2^2 * 3^2, 1
so 28 i guess??

12665
@The_Loser said:
natural no of solution a+b+c =
since 0 is not allowed here so the solution for

a+b+c = 12 is given by (12-1)C(3-1) = 11C2
a+b+c = 11 is given by (11-1)C(3-1) = 10C2
.
.
.
a+b+c = 3 is given by (3-1)C(3-1) = 2C2

So the total solutions = C(11,2) + C(10,2) + C(9,2)............C(2,2)

= 55 + 45 + 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 220

ATDH.
12666
@anilapex said:
Find the no.of factors of 1080 excluding the factors, which are perfect squares.a) 28b) 29c) 30 d) 31Do share the approach.
1080 = 135*8 = 5*27*8 = (2^3)*(3^3)*(5^1)

hence total number of factors = (3+1)*(3+1)*(1+1) = 32

for find the square factors we write 1080 as 4*9*(2*3*5) .

now square factors can come only by using 4 and 9 hence number of square factors = (1+1)*(1+1) = 4

hence required no. = 32 - 4 = 28.

ATDH.
12667
@anytomdickandhary said:
since 0 is not allowed here so the solution for a+b+c = 12 is given by (12-1)C(3-1) = 11C2a+b+c = 11 is given by (11-1)C(3-1) = 10C2...a+b+c = 3 is given by (3-1)C(3-1) = 2C2So the total solutions = C(11,2) + C(10,2) + C(9,2)............C(2,2)= 55 + 45 + 36 + 28 + 21 + 15 + 10 + 6 + 3 + 1 = 219ATDH

evath fyn, bt final summation = 220..
12668
@The_Loser said:
evath fyn, bt final summation = 220..
yes answer is 220 indeed.... I made an error in adding up the numbers in final step.


ATDH.
12669
@anytomdickandhary said:
yes answer is 220 indeed.... I made an error in adding up the numbers in final step.ATDH.
i forgt to consider natural number solution. was considering zero as well ( as in natural numbers)
12670
@veertamizhan said:
A builds a wall in 50 days. B demolishes in 60 days. Both work alternative days. The day when the wall will be first complete?
@hedonistajay said:
148 days ...
shouldnt it be 599 days..?

Assuming total work to be of 300 units.. A works at 6 units/ day B at 5U/day.. so in 2 days 1 unit.. so for 300 units 600 days.. but the wall will be complete one day before so 599..?

Edit : it should be 593 days.. See ATDH sir's reply below..
12671

6 indian & 8 americans sit across circular table sch tat no 2 indians sit together. No f ways?

12672
@The_Loser said:
6 indian & 8 americans sit across circular table sch tat no 2 indians sit together. No f ways?
7! * 8C6 * 6!
12673
@veertamizhan said:
A builds a wall in 50 days. B demolishes in 60 days. Both work alternative days. The day when the wall will be first complete?
say wall is 300 bricks ( LCM of 50 and 30)

A places (300/50) = 6 bricks per days
B demolishes (300/60) = 5 bricks per day

So in a block of 2 days net bricks laid = (6-5) = 1

after 294 bricks are laid..... next day A will put 6 bricks and wall will be complete..

so 2*294 + 1 = 589 days

ATDH.
12674

\

@The_Loser said:
6 indian & 8 americans sit across circular table sch tat no 2 indians sit together. No f ways?
7!*8c6*6!??
12675

total 9 ple= 4 gujraati & 5 marwari.

2 gujratis are vegie, & rest NV.
2 marwaris Vegie & rest NV.
All have to sit in row. sch tat no 2 marwadis sit together. further all Vegie need to sit togther. No f ways
12676
@ravi.theja said:
@pavimai270??? 210+60.. Linear ==> 7 ways u select for H_J and arrange them in 14 diff ways... 1. H_J 2. J_H... and reaming 6 postns 2 ppl ===> 6c2 =15 ways..so M= 14*15 =210N => selectng a position for H_J in circular is 8/2 ways = 4 ways.. circular (N-1)/2 ways...nw selectng 2 out of 6 is 6c2 ways ==> 15*4=60 =nM+N=270??
i dont know the oa..question is from test funda
12677

A and B start running simultaneously on a circular track from point O in the same direction .if the ratios of the speed is 6:1 respectively ,then how many times A is ahead of B by a quarter of the length of the track before they meet at O for the first time?

4
5
7
8
12678

the sum of the first 30 terms of an AP is4500.the ratio of the first 20 terms and the last ten terms is 4:5.Find the first term???????????

6.5
7.5
6
5
12679
@The_Loser said:
total 9 ple= 4 gujraati & 5 marwari.
2 gujratis are vegie, & rest NV.
2 marwaris Vegie & rest NV.
All have to sit in row. sch tat no 2 marwadis sit together. further all Vegie need to sit togther. No f ways
Only one combination possible
MGMGMGMGM
Select 4 seats for veggies=9C4
Combination for Veg Marvari=2!
Combination for Veg gujrati=2!
Combination for NonVeg Marvari=3!
Combination for Non veg Gujrati=2!
9c4*2!*2!*3!*2!=6048??
12680
@pavimai said:
the sum of the first 30 terms of an AP is4500.the ratio of the first 20 terms and the last ten terms is 4:5.Find the first term???????????6.57.565
15(2a+29d)=4500
2a+29d=300

10(2a+19d)/[15(2a+29d)-10(2a+19d)]=4/5

==> 3/2(2a+29d)/(2a+19d)=5/4+1=9/4

==> 3/2*300*4/9=2a+19d
==> 2a+19d=200

=> 10d=100 or d=10
=> 2a=300-290=10
=> a=5
12681
@pavimai said:
A and B start running simultaneously on a circular track from point O in the same direction .if the ratios of the speed is 6:1 respectively ,then how many times A is ahead of B by a quarter of the length of the track before they meet at O for the first time?4578
A and B will meet 5 times at L/6,2L/6.....5L/6... before meeting at O [starting point ]

except in the case of first meet i.e., at L/6 all other 4 instances A is ahead of B by a quarter of the length of the track Ans : 4??