Latest Elitmus Aptitude Question SOLUTION: New public school have a circular layout.the school has teachers specializing in various subjects. All classroom of the school are equally spaced apart and located along its perime

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(#M40010791) ELITMUS Numerical Ability question LR Keep an EYE

New public school have a circular layout.the school has teachers specializing in various subjects.
All classroom of the school are equally spaced apart and located along its perimeter.Each Teacher needs
four classes in day.there is strange rule.the first and last class has to be in the same class room.the
other two classes have to be at two other distinct class rooms.



Answer the following

a:) Bharti is a history teacher.in addition to above rule of the school she teaches exactly one pair of
successive classes in adjacent classrooms.how many distinct trips to classes rooms are possible for Bharti
if there are 12 classroom in school.


1;) 120
2:) 96
3:) 576
4:) 496

b:) Ram is a Math Teacher.he never teaches two successive classes in adjacent classrooms.how many distinct trip
to classroom are possible for ram if there are 9 classroom in the school.

1:) 72
2:) 324
3:) 30
4:) 180

Asked In Elitmus sandeep pandey (11 years ago)

Unsolved Read Solution (9)

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Read Solution (9) :

vineet yadav  (11 years ago)

a) possibilities for 1st and final position=12
if 1st and 2nd classes are adjacent then possibilities for 1st class=12
2nd class=2
3rd class=8
12*2*8=192
if 2nd and 3rd classes are adjacent then possibilities for 1st class=12
2nd class=2 then 3rd class=1 when they are next to next for 1st class
2nd class=7 then 3rd class=2
total=12*2*1+12*7*2=192
if 3rd and last classes are adjacent then for 1st class=12
2nd class=8
3rd class=2
total=12*8*2=192
total cmbinations possible=192+192=192=576
3 is correct......

Is this solution Helpfull? Yes (24) | No (12)  

saroj singh  (8 years ago)

Case 1 : Only first class and second class are held in adjacent class rooms

First class and fourth class can be in any of the 12 classrooms (12 ways)
Second class can be in any of the two adjacent class rooms of first class (2 ways)
Third class room can be any of the 8 class rooms which are not adjacent to second class room or fourth class room (8 ways)

Total distinct trips possible in this case are 12×2×8 = 192

Case 2: Only second class and third class are in adjacent class rooms

Second class can be in any of the 12 classrooms (12 ways)
Third class can be in any of the the two adjacent class rooms (2 ways)
First class and fourth class can be in any of the the 8 class rooms (which are not adjacent to second and third classes)

Total distinct trips possible in this case are 12×2×8= 192

Case 3: Only third class and fourth class are in adjacent class rooms

Third class can be in any of the 12 classrooms (12 ways)
Fourth and first class can be in any of the the two adjacent class rooms (2 ways)
Second class can be in any of the the 8 class rooms (which are not adjacent to first and third classes)

Total distinct trips possible in this case are 12×2×8= 192

Required number of distinct trips to classes rooms = 192+192+192 = 576

Is this solution Helpfull? Yes (11) | No (1)  

vignesh  (11 years ago)

for one adjacent combination there are 10 possibilities.there are 10 such combinations possible.so therefore 120 combinations are possibles.

Is this solution Helpfull? Yes (8) | No (23)  

Ayush  (10 years ago)

b) 1st class =9 possibilities
2nd class=4 possibilities ( dont take 6 as it will arrive you with a controvertial case where the 2nd adjacent class to the first is selected )
Now 3rd class =3 possibilities (total =9*4*3)

Now, taking left out case
1st class=9
2nd class =2 (left out classes in first case)
3rd class=4
total =9*2*4
adding , we get=180

Is this solution Helpfull? Yes (5) | No (6)  

vineet yadav  (11 years ago)

b) possibilies of 1st and 4th class=9
possibilities of 2nd class=2 then 3rd=4 when 2nd class is next to next for 1st class
possibilities of 2nd class=4 then 3rd=3 when 2nd class not follows above case
total=9*2*4+9*4*3=180
4 is correct

Is this solution Helpfull? Yes (4) | No (14)  

Karan KApur  (10 years ago)

(a)now consider a circular ring and have 12 bixes equally spaced along its circumference now first class can be any where on 12 boxes.the second class has to be the next one and the now the major point the 3rd class has 9 options as it cannot be the first class neither the class next to and previous to first class and it cannot be next to the second class room so 4 classes in total wiped out
last class will be the first only hence 12*1*8*1=96.

Is this solution Helpfull? Yes (4) | No (5)  

lokesh  (9 years ago)

B)
consider the nine classes
1
9 2

8 3


7 4

6 5


lets take an instance
first class is 1 (out of 9)(9 ways of selecting any one)
second class cannot be 2 & 9
so,
if second class is 8 (out of 2 & 8) (2 ways )
then third class can be 6 ,5 ,4 or 3 (4 ways of selecting third class)
and fourth class is same as the first.

which gives = 9*2*4=72

or if,
second class is 7 ( out of 7,6,5,4) (4 ways of selecting second class)
then third class can be 3,5 or 5( 3 ways of selecting 3rd class)

which gives = 9*4*3 = 108

distinct trips = 108 + 72 = 180

Is this solution Helpfull? Yes (4) | No (1)  

Swaraj Behera  (11 years ago)

there are 6 differnet class room which can be arrange in 6*6=36;
9 different 1st class n last class can be taken then..
the total distinct trip=36*9=324

Is this solution Helpfull? Yes (2) | No (10)  

suhaib  (10 years ago)

120 is the answer
http://www.pagalguy.com/forums/quantitative/official-quant-thread-cat-2013-t-88456/p-16907230/r-16908009

Is this solution Helpfull? Yes (2) | No (4)  

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