some problems..

.1.ABCD is a quadrilateral such that the bisectors of the four interior angles meet at a point O is Ð (AOB)= 57degree , calculate angle Ð(COD) [in degrees]

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.2. In the given figure , AD and BC are two chords extended to meet at E. if Ð (BAD) = 60 degree and CE = DE, which of the following is true

  • a.BE > 2AD
  • b.BE = 2AD
  • c.BE < 2AD
  • d.None
  • --------------------------------
  • 3. How many ordered pairs of integers (a, b), are there such that their product is a positive integer less than 1000?
  • --------------------------------------
  • 4.Chris and his wife invited a total of 10 families on their marriage anniversary. While the host family had just the two members, each family invited consisted of four members. If every person in the party shook hands with every other person belonging to a different family exactly once, then find the number of handshakes that took in the party.

soln, 4

total no. of persons in the party = 10X4 + 2

                                                         = 42

let everyone is shaking hand to other including the member of his own family

then the total no. of handshakes = 41+40+39+38+ ----+3+2+1 = 861

no. of handshakes between family members = 10X5 +1 ( 1 for the host family)

                                                                                  = 51

therefore total n of handsakes = 861- 51 = 810

soln. 1

Let angle A = 2a, angle B = 2b, angle C = 2c and angle d = 2d

we know that 2a +2b+2c+2d = 360

                           a+b+c+d = 180

                          c+ b = 180 - (a+d)                                   

for the triangle AOB   a+d+57= 180   i.e. a+d = 123

for the triangle COD   c+b+x  =180  (let angle COD = x )

                                    180 - ( a+d) +x= 180

                                    180 - 123 +x = 180

 therefore     x = 123

                    

hi madam, i think it must b

hi madam,

i think it must b 861-61=800

cos in 10 families ,the no.of handshakes between them will b

10*(3+2+1)=60 and

the couple willl account for another 1 ..

so its 61 not 51.

If I'm right can u reward me with

 

Its 800 in deed.First lets

Its 800 in deed.

First lets consider only the 10 families then total number of handshakes
= 4 x 36 + 4 x 32 + . . . . . . . . + 4
= 16 x [ 9 + 8 + . . . . . . . . . . + 1]
= 16 x 5 x 9 = 720
Now host’ll make 40 + 40 = 80 handshakes
Total number of hand shakes = 800

 

SOLUTION 4

hi ......the solution is 800 and not 810...
it can be easily with help of permut and combination

Hey Nilay Could you please

Hey Nilay Could you please explain how to solve it using combination principles

Dear all,

Sol. 2 :  AD x DE = BC x CE

             but DE = CE ; AD = BC or AE = BE

             Tringle ABE is  Isosceles ; Ang BAE = Ang ABE = 600 ; Ang AEB = 600

             Triangle ABE is Equilateral; AB = BE = AE or BE = ?

any one try further

Dear all,

Sol. 2 :  AD x DE = BC x CE

             but DE = CE ; AD = BC or AE = BE

             Tringle ABE is  Isosceles ; Ang BAE = Ang ABE = 600 ; Ang AEB = 600

             Triangle ABE is Equilateral; AB = BE = AE or BE = ?

any one try further

Thanks

Thanks for making out my mistake, yes your ans is correct, but I am going o reward you with that HA..HA..HA..

P&C approach...

Hi guys,
I am new to CAT4MBA...had been participating in forums of other web sites...thought of trying this site too... :)

total members in the party are 40 guests + 2 hosts....
if there is no restriction total handshakes among guests wld be...

40C2 = 780

every guest handshaking with every host = 80

so total 860..

now each family members among them do not handshake...

so 4C2 x 10 = 60 handshakes are extra...

so total handshakes = 860 - 60 = 800

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