How many guests were present at the party?

Question

Competition Mode Question

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once
(2) If there were half as many guests present, they could sit in a row in 120 different ways

ArunSharma12 · Answer

How many guests were present at the party?

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once
(2) If there were half as many guests present, they could sit in a row in 120 different ways

statement 1:
 n_C_2=45 
 n(n-1)=90 
n =10; sufficient

statement 2:
x! = 120; x = 5
 \frac{n}{2}=5 ; n =10
sufficient
Ans: D

freedom128 · Answer

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once 
 If there were 2 guests, there would be 1 handshake.
 If there were 3 guests, there would be (2+1)= 3 handshakes. 
 If there were 4 guests, there would be (3+2+1)= 6 handshakes.
.....
  If there are  10 guests , there will be (10+9+...+2+1)= 45 handshakes. 
SUFFICIENT

(2) If there were half as many guests present, they could sit in a row in 120 different ways 
 If there were 2 guests, the guests could sit in a row in 2!=2 ways.
 If there were 3 guests, the guests could sit in a row in 3!=6 ways.
.....
 If there are 5 guests, the guests could sit in a row in 5!=120 ways.
 THUS,  if 5 is half of the number of guests, then  the number of guests is 10.  
SUFFICIENT

FINAL ANSWER IS (D)

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GMATinsight · Answer

Question : Number of guests in Party = ?

STatement (1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once

Total Number of handshakes with n people can be written in three ways

Method 1 : if total people = n then First person will have n-1 people to shake hands with i.e. his handshakes = n-1
Second person has shook hand with first so second will shake hands with n-2 people and similarly the second last person will have 1 person to shake hands with
i.e. Total handshakes = 1+2+3+4+....+(n-1) = 45
i.e. n = 10

Method 2 : Everyone of n people has (n-1) other people to shake hands with so total handshakes = n*(n-1)
but every handshake has been counted twice e.g. first shakes hand with third when his handshakes are counted and third person shakes hand with first when his handshakes are counted
hence total unique handshakes = n*(n-1)/2 = 45 i.e. n = 10

Method 3 : We need two persons to cause one handshake
i.e. Total handshakes = total ways of choosing 2 persons out of n  = nC_2 = 45 
i.e. n = 10

SUFFICIENT

Statement (2) If there were half as many guests present, they could sit in a row in 120 different ways 
If total guests = n
then half the guests = n/2
total ways of arranging n/2 people in a row = (n/2)! = 120 (given)
but 5! = 120
i.e. n/2 = 5 i.e. n = 10
SUFFICIENT

Answer: Option D

pabpinor · Answer

(1) Tells us that there is a certain number of handshakes and that all the guests handshaked with everyone, so we would need to reverse use a combination of xC2=45. It is sufficient.

(2) also gives us another way of calculating the solution by telling us that (2x)!=120. It is also sufficient.

As both are sufficient, the right answer is D.

Regards,
Pablo

exc4libur · Answer

(1) sufic
x!/(x-2)!2!=45, x!/(x-2)!=90
x(x-1)(x-2)!/(x-2)!=90
x(x-1)=90, product of consecutive,
10(9)=90, x=10

(2) sufic
y=x/2
y!=120
3!=6, 4!=24, 5!=120
y=5, x=10

Ans (D)

preparefore · Answer

(1) This is an arithmetic progression and we can use the formula ((a1+an)n)/2. ((1+x)x)/2=45 . It is obvious that we have just one unknown value. So , we can solve it. (Sufficient)
(2) This is a simple factorial (n!=120 and x=2n)and we don't need to solve it. (Sufficient)
Hence: D

Posted from my mobile device

Archit3110 · Answer

#1

There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once

nc2=45
or say
n*(n-1)*(n-2)/2 * ( n-2) = 45
n^2-n-90=0
n=10,-9
n= 10
sufficient
#2
If there were half as many guests present, they could sit in a row in 120 different ways
n! = 120
which is when n=5
so total guest = 10
sufficeint
IMO D

How many guests were present at the party?

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once
(2) If there were half as many guests present, they could sit in a row in 120 different ways

globaldesi · Answer

How many guests were present at the party?

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once.

thus
 \frac{n(n-1)}{2 }= 45 
or n(n-1) =90
thus n =10
sufficient

(2) If there were half as many guests present, they could sit in a row in 120 different ways
(n/2)! = 120
hence n/2 = 5
or n = 10

sufficient

Each is sufficient
D

unraveled · Answer

How many guests were present at the party?

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once
Let n be the number of guests. 
Total handshakes =  \frac{n * (n-1)}{2}  = 45 
 n^2 - n - 90 = 0 
n = 10 
SUFFICIENT.

(2) If there were half as many guests present, they could sit in a row in 120 different ways
Let half the number of guests = m 
Ways to sit in a row = m! = 120
m = 5
2m = 10
SUFFICIENT.

Answer D.

minustark · Answer

How many guests were present at the party?

(1) There were 45 handshakes if the guests shook hands with each other and every 2 guests shook hands exactly once
(2) If there were half as many guests present, they could sit in a row in 120 different ways.

1) Total number of guests = n. nC2 = 45. or, n!/(n-2)!2! =45 or, n(n-1)/2=45 So n is 10. Sufficient
2) (n/2)! = 120. Since 120 =5!, so there was 10 guests. Sufficient.
D is the answer.

bumpbot · Answer

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How many guests were present at the party?

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