Last visit was: 18 Jul 2024, 10:12 It is currently 18 Jul 2024, 10:12
Toolkit
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

# How many guests were present at the party?

SORT BY:
Tags:
Show Tags
Hide Tags
Math Expert
Joined: 02 Sep 2009
Posts: 94404
Own Kudos [?]: 641999 [4]
Given Kudos: 85997
Director
Joined: 25 Oct 2015
Posts: 516
Own Kudos [?]: 896 [4]
Given Kudos: 74
Location: India
GMAT 1: 650 Q48 V31
GMAT 2: 720 Q49 V38 (Online)
GPA: 4
Director
Joined: 30 Sep 2017
Posts: 943
Own Kudos [?]: 1284 [2]
Given Kudos: 402
GMAT 1: 720 Q49 V40
GPA: 3.8
GMAT Club Legend
Joined: 08 Jul 2010
Status:GMAT/GRE Tutor l Admission Consultant l On-Demand Course creator
Posts: 6020
Own Kudos [?]: 13803 [1]
Given Kudos: 125
Location: India
GMAT: QUANT+DI EXPERT
Schools: IIM (A) ISB '24
GMAT 1: 750 Q51 V41
WE:Education (Education)
Re: How many guests were present at the party? [#permalink]
1
Kudos
Quote:
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

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

Senior Manager
Joined: 05 Aug 2019
Posts: 400
Own Kudos [?]: 208 [1]
Given Kudos: 132
Location: Spain
Concentration: Strategy, Technology
GMAT 1: 650 Q50 V28 (Online)
GMAT 2: 700 Q49 V37
GPA: 3.23
WE:General Management (Real Estate)
Re: How many guests were present at the party? [#permalink]
1
Kudos
(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
SVP
Joined: 24 Nov 2016
Posts: 1712
Own Kudos [?]: 1360 [0]
Given Kudos: 607
Location: United States
Re: How many guests were present at the party? [#permalink]
Quote:
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) 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)
Intern
Joined: 04 Dec 2019
Posts: 2
Own Kudos [?]: 1 [1]
Given Kudos: 6
Re: How many guests were present at the party? [#permalink]
1
Kudos
(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
GMAT Club Legend
Joined: 18 Aug 2017
Posts: 7998
Own Kudos [?]: 4236 [0]
Given Kudos: 243
Location: India
Concentration: Sustainability, Marketing
GMAT Focus 1:
545 Q79 V79 DI73
GPA: 4
WE:Marketing (Energy and Utilities)
Re: How many guests were present at the party? [#permalink]
#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
VP
Joined: 28 Jul 2016
Posts: 1199
Own Kudos [?]: 1751 [1]
Given Kudos: 67
Location: India
Concentration: Finance, Human Resources
Schools: ISB '18 (D)
GPA: 3.97
WE:Project Management (Investment Banking)
Re: How many guests were present at the party? [#permalink]
1
Kudos
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
CEO
Joined: 07 Mar 2019
Posts: 2620
Own Kudos [?]: 1868 [1]
Given Kudos: 763
Location: India
WE:Sales (Energy and Utilities)
Re: How many guests were present at the party? [#permalink]
1
Kudos
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.

Senior Manager
Joined: 14 Jul 2019
Status:Student
Posts: 475
Own Kudos [?]: 372 [2]
Given Kudos: 52
Location: United States
Concentration: Accounting, Finance
GMAT 1: 650 Q45 V35
GPA: 3.9
WE:Education (Accounting)
Re: How many guests were present at the party? [#permalink]
2
Kudos
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.
Non-Human User
Joined: 09 Sep 2013
Posts: 34014
Own Kudos [?]: 852 [0]
Given Kudos: 0
Re: How many guests were present at the party? [#permalink]
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
Re: How many guests were present at the party? [#permalink]
Moderator:
Math Expert
94402 posts