GMAT Question of the Day: Daily via email | Daily via Instagram New to GMAT Club? Watch this Video

 It is currently 28 Mar 2020, 07:59 ### 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.  # At the first meeting of the amputees' support group

Author Message
TAGS:

### Hide Tags

VP  P
Joined: 07 Dec 2014
Posts: 1239
At the first meeting of the amputees' support group  [#permalink]

### Show Tags 00:00

Difficulty:   65% (hard)

Question Stats: 56% (02:04) correct 44% (02:53) wrong based on 37 sessions

### HideShow timer Statistics

At the first meeting of the amputees’ support group, there were
93 fewer handshakes than there would have been if everyone
had at least one arm and shook hands once with everyone else.
The people with less than one arm made up 1/11 of the group.
How many of them were there?

A. 2
B. 3
C. 4
D. 5
E. 6

Source: me
Intern  Joined: 28 Jul 2016
Posts: 10
Re: At the first meeting of the amputees' support group  [#permalink]

### Show Tags

--> If we assume there are 2 people with less than one arm:
ie 2 people out of total 22.
So total handshakes by these two people= [2*20 + (2*1)/2]=[40+1]=41 >>> Not correct we need 93 handshakes

--->If we assume, there are 3 people:
ie 3 out of total 33.
So handshakes by these 3 persons= [3*30 + (3*2)/2]= [90+3]= 93 >>> Correct answer B
VP  P
Joined: 07 Dec 2014
Posts: 1239
Re: At the first meeting of the amputees' support group  [#permalink]

### Show Tags

algebraic approach:
let x=number of people with less than one arm
(11x)(11x-1)/2-(10x)(10x-1)/2=93
x=3
Intern  Joined: 22 Sep 2013
Posts: 4
At the first meeting of the amputees' support group  [#permalink]

### Show Tags

gracie wrote:
At the first meeting of the amputees’ support group, there were
93 fewer handshakes than there would have been if everyone
had at least one arm and shook hands once with everyone else.
The people with less than one arm made up 1/11 of the group.
How many of them were there?

A. 2
B. 3
C. 4
D. 5
E. 6

Source: me

*editted* didnt wanna leave bad info i posted to confuse anyone :/

Originally posted by SlippinJimmy on 16 Aug 2016, 10:49.
Last edited by SlippinJimmy on 16 Aug 2016, 19:17, edited 1 time in total.
Manager  Joined: 04 Jan 2014
Posts: 114
GMAT 1: 660 Q48 V32
GMAT 2: 630 Q48 V28
GMAT 3: 680 Q48 V35
Re: At the first meeting of the amputees' support group  [#permalink]

### Show Tags

thomasw713 wrote:
gracie wrote:
At the first meeting of the amputees’ support group, there were
93 fewer handshakes than there would have been if everyone
had at least one arm and shook hands once with everyone else.
The people with less than one arm made up 1/11 of the group.
How many of them were there?

A. 2
B. 3
C. 4
D. 5
E. 6

Source: me

perfect example to use the units digit to get the answer quick.

let's assume the answer is A, then (22*21)/2 - (20*19)/2 = 93?

we can actually ignore the second part (after the minus) of the equation because it will always end with a units digit of 0.
the first part will have a units digit of 1 --> (units digit of 2 * units digit of 1) / 2

for the rest

B) (33*32) / 2 --> (units digit 3 * units digit 2) / 2 = 3 (success)
C) (44*43) / 2 --> (units digit 4 * units digit 3) / 2 = 6 (fail)
D) (55*54) / 2 --> (units digit 5 * units digit 4) / 2 = 0 (fail)
E) (66*65) / 2 --> (units digit 6 * units digit 5) / 2 = 0 (fail)

Interesting approach, but I don't think the unit digit is correct for B, D and E. The even number will become half and that reduced number should be used to get the unit digit.

B) (33*32) / 2 --> (units digit 3 * units digit 6) = 8
D) (55*54) / 2 --> (units digit 5 * units digit 7) = 5
E) (66*65) / 2 --> (units digit 3 * units digit 5) = 5
Intern  Joined: 22 Sep 2013
Posts: 4
At the first meeting of the amputees' support group  [#permalink]

### Show Tags

Donnie84 wrote:
thomasw713 wrote:
gracie wrote:
At the first meeting of the amputees’ support group, there were
93 fewer handshakes than there would have been if everyone
had at least one arm and shook hands once with everyone else.
The people with less than one arm made up 1/11 of the group.
How many of them were there?

A. 2
B. 3
C. 4
D. 5
E. 6

Source: me

perfect example to use the units digit to get the answer quick.

let's assume the answer is A, then (22*21)/2 - (20*19)/2 = 93?

we can actually ignore the second part (after the minus) of the equation because it will always end with a units digit of 0.
the first part will have a units digit of 1 --> (units digit of 2 * units digit of 1) / 2

for the rest

B) (33*32) / 2 --> (units digit 3 * units digit 2) / 2 = 3 (success)
C) (44*43) / 2 --> (units digit 4 * units digit 3) / 2 = 6 (fail)
D) (55*54) / 2 --> (units digit 5 * units digit 4) / 2 = 0 (fail)
E) (66*65) / 2 --> (units digit 6 * units digit 5) / 2 = 0 (fail)

Interesting approach, but I don't think the unit digit is correct for B, D and E. The even number will become half and that reduced number should be used to get the unit digit.

B) (33*32) / 2 --> (units digit 3 * units digit 6) = 8
D) (55*54) / 2 --> (units digit 5 * units digit 7) = 5
E) (66*65) / 2 --> (units digit 3 * units digit 5) = 5

doh, you are absolutely right. wires in the brain must have gotten crossed :/ At the first meeting of the amputees' support group   [#permalink] 16 Aug 2016, 19:16
Display posts from previous: Sort by

# At the first meeting of the amputees' support group  