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At the first meeting of the amputees' support group

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At the first meeting of the amputees' support group  [#permalink]

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New post 28 Jul 2016, 15:00
00:00
A
B
C
D
E

Difficulty:

  65% (hard)

Question Stats:

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

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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
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Re: At the first meeting of the amputees' support group  [#permalink]

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New post 28 Jul 2016, 23:46
--> 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
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Re: At the first meeting of the amputees' support group  [#permalink]

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New post 29 Jul 2016, 06:13
algebraic approach:
let x=number of people with less than one arm
(11x)(11x-1)/2-(10x)(10x-1)/2=93
x=3
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At the first meeting of the amputees' support group  [#permalink]

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New post Updated on: 16 Aug 2016, 19:17
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.
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Re: At the first meeting of the amputees' support group  [#permalink]

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New post 16 Aug 2016, 19:03
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
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At the first meeting of the amputees' support group  [#permalink]

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New post 16 Aug 2016, 19:16
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 :/
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At the first meeting of the amputees' support group   [#permalink] 16 Aug 2016, 19:16
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