a greenhorn statistician

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Fri Mar 05, 2010 3:56 am

The total number of goals posted by three strikers, Amy, Bobby, and Cameron were to be presented as a sum of pairs by a greenhorn statistician. The statistician instead fed a bad command that added the product of goals by a pair along with the correct entry. This resulted in a report that read Amy and Bobby together posted 69 goals, Bobby and Cameron together posted 49 goals, and Cameron and Amy together posted 34 goals. How many goals did the trio really posted as a total?
(A) 76
(B) 57
(C) 38
(D) 22
(E) 19

Your pointer on the wordings is for ever and a day, implored.

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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Source: Beat The GMAT — Problem Solving | Fri Mar 05, 2010 3:56 am

rohan_vus: Master | Next Rank: 500 Posts; Posts: 189; Joined: Thu Apr 03, 2008 2:03 pm; Location: USA; Thanked: 21 times

by rohan_vus » Fri Mar 05, 2010 4:45 am

From wordings above , this is what i understand as my eqns show below

Let no of scores by Amy be a , Bobby be b and Cameron be c
ab+a+b = 69 --(1)
bc +b+c = 49 --(2)
ac+a+c = 34 ---(3)-->This implies a and c have to be even
If a and c are even then b must be odd either from (1) or (2)

So total goals a+b+c must be odd.
Choices B and E are only possibilities

Start with lower value ,i.e , 19
a could be 6 and c could be 4.. this gives b as 9

using , these values all above eqns are satisfied , so E should be the answer

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harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Fri Mar 05, 2010 4:53 am

sanju09 wrote:The total number of goals posted by three strikers, Amy, Bobby, and Cameron were to be presented as a sum of pairs by a greenhorn statistician. The statistician instead fed a bad command that added the product of goals by a pair along with the correct entry. This resulted in a report that read Amy and Bobby together posted 69 goals, Bobby and Cameron together posted 49 goals, and Cameron and Amy together posted 34 goals. How many goals did the trio really posted as a total?
(A) 76
(B) 57
(C) 38
(D) 22
(E) 19

Your pointer on the wordings is for ever and a day, implored.

My guess is that the term "greenhorn statistician" is only given to confuse the people(just a mental blockage)

It is apparent that B>A>C

Let the extra addition be x.
So, A+B+x= 69 -(1)
B+C+x =49 -(2)
Subtracting (2) from (1) we get
A - C =20 ( A > 20) -(4)
Also, C+A+x = 34 -(3)
Adding 3 and 4,we get
2A + x = 54 or x = 54 -2A ( A is greater than 20 so x is smaller than 14)
Adding 1,2,3
we get 2(A + B + C) +3x = 152
or A+B+C = 76 - 3x/2 [option 1 is ruled out]

Since x is smaller than 14 so 3x/2 is smaller than 21
So 76 - 3x/2 is greater than 55.
[spoiler]
Hence,by elimination method answer is 57 (B)[/spoiler]

It takes time and effort to explain, so if my comment helped you please press Thanks button

Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Fri Mar 05, 2010 4:57 am

harsh.champ wrote:
sanju09 wrote:The total number of goals posted by three strikers, Amy, Bobby, and Cameron were to be presented as a sum of pairs by a greenhorn statistician. The statistician instead fed a bad command that added the product of goals by a pair along with the correct entry. This resulted in a report that read Amy and Bobby together posted 69 goals, Bobby and Cameron together posted 49 goals, and Cameron and Amy together posted 34 goals. How many goals did the trio really posted as a total?
(A) 76
(B) 57
(C) 38
(D) 22
(E) 19

Your pointer on the wordings is for ever and a day, implored.
My guess is that the term "greenhorn statistician" is only given to confuse the people(just a mental blockage)

It is apparent that B>A>C

Let the extra addition be x.
So, A+B+x= 69 -(1)
B+C+x =49 -(2)
Subtracting (2) from (1) we get
A - C =20 ( A > 20) -(4)
Also, C+A+x = 34 -(3)
Adding 3 and 4,we get
2A + x = 54 or x = 54 -2A ( A is greater than 20 so x is smaller than 14)
Adding 1,2,3
we get 2(A + B + C) +3x = 152
or A+B+C = 76 - 3x/2 [option 1 is ruled out]

Since x is smaller than 14 so 3x/2 is smaller than 21
So 76 - 3x/2 is greater than 55.
[spoiler]
Hence,by elimination method answer is 57 (B)[/spoiler]

Why did you add the same x with all three pairs?

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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harsh.champ: Legendary Member; Posts: 1132; Joined: Mon Jul 20, 2009 3:38 amFollowed by: 6 members; Location: India; Thanked: 64 times; GMAT Score:760

by harsh.champ » Fri Mar 05, 2010 5:11 am

Didn't get the wordings.
You can use x ,y,z or ab,bc,cd instead.

It takes time and effort to explain, so if my comment helped you please press Thanks button

Just because something is hard doesn't mean you shouldn't try,it means you should just try harder.

"Keep Walking" - Johnny Walker

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ajith: Legendary Member; Posts: 1275; Joined: Thu Sep 21, 2006 11:13 pmFollowed by: 2 members; Location: Arabian Sea; Thanked: 125 times

by ajith » Fri Mar 05, 2010 5:12 am

sanju09 wrote:The total number of goals posted by three strikers, Amy, Bobby, and Cameron were to be presented as a sum of pairs by a greenhorn statistician. The statistician instead fed a bad command that added the product of goals by a pair along with the correct entry. This resulted in a report that read Amy and Bobby together posted 69 goals, Bobby and Cameron together posted 49 goals, and Cameron and Amy together posted 34 goals. How many goals did the trio really posted as a total?
(A) 76
(B) 57
(C) 38
(D) 22
(E) 19

Say, Amy scored 'a' goals , Bobby scored 'b' goals, and Cameron scored 'c' goals
a+b+ab = 69
b+c +bc = 49
a+c +ac = 34

a+b+ab+1 = 70
b+c +bc +1 = 50
a+c +ac +1 = 35

(a+1)(b+1) = 70
(b+1) (c+1) =50
(a+1) (c+1) = 35
Solving,

a+1 =7
b+1 =10
c+1 =5

a=6 b=9 c =4

[spoiler]a+b+c =19, E[/spoiler]

Always borrow money from a pessimist, he doesn't expect to be paid back.

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sanju09: GMAT Instructor; Posts: 3650; Joined: Wed Jan 21, 2009 4:27 amFollowed by: 80 members; Location: India; Thanked: 267 times; GMAT Score:760

by sanju09 » Fri Mar 05, 2010 5:27 am

ajith wrote:
sanju09 wrote:The total number of goals posted by three strikers, Amy, Bobby, and Cameron were to be presented as a sum of pairs by a greenhorn statistician. The statistician instead fed a bad command that added the product of goals by a pair along with the correct entry. This resulted in a report that read Amy and Bobby together posted 69 goals, Bobby and Cameron together posted 49 goals, and Cameron and Amy together posted 34 goals. How many goals did the trio really posted as a total?
(A) 76
(B) 57
(C) 38
(D) 22
(E) 19

Say, Amy scored 'a' goals , Bobby scored 'b' goals, and Cameron scored 'c' goals
a+b+ab = 69
b+c +bc = 49
a+c +ac = 34

a+b+ab+1 = 70
b+c +bc +1 = 50
a+c +ac +1 = 35

(a+1)(b+1) = 70
(b+1) (c+1) =50
(a+1) (c+1) = 35
Solving,

a+1 =7
b+1 =10
c+1 =5

a=6 b=9 c =4

[spoiler]a+b+c =19, E[/spoiler]

That's it!! But rohan_vus' effort had been marvelous in the absence of this master idea. Well done ajith, and you rock rohan_vus.

The mind is everything. What you think you become. -Lord Buddha

Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001

www.manyagroup.com

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