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

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]

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?

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

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]

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.