GMAT Club Around The World!!! All of the 121 baseball players particip

All of the 121 baseball players participating in a celebrity golf outing were born in either Asia, Australia, North America, or South America. If there are 25 percent fewer players that were born in Australia than were born in Asia, how many of the participating players were born in North America?


(1) The number of players who were born in South America is 40 percent greater than the number of players who were born in Asia.

(2) The ratio of the number of players who were born in North America to the number of players who were born in South America is 29:14.


solution analysis:
we have total = 121

now we have australia = 3/4 asia

statement 1:sa = 7/5 asia

so we get asia + 3/4 asia + 7/5 asia + na = 121 so 2 variables asia = x and na = y

we get eq as x+ 3/4 x+7/5 x+ y =121

we simplify to get 63x+20y = 121*20

by diophantine eq analysis we can have 1 or 2 positive integer solution

now lets see 1 trivial solution x = 0 and y = 121 now by using exchange method positive integer solution will be
x = 20 y = 58 all others will make 1 negative

so the answer is y = 58 or na born = 58 from statement 1 so its sufficient


statement 2: na = 29/14 sa and au = 3/4 asia

so we get x+3/4 x+ 29/14y + y =121

so simplifying we get 49x + 86y = 121 *28
but we see here that y has to be multiple of 7 , so giving some hit and trial withy y as multiples of 7 we
get x = 20 and y = 28 which is the only solution otherwise we get 1 negative solution using diophantine eq

so if sa = 28 then na= 29/14 *28 = 58 which is same as statement 1

so statement 2 is also sufficient

so answer is d) each statement alone is sufficient to answer this question

Let number of people in

Asia=a
Australia=b
North America=c
South America=d

a+b+c+d=121

We need the value of c
Since its number of people, a,b,c,d will be integers

Also, b=.75a
So, 1.75a+c+d=121.......eq1

(1) The number of players who were born in South America is 40 percent greater than the number of players who were born in Asia.
d=1.4a
1.75a+c+1.4a=121 (from eq1)
3.15a+c=121

For 3.15a to be an integer "a" has to a multiple of 20

3.15*20=63
3.15*40=126 (since total number of people is 121 we ca neglect this and all other values)
So, 63+c=121; c=58

Sufficient


(2) The ratio of the number of players who were born in North America to the number of players who were born in South America is 29:14.

c:d=29:14
(14/29)c=d

1.75a + c + (14/29)c = 121
50.75a+43c=3509

Again, for 50.75a to be an integer "a" has to a multiple of 20

50.75*20=1015
50.75*40=2030
50.75*60=3045
50.75*80=4060 (since total number of people is 3509 we can neglect this and all other values)

When a= 40 or 60, c will be a fraction and we need an integer. So, we can discard these values.

1015+43c=3509
43c=2494
c=58

Sufficient

IMO Option D

IMO C

total = 121
born: \({A_{s}, A_{u}, NA, SA}\)
\(A_{u} = 0.75*A_{s}\)
NA = ?

(1) The number of players who were born in South America is 40 percent greater than the number of players who were born in Asia.
\(SA = 1.4* A_{s}\)
\(121 = A_{s} + A_{u} + NA + SA\)
\(121 = A_{s} + 0.75*A_{s} + NA + 1.4* A_{s}\)
two variables and one equation
Not sufficient

(2) The ratio of the number of players who were born in North America to the number of players who were born in South America is 29:14.
NA:SA = 29:14
more than one variable in one equation
Not sufficient

combining 1 and 2
\(121 = A_{s} + 0.75*A_{s} + NA + 1.4* A_{s}\)
NA:SA = 29:14
\(SA = \frac{14*NA}{29}\)
\(SA = 1.4* A_{s}\)
\(\frac{14*NA}{29} = 1.4* A_{s}\)
\(A_{s} = \frac{10*NA}{29}\)
=> \(121 = \frac{10*NA}{29} + 0.75*\frac{10*NA}{29} + NA + 1.4* \frac{10*NA}{29}\)
only 1 unknown -
SUFFICIENT

All of the 121 baseball players participating in a celebrity golf outing were born in either Asia, Australia, North America, or South America. If there are 25 percent fewer players that were born in Australia than were born in Asia, how many of the participating players were born in North America?

Given, Asia= x, Aus= 0.75*x

Stat1: The number of players who were born in South America is 40 percent greater than the number of players who were born in Asia.
Let, NA= y and given, SA= 1.4*x
So, x + 0.75*x + y + 1.4*y = 121 or, 3.15*x + y= 121 or, 63*x +20*y = 2420, so, we can have (x,y)= (20,58) or (0,121). Atleast one player would be there. So, (x,y)= (20,58) Sufficient.

Stat2: The ratio of the number of players who were born in North America to the number of players who were born in South America is 29:14.
Additional info, NA= 29*y , SA = 14*y
So, x + 0.75*x + 29*y + 14*y = 121 or, 1.75*x + 43*y= 121 or, 7*x +172*y = 484, so, we can have (x,y)= (20,2) only. Sufficient.

So, I think D.

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