Princeton Review
How many baseball cards do Keith, Pat, and Steve own in total?
1. Keith and Pat together own half as many baseball cards as Steve does.
2. Keith and Steve together own 109 baseball cards, and Pat and Steve together own 126 baseball cards.
OA C
How many baseball cards do Keith, Pat, and Steve own in
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Your Answer
A
B
C
D
E
Global Stats
Statement 1 : Keith Pat Steve
x 2 x
Total = 3x
Not enough information to solve for x
Statement 1 not sufficient
Statement 2 : Keith. Pat. Steve
x 109-x
126-(109-x)
= 17+ x
Total = x + 17 + x + 109 - x
= 126 + x
Not enough information to solve for x
Statement 2 not sufficient
Statement 1 and 2 together :
Keith. Pat. Steve
x. 17+x. 109-x
Keith+ pat = 2*steve
x + 17+ x = 2*(109-x)
Can be solved for x
Statement 1 and 2 together are sufficient
Hence C.
x 2 x
Total = 3x
Not enough information to solve for x
Statement 1 not sufficient
Statement 2 : Keith. Pat. Steve
x 109-x
126-(109-x)
= 17+ x
Total = x + 17 + x + 109 - x
= 126 + x
Not enough information to solve for x
Statement 2 not sufficient
Statement 1 and 2 together :
Keith. Pat. Steve
x. 17+x. 109-x
Keith+ pat = 2*steve
x + 17+ x = 2*(109-x)
Can be solved for x
Statement 1 and 2 together are sufficient
Hence C.
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Question stem asks for the sum of K+S+P.
Statement 1) gives you only something like a ratio : K+P=1/2 S --> Not sufficient
Statement 2) gives you only K+S=109 P+S=126 --> 3 variables, two equations --> Not sufficient.
Together they yield: one ratio between K+P and S ; K+S=109; P+S=126. Now you can rearrange K+S=109 to K=109-S . & P+S=126 to P=126-S
Now you can express both K and P in times of S and hence you can insert them in equation ratio from 1 and you have one equation with just one variable left : S. Hence, you can solve for S and since you know how S+P depend on S you know that you can calculate the sum of K+P+S
Hence, C
Statement 1) gives you only something like a ratio : K+P=1/2 S --> Not sufficient
Statement 2) gives you only K+S=109 P+S=126 --> 3 variables, two equations --> Not sufficient.
Together they yield: one ratio between K+P and S ; K+S=109; P+S=126. Now you can rearrange K+S=109 to K=109-S . & P+S=126 to P=126-S
Now you can express both K and P in times of S and hence you can insert them in equation ratio from 1 and you have one equation with just one variable left : S. Hence, you can solve for S and since you know how S+P depend on S you know that you can calculate the sum of K+P+S
Hence, C