Carcass wrote:
Chris gave Jane x cards. He gave Betty one card more than he gave Jane and he gave Paul two cards fewer than he gave Betty. In terms of x, how many cards did Chris give Betty, Jane, and Paul altogether?
(A) \(3x + 1\)
(B) \(3x\)
(C) \(3x - 1\)
(D) \(x - 1\)
(E) \(\frac{x}{3}\)
I solved in different way. Let me know whether this is correct.
Let x be 9 , looking at options, decided to go for this.
options will be:
a.10
b.27
c.26
d.8
e.3
When we have given info that Chris gave x cards to Jane, i.e. 9 cards as per our case then we can eliminate a,d and e options.
Take option b as it bigger number than c.
Chris has 27 cards = 9(Jane) + one more than to Betty than Jane(10) + two cards fewer than Paul than Betty (8).
27 = 27.
Hence B.
close
75% (00:55) correct
