Amy's sells kale at x dollar per pound for the first 20 pounds and .8x

Amy’s price — 20x+ 0.8x*a
Lucy’s price — 14x + 0.9x*b
—> 20x+ 0.8x*a ≥ 14x+ 0.9x*b

Let’s find the minimum number of (20 + a ) :

20x + 0.8x*a = 14x + 0.9x*b
6x = x( 0.9b —0.8a)
—> \(a = \frac{(9b —60)}{8} \)— must be integer

If \(b= 12\) , then \( a = \frac{(108–60)}{8 }= 6 \)
—> \(20 + a = 26\)

The answer is C

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Solution

• Let us assume that the number of pounds over 15 is n.

o So, the total number of pounds \(= 15+ n\)
• So, the total price at which Amy's sells (15 +n) pounds of kale \(= 20*x + (15+n-20) *0.8*x …. Eq.(i)\)
• Now, the total price at which Lucia's sells (15 +n) pounds of kale \(= 14 *x + (15+n -14) *0.9*x …………. Eq. (ii)\)
• From Eq.(i) and Eq.(ii) we can say that deal offered by Amy’s to become a equal or better than Lucia’s:

o \(20x + (15+n-20) * 0.8x ≤ 14 x + (15+n -14) *0.9x\)
o Or, \(20x + (n-5) *0.8x ≤ 14x + (1+n) *0.9x\)
o Or, \(6x + 0.8nx – 4x ≤ 0.9nx+ 0.9x\)
o Or, \(1.1x ≤ 0.1nx\)
o Or, \(11 ≤ n \)

So, for minimum 11 pounds over 15, Amy’s will be a better deal.
Hence total number of pounds = 15 +11 = 26
Thus, the correct answer is Option C.

The question is confusing on two counts:
The first, as explained by IanStewart, is that it can be logically interpreted to mean the solver has to find the minimum number of pounds that a buyer has to buy IN EXCESS OF 15 POUNDS so that he gets an equal or better deal from Amy than from Lucia.
The second is the condition in the stem which states that "Amy's becomes an equal OR better deal". According to this, the correct answer could be either C (26 lbs) (the buyer spends the same amount of money whether he buys from Amy or Lucia) or D (27 lbs) ( where he spends less money when he buys from Amy).

Corrigendum: Sorry, my second point is incorrect - I overlooked the "minimum number of pounds" stipulation. First point stands though.