Ezra pays $0.75 each time he takes a public transport to reach his sch

0.75*2*k > 13 + 0.60k

Or, 1.50 k > 13 + 0.60k

Or, 0.90k > 13

Or, k > 14

Among the given option only (C) fits in , Answer must be (C)

Current plan expense per day ; 0.75*2 ; 1.5$*k
Another plan ; 13+0.3*2 ; 13+0.6*k

using answer options we can find when saving money will start

K----CP------NP(13$)
10----15-----19
12----18-----20.2
15---22.5----22 ; YES she starts to save 0.5$ per trip
IMO C ; 15 is sufficient


Ezra pays $0.75 each time he takes a public transport to reach his school. Students who frequently use this service may instead purchase token each month for $13 and then pay $0.30 per each ride. If Ezra need to take the bus twice a day for K days next month and will not use the bus on any other day, what is the least possible value of K for which Ezra will save money by using the token?

(A) 10
(B) 12
(C) 15
(D) 18
(E) 28

current cost - token cost > 0
0.75*2*k - (13+0.3*2*k) > 0
1.5k - 13 - 0.6k > 0
0.9k > 13
k > 14… = 15 days

Answer (C)

$0.75 * 2 =$1.5 per day
—> If Ezra has a token ($13 for a month) for the next month, she will spend $0.6 per day for K days.

In order to save money,
—>13+ 0.6K must be less than 1.5K

13+0.6K< 1.5K
0.9K> 13
K> 130/9= 14.4
K>14.4
—> the least possible value of K is 15
The answer is C

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We can create the following inequality:

0.75(2K) > 13 + 0.3(2K)

1.5K > 13 + 0.6K
0.9K > 13

K > 14.4

Answer: C

You can look at the difference:
Cost each day without pass = 1.5$
Cost each day with pass = 0.6$

The difference is 0.9$.

Now we need to recover the cost of 13$. So in 14 days we would have recovered 12.6$. So on the 15th day we will profit from the ownership of the pass.

Hence C.