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Greg's long - distance plan charges him $.5 for first 4 mins of a call

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Greg's long - distance plan charges him $.5 for first 4 mins of a call  [#permalink]

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New post 29 Sep 2019, 21:00
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Greg's long - distance plan charges him $.5 for first 4 mins of a call and $.07 per minute beyond. Greg makes a call for x minutes, how many minutes was the call?

(1) Last 7 minutes of Greg's call costed him $.22 less than the first 7 minutes of the call
(2) The call did not cost more than $1.05

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Re: Greg's long - distance plan charges him $.5 for first 4 mins of a call  [#permalink]

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New post 29 Sep 2019, 22:57
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(1) Last 7 minutes of Greg's call costed him $.22 less than the first 7 minutes of the call

This information is not sufficient, as we don’t know how many calls Greg has made in between first and last 7 calls.

Minimum number of minutes are 11, as 10 minutes would give the difference between first 7 minutes = 0.5 + 3*0.07 = $0.71 and last 7 minutes = 0.5/4 + 6*0.07 = $0.545 as $0.165

So, possible number of minutes are 11, 12, 13, 14, 15, 16, . . . .
—> Insufficient

(2) The call did not cost more than $1.05
So, total cost can be less than or equal to $1.05 —> Insufficient

Combining (1) & (2),
Cost of 11 minutes = 0.5 + 7*0.07 = $0.99 —> Possible
Cost of 12 minutes = 0.5 + 8*0.07 = $1.06 —> Not Possible

So, we get a unique number of minutes = 11
—> Sufficient

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Re: Greg's long - distance plan charges him $.5 for first 4 mins of a call  [#permalink]

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New post 29 Sep 2019, 23:37
Total charges can be calculated as = $ (0.5 + 0.07 * y) where y is number of minutes beyond 4 minutes.

(1) Last 7 minutes of Greg's call costed him $.22 less than the first 7 minutes of the call
So Call minutes = 7 + 7 = 14 minutes. Verifying we have
0.5 + 0.07 * 3 - 0.22 = 0.07 * 7
0.5 - 0.01 = 0.49
LHS = RHS

SUFFICIENT.

(2) The call did not cost more than $1.05
0.5 + 0.07 * y ≤ 1.05 where x = 4 + y
Since y = 0 or y < 8, ‘x’ can be
0 < x < 12 as call may cost from $0.5 to $1.05 including both of which satisfies the condition.

INSUFFICIENT.

Answer (A).
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Re: Greg's long - distance plan charges him $.5 for first 4 mins of a call  [#permalink]

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New post 29 Sep 2019, 23:39
Cost per minute for first 4 minutes = $0.5, and every extra minute beyond 4 minutes costs $0.07 per minute.
we are to determine how many minutes Greg's call lasted.

From statement 1, we know that the last 7 minutes of his call costs him $0.22 less than the first seven minutes.
This statement is clearly insufficient because it only tells us that Greg's call lasted more than 14 minutes. Which means Greg's call could be 15 minutes, 20 minutes, in fact infinite possibilities satisfy the criteria given in 1.

Statement 2 states that the total call costs is not more than $1.05. This is also insufficient. Greg could have spoken for a minute which would have cost him $0.5 or two minutes which could cost him $1.

1+2.
from 1, we know that the length of the call is more than 14 minutes. However, there is no way that a call that is more than 14 minutes can cost less than $1.05. Even the cost of the first three minutes would cost $1.5, which is more than the requirement in statement 2, and less than the required time per statement 1. Both statements taken together is still not sufficient to narrow down on the total length of time for Greg's phone call.

The asnwer is E in my view.
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Re: Greg's long - distance plan charges him $.5 for first 4 mins of a call  [#permalink]

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New post 30 Sep 2019, 02:34
Greg's long - distance plan charges him $.5 for first 4 mins of a call and $.07 per minute beyond. Greg makes a call for x minutes, how many minutes was the call?

Total call time = x, remaining duration call time = x-4
Cost of first 4 min. = 4*0.5 = $2,
Cost of remaining minutes of call = 0.07 * x-4
Total cost = $ 2 + 0.07 *x-4.
Need to find unique value of x
1) Last 7 minutes of Greg's call costed him $.22 less than the first 7 minutes of the call
x-7= 2+0.21 -0.22
Linear equation with one unknown. Hence, sufficient.
2) The call did not cost more than $1.05
There can be multiple values of x as higher limit is given as it can be 4 minutes to 11 minutes . So, insufficient.
Ans. A.
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Re: Greg's long - distance plan charges him $.5 for first 4 mins of a call  [#permalink]

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New post Updated on: 30 Sep 2019, 21:36
1
(1) Last 7 minutes of Greg's call costed him $.22 less than the first 7 minutes of the call


for 1st 7 minutes it costs....a constant amount of $0.71 which is $0.22 more than last 7 minutes which costs him $0.49
SO any 7 min after 1st 4 min it will costs him $0.49..using which we cannot calculate total time taken so........ insuff

(2) The call did not cost more than $1.05 .....clearly insuff

combining,....for the call cost not to exceed $1.05, total time must be less than or equal to 10min...

so total time of 10min satiesfies both conditions



OA:C

Originally posted by madgmat2019 on 30 Sep 2019, 04:20.
Last edited by madgmat2019 on 30 Sep 2019, 21:36, edited 1 time in total.
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Re: Greg's long - distance plan charges him $.5 for first 4 mins of a call  [#permalink]

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New post 30 Sep 2019, 05:19
1
Quote:
Greg's long - distance plan charges him $.5 for first 4 mins of a call and $.07 per minute beyond. Greg makes a call for x minutes, how many minutes was the call?

(1) Last 7 minutes of Greg's call costed him $.22 less than the first 7 minutes of the call
(2) The call did not cost more than $1.05


(1) Last 7 minutes of Greg's call costed him $.22 less than the first 7 minutes of the call: insufic.
Last 7 mins = First 7 mins - 0.22… First - Last = 0.22
First 7 mins {0 to 7 mins}: 0.5+3mins*(0.07)=0.71
Last 7 mins {4 to 11 mins}: 7mins*(0.07)=0.49
Last 7 mins {5 to 12 mins}: 7mins*(0.07)=0.49
Last 7 mins {6 to 13 mins}: 7mins*(0.07)=0.49
0.71 - 0.49 = 0.22…
There could be any number of minutes in between First and Last,… x ≥ 11;

(2) The call did not cost more than $1.05: $0≤cost≤$1.05, insufic.

(1&2): sufic.
min(x=11)≤0.5+(x-4)(0.07)≤1.05
0.5+(0.07)(11-4)<0.5+(x-4)(0.07)≤1.05
0.99≤0.5+(x-4)(0.07)≤1.05
0.49≤(x-4)(0.07)≤0.55
7≤x-4≤55/7… 11≤x≤55/7+4… 11≤x≤83/7…11≤x<~12…x=11

Answer (C)
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Greg's long - distance plan charges him $.5 for first 4 mins of a call  [#permalink]

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New post 30 Sep 2019, 15:52
1
given
total cost for call
0.5+(x-4)*0.07
find value of x
(1) Last 7 minutes of Greg's call costed him $.22 less than the first 7 minutes of the call
first 7 mins value ; 0.5+3*0.07 ; $.71
last 7 mins value ; .71-.22 ; $.49 ; i.e total mins ; .49/.7 ; .7 mins
last 7 mins can also be from 6 min to 13 mins
5 min to 12 mins so on insufficient


#2 The call did not cost more than $1.05
0.5+(x-4)*0.07 <=1.05
0.5+.07x-.28<=1.05
.07x<=.83
x<=11.85
total mins value is not unique
insufficient
from 1 &2 ; the total mins x=11
IMO C

Greg's long - distance plan charges him $.5 for first 4 mins of a call and $.07 per minute beyond. Greg makes a call for x minutes, how many minutes was the call?

(1) Last 7 minutes of Greg's call costed him $.22 less than the first 7 minutes of the call
(2) The call did not cost more than $1.05
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Greg's long - distance plan charges him $.5 for first 4 mins of a call   [#permalink] 30 Sep 2019, 15:52
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