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# If a phone call costs a cents for the first minute and a/3 cents

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Joined: 01 Sep 2010
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If a phone call costs a cents for the first minute and a/3 cents  [#permalink]

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05 Oct 2012, 16:56
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Difficulty:

15% (low)

Question Stats:

81% (02:07) correct 19% (02:40) wrong based on 186 sessions

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If a phone call costs a cents for the first minute and a/3 cents for each additional half-minute, how much a b minute phone call cost, in cents ?

A. $$\frac{(2a+2ab)}{27}$$

B. $$\frac{(a+2ab)}{3}$$

C. $$\frac{(a+ab)}{300}$$

D. $$\frac{(a+ab)}{3}$$

E. $$\frac{ab}{3}$$

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Re: If a phone call costs a cents for the first minute and a/3 cents  [#permalink]

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05 Oct 2012, 22:19
2
First, let's set up an equation:

No matter how long the phone call is, we always have to pay a cents for the first minute. Every minute beyond the first minute costs 2a/3 cents.

Cost of phone call = a + (b-1) * 2a/3

Now we have to simplify this equation.

3a/3 + (b-1) * 2a/3
(3a + (b-1) * 2a) / 3
(3a + 2ab - 2a) / 3
(a + 2ab) / 3
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Re: If a phone call costs a cents for the first minute and a/3 cents  [#permalink]

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10 Dec 2012, 04:59
1
Cost = (1)(a) + ((b-1)/(1/2))(a/3)
Cost = a + (2b - 2)(a/3)
Cost = a + (2ab -2a)/3
Cost = (3a - 2a + 2ab) / 3
Cost = (a + 2ab)/3

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Re: If a phone call costs a cents for the first minute and a/3 cents  [#permalink]

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17 Dec 2014, 06:42
Could someone please help with this question? I thought "a b-minute" could also be half-a-minute in which case it costs a/2 cents. IOW, the cost of "b" minutes is ab, no?
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If a phone call costs a cents for the first minute and a/3 cents  [#permalink]

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17 Dec 2014, 06:55
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Blackbox wrote:
If a phone call costs a cents for the first minute and a/3 cents for each additional half-minute, how much a b minute phone call cost, in cents ?

Could someone please help with this question? I thought "a b-minute" could also be half-a-minute in which case it costs a/2 cents. IOW, the cost of "b" minutes is ab, no?

A call less than or equal to 1 minute costs a cents. So, if b = 1/2 minutes the cost would still be a cents.

Easy way would be to just plug numbers, say a = 1 cent and b = 1 minute what would be the cost of a call then? It would cost 1 cent. Plug a = b = 1 and check for an option which would give 1. Only B fits.
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Re: If a phone call costs a cents for the first minute and a/3 cents  [#permalink]

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17 Dec 2014, 07:06
Nice! It is much easier to sub-in values and understand. Thanks
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Re: If a phone call costs a cents for the first minute and a/3 cents  [#permalink]

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05 Jan 2015, 08:19
This is what I though after reading this:

First minute: a --> this actually means "a" times the number of minutes; in this case a*1.
Extra minutes: (2a)/3 --> a/3 for half a minute, so double this for a whole minute.
Based on the logic we used above to calculate the first minute, which is "a", the following minutes will be (2a)/3 times "b", because "b" is the number of minutes the person spoke.

Adding these: a + (2a)/3 times b, so a + (2ab)/3 --> ***if you write down the fractions horizontally it makes much more sense, because you don't use the parenthesis.
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Re: If a phone call costs a cents for the first minute and a/3 cents  [#permalink]

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02 Mar 2018, 00:15
Cost = (1)(a) + ((b-1)/(1/2))(a/3)
Cost = a + (2b - 2)(a/3)
Cost = a + (2ab -2a)/3
Cost = (3a - 2a + 2ab) / 3
Cost = (a + 2ab)/3
Re: If a phone call costs a cents for the first minute and a/3 cents   [#permalink] 02 Mar 2018, 00:15
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