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If Janice was 25 years old z years ago and Lisa will be 23

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rxs0005
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If Janice was 25 years old z years ago and Lisa will be 23 [#permalink] New post   Updated on: 21 Nov 2010, 08:48
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If Janice was 25 years old z years ago and Lisa will be 23 years old in p years, what was the average (arithmetic mean) of their ages 5 years ago?

(z + p)/2

(z - p + 38)/4

(z - p + 28)/4

(z + p + 48)/2

(z - p + 38)/2
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E

Originally posted by rxs0005 on 21 Nov 2010, 06:48.
Last edited by rxs0005 on 21 Nov 2010, 08:48, edited 1 time in total.
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padukaprasad
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Re: Airthmetic PS [#permalink] New post   21 Nov 2010, 07:30
For the sake of simplicity let us assume z=10 and p=13

So Janice's current age = 25+10=35
Lisa's current age= 23-13=10

Janice's age before 5 years = 35-5=30
Lisa's age before 5 years=10-5=5

Average age =35/2=17.5

Now substituting the values of z and p in the options a) ruled out(as z+p/2 is not equal to 17.5)
b) ruled out
c) ruled out
d) ruled out

None of the options are satisfying the condition..so I feel that either b can be the correct answer if the option is (z-p+38)/2.
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rxs0005
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Re: Airthmetic PS [#permalink] New post   21 Nov 2010, 08:16
can someone do this algebraically i cant seem to quite get any of the AC using algebra
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Re: Airthmetic PS [#permalink] New post   21 Nov 2010, 08:22
rxs0005 wrote:
If Janice was 25 years old z years ago and Lisa will be 23 years old in p years, what was the average (arithmetic mean) of their ages 5 years ago?

(z + p) / 2

( z - p +38) / 4

( z - p +28) /4

( z + p +48) /2

( z - p +38) / 3


Let me try...
Current age of Janice =z+25
Current age of Lisa=23-p
Age of Janice 5 years ago =z+20
Age of Lisa 5 years ago=23-p-5=18-p

Avearage(Arithmetic mean)=(z+20+18-p)/2=(z-p+38)/2

I feel that the option B is wrong..It should be (z-p+38)/2 as I had derived before...
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Re: Airthmetic PS [#permalink] New post   21 Nov 2010, 08:25
Expert Reply
rxs0005 wrote:
If Janice was 25 years old z years ago and Lisa will be 23 years old in p years, what was the average (arithmetic mean) of their ages 5 years ago?

(z + p) / 2

( z - p +38) / 4

( z - p +28) /4

( z + p +48) /2

( z - p +38) / 3


Janice was 25 years old z years ago --> \(J-z=25\), so \(J=25+z\);
Lisa will be 23 years old in p years --> \(L+p=23\), so \(L=23-p\);

5 years ago Janice was \(J-5=25+z-5=20+z\) and Lisa was \(L-5=23-p-5=18-p\), so the average (arithmetic mean) of their ages was \(\frac{20+z+18-p}{2}=\frac{38+z-p}{2}\).

No correct answer among answer choices.
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Re: Airthmetic PS [#permalink] New post   21 Nov 2010, 08:48
thanks i had a typo AC E had 2 in the denominator
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Re: Airthmetic PS [#permalink] New post   21 Nov 2010, 08:58
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rxs0005 wrote:
If Janice was 25 years old z years ago and Lisa will be 23 years old in p years, what was the average (arithmetic mean) of their ages 5 years ago?

(z + p) / 2

( z - p +38) / 4

( z - p +28) /4

( z + p +48) /2

( z - p +38) / 2


I would still suggest, the fastest way to do it is by assuming values for z and p. Say z = 0 so Janice is 25 yrs old now. p = 0 so Lisa is 23 yrs old now. They were 20 and 18 respectively 5 yrs back so their average age was 19. So you are looking for the option that gives you 19. (When you are putting values, make sure you are very clear about what value you want from the options)
Put z = 0, p =0, you get 19 from option (E).
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If Janice was 25 years old z years ago and Lisa will be 23 [#permalink] New post   10 Dec 2015, 07:42
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rxs0005 wrote:
If Janice was 25 years old z years ago and Lisa will be 23 years old in p years, what was the average (arithmetic mean) of their ages 5 years ago?

(z + p)/2

(z - p + 38)/4

(z - p + 28)/4

(z + p + 48)/2

(z - p + 38)/2


Since Janice was 25 years old z years ago,
Hence current Age of Janice = 25 + z
Age 5 years before = 20 + z (i)

Since Lisa will be 23 years old in p years.
Hence current age of Lisa = 23 - p
Age 5 years before = 18 - p (ii)

Average of (i) and (ii) = \(\frac{{20 + z + 18 - p}}{2}\) = \(\frac{{z - p + 38}}{2}\)

Option E
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Re: If Janice was 25 years old z years ago and Lisa will be 23 [#permalink] New post   10 Dec 2015, 09:02
Quote:
what was the average (arithmetic mean) of their ages 5 years ago?


Avergae (arithmetic mean) = \(\frac{{(18-p) +( 20+z)}}{2}\)

Avergae (arithmetic mean) = \(\frac{(38+z-p)}{2}\)

Hence IMHO (E)
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Re: If Janice was 25 years old z years ago and Lisa will be 23 [#permalink] New post   04 Dec 2016, 02:56
I solved this question using algebra, but as others have suggested, choosing smart numbers is a better approach.
Let z=0 and p=0. With this extreme case, can we eliminate choices?
This means Janice is 25 years now while Lisa is 23 years now. 5 years ago, Janice was 20 and Lisa was 18. Ok average = 38/2. Looking at the choices and substituting zero values for both p and z, we can see E work! . Rest others dont work at all
So E
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Re: If Janice was 25 years old z years ago and Lisa will be 23 [#permalink] New post   26 Mar 2022, 20:07
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Re: If Janice was 25 years old z years ago and Lisa will be 23 [#permalink]
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