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Joan, Kylie, Lillian, and Miriam all celebrate their birthda

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Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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20 Apr 2008, 12:27
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Difficulty:

45% (medium)

Question Stats:

71% (01:58) correct 29% (02:09) wrong based on 908 sessions

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Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today. Joan is 2 years younger than Kylie, Kylie is 3 years older than Lillian, and Miriam is one year older than Joan. Which of the following could be the combined age of all four women today?

A. 51
B. 52
C. 53
D. 54
E. 55

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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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11 Apr 2014, 02:54
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Re: PS: 4 birthdays  [#permalink]

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21 Apr 2008, 16:01
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Some time ago I read in Kaplan... that if a question has the phrase "Which of the following..." (verbal or Quant) try the answer choices from the bottom (from E up)...

That's how I got in my very second choice (D) I got the answer without attempting the other 3...

Just thought I'd share that piece of info...

If u guys already know this... discard this message
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Re: PS: 4 birthdays  [#permalink]

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20 Apr 2008, 13:10
2

total- x+x+1+x+2+x+3
total= 4x+6
birthdays at the same day, so total = integer
54 -6 = 48, which is divisible by 4
so, x= 12
4*12 +6 = 54
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Re: PS: 4 birthdays  [#permalink]

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20 Apr 2008, 18:02
4
1
chineseburned wrote:
Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today. Joan is 2 years younger than Kylie, Kylie is 3 years older than Lillian, and Miriam is one year older than Joan. Which of the following could be the combined age of all four women today?
A. 51
B. 52
C. 53
D. 54
E. 55

D

K = J + 2
K = L + 3
M = J + 1
Total = K + J + M + L = K + K - 2 + K - 2 + 1 + K - 3 = 4K - 6

I tried out all numbers and only 54 works as an integer of K.
(54 + 6)/4 = K = integer
The rest give fractions.
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Re: PS: 4 birthdays  [#permalink]

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20 Apr 2008, 18:23
2
the way i did is very similar to Bkk's approach..

j=k-2
k=L+3
m=1+J
L=L

j=L+1; M=L+2

total sum of ages= (L+1+L+2+L+3+L=4L+6

then i subtract 6 from all the answer choices..the resulting number should be divisible by 4..

54 is the only one

D it is..
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Re: PS: 4 birthdays  [#permalink]

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21 Apr 2008, 11:16
1
I'd say 54 as well...

J= K-2 = K-2
K= L+3 = K
L= K-3 = K-3
M= J+1 = K-1

Then, I gave K=15 so 15+14+13+12= 54
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Re: PS: 4 birthdays  [#permalink]

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21 Apr 2008, 14:51
1
D

J = K - 2
K = 3 + L
M = 1 + J

L = 12
K = 15 27
J = 13 40
M = 14 54
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Re: PS: 4 birthdays  [#permalink]

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21 Apr 2008, 18:46
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axl_oz wrote:
Just thought I'd share that piece of info...

If u guys already know this... discard this message

I have just be informed, thanks!
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Re: PS: 4 birthdays  [#permalink]

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21 May 2008, 12:16
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i was just thinking..if you are in a tight spot and have to quickly think..maybe you have 3 questions left and only 1 minute left to do em..

you know that one of the girls is 2 years younger and one is 3 years older..

so then of the answer choice..1 of these choices must be divisble by 3..now look at 2..the answer choice has to be even..so 54 is the best choice..51 is close but its not divisible by 2..
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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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11 Apr 2014, 00:24
Let present age of Kylie = x, then present ages of....

Joan = x-2

Lillian = x-3

Miriam = x-1

Total age = x + x-1 + x-2 + x-3

= 4x - 6

Option A, C, E are odd, hence can be discarded

54 + 6 = 60 which is divisible by 4

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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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11 Apr 2014, 07:27
My strategy is Word Problem ,By plugging in
And forming the Equation k = x , J = x -2 , L = x - 3 , M = x - 2 + 1
So x+ x - 2 + x- 3 + x - 2 + 1 = 4x - 6
4x-6 = Total By Using Options x = 15
So Total is 54
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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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13 Apr 2014, 04:48
Assume L =x
then K= x+3
j=x+1
and M= x+2 .

Total Sum = x+x+1+x+2+x+3= 4x+6
Substitute X =12 (random guess) then we can find 54 is answer
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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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29 Jul 2015, 17:53
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2
Hi All,

The first part of this question requires translating sentences into formulas:

J + 2 = K
L + 3 = K
J + 1 = M

Even if you translated these formulas differently (but correctly), you still have a relationship among the 4 variables that needs to be discovered. You can TEST VALUES to figure out the relationship:

If K = 10, then
J = 8
L = 7
M = 9

Now we know that the 4 values are consecutive.

To solve, you can either use "brute-force" (since the answers are small) or algebra:

Brute-Force:
11 + 12 + 13 + 14 = 50 NOT ENOUGH BASED ON THE ANSWERS
12 + 13 + 14 + 15 = 54 WINNER

Algebra:
Call the youngest person X
X + (X+1) + (X+2) + (X+3) = 4X + 6
If X = 11, total = 50 NOT ENOUGH BASED ON THE ANSWERS
If X = 12, total = 54 WINNER

Either way, you have the correct answer.

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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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17 Aug 2015, 02:40
chineseburned wrote:
Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today. Joan is 2 years younger than Kylie, Kylie is 3 years older than Lillian, and Miriam is one year older than Joan. Which of the following could be the combined age of all four women today?

A. 51
B. 52
C. 53
D. 54
E. 55

let the age of Lillian is L
Kylie's age is L+3
Joan's age is L+1
Miriam's age is L+2
combined age is 4L+6, which is even
The first part is a multiple of 4
44+6 or 48+6 or 52+6
Hope, you got it
The correct option is D
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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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23 Aug 2016, 03:27
Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today. Joan is 2 years younger than Kylie, Kylie is 3 years older than Lillian, and Miriam is one year older than Joan. Which of the following could be the combined age of all four women today?

A. 51
B. 52
C. 53
D. 54
E. 55

J =k-2
K= L+3 => J =L+3-2
M = J+1

J+K+L +M = J+J+2+J-1+J+1
4J+2 so D
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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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10 Jan 2017, 05:46
Total age = x + x-1 + x-2 + x-3

= 4x - 6

After arriving at this stage can this be said that all values except 54 will yield a decimal value of x and as age cannot be in decimal the answer is 54?
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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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12 Jan 2017, 16:37
agold wrote:
D

J = K - 2
K = 3 + L
M = 1 + J

L = 12
K = 15 27
J = 13 40
M = 14 54

agold, How do you know L=12?
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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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04 Sep 2017, 14:34
While I used the same approach (reduced J: J+(J+2)+((J+2)-3)+J+1 = 4J+2)

A shortcut to know which numbers to test would be that you know the number must be even - so only 52 or 54

Another way to look at this is that based off of the age differences, you have J and K being an even number apart. They are either both even or both odd.

K and L are an odd number apart, so L is the opposite of K.
J and M are an odd number apart, so M is the opposite of J.

This means the sum of ages would be 2 odd + 2 even numbers, which will always come out as even.
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Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda  [#permalink]

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19 Oct 2017, 18:01
1
chineseburned wrote:
Joan, Kylie, Lillian, and Miriam all celebrate their birthdays today. Joan is 2 years younger than Kylie, Kylie is 3 years older than Lillian, and Miriam is one year older than Joan. Which of the following could be the combined age of all four women today?

A. 51
B. 52
C. 53
D. 54
E. 55

the problem is solved as soon as the data, given, are elegantly simplified

given
j = k -2
k = L + 3
m = j + 1

So

j = k-2 so far so good
k = L +3
L = k - 3 now manageable
m = j + 1
or m = k-2 + 1 putting the value of j
or m = k - 1

now we are in business

j + k + L + m
= k -2 + k + k-3 + k-1
=4k - 6

only D fits in
Re: Joan, Kylie, Lillian, and Miriam all celebrate their birthda &nbs [#permalink] 19 Oct 2017, 18:01

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