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Re: In three years, Janice will be three times as old as her daughter. Six
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15 Aug 2016, 06:50

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Bunuel wrote:

In three years, Janice will be three times as old as her daughter. Six years ago, her age was her daughter’s age squared. How old is Janice?

A. 18 B. 36 C. 40 D. 42 E. 45

msk0657's solution is great. Here's an approach that uses 1 variable:

Let J = Janice's PRESENT AGE So, 6 years ago, Janice's age WAS J - 6 Also, in 3 years, Janice's age WILL BE J + 3

Six years ago, Janice's age was her daughter’s age squared. 6 years ago, Janice's age WAS J - 6 So, 6 years ago, the DAUGHTER'S age was √(J - 6)

If the DAUGHTER'S age was √(J - 6)6 years ago, then to find her age 3 years from now, we must add 9 to her age 6 years ago. In other words, √(J - 6) + 9 = the DAUGHTER'S age in 3 years

In three years, Janice will be three times as old as her daughter √(J - 6) + 9 = the DAUGHTER'S age in 3 years J + 3 = Janice's age in 3 years At this point, Janice's age is 3 times her daughter's age. So, to make the two values EQUAL, we must multiply the smaller value (the daughter's age) by 3 We get: 3[√(J - 6) + 9] = J + 3 ......now solve for J Divide both sides by 3 to get: √(J - 6) + 9 = (J/3) + 1 Subtract 9 from both sides to get: √(J - 6) = (J/3) - 8

IMPORTANT: At this point, you might find it easier to plug in the answer choices and see which one satisfies the above equation... Notice that the correct value of J will yield an INTEGER value when plugged into √(J - 6) AND it will yield an INTEGER value when plugged into (J/3) We can quickly see that only answer choice D (J = 42) will yield integer values for √(J - 6) AND (J/3)

Re: In three years, Janice will be three times as old as her daughter. Six
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26 Aug 2017, 06:35

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Bunuel wrote:

In three years, Janice will be three times as old as her daughter. Six years ago, her age was her daughter’s age squared. How old is Janice?

A. 18 B. 36 C. 40 D. 42 E. 45

One of my students (username: aashaybaindur) just pointed out a super fast way to solve this question:

GIVEN: SIX YEARS AGO, Janice's age was her daughter’s age squared.

Take each answer choice (Janice's PRESENT age) and subtract 6 years to see which one(s) yield(s) the square of an integer. A. 18 - 6 = 12. Since 12 is NOT the square of an integer, ELIMINATE A B. 36 - 6 = 30. Since 30 is NOT the square of an integer, ELIMINATE B C. 40 - 6 = 34. Since 34 is NOT the square of an integer, ELIMINATE C D. 42 - 6 = 36. Since 36 IS the square of an integer, KEEP D E. 45 - 6 = 39. Since 39 is NOT the square of an integer, ELIMINATE E

Re: In three years, Janice will be three times as old as her daughter. Six
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01 Sep 2018, 10:07

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Let the daughter's age after 3 years be D ; The corresponding age of Janice = 3D 6 years ago (from the present) daughter's age = d-9 ; Janice's age = (d-9)^2

(d-9)^2 + 9 = 3d (d-6) (d-15) = 0 d = 6 or 15

Janice's present age = 3d - 3 = 15 or 42

15 is not present in the options, thus 42.

gmatclubot

Re: In three years, Janice will be three times as old as her daughter. Six &nbs
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01 Sep 2018, 10:07