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# Hunter was one fifth as old as Erica, x years ago. In x years, Erica

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Math Expert
Joined: 02 Sep 2009
Posts: 43377
Hunter was one fifth as old as Erica, x years ago. In x years, Erica [#permalink]

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17 Sep 2017, 02:24
00:00

Difficulty:

35% (medium)

Question Stats:

77% (02:38) correct 23% (02:15) wrong based on 43 sessions

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Hunter was one fifth as old as Erica, x years ago. In x years, Erica will be twice as old as Hunter. What is the ratio of Hunter's current age to Erica's current age?

A. 11:15
B. 3:7
C. 5:13
D. 9:17
E. 7:23
[Reveal] Spoiler: OA

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Hunter was one fifth as old as Erica, x years ago. In x years, Erica [#permalink]

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17 Sep 2017, 06:24
Let age of Erica after x years be E and Hunter's age be H.

From questions stem, E = 2H

Also,$$H - 2x = \frac{1}{5}(E - 2x) => 5H - 10x = E - 2x$$

Substituting the value of E, we will get $$3H = 8x => \frac{H}{x}= \frac{8}{3}$$

If H = 8, E = 16
2x(6) years ago, Hunter's age is 2 and that is one-fifth the age of Erica!

Since, we are asked for their current ages, it must be (8-3:16-3) or 5:13(Option C)
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Joined: 02 Jul 2017
Posts: 287
GMAT 1: 730 Q50 V38
Hunter was one fifth as old as Erica, x years ago. In x years, Erica [#permalink]

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17 Sep 2017, 10:10
Hunter was one fifth as old as Erica, x years ago.
=> $$(H-x)= \frac{1}{5}(E-x)$$
=> $$5H-5x= E-x$$
=> $$5H =E+4x$$

In x years, Erica will be twice as old as Hunter.
=> $$2(H+x)= E+x$$
=> $$2H+2x=E+x$$
=> $$2H=E-x$$

What is the ratio of Hunter's current age to Erica's current age: $$\frac{H}{E}$$????

=>$$5H =E+4x$$
=> $$x= \frac{5H-E}{4}$$

as $$2H=E-x$$ => $$2H=E-\frac{5H-E}{4}$$ => $$13H =5E$$
$$\frac{H}{E} = \frac{5}{13}$$

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Joined: 22 May 2016
Posts: 1260
Hunter was one fifth as old as Erica, x years ago. In x years, Erica [#permalink]

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19 Sep 2017, 17:17
Bunuel wrote:
Hunter was one fifth as old as Erica, x years ago. In x years, Erica will be twice as old as Hunter. What is the ratio of Hunter's current age to Erica's current age?

A. 11:15
B. 3:7
C. 5:13
D. 9:17
E. 7:23

x years ago, Hunter was 1/5 as old as Erica

$$H - x = \frac{1}{5}(E - x)$$
$$5(H - x) = E - x$$
$$5H - 5x = E - x$$
$$5H = E + 4x$$ ······ (A)

In x years, Erica will be twice as old as Hunter

$$2(H + x) = E + x$$
$$2H + 2x = E + x$$
$$2H = E - x$$ ······· (B)

Looking at (A) and (B) . . . Multiply (B) by 4 (to eliminate x). Add it to (A).

$$8H = 4E - 4x$$
$$5H = E + 4x$$
----------------------
$$13H = 5E$$

Ratio of Hunter's current age to Erica's current age?

$$13H = 5E$$

$$\frac{H}{E} = \frac{5}{13}$$

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Hunter was one fifth as old as Erica, x years ago. In x years, Erica   [#permalink] 19 Sep 2017, 17:17
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