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Five years ago, Beth's age was three times that of Amy. Ten years ago,

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Bunuel
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Five years ago, Beth's age was three times that of Amy. Ten years ago, [#permalink] New post  30 May 2019, 02:48
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Five years ago, Beth's age was three times that of Amy. Ten years ago, Beth's age was one half that of Chelsea. If C represents Chelsea's current age, which of the following represents Amy's current age?

(A) C/6 + 5
(B) 2C
(C) (C - 10)/3
(D) 3C - 5
(E) (5/3)C - 10
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Re: Five years ago, Beth's age was three times that of Amy. Ten years ago, [#permalink] New post  30 May 2019, 02:57
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Bunuel wrote:
Five years ago, Beth's age was three times that of Amy. Ten years ago, Beth's age was one half that of Chelsea. If C represents Chelsea's current age, which of the following represents Amy's current age?

(A) C/6 + 5
(B) 2C
(C) (C - 10)/3
(D) 3C - 5
(E) (5/3)C - 10


Let Amy's current age= A; Beth's=B; Chelsea's=C
Given,
B-5 = 3*(A-5)
=> B=3A-10 -- equation [1]

Also given,
B-10 = 1/2 * (C-10)
=> B=C/2 +5 -- equation [2]

From equation [1] & [2],
3A-10= C/2 +5
=> A= C/6 +5

Hence, option A.
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Five years ago, Beth's age was three times that of Amy. Ten years ago, [#permalink] New post  30 May 2019, 03:27
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Bunuel wrote:
Five years ago, Beth's age was three times that of Amy. Ten years ago, Beth's age was one half that of Chelsea. If C represents Chelsea's current age, which of the following represents Amy's current age?

(A) C/6 + 5
(B) 2C
(C) (C - 10)/3
(D) 3C - 5
(E) (5/3)C - 10


Let the age of Amy 5 years ago = x
Current Age of Chelsea = C


Ten years ago, Beth's age was one half that of Chelsea.
--> (3x - 5) = 1/2(C - 10)
--> 3x - 5 = C/2 - 5
--> X = C/6

Current Age of Amy = X + 5 = C/6 + 5

IMO Option A

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Re: Five years ago, Beth's age was three times that of Amy. Ten years ago, [#permalink] New post  30 May 2019, 11:10
Let Beth`s age be B years. Let Amy`s age be A years. Chelsea`s age C years. 5 years ago : B-5 = 3(A-5). or, B-5 = 3A-15
or, B= 3A-10 -----(1). Ten years ago : B-10 = 1/2(C-10) ---(2) . Substituting B=3A-10 in equation (2) we get 3A -10-10 = 1/2(C-10). or, 3A-20 = 1/2(C-10). or, 6A-40 = C-10. or, 6A= C+30. or A = C/6+5. Ans: A
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Re: Five years ago, Beth's age was three times that of Amy. Ten years ago, [#permalink] New post  02 Jun 2019, 17:16
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Bunuel wrote:
Five years ago, Beth's age was three times that of Amy. Ten years ago, Beth's age was one half that of Chelsea. If C represents Chelsea's current age, which of the following represents Amy's current age?

(A) C/6 + 5
(B) 2C
(C) (C - 10)/3
(D) 3C - 5
(E) (5/3)C - 10


Let’s let a, b, and c represent the current ages of Amy, Beth, and Chelsea, respectively. We can create the equations:

b - 5 = 3(a - 5)

b - 5 = 3a - 15

b = 3a - 10

and

b - 10 = (1/2)(c - 10)

2b - 20 = c - 10

2b - 10 = c

Substituting, we have:

2(3a - 10) - 10 = c

6a - 20 - 10 = c

6a - 30 = c

a - 5 = c/6

a = c/6 + 5

Answer: A
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Re: Five years ago, Beth's age was three times that of Amy. Ten years ago, [#permalink]
02 Jun 2019, 17:16
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