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problem on ages

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05 Jul 2010, 11:03
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75% (02:54) correct 25% (00:00) wrong based on 5 sessions

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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)(5c/3)-10
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05 Jul 2010, 11:13
dineshboinapalli 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)(5c/3)-10

Were did you get this problem from?

I worked it out algabreically and got
$$A = \frac{(C+30)}{6}$$

I then tried to plug in numbers to make sure I didn't make any stupid mistakes.
If B = 20, then
A = 5+5 = 10
C = 30

I then plugged in these values and none of the answers were correct. As I suspected, my algebraic answer was correct
$$A = \frac{(C+30)}{6} = \frac{(30+30)}{6} = 10$$

I believe that these answer choices are all incorrect.
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05 Jul 2010, 11:16
Also, I believe this should be in the Quant forum under the Problem Solving sub-forum.

Mods - would you mind moving this to get better visibility?
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05 Jul 2010, 18:49
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)(5c/3)-10
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05 Jul 2010, 19:27
Hi,

we have to translate the English into algebraic expressions that reflect the same meaning.

Quote:
five years ago,Beth's age was three times that of Amy.

How old is Beth today?...Let's call it "B"

Then, how old was Beth five years ago?..."B-5"

Similarly, how old was Amy five years ago?..."A-5"

At this point in time, Beth's age was three times that of Amy. So:

B-5 = 3*(A-5) (1)

Similarly, this sentence:

Quote:
Ten years ago ,Beth's age was one half that of Chelsea.

yields:

B-10 = (C-10)/2 (2)

We want to solve for A in terms of C. So, we want to get rid of B. The easiest way is to equate the two equations. But the left-hand sides of each equation are different. So, just subtract 5 from both sides of (1) (OR add 5 to both sides of (2)):

B-5 -5 = 3*(A-5) - 5
B-10 = 3*(A-5) -5

Because B-10 = B-10, we have:

3*(A-5) - 5 = (C-10)/2

Solving for A, we have:

A = C/6 + 5

Choose A!
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05 Jul 2010, 19:35

The solution I got is equivalent to answer A.
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05 Jul 2010, 20:14
Hi dinesh,

did you double post this? Seems there's two threads on this same topic started by the same OP at teh same time! Anyways, I've just copied and pasted the explanation I gave in the other thread. Hope it helps!

we have to translate the English into algebraic expressions that reflect the same meaning.

Quote:
five years ago,Beth's age was three times that of Amy
.

How old is Beth today?...Let's call it "B"

Then, how old was Beth five years ago?..."B-5"

Similarly, how old was Amy five years ago?..."A-5"

At this point in time, Beth's age was three times that of Amy. So:

B-5 = 3*(A-5) (1)

Similarly, this sentence:

Quote:
Ten years ago ,Beth's age was one half that of Chelsea.

yields:

B-10 = (C-10)/2 (2)

We want to solve for A in terms of C. So, we want to get rid of B. The easiest way is to equate the two equations. But the left-hand sides of each equation are different. So, just subtract 5 from both sides of (1) (OR add 5 to both sides of (2)):

B-5 -5 = 3*(A-5) - 5
B-10 = 3*(A-5) -5

Because B-10 = B-10, we have:

3*(A-5) - 5 = (C-10)/2

Solving for A, we have:

A = C/6 + 5

Choose A!
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06 Jul 2010, 05:47
I started with Chelsey. Her age 10 years ago was C-10, so Beth's age 10 years ago was (c-10)/2. Next stage: Beth's age 5 years ago was (c-10)/2+5, and Amy's age 5 years ago was ((c-10)/2+5)/3. Now: Amy's age is ((c-10)/2+5)/3+5.
Now we should simplify the algebraic expression:
((c-10)/2+10/2)/3+5=c/6+5
Re: problem on ages   [#permalink] 06 Jul 2010, 05:47
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