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Which of the following is equivalent to 1 + 3/(b+2)

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Which of the following is equivalent to 1 + 3/(b+2)  [#permalink]

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New post Updated on: 29 Aug 2018, 15:28
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Which of the following is equivalent to \(\frac{1+\frac{3}{(b+2)}}{1 + \frac{7}{(b-2)}}\) for all values of \(b\) for which the expression is defined?

A) \(\frac{b-2}{b+3}\)

B) \(\frac{b-1}{b+1}\)

C) \(\frac{b+3}{b-1}\)

D) \(\frac{b-2}{b+2}\)

E) \(\frac{b-3}{b-2}\)


I have a problem from Target Test prep. I cant wrap my head around this. It asks me to what is equivalent for all values of b for which the expression is defined. Thank you big time in advance.

Originally posted by raycal84 on 29 Aug 2018, 13:58.
Last edited by generis on 29 Aug 2018, 15:28, edited 1 time in total.
Edited and formatted the question, renamed the topic, added answer choices
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Re: Which of the following is equivalent to 1 + 3/(b+2)  [#permalink]

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New post 29 Aug 2018, 14:48
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raycal84 wrote:
I have a problem from Target Test prep. I cant wrap my head around this. It asks me to what is equivalent for all values of b for which the expression is defined. Thank you big time in advance.


1+ 3/B+2
_________

1+7/b-2


Can you share the answer choices with us?
It's quite possible that the best (fastest) solution can be found by testing the answer choices.

Cheers,
Brent
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Re: Which of the following is equivalent to 1 + 3/(b+2)  [#permalink]

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New post 29 Aug 2018, 14:56
Please see attached.
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Re: Which of the following is equivalent to 1 + 3/(b+2)  [#permalink]

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New post 29 Aug 2018, 15:06
1
raycal84 wrote:
I have a problem from Target Test prep. I cant wrap my head around this. It asks me to what is equivalent for all values of b for which the expression is defined. Thank you big time in advance.


1+ 3/B+2
_________

1+7/b-2



Numerator of the given expression:-
\(1+\frac{3}{b+2}\) can be simplified by LCM method,
=\(\frac{b+2+3}{b+2}\)
=\(\frac{b+5}{b+2}\)

Denominator of the given expression:-
\(1+\frac{7}{b-2}\) can be simplified by LCM method,
=\(\frac{b-2+7}{b-2}\)
=\(\frac{b+5}{b-2}\)

Now , given expression=\(\frac{Numerator}{Denominator}\)=\(\frac{\frac{b+5}{b+2}}{\frac{b+5}{b-2}}\)

=\(\frac{b+5}{b+2}*\frac{b-2}{b+5}\)
(b+5) is cancelled out.
=\(\frac{b-2}{b+2}\)

Ans. (D)
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Re: Which of the following is equivalent to 1 + 3/(b+2)  [#permalink]

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New post 29 Aug 2018, 15:19
1
PKN wrote:
raycal84 wrote:
I have a problem from Target Test prep. I cant wrap my head around this. It asks me to what is equivalent for all values of b for which the expression is defined. Thank you big time in advance.


1+ 3/B+2
_________

1+7/b-2



Numerator of the given expression:-
\(1+\frac{3}{b+2}\) can be simplified by LCM method,
=\(\frac{b+2+3}{b+2}\)
=\(\frac{b+5}{b+2}\)

Denominator of the given expression:-
\(1+\frac{7}{b-2}\) can be simplified by LCM method,
=\(\frac{b-2+7}{b-2}\)
=\(\frac{b+5}{b-2}\)

Now , given expression=\(\frac{Numerator}{Denominator}\)=\(\frac{\frac{b+5}{b+2}}{\frac{b+5}{b-2}}\)

=\(\frac{b+5}{b+2}*\frac{b-2}{b+5}\)
(b+5) is cancelled out.
=\(\frac{b-2}{b+2}\)

Ans. (D)


ahhh. Thank you so much. I got to the second to last part on my own, but i multiplied the "b" and 5 and -2.

So i ended up with 2b-10 over 2b+10. So why would that be the answer?i mean I see why yours is right, but why couldnt we multiply the Bs?
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Re: Which of the following is equivalent to 1 + 3/(b+2)  [#permalink]

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New post 29 Aug 2018, 15:20
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raycal84 wrote:
Please see attached.


Another approach is to test a value of b

We're looking for an expression that is equivalent to the given expression.
So, if we find the value of the given expression for a certain value of b, then the correct answer choice must also equal the same value for the same value of b.

So, let's see what happens when b = 3

Given expression: [1 + 3/(b + 2)]/[1 + 7/(b - 2)] = [1 + 3/(3 + 2)]/[1 + 7/(3 - 2)]
= [1 + 3/5]/[1 + 7/1]
= [8/5]/[8]
= 8/5 x 1/8
= 1/5

So, when b = 3, the given expression evaluates to be 1/5
So, the correct answer choice must also evaluate to be 1/5 when we plug in b = 3

Check the answer choices....
A) (b - 2)/(b + 3) = (3 - 2)/(3 + 3) = 1/6. No good. We want 1/5. ELIMINATE.
B) (b - 1)/(b + 1) = (3 - 1)/(3 + 1) = 2/4 = 1/2. No good. We want 1/5. ELIMINATE.
C) (b + 3)/(b - 1) = (3 + 3)/(3 - 1) = 6/2 = 3. No good. We want 1/5. ELIMINATE.
D) (b - 2)/(b + 2) = (3 - 2)/(3 + 2) = 1/5. GREAT!!! Keep
E) (b - 3)/(b - 2) = (3 - 3)/(3 - 2) = 0/1. No good. We want 1/5. ELIMINATE.

Answer: D

Cheers,
Brent
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Re: Target Test Prep Quiz Problem.  [#permalink]

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New post 29 Aug 2018, 15:31
raycal84 wrote:
PKN wrote:
raycal84 wrote:
I have a problem from Target Test prep. I cant wrap my head around this. It asks me to what is equivalent for all values of b for which the expression is defined. Thank you big time in advance.


1+ 3/B+2
_________

1+7/b-2



Numerator of the given expression:-
\(1+\frac{3}{b+2}\) can be simplified by LCM method,
=\(\frac{b+2+3}{b+2}\)
=\(\frac{b+5}{b+2}\)

Denominator of the given expression:-
\(1+\frac{7}{b-2}\) can be simplified by LCM method,
=\(\frac{b-2+7}{b-2}\)
=\(\frac{b+5}{b-2}\)

Now , given expression=\(\frac{Numerator}{Denominator}\)=\(\frac{\frac{b+5}{b+2}}{\frac{b+5}{b-2}}\)

=\(\frac{b+5}{b+2}*\frac{b-2}{b+5}\)
(b+5) is cancelled out.
=\(\frac{b-2}{b+2}\)

Ans. (D)


ahhh. Thank you so much. I got to the second to last part on my own, but i multiplied the "b" and 5 and -2.

So i ended up with 2b-10 over 2b+10. So why would that be the answer?i mean I see why yours is right, but why couldnt we multiply the Bs?


Hi raycal84,

You have to see the answer options(especially the format of answer choices). Moreover, in GMAT, answer options are always in reduced factor form. ( I mean 10/12 would be written as 5/6 etc.)

I am not sure how you have arrived at \(\frac{2b-10}{2b+10}\)

Though multiplication is not required(since same terms in numerator and denominator get cancelled out), let's multiply.
1) \((b+5)(b-2)=b*b-b*2+5*b-5*2=b^2-2b+5b-10=b^2+3b-10\)
2) \((b-5)(b+2)=b*b+b*2-5*b-5*2=b^2+2b-5b-10=b^2-3b-10\)

Hope it helps.
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Which of the following is equivalent to 1 + 3/(b+2)  [#permalink]

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New post 29 Aug 2018, 15:32
1
raycal84 wrote:
Which of the following is equivalent to \(\frac{1+\frac{3}{(b+2)}}{1 + \frac{7}{(b-2)}}\) for all values of \(b\) for which the expression is defined?

A) \(\frac{b-2}{b+3}\)

B) \(\frac{b-1}{b+1}\)

C) \(\frac{b+3}{b-1}\)

D) \(\frac{b-2}{b+2}\)

E) \(\frac{b-3}{b-2}\)


I have a problem from Target Test prep. I cant wrap my head around this. It asks me to what is equivalent for all values of b for which the expression is defined. Thank you big time in advance.

Calculate the numerator and denominator separately.

NUMERATOR

\(1 + \frac{3}{(b+2)}= (\frac{1}{1} + \frac{3}{(b+2)})\)

Common denominator: \((b+2)\)

\((\frac{b+2}{b+2})*1 + \frac{3}{b+2} = (\frac{b+2}{b+2} + \frac{3}{b+2}) =\frac{b+2+3}{b+2}=\frac{b+5}{b+2}\)

DENOMINATOR - common denominator: \((b-2)\)

\((\frac{b-2}{b-2})*1+\frac{7}{b-2}=(\frac{b-2}{b-2} + \frac{7}{b-2})=(\frac{b-2+7}{b-2})=\frac{b+5}{b-2}\)

Put them together

\(\frac{\frac{b+5}{b+2}}{\frac{b+5}{b-2}}= (\frac{b+5}{b+2}*\frac{b-2}{b+5})=\frac{b-2}{b+2}\)

(The (\(b+5\)) factors out)

Answer D

P.S. Many people use a shortcut for adding TWO fractions. (In other words, I did not solve as I demonstrated.) I have no idea why many people are not taught the shortcut. I was not. At times the shortcut saves 20 seconds. I am not worried that I will forget how to find the LCM (that issue is raised on the site below by a commentator).
You can find the shortcut here.
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Re: Which of the following is equivalent to 1 + 3/(b+2)  [#permalink]

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New post 30 Aug 2018, 15:45
Ok i see my mistake on the last question, but I still keep running into this problem. Work attached. What am i missing?
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Re: Which of the following is equivalent to 1 + 3/(b+2)  [#permalink]

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New post 30 Aug 2018, 15:46
raycal84 wrote:
Ok i see my mistake on the last question, but I still keep running into this problem. Work attached. What am i missing?


Work is now attached.
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Which of the following is equivalent to 1 + 3/(b+2)  [#permalink]

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New post 30 Aug 2018, 19:34
raycal84 wrote:
Ok i see my mistake on the last question, but I still keep running into this problem. Work attached. What am i missing?

raycal84

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