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If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.

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If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post Updated on: 05 Oct 2013, 05:59
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If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7=

A. 1/a+ 1/8
B. 1/8 - 1/a
C. 1/8-a
D. 3/7(8 – a)
E. 9/49 (8 – a)

Originally posted by gorden on 11 Nov 2008, 17:28.
Last edited by Bunuel on 05 Oct 2013, 05:59, edited 1 time in total.
Edited the question, added the OA and moved to PS forum.
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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 05 Oct 2013, 06:03
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Mansi12 wrote:
Hi,

Can anyone help me with this one. I got stuck at -
x=(24-3a)/7
=> x= 3(8-a)/7

how do we get to 3x/7 ?


If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7=

A. 1/a+ 1/8
B. 1/8 - 1/a
C. 1/8-a
D. 3/7(8 – a)
E. 9/49 (8 – a)

\(\frac{7x}{3} + a = 8\) --> \(\frac{7x}{3}= 8-a\).

Now, multiply both sides by \(\frac{3}{7}*\frac{3}{7}\) in order to get \(\frac{3x}{7}\) on the LHS:

\(\frac{3}{7}*\frac{3}{7}*\frac{7x}{3}=\frac{3}{7}*\frac{3}{7}( 8-a)\).

\(\frac{3}{7}*x=\frac{9}{49}( 8-a)\).

Answer: E.

Hope it helps.
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Re: a 8---27  [#permalink]

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New post 11 Nov 2008, 19:01
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gorden wrote:
If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7=

A. 1/a+ 1/8
B. 1/8 - 1/a
C. 1/8-a
D. 3/7(8 – a)
E. 9/49 (8 – a)


E.
assuming that the original eqn (7x/3) + a = 8

7x + 3a = 24
7x = 24 -3a

x = (24 - 3a )/7 = 3(8 -a)/7

i.e 3x/7 = 9(8-a)/49
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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 05 Oct 2013, 05:29
Hi,

Can anyone help me with this one. I got stuck at -
x=(24-3a)/7
=> x= 3(8-a)/7

how do we get to 3x/7 ?
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If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 30 Nov 2014, 17:15
Exercise raises 7x/3 + a = 8 and a > 8

And express 3x/7 according to request.

One suggestion is to form 3x/7, from equal 7x/3 + a = 8

In the last ogualdad, clearing 7x/3, is:

7x/3 = 8 - a

Here we can form in the ladi left 3x/7 equality, simply multiply the ogualdad by 9/49, so must

(9/49)(7x/3) = (9/49) (8 - a)

Simplifying on the left side of the equality is:

7x/3 = (9/49) (8 - a)

It is requested.

Answer E

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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 30 Nov 2014, 23:18
gorden wrote:
If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7=

A. 1/a+ 1/8
B. 1/8 - 1/a
C. 1/8-a
D. 3/7(8 – a)
E. 9/49 (8 – a)


What's the need to specifically say a > 8 ? Would it make any difference?
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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 31 Jan 2015, 00:17
1
Bunuel wrote:
Mansi12 wrote:
Hi,

Can anyone help me with this one. I got stuck at -
x=(24-3a)/7
=> x= 3(8-a)/7

how do we get to 3x/7 ?


If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7=

A. 1/a+ 1/8
B. 1/8 - 1/a
C. 1/8-a
D. 3/7(8 – a)
E. 9/49 (8 – a)

\(\frac{7x}{3} + a = 8\) --> \(\frac{7x}{3}= 8-a\).

Now, multiply both sides by \(\frac{3}{7}*\frac{3}{7}\) in order to get \(\frac{3x}{7}\) on the LHS:

\(\frac{3}{7}*\frac{3}{7}*\frac{7x}{3}=\frac{3}{7}*\frac{3}{7}( 8-a)\).

\(\frac{3}{7}*x=\frac{9}{49}( 8-a)\).

Answer: E.

Hope it helps.


Hi,

Thank you.

After 7x/3 = 8 - a, why can we not take the reciprocal of both sides?

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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 31 Jan 2015, 05:28
1
1
thorinoakenshield wrote:
Bunuel wrote:
Mansi12 wrote:
Hi,

Can anyone help me with this one. I got stuck at -
x=(24-3a)/7
=> x= 3(8-a)/7

how do we get to 3x/7 ?


If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7=

A. 1/a+ 1/8
B. 1/8 - 1/a
C. 1/8-a
D. 3/7(8 – a)
E. 9/49 (8 – a)

\(\frac{7x}{3} + a = 8\) --> \(\frac{7x}{3}= 8-a\).

Now, multiply both sides by \(\frac{3}{7}*\frac{3}{7}\) in order to get \(\frac{3x}{7}\) on the LHS:

\(\frac{3}{7}*\frac{3}{7}*\frac{7x}{3}=\frac{3}{7}*\frac{3}{7}( 8-a)\).

\(\frac{3}{7}*x=\frac{9}{49}( 8-a)\).

Answer: E.

Hope it helps.


Hi,

Thank you.

After 7x/3 = 8 - a, why can we not take the reciprocal of both sides?

TO


You mean why don't we write 3/(7x) = 1/(8-a) ? What does it give?
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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 31 Jan 2015, 14:01
1
Hi All,

This question can be solved by TESTing VALUES, although it does come with a "twist" that will require that you think about which type of values would make the math easy to deal with.

We're told that 7X/3 + A = 8 and that A > 8. We're asked for the value of 3X/7.

Since A > 8......7X/3 will have to be NEGATIVE.

To eliminate the fraction, I'm going to TEST...
X = -3

7(-3)/3 = -7
-7 + A = 8
A = 15

So, with X = -3 and A = 15, the answer to the question is 3(-3)/7 = -9/7

Answer A: 1/15 + 1/8 is positive. NOT a match
Answer B: 1/8 - 1/15 is positive. NOT a match
Answer C: 1/8 - 15 is TOO negative. NOT a match
Answer D: (3/7)(8-15) = (3/7)(-7) = -3 NOT a match
Answer E: (9/49)(8-15) = (9/49)(-7) = -9/7 This IS a MATCH

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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 31 Jan 2015, 16:47
1
Hi,

Thank you.

After 7x/3 = 8 - a, why can we not take the reciprocal of both sides?

TO


Bunuel: You mean why don't we write 3/(7x) = 1/(8-a) ? What does it give?

Yes. Will this be mathematically illegal? I am sorry if the question sounds very basic.

TO

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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 01 Feb 2015, 04:17
thorinoakenshield wrote:
Hi,

Thank you.

After 7x/3 = 8 - a, why can we not take the reciprocal of both sides?

TO


Bunuel: You mean why don't we write 3/(7x) = 1/(8-a) ? What does it give?

Yes. Will this be mathematically illegal? I am sorry if the question sounds very basic.

TO



The question asks to find the value of 3x/7 in terms of a, not the value of 3/(7x). You should read questions more carefully. Also, I think you need to brush up fundamentals and only then to attempt questions. Here are the books you might find helpful: best-gmat-math-prep-books-reviews-recommendations-77291.html
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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 15 Nov 2017, 11:54
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Please spend some time to write the answer choices correctly.

9/49 (8 – a) can be misread as \(\frac{9}{49(8-a)}\) and is not the same as \(\frac{9}{49} * (8-a)\).
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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.  [#permalink]

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New post 10 Feb 2019, 10:26
gorden wrote:
If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7=

A. 1/a+ 1/8
B. 1/8 - 1/a
C. 1/8-a
D. 3/7(8 – a)
E. 9/49 (8 – a)


7x/3 + a = 8
7x + 3a = 24

a > 8, a = 9

7x = -3
x = -3/7

3x/7 = -9/49 -> target value

We can get this from only E
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Re: If 7x/3 + a = 8 and a > 8, then, in terms of a, 3x/7= A.   [#permalink] 10 Feb 2019, 10:26
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