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Math Expert V
Joined: 02 Sep 2009
Posts: 57155
The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then what is the value of a^2 - c^2?

(1) a – b = 3
(2) (a + b)/(a - b)= 7

Kudos for a correct solution.

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The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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4
Answer is C

Statement 1 : a-b=3 so b=a-3 and b^2 = a^2+9-6a substituting in given eq we get a^2-c^2=17-9+6a

Statement 2 : a+b/a-b= 7 on solving and substituting in given equation we get c^2=9/16a^2-17 so not sufficient

(1)+(2)

On solving together we get two values of a = 12 b = 9 so a^2-C^2=144-64

so C

Got Confused while combining 1 and 2 first time so edited

Originally posted by RAHULSAINI on 23 Feb 2015, 03:58.
Last edited by RAHULSAINI on 23 Feb 2015, 04:24, edited 1 time in total.
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Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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Quote:
The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then what is the value of a^2 - c^2?

(1) a – b = 3
(2) (a + b)/(a - b)= 7

We are given that b^2 - c^2 = 17
We are asked to find out the value of (a^2 - c^2)

Statement 1 says (a - b) = 3
=> a = b + 3
=> a^2 - c^2 = b^2 + 6b + 9 - c^2 = 17 + 6b + 9
This statement alone is not sufficient.

Statement 2 says (a + b)/(a - b) = 7
=> a + b = 7a - 7b
=> 8b = 6a
=> a = 4b/3
=> a^2 - c^2 = 16b^2/9 - c^2 = 7b^2/9 +17
This statement alone is not sufficient.

When we combine the two we get
a = b + 3 and a = 4b/3
=> 4b/3 = b + 3
=> b = 9
=> a = b + 3 = 12
=> c = sqrt(b^2 - 17) = sqrt(81 - 17) = sqrt(64) = 8
=> a^2 - c^2 = 12^2 - 8^2 = 80.

So, we have found out the answer by using both the statements together.
Correct answer option : C
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Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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(a+b)(a-b)=17

both statements don't get us close to the value of C
in order to get value of c we have to know value of A or B for that we can use both statements and get value of A and B to get value of C

b^2-c^2=17>>(a-3)^2-c^2=17
and a+b=21
a=12,b=9
with this we get c=+-8 we know they are positive so 8

hence answer is C
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GMAT 1: 780 Q51 V46 Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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1
Hi ,
(b+c)(b-c)=17, what is the value of (a+c)(a-c)?
Statement 1 is insufficient:
This does not say anything c. Also a-b=3 will have many values for a and b.
Statement 2 is also insufficient:
This again does not say anything about c. Also (a + b)/(a - b)= 7 has many values.
Together sufficient:
(a+b)=21 and a-b=3
Solving we get a=12 and b=9
If we subs in b^2 - c^2 = 17 we get c^2=8.
So we know values of a and c.
So together sufficient.
Answer is C.
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GMAT 1: 710 Q49 V38 GMAT 2: 760 Q48 V47 Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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I think the answer should be D. b^2-c^2 = 17, leaves us with two b = 9 and C = 8. These are the only two possible positive numbers whose squares differ by 17.
a. Substituting for b, we get a, and hence sufficient.
b. Again substituting for b, we get the value for A. Sufficient.
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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Hi nphatak,

The prompt does NOT state that A, B and C are integers. Knowing that they might be non-integers, are there any OTHER values that you can come up with for B and C.....?

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Math Expert V
Joined: 02 Sep 2009
Posts: 57155
The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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Bunuel wrote:
The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then what is the value of a^2 - c^2?

(1) a – b = 3
(2) (a + b)/(a - b)= 7

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

Let X = a^2 - c^2, the value we are seeking. Notice if we subtract the first equation in the prompt from this equation, we get a^2 - b^2 = X – 17. In other words, if we could find the value of a^2 - b^2, then we could find the value of X.

Statement #1: a – b = 3

From this statement alone, we cannot calculate a^2 - b^2, so we can’t find the value of X. Statement #1, alone and by itself, is insufficient.

Statement #2: (a + b)/(a - b) = 7

From this statement alone, we cannot calculate a^2 - b^2, so we can’t find the value of X. Statement #2, alone and by itself, is insufficient.

Statements #1 & #2 combined: Now, if we know both statements are true, then we could multiple these two equations, which cancel the denominator, and result in the simple equation a + b = 21. Now, we have the numerical value of both (a – b) and (a + b), so from the difference of two squares formula, we can figure out a^2 - b^2, and if we know the numerical value of that, we can calculate X and answer the prompt. Combined, the statements are sufficient.

Answer = C
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Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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Hi All,

Given That a,b,c are positive, The only possible solution for
b^2 - c^2 = 17 is b = 9 and c = 8, (Correct me if i am wrong).
So either of the statements are suff. to get a value.
So I think the answer is D.
Please comment.

Thanks,
Math Expert V
Joined: 02 Sep 2009
Posts: 57155
Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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rmselva wrote:
Hi All,

Given That a,b,c are positive, The only possible solution for
b^2 - c^2 = 17 is b = 9 and c = 8, (Correct me if i am wrong).
So either of the statements are suff. to get a value.
So I think the answer is D.
Please comment.

Thanks,

Please read the whole thread: the-numbers-a-b-and-c-are-all-positive-if-b-2-c-2-17-then-wha-193643.html#p1491552
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Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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EMPOWERgmatRichC wrote:
Hi nphatak,

The prompt does NOT state that A, B and C are integers. Knowing that they might be non-integers, are there any OTHER values that you can come up with for B and C.....?

GMAT assassins aren't born, they're made,
Rich

Dear , can you I tried to find examples for A,B as non-integer numbers in form (A+B)(A-B)=17 but i did not found?

can you give me examples
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Math Expert V
Joined: 02 Aug 2009
Posts: 7755
The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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2
Dear , can you I tried to find examples for A,B as non-integer numbers in form (A+B)(A-B)=17 but i did not found?

can you give me examples

hi 23a2012
you can have lot of examples ...
let a+b=17/2.. a-b=2... so (a+b)(a-b)=2*17/2=17..
there are two numbers a and b and two eq, so u can find the answer..
a+b=17/2..
a-b=2 or a=b+2.
put the value in above equation..
b+2+b=17/2..
2*b=17/2 - 2 = 13/2.... so b = 13/4 and a= b+2=13/4+2=21/4....
similarly a+b=17/3.. a-b=3... and so on
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Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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chetan2u wrote:
Dear , can you I tried to find examples for A,B as non-integer numbers in form (A+B)(A-B)=17 but i did not found?

can you give me examples

hi 23a2012
you can have lot of examples ...
let a+b=17/2.. a-b=2... so (a+b)(a-b)=2*17/2=17..
there are two numbers a and b and two eq, so u can find the answer..
a+b=17/2..
a-b=2 or a=b+2.
put the value in above equation..
b+2+b=17/2..
2*b=17/2 - 2 = 13/2.... so b = 13/4 and a= b+2=13/4+2=21/4....
similarly a+b=17/3.. a-b=3... and so on

Dear, Thank you for clarification.
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Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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2
Bunuel wrote:
Bunuel wrote:
The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then what is the value of a^2 - c^2?

(1) a – b = 3
(2) (a + b)/(a - b)= 7

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

The numbers a, b, c, and d are all positive, and J = a/b and K = c/d. Is J greater than K?

Statement #1: a < b, c < d, and c > a

The first two inequalities tells us that both J and K are fractions less than one. From this and third inequality, we can draw no conclusion about which is bigger. For example, J = 9/10 and K = 11/50 satisfy all three conditions, and J > K, while J = 7/100 and K = 9/00 satisfy all three conditions, and J < K. This statement can produce either answer to the prompt question. Therefore, this statement, alone and by itself, is insufficient.

Statement #2: a + 7 = c and b + 4 = d

We need to forget everything associated with statement #1. Now we are focused purely on the issue of adding to the numerator and the denominator. Understanding this question depends very much on the concept discussed in that blog article. As you will see in that post, adding 7 to the numerator and 4 to the denominator will have the net effect of moving the fraction closer to 7/4 on the number line. If the original fraction is less than 7/4, for example, J = 1/2, then this addition, in moving the fraction closer to 7/4, makes the fraction bigger — K = 7/6 is greater than J = 1/2. BUT, if the original fraction is more than 7/4, for example, J = 25/4 = 6.25, then this addition, in moving the fraction closer to 7/4, makes the fraction smaller — K = 32/8 = 4 is less than J = 1/2. This statement can produce either answer to the prompt question. Therefore, this statement, alone and by itself, is insufficient.

Statements #1 & #2 combined:

The first statement tells us that J < 1 and K < 1. The second statement, in adding 7 to the numerator and 4 to the denominator will have the net effect of moving the K closer than J to 7/4 on the number line. Any fraction that is less than one gets bigger if it moves closer to 7/4. Therefore, J < K. Combined, the statements are sufficient.

Answer = C

Apoogies for finding a mistake
The solution you have provided doesn't match with the question stem.From where did the info about d came from
Math Expert V
Joined: 02 Sep 2009
Posts: 57155
Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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rhine29388 wrote:
Bunuel wrote:
Bunuel wrote:
The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then what is the value of a^2 - c^2?

(1) a – b = 3
(2) (a + b)/(a - b)= 7

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

The numbers a, b, c, and d are all positive, and J = a/b and K = c/d. Is J greater than K?

Statement #1: a < b, c < d, and c > a

The first two inequalities tells us that both J and K are fractions less than one. From this and third inequality, we can draw no conclusion about which is bigger. For example, J = 9/10 and K = 11/50 satisfy all three conditions, and J > K, while J = 7/100 and K = 9/00 satisfy all three conditions, and J < K. This statement can produce either answer to the prompt question. Therefore, this statement, alone and by itself, is insufficient.

Statement #2: a + 7 = c and b + 4 = d

We need to forget everything associated with statement #1. Now we are focused purely on the issue of adding to the numerator and the denominator. Understanding this question depends very much on the concept discussed in that blog article. As you will see in that post, adding 7 to the numerator and 4 to the denominator will have the net effect of moving the fraction closer to 7/4 on the number line. If the original fraction is less than 7/4, for example, J = 1/2, then this addition, in moving the fraction closer to 7/4, makes the fraction bigger — K = 7/6 is greater than J = 1/2. BUT, if the original fraction is more than 7/4, for example, J = 25/4 = 6.25, then this addition, in moving the fraction closer to 7/4, makes the fraction smaller — K = 32/8 = 4 is less than J = 1/2. This statement can produce either answer to the prompt question. Therefore, this statement, alone and by itself, is insufficient.

Statements #1 & #2 combined:

The first statement tells us that J < 1 and K < 1. The second statement, in adding 7 to the numerator and 4 to the denominator will have the net effect of moving the K closer than J to 7/4 on the number line. Any fraction that is less than one gets bigger if it moves closer to 7/4. Therefore, J < K. Combined, the statements are sufficient.

Answer = C

Apoogies for finding a mistake
The solution you have provided doesn't match with the question stem.From where did the info about d came from

Thank you for noticing this. Edited. +1.
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The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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Bunuel wrote:
Bunuel wrote:
The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then what is the value of a^2 - c^2?

(1) a – b = 3
(2) (a + b)/(a - b)= 7

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

Let X = a^2 - c^2, the value we are seeking. Notice if we subtract the first equation in the prompt from this equation, we get a^2 - b^2 = X – 17. In other words, if we could find the value of a^2 - b^2, then we could find the value of X.

Statement #1: a – b = 3

From this statement alone, we cannot calculate a^2 - b^2, so we can’t find the value of X. Statement #1, alone and by itself, is insufficient.

Statement #2: (a + b)/(a - b) = 7

From this statement alone, we cannot calculate a^2 - b^2, so we can’t find the value of X. Statement #2, alone and by itself, is insufficient.

Statements #1 & #2 combined: Now, if we know both statements are true, then we could multiple these two equations, which cancel the denominator, and result in the simple equation a + b = 21. Now, we have the numerical value of both (a – b) and (a + b), so from the difference of two squares formula, we can figure out a^2 - b^2, and if we know the numerical value of that, we can calculate X and answer the prompt. Combined, the statements are sufficient.

Answer = C

Hi Bunuel,
I don't know why I don't get the highlighted part above. I get the arithmetic behind it but fail to understand why we can do this. I am familiar with systems of equations where you add and subtract equations to eliminate terms. I have a feeling this is very simple but I am just not seeing it. Please help!
Thanks
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GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha  [#permalink]

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renept wrote:
Bunuel wrote:
Bunuel wrote:
The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then what is the value of a^2 - c^2?

(1) a – b = 3
(2) (a + b)/(a - b)= 7

Kudos for a correct solution.

MAGOOSH OFFICIAL SOLUTION:

Let X = a^2 - c^2, the value we are seeking. Notice if we subtract the first equation in the prompt from this equation, we get a^2 - b^2 = X – 17. In other words, if we could find the value of a^2 - b^2, then we could find the value of X.

Statement #1: a – b = 3

From this statement alone, we cannot calculate a^2 - b^2, so we can’t find the value of X. Statement #1, alone and by itself, is insufficient.

Statement #2: (a + b)/(a - b) = 7

From this statement alone, we cannot calculate a^2 - b^2, so we can’t find the value of X. Statement #2, alone and by itself, is insufficient.

Statements #1 & #2 combined: Now, if we know both statements are true, then we could multiple these two equations, which cancel the denominator, and result in the simple equation a + b = 21. Now, we have the numerical value of both (a – b) and (a + b), so from the difference of two squares formula, we can figure out a^2 - b^2, and if we know the numerical value of that, we can calculate X and answer the prompt. Combined, the statements are sufficient.

Answer = C

Hi Bunuel,
I don't know why I don't get the highlighted part above. I get the arithmetic behind it but fail to understand why we can do this. I am familiar with systems of equations where you add and subtract equations to eliminate terms. I have a feeling this is very simple but I am just not seeing it. Please help!
Thanks

Hi renept,

I think that you might find it easier to think of that math as 'substitution' instead of multiplication.

Fact 1 tells us that (a - b) = 3
Fact 2 gives us another equation: (a + b)/(a - b) = 7

Combining the two Facts, notice how (a - b) appears in both equations? We can 'substitute' the value of (a - b) into the second equation, which would give us...

(a + b)/3 = 7
(a + b) = 21

GMAT assassins aren't born, they're made,
Rich
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www.empowergmat.com/ Re: The numbers a, b, and c are all positive. If b^2 - c^2 = 17, then wha   [#permalink] 17 Feb 2019, 14:38
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