GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 21 Apr 2019, 07:22

### GMAT Club Daily Prep

#### Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

# Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (

Author Message
TAGS:

### Hide Tags

Manager
Status: Manager
Joined: 02 Nov 2018
Posts: 178
Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

11 Apr 2019, 13:32
3
00:00

Difficulty:

45% (medium)

Question Stats:

74% (02:13) correct 26% (01:46) wrong based on 19 sessions

### HideShow timer Statistics

Let a and b be relatively prime integers with a > b > 0 and $$\frac{a^3 −b^3}{(a−b)^3}$$ =$$\frac{73}{3}$$ What is a − b?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

_________________
Give a kudos if u find my post helpful. kudos motivates active discussions

Intern
Joined: 31 Aug 2016
Posts: 5
Location: India
Concentration: Finance, Economics
WE: Analyst (Consulting)
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

11 Apr 2019, 14:25
Can someone explain me how to actually solve this.
I got "C" because I assumed (a-b) should be an integer and multiple of 3 because of 3 in denominator of 73/3

Manager
Joined: 21 Feb 2019
Posts: 106
Location: Italy
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

11 Apr 2019, 15:23
1
vedavyas003 wrote:
Can someone explain me how to actually solve this.
I got "C" because I assumed (a-b) should be an integer and multiple of 3 because of 3 in denominator of 73/3

I think the "logic way" is the best here since algebraic method is too long to answer to the question in less than 2 minutes.

Just rewrite it as: $$\frac{(a-b)(a^2 + ab + b^2)}{{(a-b)^3}} = \frac{73}{3}$$.

Taking into account the value of the denominator (3), you should discard all options but C, since the other ones will never give a 3 when simplified with something above the fraction.

However, I'm interested if there is a more methodical approach enough fast to deal with.
_________________
If you like my post, Kudos are appreciated! Thank you.

MEMENTO AUDERE SEMPER
Intern
Joined: 04 Apr 2019
Posts: 2
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

15 Apr 2019, 18:27
Can someone please post solution to this problem?
Intern
Joined: 19 Oct 2018
Posts: 33
Location: India
GPA: 3.25
WE: Engineering (Other)
Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

15 Apr 2019, 19:34
2
(a-b) (a^2+b^2+ab)/(a-b) (a^2+b^2-2ab)=73/3
1+(3ab/a^2+b^2-2ab)=73/3
ab/(a-b)^2=70/9
as a and b are relatively prime, there will be no common factor between their difference and product
we can further find their values
a=10, b=7
Intern
Joined: 04 Apr 2019
Posts: 2
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

16 Apr 2019, 03:15
nick1816 wrote:
(a-b) (a^2+b^2+ab)/(a-b) (a^2+b^2-2ab)=73/3
1+(3ab/a^2+b^2-2ab)=73/3
ab/(a-b)^2=70/9
as a and b are relatively prime, there will be no common factor between their difference and product
we can further find their values
a=10, b=7

Thanks, but how did you get to ab=70?

Posted from my mobile device
Intern
Joined: 18 Jan 2019
Posts: 27
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

16 Apr 2019, 05:34
lucajava wrote:
vedavyas003 wrote:
Can someone explain me how to actually solve this.
I got "C" because I assumed (a-b) should be an integer and multiple of 3 because of 3 in denominator of 73/3

I think the "logic way" is the best here since algebraic method is too long to answer to the question in less than 2 minutes.

Just rewrite it as: $$\frac{(a-b)(a^2 + ab + b^2)}{{(a-b)^3}} = \frac{73}{3}$$.

Taking into account the value of the denominator (3), you should discard all options but C, since the other ones will never give a 3 when simplified with something above the fraction.

However, I'm interested if there is a more methodical approach enough fast to deal with.

can you please explain this with more detail?
Manager
Joined: 27 Oct 2017
Posts: 52
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

17 Apr 2019, 10:21
nick1816 wrote:
(a-b) (a^2+b^2+ab)/(a-b) (a^2+b^2-2ab)=73/3
1+(3ab/a^2+b^2-2ab)=73/3
ab/(a-b)^2=70/9
as a and b are relatively prime, there will be no common factor between their difference and product
we can further find their values
a=10, b=7

Can you please explain it in more detail?
Director
Joined: 13 Mar 2017
Posts: 722
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

17 Apr 2019, 10:37
nick1816 wrote:
(a-b) (a^2+b^2+ab)/(a-b) (a^2+b^2-2ab)=73/3
1+(3ab/a^2+b^2-2ab)=73/3
ab/(a-b)^2=70/9
as a and b are relatively prime, there will be no common factor between their difference and product
we can further find their values
a=10, b=7

Your solution is really awesome and well thought of.
_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.

MBA Social Network : WebMaggu

Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)

What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".
Director
Joined: 13 Mar 2017
Posts: 722
Location: India
Concentration: General Management, Entrepreneurship
GPA: 3.8
WE: Engineering (Energy and Utilities)
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

17 Apr 2019, 10:52
2
Let a and b be relatively prime integers with a > b > 0 and $$\frac{a^3 −b^3}{(a−b)^3}$$ =$$\frac{73}{3}$$ What is a − b?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

a and b are relatively prime and a>b>0.

$$\frac{a^3 −b^3}{(a−b)^3}$$ =$$\frac{73}{3}$$
$$\frac{(a-b)(a^2 +ab + b^2)}{(a−b)^3}$$ =$$\frac{73}{3}$$
Since a>b>0 So, a-b=/=0
$$\frac{(a^2 + ab + b^2)}{(a−b)^2}$$ =$$\frac{73}{3}$$
$$\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}$$ =$$\frac{73}{3}$$
$$\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}-1$$ =$$\frac{73}{3}-1$$
$$\frac{(3ab)}{(a^2 - 2ab + b^2)}$$ =$$\frac{70}{3}$$
$$\frac{(ab)}{(a-b)^2}$$ =$$\frac{70}{9}$$
$$\frac{(ab)}{(a-b)^2}$$ =$$\frac{70x^2}{(3x)^2}$$

so, a-b = 3x and ab = 70x^2
At x =1, a-b = 3, ab = 70 -> (3+b)b= 70 ->b^2 +3b -70 = 0 -> (b+10)(b-7) = 0 -> b= 7, a = 10 , a-b = 3
At x= 2, a-b = 6 , ab = 70*2^2, a= 20, b=14

and so on... But a and b are co-prime and hence a-b =3 , ab = 70, a = 10 & b=7.

_________________
CAT 2017 (98.95) & 2018 (98.91) : 99th percentiler
UPSC Aspirants : Get my app UPSC Important News Reader from Play store.

MBA Social Network : WebMaggu

Appreciate by Clicking +1 Kudos ( Lets be more generous friends.)

What I believe is : "Nothing is Impossible, Even Impossible says I'm Possible" : "Stay Hungry, Stay Foolish".
Manager
Joined: 27 Oct 2017
Posts: 52
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (  [#permalink]

### Show Tags

17 Apr 2019, 11:10
shashankism wrote:
Let a and b be relatively prime integers with a > b > 0 and $$\frac{a^3 −b^3}{(a−b)^3}$$ =$$\frac{73}{3}$$ What is a − b?

(A) 1

(B) 2

(C) 3

(D) 4

(E) 5

a and b are relatively prime and a>b>0.

$$\frac{a^3 −b^3}{(a−b)^3}$$ =$$\frac{73}{3}$$
$$\frac{(a-b)(a^2 +ab + b^2)}{(a−b)^3}$$ =$$\frac{73}{3}$$
Since a>b>0 So, a-b=/=0
$$\frac{(a^2 + ab + b^2)}{(a−b)^2}$$ =$$\frac{73}{3}$$
$$\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}$$ =$$\frac{73}{3}$$
$$\frac{(a^2 + ab + b^2)}{(a^2 - 2ab + b^2)}-1$$ =$$\frac{73}{3}-1$$
$$\frac{(3ab)}{(a^2 - 2ab + b^2)}$$ =$$\frac{70}{3}$$
$$\frac{(ab)}{(a-b)^2}$$ =$$\frac{70}{9}$$
$$\frac{(ab)}{(a-b)^2}$$ =$$\frac{70x^2}{(3x)^2}$$

so, a-b = 3x and ab = 70x^2
At x =1, a-b = 3, ab = 70 -> (3+b)b= 70 ->b^2 +3b -70 = 0 -> (b+10)(b-7) = 0 -> b= 7, a = 10 , a-b = 3
At x= 2, a-b = 6 , ab = 70*2^2, a= 20, b=14

and so on... But a and b are co-prime and hence a-b =3 , ab = 70, a = 10 & b=7.

Thanks a lot..

Posted from my mobile device
Re: Let a and b be relatively prime integers with a > b > 0 and a 3 −b 3 (   [#permalink] 17 Apr 2019, 11:10
Display posts from previous: Sort by