Last visit was: 23 Apr 2026, 10:14 It is currently 23 Apr 2026, 10:14
Close
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
Your Progress

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
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Close
Request Expert Reply
Confirm Cancel
User avatar
monsoon1
Joined: 23 Sep 2008
Last visit: 05 Jun 2015
Posts: 22
Own Kudos:
167
 [117]
Given Kudos: 137
Posts: 22
Kudos: 167
 [117]
13
Kudos
Add Kudos
103
Bookmarks
Bookmark this Post
Most Helpful Reply
User avatar
Vips0000
User avatar
Current Student
Joined: 15 Sep 2012
Last visit: 02 Feb 2016
Posts: 521
Own Kudos:
1,313
 [42]
Given Kudos: 23
Status:Done with formalities.. and back..
Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE:Information Technology (Computer Software)
Products:
Schools: Olin - Wash U - Class of 2015
Posts: 521
Kudos: 1,313
 [42]
29
Kudos
Add Kudos
13
Bookmarks
Bookmark this Post
User avatar
mau5
User avatar
Verbal Forum Moderator
Joined: 10 Oct 2012
Last visit: 31 Dec 2024
Posts: 478
Own Kudos:
3,386
 [20]
Given Kudos: 141
Posts: 478
Kudos: 3,386
 [20]
10
Kudos
Add Kudos
10
Bookmarks
Bookmark this Post
General Discussion
User avatar
Vips0000
User avatar
Current Student
Joined: 15 Sep 2012
Last visit: 02 Feb 2016
Posts: 521
Own Kudos:
1,313
 [17]
Given Kudos: 23
Status:Done with formalities.. and back..
Location: India
Concentration: Strategy, General Management
Schools: Olin - Wash U - Class of 2015
WE:Information Technology (Computer Software)
Products:
Schools: Olin - Wash U - Class of 2015
Posts: 521
Kudos: 1,313
 [17]
8
Kudos
Add Kudos
9
Bookmarks
Bookmark this Post
monsoon1
If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4
(2) ab ≤ 15

given is c= ab/(a+b)
thus, since c is integer, ab/a+b must be an integer.

statement 1: b ≤ 4
No information can be drawn. Not sufficient
statement 2: ab ≤ 15
Only possible value of ab, such that ab/a+b is integer could be when a=2,b=2. Thus, c=1.
Sufficient.

Ans B it is.
User avatar
monsoon1
Joined: 23 Sep 2008
Last visit: 05 Jun 2015
Posts: 22
Own Kudos:
167
 [1]
Given Kudos: 137
Posts: 22
Kudos: 167
 [1]
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Vips0000
monsoon1
If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4
(2) ab ≤ 15

given is c= ab/(a+b)
thus, since c is integer, ab/a+b must be an integer.

statement 1: b ≤ 4
No information can be drawn. Not sufficient
statement 2: ab ≤ 15
Only possible value of ab, such that ab/a+b is integer could be when a=2,b=2. Thus, c=1.
Sufficient.

Ans B it is.


How did you find that only these values would satisfy?
Did you test several numbers?

The question also doesn't give us the clue whether the numbers are the same or different.So, we have to test many numbers.right?

Can you please show the steps or any other way to get to the correct answer?
avatar
Nilohit
Joined: 03 Sep 2013
Last visit: 06 Apr 2023
Posts: 8
Own Kudos:
2
 [1]
Given Kudos: 2
Posts: 8
Kudos: 2
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I have a question here. If the solution given by Vips0000 is correct and to me it seems perfect, then neither of the statements is actually needed to solve the question there can be only one combination possible from the question stem itself; a=b=2 and consequently c=1. In such a scenario shouldn't the answer be D, since both statements independently will lead us to the answer.
avatar
hsarci
Joined: 04 Oct 2013
Last visit: 25 May 2021
Posts: 2
Own Kudos:
Posts: 2
Kudos: 4
Kudos
Add Kudos
Bookmarks
Bookmark this Post
I can't see how you did the following:


Vips0000
monsoon1


we have\(c= ab/(a+b)\)
or \(c = \frac{1}{(1/a+1/b)}\)



Could you explain? thanks.


Edit: Think I got it now. Reciprocal of both sides of the original equation? or is there a different way?
avatar
hsarci
Joined: 04 Oct 2013
Last visit: 25 May 2021
Posts: 2
Own Kudos:
4
 [4]
Posts: 2
Kudos: 4
 [4]
3
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
Nilohit
I have a question here. If the solution given by Vips0000 is correct and to me it seems perfect, then neither of the statements is actually needed to solve the question there can be only one combination possible from the question stem itself; a=b=2 and consequently c=1. In such a scenario shouldn't the answer be D, since both statements independently will lead us to the answer.


Well, lets say the ab<or= 15 restriction was not there. Then if a=10 and b=10 then ab/(a+b) would equal 100/20...which is an integer.

The only possible values when ab<or=15 are 2 and 2.


but for the b<or=4 statement, you could still have a case where say b=4 and a=12, and ab/a+b=48/16=3. Then you would have 2 (or more) possible values for a and b if you also include the possible value set of "a=2 and b=2".


Vips0000 reasoning isn't actually perfect. In the case of \(\frac{1}{1/a+1/b}\) the denominator does not have to be "1". \(\frac{1}{1/a+1/b}\) could equal \(\frac{1}{1/4+1/12}\) which reduces to \(\frac{1}{4/12}\) then \(1*\frac{12}{4}\) which equals 3
User avatar
jlgdr
Joined: 06 Sep 2013
Last visit: 24 Jul 2015
Posts: 1,302
Own Kudos:
2,976
 [1]
Given Kudos: 355
Concentration: Finance
Posts: 1,302
Kudos: 2,976
 [1]
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
monsoon1
If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4
(2) ab ≤ 15

Updating with new solution by jlgdr

OK so we have that ab / a+b = C is an integer. Therefore let's hit the first statement.

Statement 1 says that b<=4. We have two choices here (actually 3). Let's begin with (2,2) C would equal 2. Now if we pick (4,4), C is again two so same answer. But if we pick (6,6) then C= 3. So not sufficient.

From statement 2 we know that ab<=15. Hence ab has to be 2,2 since both a,b are positive integers and of course C = 2.

Therefore B stands

Gimme kudos
Cheers
J
avatar
JepicPhail
Joined: 17 Oct 2013
Last visit: 27 Feb 2014
Posts: 4
Own Kudos:
Given Kudos: 104
Posts: 4
Kudos: 5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Doesn't mau5's solution, if correct, bring us back to Nilohit's point of 'why do we need know statement 1 and 2 if we already know C equals 1'? Regardless, this is definitely a challenging question. I would like to know if there is a shortcut to this problem, especially for statement #2. Just plugging in some numbers would be impossible given the time limit...
User avatar
mau5
User avatar
Verbal Forum Moderator
Joined: 10 Oct 2012
Last visit: 31 Dec 2024
Posts: 478
Own Kudos:
3,386
 [1]
Given Kudos: 141
Posts: 478
Kudos: 3,386
 [1]
Kudos
Add Kudos
Bookmarks
Bookmark this Post
JepicPhail
Doesn't mau5's solution, if correct, bring us back to Nilohit's point of 'why do we need know statement 1 and 2 if we already know C equals 1'? Regardless, this is definitely a challenging question. I would like to know if there is a shortcut to this problem, especially for statement #2. Just plugging in some numbers would be impossible given the time limit...

I don't understand how you can get the answer from the First fact statement.

Also, how can you get that c=1, without fact statement 2?
avatar
JepicPhail
Joined: 17 Oct 2013
Last visit: 27 Feb 2014
Posts: 4
Own Kudos:
Given Kudos: 104
Posts: 4
Kudos: 5
Kudos
Add Kudos
Bookmarks
Bookmark this Post
mau5
JepicPhail
Also, how can you get that c=1, without fact statement 2?

Oh, I see now why you need 15. \(sqrt(15)\) is less than 4, so C is less than or equal to something like 1/2, 2/2, 3/2, etc... and since only integer here is 1, C equals 1. :o
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,441
Own Kudos:
79,396
 [7]
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,441
Kudos: 79,396
 [7]
6
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
JepicPhail
Doesn't mau5's solution, if correct, bring us back to Nilohit's point of 'why do we need know statement 1 and 2 if we already know C equals 1'? Regardless, this is definitely a challenging question. I would like to know if there is a shortcut to this problem, especially for statement #2. Just plugging in some numbers would be impossible given the time limit...

Actually, this is not correct. c needn't be 1 in every case.

Take a = 2, b = 2. In this case c = 1

Take a = 4, b = 4. In this case,
1/4 + 1/4 = 1/2
c = 2

Take a = 3, b = 6. In this case,
1/3 + 1/6 = 1/2
c = 2

Take a = 15, b = 30. In this case
1/15 + 1/30 = 1/10
etc

Basically, you have to look for values such that when the numerators add up, the sum is divisible by the denominator. a and b cannot be 1 since we need the sum to be less than 1.

Statement 1: b <= 4
This gives you different values of c. c could be 1 or 2. Not sufficient.

Statement 2: ab <= 15
a and b could be 2 each. There is no other set of values. Try the small pairs (2, 4), (3, 3).
Hence (B) alone is sufficient.
(Note the algebraic solution provided by mau5 for statement 2.)
avatar
Rohan_Kanungo
Joined: 10 Dec 2013
Last visit: 07 Nov 2016
Posts: 14
Own Kudos:
Given Kudos: 7
Location: India
Concentration: Technology, Strategy
Schools: ISB '16 (S)
GMAT 1: 710 Q48 V38
GPA: 3.9
WE:Consulting (Consulting)
Schools: ISB '16 (S)
GMAT 1: 710 Q48 V38
Posts: 14
Kudos: 10
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Quote:
Statement 2: ab <= 15
a and b could be 2 each. There is no other set of values. Try the small pairs (2, 4), (3, 3).
Hence (B) alone is sufficient.

What if a = 1 and b =1, it would still satisfy all the conditions i.e. ab<15 and c would be an integer only. Doesnt this gives 2 solutions for statement 2 as well??
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,441
Own Kudos:
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,441
Kudos: 79,396
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Rohan_Kanungo
Quote:
Statement 2: ab <= 15
a and b could be 2 each. There is no other set of values. Try the small pairs (2, 4), (3, 3).
Hence (B) alone is sufficient.

What if a = 1 and b =1, it would still satisfy all the conditions i.e. ab<15 and c would be an integer only. Doesnt this gives 2 solutions for statement 2 as well??

If a = 1, b = 1,

1 + 1 = 1/c
c = 1/2
c is not an integer in this case.
User avatar
sharkr
Joined: 12 Jan 2017
Last visit: 04 Sep 2024
Posts: 8
Own Kudos:
Given Kudos: 27
Location: United States
GMAT 1: 740 Q50 V40
GMAT 1: 740 Q50 V40
Posts: 8
Kudos: 12
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma

Basically, you have to look for values such that when the numerators add up, the sum is divisible by the denominator. a and b cannot be 1 since we need the sum to be less than 1.

What are the numerators, the sum, and the denominator referring to?

Thanks in advance.
User avatar
KarishmaB
Joined: 16 Oct 2010
Last visit: 23 Apr 2026
Posts: 16,441
Own Kudos:
79,396
 [2]
Given Kudos: 484
Location: Pune, India
Expert
Expert reply
Active GMAT Club Expert! Tag them with @ followed by their username for a faster response.
Posts: 16,441
Kudos: 79,396
 [2]
1
Kudos
Add Kudos
1
Bookmarks
Bookmark this Post
sharkr
VeritasPrepKarishma

Basically, you have to look for values such that when the numerators add up, the sum is divisible by the denominator. a and b cannot be 1 since we need the sum to be less than 1.

What are the numerators, the sum, and the denominator referring to?

Thanks in advance.


1/a + 1/b = 1/c

You have to look for values of a and b (denominators) such that when you add 1/a and 1/b, the numerator you get is divisible by the denominator of the fraction (which you get after adding)

e.g. if a = 3, b = 6

1/a + 1/b = 1/3 + 1/6 = (2 + 1)/6 = 3/6
The numerator is 3 and the denominator is 6. The numerator is divisible by the denominator such that you get 1 in the numerator after cancelling.

1/a + 1/b = 1/3 + 1/6 = 1/2

When you get 1 in the numerator, the denominator of the resulting fraction is the value of c (i.e. 2 above)
User avatar
sharkr
Joined: 12 Jan 2017
Last visit: 04 Sep 2024
Posts: 8
Own Kudos:
Given Kudos: 27
Location: United States
GMAT 1: 740 Q50 V40
GMAT 1: 740 Q50 V40
Posts: 8
Kudos: 12
Kudos
Add Kudos
Bookmarks
Bookmark this Post
VeritasPrepKarishma
sharkr
VeritasPrepKarishma

Basically, you have to look for values such that when the numerators add up, the sum is divisible by the denominator. a and b cannot be 1 since we need the sum to be less than 1.

What are the numerators, the sum, and the denominator referring to?

Thanks in advance.


1/a + 1/b = 1/c

You have to look for values of a and b (denominators) such that when you add 1/a and 1/b, the numerator you get is divisible by the denominator of the fraction (which you get after adding)

e.g. if a = 3, b = 6

1/a + 1/b = 1/3 + 1/6 = (2 + 1)/6 = 3/6
The numerator is 3 and the denominator is 6. The numerator is divisible by the denominator such that you get 1 in the numerator after cancelling.

1/a + 1/b = 1/3 + 1/6 = 1/2

When you get 1 in the numerator, the denominator of the resulting fraction is the value of c (i.e. 2 above)

Thanks for the reply.
For clarification, don't you mean the denominator (ie. 6) is divisible by the numerator (ie. 3)?
3 / 6 = 0*6 + 3 whereas 6/3 = 3*2 (no remainder).
User avatar
BillyZ
User avatar
Current Student
Joined: 14 Nov 2016
Last visit: 24 Jan 2026
Posts: 1,135
Own Kudos:
22,610
 [3]
Given Kudos: 926
Location: Malaysia
Concentration: General Management, Strategy
GMAT 1: 750 Q51 V40 (Online)
GPA: 3.53
Products:
3
Kudos
Add Kudos
Bookmarks
Bookmark this Post
monsoon1
If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4
(2) ab ≤ 15

Source : Manhattan Advanced Quant Question No. 3

OFFICIAL SOLUTION



Since a, b, and c are positive integers, 1/a, 1/b, and 1/c are each less than or equal to 1. Also, 1/a and 1/b must both be less than 1/c, implying that a and b must both be greater than c. Furthermore, either 1/a or 1/b must be no less than ½ of 1/c, because if both fractions are less than ½ of 1/c, the sum will be less than 1/c , which implies that either a or b must be less than or equal to 2c.

These implications, along with the integer constraints and the given equation, greatly reduce the number of possible values for a, b, and c. We should make a comprehensive list for the first few c values. A good approach is to work backwards from the target value of c (= 1, 2, 3, etc.) and try to find integer values of a and b that fit the equation. There are only a few possibilities in each case. One pair that always works is making both a and b equal to 2c. Also, if we make a equal to c + 1, then there is always an integer value for b (which winds up equaling ac or c(c + 1), as we can show by a little algebra).

This question can be rephrased to “Which (a, b) pairs listed above are valid, and what is the resulting value of c?”

(1) INSUFFICIENT: If b ≤ 4, then valid (a, b) pairs are (2, 2) and (6, 3) and (4, 4) and (12, 4). This implies that c could be 1, 2, or 3.

(2) SUFFICIENT: If ab ≤ 15, then the only valid (a, b) pair is (2, 2) and c must be 1.

By the way, fractions of the form 1/integer are known as Egyptian fractions, because they were used first in ancient Egypt.

The correct answer is B.
Attachments

Untitled.jpg
Untitled.jpg [ 42.44 KiB | Viewed 51210 times ]

User avatar
KARISHMA315
Joined: 09 May 2016
Last visit: 01 Jan 2024
Posts: 51
Own Kudos:
Given Kudos: 12
Location: United States (NY)
Concentration: Finance, General Management
GMAT 1: 710 Q49 V38
WE:Information Technology (Consulting)
GMAT 1: 710 Q49 V38
Posts: 51
Kudos: 55
Kudos
Add Kudos
Bookmarks
Bookmark this Post
monsoon1
If a, b, and c are positive integers such that 1/a + 1/b = 1/c, what is the value of c?

(1) b ≤ 4
(2) ab ≤ 15



Is there any algebric approach to be sure that B is sufficient
 1   2   
Moderators:
Math Expert
109782 posts
498 posts
212 posts