Last visit was: 19 Nov 2025, 09:06 It is currently 19 Nov 2025, 09:06
Home
GMAT
Messages
Tests
Schools
Events
Rewards
Chat
My Profile
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.
Go to My Error Log Learn more
Close
Request Expert Reply
Confirm Cancel
Back to Forum
Create Topic
   1   2 
555-605 Level|   Algebra|   Word Problems|         
User avatar
kornn
User avatar
Joined: 28 Jan 2017
Last visit: 18 Dec 2021
Posts: 357
Own Kudos:
93
Given Kudos: 832
Posts: 357
Kudos: 93
Martha bought several pencils. If each pencil was either a 06 Jul 2020, 05:03
Kudos
Add Kudos
Bookmarks
Bookmark this Post
IanStewart
Bull78
Martha bought several pencils. If each pencil was either a 23 cent pencil or a 21 pencil, how many 23 cent pencils did martha buy?

(1) Martha bought a total of 6 pencils

(2) The total value of the pencils Martha bought was 130 cents.

You can also look at this as follows: if she bought 7 pencils, she would spend at least 7*0.21 = $1.49. If she bought 5 pencils, she would spend at most 5*0.23 = $1.15. So if she spent $1.30, as we learn from Statement 2 alone, she must have bought exactly 6 pencils. So Statement 2 is sufficient alone (once we know that she bought 6 pencils, there can only be one combination of $0.23 and $0.21 pencils that give a total cost of $1.30, since the more $0.23 pencils she buys, the greater the cost will be).
Show more
Dear IanStewart,

Did you go through the process of trial and error there?
Are there any quicker way?
User avatar
IanStewart
User avatar
GMAT Tutor
Joined: 24 Jun 2008
Last visit: 18 Nov 2025
Posts: 4,145
Own Kudos:
10,988
 [2]
Given Kudos: 99
Expert
Expert reply
Posts: 4,145
Kudos: 10,988
 [2]
Martha bought several pencils. If each pencil was either a 06 Jul 2020, 05:10
2
Kudos
Add Kudos
Bookmarks
Bookmark this Post
varotkorn

Did you go through the process of trial and error there?
Are there any quicker way?
Show more

It's DS, so there's no reason to find the actual answer to the question - you only need to know that there is one and only one answer. If she bought six pencils, some of which cost 23 cents, some of which cost 21 cents, only one combination can possibly cost $1.30, because each possible combination will have a different total price (if you replace a 21 cent pencil with a 23 cent one, the cost goes up).
User avatar
GMATGuruNY
User avatar
Joined: 04 Aug 2010
Last visit: 18 Nov 2025
Posts: 1,344
Own Kudos:
3,796
 [1]
Given Kudos: 9
Schools:Dartmouth College
Expert
Expert reply
Posts: 1,344
Kudos: 3,796
 [1]
Martha bought several pencils. If each pencil was either a 04 Feb 2021, 08:25
1
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Mck2023
how can i work it with speed?
Show more

Consider the following equation:
2x + 3y = 30.

If x and y are nonnegative integers, the following solutions are possible:
x=15, y=0
x=12, y=2
x=9, y=4
x=6, y=6
x=3, y=8
x=0, y=10

Notice the following:
The value of x changes in increments of 3 (the coefficient for y).
The value of y changes in increments of 2 (the coefficient for x).
This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers.

Onto the problem at hand:

Quote:
Martha bought several pencils. If each pencil was either a 23 cent pencil or a 21 pencil, how many 23 cent pencils did martha buy?

(1) Martha bought a total of 6 pencils

(2) The total value of the pencils Martha bought was 130 cents.
Show more

Statement 1 is clearly INSUFFICIENT.

Statement 2 implies the following:
23x + 21y = 130

Clearly, this equation must yield at least one viable integer combination for x and y.
As noted above:
Once a viable option for x has been determined, the value of x may change only in increments of 21 (the coefficient for y).
BUT SUCH A CHANGE IS NOT POSSIBLE.
If the value of x increases by 21, then the value of 23x will exceed the sum of 130.
If the value of x decreases by 21, then the value of 23x will be negative.
Implication:
ONLY ONE INTEGER COMBINATION for x and y will satisfy the equation.
SUFFICIENT.

Show Spoiler
B

Using this line of reasoning, we can determine the correct answer with virtually no work at all.
User avatar
AnishPassi
User avatar
Joined: 16 Jul 2014
Last visit: 15 Nov 2025
Posts: 112
Own Kudos:
661
Given Kudos: 18
Status:GMAT Coach
Affiliations: The GMAT Co.
Concentration: Strategy
Schools: IIMA  (A)
GMAT 1: 760 Q50 V41
Expert
Expert reply
Schools: IIMA  (A)
GMAT 1: 760 Q50 V41
Posts: 112
Kudos: 661
Martha bought several pencils. If each pencil was either a 05 Feb 2022, 09:33
Kudos
Add Kudos
Bookmarks
Bookmark this Post
shash
How can this problem be solved without 2 equations. How do you in fact spot such cases?
Show more

Actually the second statement also gives us two equations. Albeit, one is not explicit.

The equation that be derived clearly from the second statement is:
23x + 21y = 130

Now, how do we have another equation here?

As IanStewart has also explained,

If the pencils cost either 21c or 23c, and the total value of the pencils was 130 cents, the total pencils had to be six. Had to be.

Here's why:

1. What's the minimum # of pencils she must have bought?
I'm thinking in terms of the expensive pencils.
I'll start with a relaxed number.
e.g. 23 x 3 = 69. Less than 130. Too few.
23 x 4 = 92. Less than 130. Too few.
23 x 5 = 115. Less than 130. Too few.
23 x 6 > 130. Wait. That's too many.
Even if all the pencils were the expensive ones and she bought five pencils, her total would have been less than 130. So, she would have bought more than 5 pencils to get a total of 130 cents.


2. What's the maximum # of pencils she must have bought?
I'm thinking in terms of the cheaper pencils now.
21 x 10 = 210. Greater than 130. Too many.
21 x 7 = 140 + 7. Greater than 130. Too many.
21 x 6 = 126. Ok, that's too few.
Even if all the pencils were the cheaper ones and she bought seven pencils, her total would have been greater than 130. So, she would have bought fewer than 7 pencils to get a total of 130 cents.

Combining these two, we know for sure that she bought six pencils total.

So, we have our second equation now:
x + y = 6
User avatar
brains
User avatar
Joined: 30 May 2017
Last visit: 22 Sep 2024
Posts: 91
Own Kudos:
132
Given Kudos: 169
Location: India
Concentration: Finance, Strategy
Schools: Darden '23 Goizueta '24 Fuqua '24 ISB '23 Kelley '24 Mendoza Ross '24 Tepper '24 Tuck '24 WBS Rotman '24
GPA: 3.73
WE:Engineering (Consulting)
Products:
My Reviews
Schools: Darden '23 Goizueta '24 Fuqua '24 ISB '23 Kelley '24 Mendoza Ross '24 Tepper '24 Tuck '24 WBS Rotman '24
Posts: 91
Kudos: 132
Martha bought several pencils. If each pencil was either a 16 Feb 2022, 03:31
Kudos
Add Kudos
Bookmarks
Bookmark this Post
This can be solved using the system of coordinate geometry. We know when an equation has a negative slope then it is called a decreasing line, in terms of Co-ord Geometry. The particular property of a decreasing line is this: for any satisfying value of (x,y), the next value of x, y will take the opposite coordinates of (x,y):

What I meant is this : We know that 23x + 21 y= 130

Now, since the constraint here is nos. of pencils has to be an integer, the nearest value will be 23 x 2 + 21 x 4 . How do I know because the unit digit will become zero and hence that will be my first iteration. other times what I do is take any value of x or y to be zero, if the constant is divisible by any of coefficient of x or y,

So we have the initial value x=2 and y =4/ We need to check if there are any other values within the range,
So the property dictates this , Since it is a decreasing line, if x increases, y has to decrease. or vice versa to satisfy the equation.

Now the next value of x will change by the coefficient of y,
So what was the coefficient of y?
It was 21. !

So x will take the value of x =2 +21 =23

Now when x =23, y will take the coefficient of x, which is 23 but in a decreasing way because you increased x. So y will be 4-23= -19, which is not possible as no of pencils cannot be negative

Similarly, if you increase y by the coefficient of x, y becomes = 4+ 23 =27, so x will have to decrease by the coefficient of y which is 21. So x will become, 2-21= -19, which is not possible again,

So both are not possible and hence statement 2 has a unique solution. Hence B.

Solve these kinds of questions with this logic, and mostly you may not fail.
Hope it helps
User avatar
Maria240895chile
User avatar
Joined: 23 Apr 2021
Last visit: 07 Jun 2023
Posts: 115
Own Kudos:
59
Given Kudos: 115
Posts: 115
Kudos: 59
Martha bought several pencils. If each pencil was either a 01 Apr 2023, 07:12
Kudos
Add Kudos
Bookmarks
Bookmark this Post
The way I thought this problem was the following.

(1) is clearly insufficient bc we only know the sum of the quantities but not the money spent.

(2) 23x+21y=130
I realized this can be solve by trying different numbers since we know x and y are integers since they are pencils and we also know they must be positive. So we can use info from (1) to help us to solve just using (2) without actually use (1). So we have a clue that x+y=6 so instead of trying a lot of numbers in (2) to get 130 we just try the pair of numbers that sum 6. (0 and 6) (1 and 5) (2 and 4) ( 3 and 3)

x=0, y =6, 23x+21y=130 --> 126=130, NO
x=6, y=0, -- >138=130, NO
x=y=3, 69+63= 132 -->132=130 NO
x=1, y=5, 23+105= 128, -->128=130 NO
x=5, Y=1, 115+21=136. --> 136=130 NO
x=2, Y=4; 46+84= 130. --> 130=130 = YES
x=4, Y=2, 92+42=134. --> 134=130 NO

so the only possible combination is x=2 and Y=4
so (2) is sufficient by itself, even though we use (1) as a clue, it is not necessary to solve the problem

IMO B
User avatar
bumpbot
User avatar
Non-Human User
Joined: 09 Sep 2013
Last visit: 04 Jan 2021
Posts: 38,587
Own Kudos:
1,079
Posts: 38,587
Kudos: 1,079
Martha bought several pencils. If each pencil was either a 13 Apr 2025, 17:46
Kudos
Add Kudos
Bookmarks
Bookmark this Post
Hello from the GMAT Club BumpBot!

Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
   1   2 
Moderators:
Bunuel
Math Expert
105389 posts
kingbucky
496 posts