Source: Veritas Prep
A store sells erasers for 0.23$ per piece and pencil for 0.11$ per piece. How many eraser and pencils did Jessica buy?
1) She bought 5 erasers.
2) She spent a total of 1.70$.
The OA is B
A store sells erasers for 0.23$ per piece and pencil for
This topic has expert replies
-
- Moderator
- Posts: 2209
- Joined: Sun Oct 15, 2017 1:50 pm
- Followed by:6 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
GMAT/MBA Expert
- Jay@ManhattanReview
- GMAT Instructor
- Posts: 3008
- Joined: Mon Aug 22, 2016 6:19 am
- Location: Grand Central / New York
- Thanked: 470 times
- Followed by:34 members
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Say Jessica bought x numbers of erasers and y x numbers of pencils.BTGmoderatorLU wrote:Source: Veritas Prep
A store sells erasers for 0.23$ per piece and pencil for 0.11$ per piece. How many eraser and pencils did Jessica buy?
1) She bought 5 erasers.
2) She spent a total of 1.70$.
The OA is B
Thus, she spent a total of 0.23x + 0.11y to buy (x + y) items.
We have to get the value of (x + y).
Let's take each statement one by one.
1) She bought 5 erasers.
No information about the numbers of pencils. Insufficient.
2) She spent a total of 1.70$.
=> 0.23x + 0.11y = 1.70
23x + 11y = 170
x = (170 - 11y)/23
x = (161 + 9 - 11y)/23
x = 161/23 + (9 - 11y)/23
x = 7 + (9 - 11y)/23
Since x and y are positive integers, 9 - 11y must be a multiple of 23. With some hit and trail, you'll find one eligible value for y = 5.
At y = 5, we have x = 7 + (9 - 11y)/23 => x = 7 + (9 - 11*5)/23 => x = 7 + (9 - 55)/23 => x = 7 + (-46/23) => x = 7 - 2 = 5.
So, one solution is x = y = 5, thus, x + y = 10. There is no need to try for higher qualified values of y since at higher values x would turn out to be negative, unqualified value.
So, the unique value of x + y = 10. Sufficient.
The correct answer: B
Hope this helps!
-Jay
_________________
Manhattan Review
Locations: Manhattan Review New Delhi | GMAT Prep Malleswaram | GRE Prep Vijayawada | Mumbai GRE Coaching | and many more...
Schedule your free consultation with an experienced GMAT Prep Advisor! Click here.
GMAT/MBA Expert
- Brent@GMATPrepNow
- GMAT Instructor
- Posts: 16207
- Joined: Mon Dec 08, 2008 6:26 pm
- Location: Vancouver, BC
- Thanked: 5254 times
- Followed by:1268 members
- GMAT Score:770
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
This question illustrates a common trap on the GMAT.
As Jay has shown, with statement 2, we're able to write the equation 23x + 11y = 170 , and in high school we learned that, if we're given 1 equation with 2 variables, we cannot find the value of either variable.
However, if we restrict the variables to positive integers within a certain range of values, then there are times when we can, indeed, find the value of a variable if we're given 1 equation with 2 variables.
Here's a very similar Official GMAT question to practice with: https://www.beatthegmat.com/why-does-th ... 11112.html
Cheers,
Brent
As Jay has shown, with statement 2, we're able to write the equation 23x + 11y = 170 , and in high school we learned that, if we're given 1 equation with 2 variables, we cannot find the value of either variable.
However, if we restrict the variables to positive integers within a certain range of values, then there are times when we can, indeed, find the value of a variable if we're given 1 equation with 2 variables.
Here's a very similar Official GMAT question to practice with: https://www.beatthegmat.com/why-does-th ... 11112.html
Cheers,
Brent
- GMATGuruNY
- GMAT Instructor
- Posts: 15539
- Joined: Tue May 25, 2010 12:04 pm
- Location: New York, NY
- Thanked: 13060 times
- Followed by:1906 members
- GMAT Score:790
Timer
00:00
Your Answer
A
B
C
D
E
Global Stats
Consider the following equation:
5x + 7y = 70.
If x and y are nonnegative integers, the following solutions are possible:
x=0, y=10
x=7, y=5
x=14, y=0.
Notice the following:
The value of x changes in increments of 7 (the coefficient for y).
The value of y changes in increments of 5 (the coefficient for x).
This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers.
Since the number of pencils can be any nonnegative value, INSUFFICIENT.
Statement 2:
23E + 11P = 170.
This equation is constrained to nonnegative integers.
Test a case that also satisfies Statement 1.
If E = 5, we get:
23*5 + 11P = 170
115 + 11P = 170
11P = 55
P = 5.
Thus, one solution for 23E + 11P = 170 is E=5 and P=5.
Since the value of E may change only in increments of 11 -- the coefficient for P -- we get the following alternate options for E:
16, 27, 38...
All of these alternate options for E will yield a sum greater than 170 and thus are not viable.
Thus, the only viable solution for 23E + 11P = 170 is E=5 and P=5.
SUFFICIENT.
The correct answer is B.
5x + 7y = 70.
If x and y are nonnegative integers, the following solutions are possible:
x=0, y=10
x=7, y=5
x=14, y=0.
Notice the following:
The value of x changes in increments of 7 (the coefficient for y).
The value of y changes in increments of 5 (the coefficient for x).
This pattern will be exhibited by any fully reduced equation that has two variables constrained to nonnegative integers.
Statement 1:BTGmoderatorLU wrote:Source: Veritas Prep
A store sells erasers for 0.23$ per piece and pencil for 0.11$ per piece. How many eraser and pencils did Jessica buy?
1) She bought 5 erasers.
2) She spent a total of 1.70$.
Since the number of pencils can be any nonnegative value, INSUFFICIENT.
Statement 2:
23E + 11P = 170.
This equation is constrained to nonnegative integers.
Test a case that also satisfies Statement 1.
If E = 5, we get:
23*5 + 11P = 170
115 + 11P = 170
11P = 55
P = 5.
Thus, one solution for 23E + 11P = 170 is E=5 and P=5.
Since the value of E may change only in increments of 11 -- the coefficient for P -- we get the following alternate options for E:
16, 27, 38...
All of these alternate options for E will yield a sum greater than 170 and thus are not viable.
Thus, the only viable solution for 23E + 11P = 170 is E=5 and P=5.
SUFFICIENT.
The correct answer is B.
Private tutor exclusively for the GMAT and GRE, with over 20 years of experience.
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3
Followed here and elsewhere by over 1900 test-takers.
I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
My students have been admitted to HBS, CBS, Tuck, Yale, Stern, Fuqua -- a long list of top programs.
As a tutor, I don't simply teach you how I would approach problems.
I unlock the best way for YOU to solve problems.
For more information, please email me (Mitch Hunt) at [email protected].
Student Review #1
Student Review #2
Student Review #3