msmith8754 wrote:If each pencil was either 23cents or 21 cents, how many 23 cents pencils were purchased?
(1) 6 pencils were bought
(2)Total value of pencils bought was 130 cents

I am stumped with this one . Thank you in advance for your feedback.

IOM C

given: P(21;23) <=> P(x;y)
st(1) x+y=6 Not Sufficient
st(2) 21x+23y=130 Not Sufficient
Combined st(1&2) two variables & two equations Sufficient

x=6-y --> 21(6-y)+23y=130, y=2; x=6-2=4
4*21 + 2*23 = 130

msmith8754 wrote:If each pencil was either 23cents or 21 cents, how many 23 cents pencils were purchased?
(1) 6 pencils were bought
(2)Total value of pencils bought was 130 cents

I am stumped with this one . Thank you in advance for your feedback.

Statement 1: 6 pencils were bought
No way to determine how many of the 6 pencils cost 23 cents.
Insufficient.

Statement 2: Total value = 130 cents
Since 6*23 = 138, we know that the number of 23 cent pencils was less than 6.
Let's try buying 5 or fewer 23 cent pencils to see whether the remaining amount could be purchased with the 21 cent pencils:
5*23 = 115. 130-115 = 15. 15 is not divisible by 21.
4*23 = 92. 130-92 = 38. 38 is not divisible by 21.
3*23 = 69. 130-69 = 61. 61 is not divisible by 21.
2*23 = 46. 130-46 = 84. 84/21 = 4. We could buy two 23 cent pencils and four 21 cent pencils.
1*23 = 23. 130-23 = 107. 107 is not divisible by 21.

Since the only viable combination requires that two 23 cent pencils be purchased, sufficient.

The correct answer is B.

This question involves a common GMAT trick. When a DS problem is restricted to positive integers -- we can't buy 1/2 a pencil or -3 pencils -- one equation might be sufficient information to solve for 2 variables. Be careful!

1st stem is definitely insufficient

2nd stem:

21x+23y=130

Typically, an equation like this cannot be solved. However, you will have sufficient data to solve under these conditions:
(1) unknowns are integers
(2) total (130) is less than the LCM of the coefficients (21 & 23) + the smaller coefficient.

That's it. Because 23 is prime, the LCM of 23 and 21 is 23*21. Since the total (130) is less than 23*21+21, there is only one possible solution. Statment (2) is sufficient.

yes, I remember Mitch showing this add-withdraw method once- hastened myself

Hi, Could someone please explain how we got this answer in more details ?

ankurmit wrote:1st stem is definitely insufficient

2nd stem:

21x+23y=130

Typically, an equation like this cannot be solved. However, you will have sufficient data to solve under these conditions:
(1) unknowns are integers
(2) total (130) is less than the LCM of the coefficients (21 & 23) + the smaller coefficient.

That's it. Because 23 is prime, the LCM of 23 and 21 is 23*21. Since the total (130) is less than 23*21+21, there is only one possible solution. Statment (2) is sufficient.