A boy goes to a supermarket and buys some pencils and erasers.

Question

A boy goes to a supermarket and buys some pencils and erasers. The cost of each pencil is $0.3 and cost of each eraser is $0.5. If he bought at least one pencil and at least one eraser and the number of pencils he bought was not four, how many pencils did he buy?

(1) He paid a total of $4.20

(2) He bought three erasers.

This is a  Data Sufficiency Butler Question

Check the links to other Butler Projects:

Data Sufficiency Butler  
  Problem Solving Butler  
  Sentence Correction Butler  
  Reading Comprehension Butler  
  Integrated Reasoning Butler

pierjoejoe · Answer

the equation to solve is

0.3 * x + 0.5 * y = K

where x is the number of pencil, y is the number of erasers. 
x != 4

the point is that x and y must be integers.

(1) 
0.3 * x + 0.5 * y = 4.2

3 * x + 5 * y = 42

3*x must be a number such that 42 - 3*x is divisible by 5, and the number y is at least 1.
so we cannot have x = 14, neither x = 4. the only option left is x = 9, in that case 3x = 27 and 42-27 is divisible by 5 -->y=3

(2)
we do not need this information

Imtiyazshaikh99 · Answer

Ans - A

S1: 3x+5y = 42
If x!= 4 then the only other option for x as an integer is 9.
Suff.

S2: y= 3
But total no given. Hence insuff.

Krunaal · Answer

Let no. of pencils be P and no. of erasers be E
Price of total Pencils = $0.3P, where P is greater than 0 and not equal to 4
Price of total Erasers = $0.5E, where E is greater than 0

Statement 1.

0.3P + 0.5E = 4.2

P and E need to be integers, and P can't be 4
Now, only when P is 9, E can assume a value of an integer i.e. 3  SUFFICIENT

Statement 2.

E = 3, but what is P?  INSUFFICIENT

Answer A.

jmaxr · Answer

Firstly, for my own sanity, I multiplied every number by 10 so I was working with integers.

From the problem stem we can conclude that the number of Pencils will be a multiple of 3 and the number of Erasers will be a multiple of 5. We also are given the number of pencils purchased cannot equal 4, or in other words the total cost of the pencils cannot equal 12.

1) This provides us a total of 42 for the combined cost of pencils and erasers. From here I simply ran through the combinations of multiples of 5 subtracted from 42 in order to find a multiple of 3. 
Visually It looked a bit like this
42=5(1)+37. Incorrect, 37 is not a multiple of 3.
42=5(2)+32. Incorrect, 32 is not a multiple of 3.
42=5(3)+27.  Possibly correct , 27 is 3x9. 
42=5(4)+22. Incorrect, 22 is not a multiple of 3.
42=5(5)+17. Incorrect, 17 is not a multiple of 3 
42=5(6)+12.  Incorrect  the number of pencils purchased cannot be 4. 3x4=12. 
42=5(7)+7. Incorrect, 7 is not a multiple of 3.

This leaves us with the only possible combination of 3 Erasers and 3 Pencils purchased. 
 Suff

2). Gives us the number of erasers purchased but without an upper limit for the total cost we have nothing to go off here as the number of pencils could be anywhere between 1 and ∞. 
 Insuff

Answer choice A

A boy goes to a supermarket and buys some pencils and erasers.

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Data Sufficiency (DS) Questions

A boy goes to a supermarket and buys some pencils and erasers.

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards