Problem Solving

Joanna bought only $0.15 stamps and $0.29 stamps.
How many $0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps
and $0.29 stamps.

The trick is the number of stamps she bought must be integers.

Let us assume the numbers of $0.15 and $0.29 stamps she bought are equal to x and y, respectively.
We need to determine x.

Statement 1: 0.15x + 0.29y = 4.40
So, 15x + 29y = 440
--> x = (440 - 29y)/15

Now, x and y must be an integer.
Also x and y cannot be equal to zero as 440 is not a multiple of either 15 or 29.
So, x and y must be positive integers.

As x must be a positive integer, (440 - 29y) must be divisible by 15 and hence by both 5 and 3.
Now, 440 is divisible by 5. So, (440 - 29y) to be divisible by 5, y must be divisible by 5.
So, possible values of y are 5, 10, 15, etc...

For y = 5, (440 - 29y) = (440 - 29*5) = 440 - 145 = 295 --> not divisible by 15
For y = 10, (440 - 29y) = (440 - 29*10) = 440 - 290 = 150 --> divisible by 15
For y = 15, (440 - 29y) = (440 - 29*15) = 440 - 435 = 5 --> not divisible by 15
For y ≥ 20, (440 - 29y) ≤ (440 - 29*20) = 440 - 580 < 0 --> x is negative

So, only possible value of y is 10.
So, x = (440 - 29y)/15 = 150/15 = 10

So, statement 1 is sufficient.


Statement 2: x can have different possible values.
So, statement 2 is not sufficient.

Answer : A

how x has different values?

like if she bought same number of stamps so .15x + .29x = 4.40

then .44x = 4.40 and x=10.

let me know if i am missing something.

[email protected] wrote:Joanna bought only $0.15 stamps and $0.29 stamps.
How many $0.15 stamps did she buy?
(1) She bought $4.40 worth of stamps.
(2) She bought an equal number of $0.15 stamps
and $0.29 stamps.

A savvy test-taker can determine the correct answer here without actually solving.
Here are the two statements, translated into math:

Statement 1: 15x + 29y = 440
Statement 2: x=y

Clearly, the two statements combined provide enough information to solve for x and y.
Eliminate E.

The two equations above are very straightforward.
Virtually every test-taker will be able to generate them easily.
This means that the correct answer CANNOT be C.
If the correct answer is C, then almost every test-taker will choose the correct answer, making the problem pointless.
Eliminate C.

Statement 2 is clearly insufficient on its own.
If all we know is that x=y, it's possible that x=y=1 or that x=y=2.
Eliminate B and D.

The correct answer is A.

One last fun fact! These sort of equations - algebraic equations in at least two variables for which we only consider integer solutions - are called Diophantine equations, and they've been entertaining and befuddling recreational mathematicians for about 2000 years.

Here's a nice rundown of the general approach to a simple Diophantine equation: https://mathforum.org/library/drmath/view/51595.html

And here's a fun set of problems (with detailed solutions!): https://www.cemc.uwaterloo.ca/events/mat ... -Solns.pdf