Joanna bought only 0.15$ stamps and 0.29$ stamps. how many 0.15$ and 0.29$ stamps did she buy?
1) she bought 4.4$ worth of stamps
2) she bought an equal number of 0.15$ stamps and 0.29$ stamps
stamps
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Let's assume Joanna bought X number of $0.15 stamps and Y number of $0.29 stamps.
Statement 1 : she bought 4.4$ worth of stamps
This implies:
(0.15)X + (0.29)Y = 4.4
15X + 29Y = 440 (multiply both sides by 100)
15X = 440 - 29Y
X = (440 - 29Y) / 15
This tells that the subtraction (440 - 29Y) will be a multiple of 15, because we can't have fractional amount of stamps.
Now list out various multiples of 15
0,15,30,45,60,75,90....
Notice that every time the number ends in either a 5 or a 0. This should mean that 440 - 29Y must also end in either a 5 or a 0
The unit digit of 440 is 0 , so the only way to subtract something from it (to get a number which ends with a 5 or 0) is to subtract a number whose unit digit ends in a 5 or 0
Now its fairly simple. 29 * 5 = 145 or 29 * 10 = 290 (9 multiplied by a 5 or a multiple of 5 or a 0 will satisfy this ,notice that only those two will give us the desired result)
440 - 145 = 295 --> Not divisible by 15
440 - 290 = 150 --> Divisible by 15
So, X = 10 and from our previous equation Y = 10
SUFFICIENT
Statement 2 : she bought an equal number of 0.15$ stamps and 0.29$ stamps
Joanna could have bought any number of both $0.15 and $0.29 stamps
NOT SUFFICIENT
Correct Answer : A
Statement 1 : she bought 4.4$ worth of stamps
This implies:
(0.15)X + (0.29)Y = 4.4
15X + 29Y = 440 (multiply both sides by 100)
15X = 440 - 29Y
X = (440 - 29Y) / 15
This tells that the subtraction (440 - 29Y) will be a multiple of 15, because we can't have fractional amount of stamps.
Now list out various multiples of 15
0,15,30,45,60,75,90....
Notice that every time the number ends in either a 5 or a 0. This should mean that 440 - 29Y must also end in either a 5 or a 0
The unit digit of 440 is 0 , so the only way to subtract something from it (to get a number which ends with a 5 or 0) is to subtract a number whose unit digit ends in a 5 or 0
Now its fairly simple. 29 * 5 = 145 or 29 * 10 = 290 (9 multiplied by a 5 or a multiple of 5 or a 0 will satisfy this ,notice that only those two will give us the desired result)
440 - 145 = 295 --> Not divisible by 15
440 - 290 = 150 --> Divisible by 15
So, X = 10 and from our previous equation Y = 10
SUFFICIENT
Statement 2 : she bought an equal number of 0.15$ stamps and 0.29$ stamps
Joanna could have bought any number of both $0.15 and $0.29 stamps
NOT SUFFICIENT
Correct Answer : A
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This question illustrates a common trap on the GMAT.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.
For statement 1, we're able to write the equation 15x + 29y = 440, 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 find the value of a variable if we're given 1 equation with 2 variables.
This common GMAT trap (along with other common traps) is addressed in the following free videos:
https://www.gmatprepnow.com/module/gmat- ... cy?id=1105
https://www.gmatprepnow.com/module/gmat- ... cy?id=1106
https://www.gmatprepnow.com/module/gmat- ... cy?id=1107
Cheers,
Brent
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For an alternate approach, check my second post here:
https://www.beatthegmat.com/to-find-the- ... 73004.html
https://www.beatthegmat.com/to-find-the- ... 73004.html
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I have worked with students based in the US, Australia, Taiwan, China, Tajikistan, Kuwait, Saudi Arabia -- a long list of countries.
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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].
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Here's a quick way that doesn't require much casework.
If we have 15x + 29y = 440, and we know that x and y are integers, we can do a little manipulation to start.
15x + 15y + 14y = 440
15x + 15y + 15y - y = 440
15*(x + 2y) - y = 440
Now let's look at multiples of 15 in the vicinity of 440. We have 15 * 30 and 15 * 31, or 450 and 465.
But if 465 is our answer, then y = 25. If y = 25, then Joanna bought 25 29¢ stamps ... but that's $7.25!
So the only possible solution is 450. This gives us
15*(x + 2y) = 450
and 15*(x + 2y) - y = 440
so y = 10. From there, x = 10, and we're done; S1 is sufficient.
If we have 15x + 29y = 440, and we know that x and y are integers, we can do a little manipulation to start.
15x + 15y + 14y = 440
15x + 15y + 15y - y = 440
15*(x + 2y) - y = 440
Now let's look at multiples of 15 in the vicinity of 440. We have 15 * 30 and 15 * 31, or 450 and 465.
But if 465 is our answer, then y = 25. If y = 25, then Joanna bought 25 29¢ stamps ... but that's $7.25!
So the only possible solution is 450. This gives us
15*(x + 2y) = 450
and 15*(x + 2y) - y = 440
so y = 10. From there, x = 10, and we're done; S1 is sufficient.