Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40 cent stamps. She spent a total of $4.20 in buying these stamps. Did she buy more than 5 stamps of any of the three values?
(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.
(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.
OA B
Source: Princeton Review
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Alice bought a certain number of 30 cent stamps, 35 cent
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BTGmoderatorDC
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Jay@ManhattanReview
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Let the number of 30 cent stamps, 35 cent stamps and 40 cent stamps be x, y and z, respectively.BTGmoderatorDC wrote:Alice bought a certain number of 30 cent stamps, 35 cent stamps and 40 cent stamps. She spent a total of $4.20 in buying these stamps. Did she buy more than 5 stamps of any of the three values?
(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.
(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.
OA B
Source: Princeton Review
Thus, we have
30x + 35y + 40z = 420
6x + 7y + 8z = 84
We have to determine where at least one of x, y and z is greater than 5.
Let's take each statement one by one.
(1) The number of 35 cent stamps and 40 cent stamps that Alice bought are equal.
=> y = z
Thus, from 6x + 7y + 8z = 84, we have
6x + 7z + 8z = 84
6x + 15z = 84
2x + 5z = 28
Case 1: Say z = y = 2,
=> x = 9 > 5. The answer is Yes.
Case 1: Say z = y = 4,
=> x = 4 < 5. The answer is No.
No unique answer. Insufficient.
(2) The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.
=> x = z and y ≤ z
Thus, from 6x + 7y + 8z = 84, we have
6z + 7y + 8z = 84
14z + 7y = 84
Case 1: Say x = z = 4
=> y = 4 < 5. The answer is No.
Case 2: Say x = z = 2
=> y = 8 > 5. But this is not a valid case since we know that y ≤ z; however, since for this case, y = 8 > z = 2.
Thus, we can conclude that x, y, and z, each is not greater than 5. Sufficient.
The correct answer: C
Hope this helps!
-Jay
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deloitte247
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Let x =number of 30 stamps and x > 0
Let y = number of 35 stamps and y > 0
Let z = number of 40 stamps and z > 0
Alice spent $4.20 in buying all the stamps
$$30x+35y+40z\ =420\ cent\ ----eqn\ i$$
Statement 1: The number of 35 cent and 40 cent stamps that Alice bought are equal.
y=z, then from eqn i,
$$30x+y\left(35+40\right)=420\ cents$$
$$30x+75y=420\ -----\ eqn\ ii$$
If x=9, and y=2; it answers eqn (ii) but Alice mighr have bought more than 5 of the 30 cent stamps.
So, if x=4 and y=4, it also answer eqn (ii) and Alice might have bought less than 5 of the 30 cent stamps. Hence, statement 1 is INSUFFICIENT because we have no unique answers.
Statement 2: The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.
$$Therefore,\ x=z,\ and\ y\le z=x,\ and\ from\ eqn\ i,$$
$$30x+40x+35y=420$$
$$70x+35y=420\ ----\ eqn\ iii$$
$$Since\ y\ge0\ and\ y\le x,\ the\ \minimum\ possible\ value\ of\ y=2.$$
$$So,\ if\ y=2,\ x=z=5.\ Therefore,\ x\ and\ z\ \le5$$
None of the three values are more than 5. Hence, statement 2 alone is SUFFICIENT.
Option B is therefore the correct answer
Let y = number of 35 stamps and y > 0
Let z = number of 40 stamps and z > 0
Alice spent $4.20 in buying all the stamps
$$30x+35y+40z\ =420\ cent\ ----eqn\ i$$
Statement 1: The number of 35 cent and 40 cent stamps that Alice bought are equal.
y=z, then from eqn i,
$$30x+y\left(35+40\right)=420\ cents$$
$$30x+75y=420\ -----\ eqn\ ii$$
If x=9, and y=2; it answers eqn (ii) but Alice mighr have bought more than 5 of the 30 cent stamps.
So, if x=4 and y=4, it also answer eqn (ii) and Alice might have bought less than 5 of the 30 cent stamps. Hence, statement 1 is INSUFFICIENT because we have no unique answers.
Statement 2: The number of 30 cent stamps and 40 cent stamps that Alice bought are equal. The number of 35 cent stamps that she bought was not more than the number of 40 cent stamps that she bought.
$$Therefore,\ x=z,\ and\ y\le z=x,\ and\ from\ eqn\ i,$$
$$30x+40x+35y=420$$
$$70x+35y=420\ ----\ eqn\ iii$$
$$Since\ y\ge0\ and\ y\le x,\ the\ \minimum\ possible\ value\ of\ y=2.$$
$$So,\ if\ y=2,\ x=z=5.\ Therefore,\ x\ and\ z\ \le5$$
None of the three values are more than 5. Hence, statement 2 alone is SUFFICIENT.
Option B is therefore the correct answer
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Ian Stewart
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Using Statement 2 alone, if she bought x stamps at 30 cents, and x stamps at 40 cents, she spent 70x cents in total just on those stamps. If she only spent 420 cents overall, then x clearly is no greater than 6. But if she also bought some 35 cent stamps, x must be strictly less than 6. And Statement 2 tells us the number of 35 cent stamps bought was less than or equal to x, so is also less than 6, and Statement 2 is sufficient alone.
Statement 1 is not sufficient, since she could have bought 2 of each (and nine of the $0.30 stamps) or 4 of each (and 4 of the $0.30 stamps).
Statement 1 is not sufficient, since she could have bought 2 of each (and nine of the $0.30 stamps) or 4 of each (and 4 of the $0.30 stamps).
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