A certain bookstore, each notepad costs x dollars and each

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At a certain bookstore, each notepad costs x dollars and each marker costs y dollars. If $10 is enough to buy 5 notepads and 3 markers, is $10 enough to buy 4 notepads and 4 markers instead?

1) Each notepad cost less than $1.
2) $10 is enough to buy 11 notepads.

OA E.

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by ceilidh.erickson » Sat Dec 01, 2018 4:41 pm

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Translate the given information:
$10 is enough to buy 5 notepads and 3 markers -->
$$5x+3y\le10$$

Translate the question:
is $10 enough to buy 4 notepads and 4 markers instead? -->
$$4x+4y\le10\ ?$$
Simplify:
$$2x+2y\le5\ ?$$

1) Each notepad cost less than $1.
If x < 1, then 5x < 5. Let's test cases to see whether this is sufficient:

Case 1:
x = $0.90
y = $0.30
5x + 3y = $4.50 + $0.90 = $5.40 --> less than $10, holds statement true
4x + 4y = $3.60 + $1.20 = $4.80 --> less than $10, answer to the question is YES.

Case 1:
x = $0.10
y = $3.00
5x + 3y = $0.50 + $9.00 = $9.50 --> less than $10, holds statement true
4x + 4y = $0.40 + $12.00 = $12.40 --> NOT less than $10, so the answer to the question is NO.
Insufficient

2) $10 is enough to buy 11 notepads.
$$11x\ \le10$$
$$x\ \le0.90$$

Both of the cases we tested for statement 1 are consistent with this information, so this must also be insufficient. And putting the statements together won't help, since both cases will still be relevant (i.e. we can find a "yes" answer or a "no" answer to the question using both statements).

The answer is E.
Ceilidh Erickson
EdM in Mind, Brain, and Education
Harvard Graduate School of Education

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by Jay@ManhattanReview » Sun Dec 02, 2018 4:41 am

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AAPL wrote:GMAT Prep

At a certain bookstore, each notepad costs x dollars and each marker costs y dollars. If $10 is enough to buy 5 notepads and 3 markers, is $10 enough to buy 4 notepads and 4 markers instead?

1) Each notepad cost less than $1.
2) $10 is enough to buy 11 notepads.

OA E.
We have 5x + 3y ≤ 10.

We have to determine whether 4x + 4y ≤ 10 or x + y ≤ 2.5

Question: Is x + y ≤ 2.5?

When we sacrifice x for y in the inequality 5x + 3y ≤ 10, we get 4x + 4y ≤ 10. Thus, for the inequality 4x + 4y ≤ 10 to be true, we must have x ≥ y; otherwise, we cannot determine conclusively whether 4x + 4y ≤ 10 or x + y ≤ 2.5.

Looking at the given inequality 5x + 3y ≤ 10, we see that if x = y, then 8x ≤ 10 => x ≤ 1.25 and y ≤ 1.25. So, here $1.25 is the average price of the items.

Let's take each statement one by one.

1) Each notepad cost less than $1.

=> x < 1. Since the average price per item = 1.25 (considering the prices of both the items equal), and x < 1, we must have y > 1.25; this means that y > x. We already concluded that we can have a unique answer only if x ≥ y. Since x < y, we do not get a unique answer. Insufficient.

2) $10 is enough to buy 11 notepads.

> x = 10/11 < 1. It calls for the same analysis as we did in Statement 1. Insufficient.

(1) and (2) together

Both the statements lead to the same conclusion. Insufficient.

The correct answer: E

Hope this helps!

-Jay
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AAPL wrote:GMAT Prep

At a certain bookstore, each notepad costs x dollars and each marker costs y dollars. If $10 is enough to buy 5 notepads and 3 markers, is $10 enough to buy 4 notepads and 4 markers instead?

1) Each notepad cost less than $1.
2) $10 is enough to buy 11 notepads.
Do not forget that there are noninteger dollar values, but all values in cents are integers!

$$\left\{ \matrix{
\,{\rm{notepads}}\,,\,\,\$ \,n\,\,{\rm{cents}}\,\,{\rm{each}}\,\,\,\left( {n = 100x} \right) \hfill \cr
\,{\rm{markers}}\,,\,\,\$ \,m\,\,{\rm{cents}}\,\,{\rm{each}}\,\,\,\left( {m = 100y} \right) \hfill \cr} \right.\,\,\,\,\,\,\,\,\,\,\,\,5n + 3m \le 1000\,\,\,\left[ {{\rm{cents}}} \right]\,\,\,\,\left( * \right)$$

$$\left( 1 \right)\,\,n < 100\,\,\,\left\{ \matrix{
\,{\rm{Take}}\,\,\left( {n,m} \right) = \left( {50\,,\,200} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\,\,\,\,\,\,\,\,\left[ {\,250 + 600 < 1000\,\,\left( * \right)\,} \right]\,\,\,\,\,\, \hfill \cr
\,{\rm{Take}}\,\,\left( {n,m} \right)\, = \left( {50,210} \right)\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \,\,\,\,\,\,\,\,\,\,\,\,\left[ {\,250 + 630 < 1000\,\,\left( * \right)\,} \right]\,\,\, \hfill \cr} \right.$$

$$\left( 2 \right)\,\,11n \le 1000\,\,\,\,\mathop \Leftrightarrow \limits^{n\,\,{\mathop{\rm int}} } \,\,\,\,n \le 90\,\,\,\left\{ \matrix{
\,\left( {{\rm{Re}}} \right){\rm{Take}}\,\,\left( {n,m} \right) = \left( {50,200} \right)\,\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{YES}}} \right\rangle \,\, \hfill \cr
\,\left( {{\mathop{\rm Re}\nolimits} } \right){\rm{Take}}\,\,\left( {n,m} \right)\, = \left( {50,210} \right)\,\,\, \Rightarrow \,\,\,\left\langle {{\rm{NO}}} \right\rangle \hfill \cr} \right.\,\,\,\,\,\,\,\,$$

$$ \Rightarrow \,\,\,\,\left( {\rm{E}} \right)$$

This solution follows the notations and rationale taught in the GMATH method.

Regards,
Fabio.
Fabio Skilnik :: GMATH method creator ( Math for the GMAT)
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