At a certain bookstore, each notepad costs x dollars and

Given: \(5x+3y\leq{10}\). Question: is \(4x+4y\leq{10}\)?

Note that both statements are telling us basically the same thing:
(1) \(x<1\);
(2) \(11x<10\) --> \(x<\frac{10}{11}\approx{0.9}\);

Now, if both notepad and marker cost very cheap then the answer would be YES but if notepad costs $0.1 and marker costs $3 then the answer would be NO. Not sufficient.

Answer: E.
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Let me put it in this way:
given :
\(5x+3y <= 10\)
asking:
is \(4x+4y <= 10\)

NOTE: in both the above cases we buy total of \(8\) items. All it is asking is, with the same amount i.e. \($10\) can i buy \(1\) more\(y\), making \(y\) to \(4\) items from \(3\)items, instead of buying \(1\) \(x\), making \(x\) to \(4\) items from \(5\) items.

Hence, to answer this question, all we need to know is is the price of \(1 y\) less than that of one \(1 x\). ==> \(is Y < X\) If this is the case, then i can safely replace 1 x with 1 y for that/leass than the amount ($10). Can't I?

stmnt1: \(x < 1\). who cares. i need to know \(is Y < X\)..Not suff.
stmnt2: 11x<10. ==> x<1. Same as above...Not suff.
stmnts 1 & 2. Both are bread slices. Can not make the sandwich with out the extra stuff....Not suff.

Answer "E".

Regards,
Murali.

Kudos?

Given: 5x + 3y <= 10 .... (I)
Question: Is 4x + 4y <=10 ?....(II)

Statement (1) x < 1.
If x is a little less than 1 but almost 1, then 5x is little less than 5. Putting in (I) above, we get that y is less than \(\frac{5}{3}\).
Now we need to put these values in equation (II) and check.
4x will be a little less than 4 and 4y will be less than 4x\(\frac{5}{3}\) i.e. 6.66. Together, 4x + 4y will be less than 10.66. It may be less than 10 or a little more than 10, hence this is not sufficient.

Statement (2) x < \(\frac{10}{11}\)
If x is a little less than 10/11 but almost 10/11, then 5x is little less than 50/11. Putting in (I) above, we get that y is less than \(\frac{20}{11}\).
Now we need to put these values in equation (II) and check.
4x will be a little less than 40/11 and 4y will be less than 4x\(\frac{20}{11}\) i.e. \(\frac{80}{11}\). Together, 4x + 4y will be less than 120/11 i.e. less than 10.9. It may be less than 10 or more, we do not know, hence this is not sufficient.

Note: You didn't actually need to calculate the data in the second statement. This is because, if we compare 5x + 3y with 4x + 4y, the x term has reduced but the y term has increased. If x is as great as possible, it will keep y small and the extra y in second statement may not matter. We tried with x a little less than 1 in first statement. In second statement, x is smaller than 10/11. Hence, if first statement was not sufficient, no way could the second statement be sufficient.

can someone explain why "If $10 is enough to buy 5 notepads and 3 markers" => the translation can't be 5x+ 3y = 10 ( what's wrong with this)

and why is it => 5x+3y < = 10

LET SAY X= 1 Y =1
THEN 5X + 3Y = 8
STILL 10 DOLLAR IS enough TO BUY 5 NOTEPADS AND 3 MARKERS.

so question wants to say that price of 5 notepads and 3 markers is less than or equal to 10 $

hope it is clear.

'is enough' means that it suffices. Whether it suffices perfectly or something is leftover is not specified. e.g. it is correct to say $10 is enough to buy the chocolate bar I want. The bar could cost $5 or $8 or $9.2 or $10 etc. I any case, $10 is enough.

Check out a detailed discussion of this question here: https://www.gmatclub.com/forum/veritas-prep-resource-links-no-longer-available-399979.html#/2013/09 ... -of-words/

I am pretty sure this wasnt intended to be a hard question.
Both statements in a way give the same info. So answer should be either D or E.
Since both x and y have changed in the question (4x+4y≤10?) from the given (5x+3y≤10) the answer has to be E

Do not forget that there are noninteger dollar values, but all values in cents are integers!

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

\(4n + 4m\,\,\mathop \le \limits^? \,\,1000\,\,\,\left[ {{\rm{cents}}} \right]\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,n + m\,\,\mathop \le \limits^? \,\,250\,\,\,\left[ {{\rm{cents}}} \right]\)

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

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

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


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

Regards,
Fabio.
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Here's one way to visualize it. It helps to put yourself into the story when the problem asks about things like shopping and prices! Our intuitions about these topics are often better than our pure math skills.

$10 is enough to buy 5 notepads and 3 markers. Imagine that you have 5 notepads and 3 markers in your shopping cart. But, you'd rather buy 4 notepads and 4 markers. Can you safely 'swap out' a notepad for a marker without going over budget?

For starters, you can definitely do that if a marker is less expensive than a notepad. Since you were under $10 before, you'll still be under $10.

On the other hand, suppose that you're already very close to your budget when you have 5 notepads and 3 markers in your cart. If a marker is more expensive than a notepad, then switching might bring you over budget.

With that in mind, let's look at the statements.

(1) says that a notepad is inexpensive. But a marker could be even cheaper - or it could be much more expensive! Insufficient.

(2) says more or less the same thing. A marker could cost as little as $0.01, or as much as $3 (or more), as long as the notepads were cheap enough. If we switch a notepad for a very cheap marker, we'd stay under budget. If we switch for an expensive one, we could go over.

Since (1) and (2) are redundant, having both of them doesn't help answer the question, either.

The answer is E.
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we can do one step more before we plug in specific numbers
5x+3y<10
whether
4x+4y<10
the change between two expression is
y-x
if y-x is big, 10 usd is not enough, if y-x is small, 10 usd is enough. we plug in the specific number now
supose y=3
from the first expression, 5x <10-3*3, x<0.2
from the second expression
y=3 can not satifies expression 2.
if y=1
then we can see that both expression are good.

so, answer is E

Dear IanStewart,

I'm not sure whether the highlighted part (written by Karishma) is valid. It seems that she subtracted 5x < 5 from (I).
I don't think that is allowed, right? (as picture attached herein) Please confirm.


Again, I'm not sure whether this is valid for the very same reason above.


I completely lost in this part. If in the second statement x is smaller, then y could be larger. Why don't we need to check? I don't get the reasoning behind it.

Looking forward to your awesome method sir

First, you cannot directly subtract inequalities that face the same way (you can, in theory, learn some useless rules about how to subtract them if they face in opposite directions, but there's no reason to learn those if you know how to add inequalities correctly), so your screenshot is giving good advice about that. But that's not what Karishma is doing. She is just working out, when notepads cost almost $1, what the maximum price of markers could be.

But if we know that 5 notepads and 3 markers cost $10, and we want the maximum possible value for the cost of 4 notepads and 4 markers, we don't want to maximize the cost of notepads. We want to minimize the cost of notepads. It's probably easier to see why that's true with simpler numbers. If I tell you x and y are positive integers, and

x + y = 10

and I ask for the maximum possible value of 2y, we don't get that by maximizing x. We get that value by minimizing x, so it contributes as little as possible to the sum "x + y = 10" and y therefore contributes as much as possible to that sum. We get the maximum value of 2y when x = 1, and then 2y = 18.

In this question, we're comparing what happens when we buy 5 notepads and 3 markers with what happens when we buy 4 notepads and 4 markers. So we're just replacing one notepad with one marker, and the total cost will go up if markers cost more than notepads. We know 5 notepads and 3 markers cost at most $10. Let's say they cost exactly $10. We'll get the maximum possible price of 4 notepads and 4 markers if markers are as expensive as possible, so if notepads are as cheap as possible. If notepads are free, then each marker costs $10/3, and 4 notepads and 4 markers can cost as much as $40/3 = $13.3333... That's the maximum possible price here of 4 markers and 4 notepads, using both Statements.

There's no need to really do any of that though - I wrote a long reply to answer your questions. If I were solving this problem, I'd just say "using both statements, markers and notepads can both be free, and then the answer is 'yes', or notepads can be free and markers can cost $3 and then the answer is 'no', so the answer is E".

Dear IanStewart,

If I were to switch the 2 conditions in question stem as highlighted:



Would the answer to the above modified question be D?
Here we would want to MAXIMIZE the cost of notepads because the number of notepads is increased from 4 to 5, right?

IMO, Karishma's approach would make more sense in this question, right?

Yes, that's exactly right.

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