Both statements are insufficient
so E

E in answer.

p+h=834

f=2p+3.5h

=2(834-h)+3.5h

=1668+1.5h

Cedagmat wrote:The original cost for paperback copy and hardback copy is $8 and $9.5, respectively. The sales price for the paperback copy and hardback copy is $10 and $13, respectively. If a total of 834 books were sold out, was the total profit greater than $2,000?
(1) More paperback copies were sold.
(2) At least 100 hardback copies were sold.
[spoiler]Answer: E[/spoiler]

S1: If more paperback copies were sold, then the number of paperbacks must be at least greater than 834/2=417. P = R-C. Thus for paperback books, there must be at least 417 sold. The profit margin is $1.5 (417) = $625. Then at most 417 hardbacks are sold at a profit margin of $3 (417)=$1251. Thus the profit will never be greater than 2,000. Sufficient.
S2: if 100 Hs were sold then $3(100)=$300 in profit. However, if $3(800) then $2,400, then the profit would be greater than $2000. Thus insufficient.

Does anyone know what I'm doing wrong here?


146. There are less than 50 books are to be divided by students. If the books are divided by 7 students, one book will be left. How many books are there?
(1) If the books are divided by 9 students, 8 books will be left.
(2) If the books are divided by 5 students, 3 books will be left.
[spoiler] Answer: A[/spoiler]

Let x = number of books.
x=7a+1

S1: x = 9b+8
set equations to equal each other to find integer possibilites
7a+1=9b+8
7a=9b+7
a=9b/7+1

pick b =7, a = 10, plug in a values. x=7(10)+1 = 71. Insufficient since question mentions no. of books <50.

S2: x=5c+3
7a+1=5c+3
a=(5c+2)/7
a works for values c at 1 and 6. Either 8 or 43. Not sufficient.

Does anyone have suggestions here?

S2: x=5c+3
7a+1=5c+3
a=(5c+2)/7
a works for values c at 1 and 8 (not 6). Either 8 or 43. Not sufficient.

E!

total books sold: 834$.

profit for paperback/book: 10-8= 2$
profit for hardback/book: 13-9.5= 3.5$.

question: sale total: over 2000$.(revenure>2000$)

1. paperback >hardback. (if all paperbacks were sold :834 x 2 insufficient)
2. at least 100 hardbacks were sold : = 100 x 3.5 + 734 x 2 insufficient)

3 : 1+ 2 insufficient:

25s were counted! teehee (we have no time to make a long math)

IMO E, neither 1 or 2 are sufficient.

Answer: Option E

I liked the methods that I read above, but I found that I went over 2 min with them (because of the math - if you have quick ways to do the multiplication, feel free to post). I ended up doing this in less than 2 by doing the following:

Let p = # paperback
Let h = # hardcovers

p + h = 834 --(a)
(10 - 8)p + (13 - 9.5)h > 2000
2p + 3.5h > 2000 --(b)
2(834 - h) + 3.5p > 2000 (substitute (a) in)
h > 332(2/3)
h >= 222 (approximation)

Basically, if h >= 222 or p < 612, we have a Yes or No answer

(1) Insufficient - p can be between 834/2~=415 and 834
(2) Insufficient - h could be below 222
(1+2) Insufficient - h could be between 100 and 222 but also between 222 and 834/2~=415

Answer: E

Hi,

I have a doubt.

In 146, S1 says: If the books are divided by 9 students, 8 books will be left.

Doesn't that mean that the number of students is at least 9? Because the books are not divided by 9(number), but by 9 students and then 8 are left.
If this is the case, the number of books should be at least 9 and 8 should not be considered.

Please clarify.

The GMAT would use friendlier numbers:

A store purchases paperback copies and hardback copies for $8 and $9.50, respectively. Each paperback copy and hardback copy is then sold for $10 and $13, respectively. If a total of 840 books were sold, was the total profit greater than $2,100?

(1) More paperback copies than hardback copies were sold.
(2) At least 100 hardback copies were sold.

This is a MIXTURE problem.

Average profit per paperback = selling price - cost = 10 - 8 = 2.
Average profit per hardback = selling price - cost = 13 - 9.5 = 3.5.
To yield a profit of $2100, average profit required for the MIXTURE of paperbacks and hardbacks = revenues/books = 2100/840 = 2.5.

To determine the required ratio of paperbacks to hardbacks, use ALLIGATION -- a great way to handle mixture problems.

Step 1: Plot the 3 averages on a number line, with the averages for the paperbacks and hardbacks on the ends and the average for the mixture in the middle.
P 2------------------2.5---------------3.5 H

Step 2: Calculate the distances between the averages.
P 2--------0.5-------2.5--------1------3.5 H

Step 3: Determine the ratio in the mixture.
The required ratio of paperbacks to hardbacks is equal to the RECIPROCAL of the distances in red.
P/H = 1/0.5 = 2/1.

Implication:
To yield a profit of $2100, for every 2 paperbacks that are sold, 1 hardback must be sold.
Since 1 of every 3 books must be a hardback, the required number of hardbacks is equal to 1/3 of the 840 books sold:
(1/3)(840) = 280.
If MORE than 280 hardbacks are sold, then the total profit will be even GREATER THAN $2100, since hardbacks yield more profit than paperbacks.

Question stem, rephrased:
Were at least 280 hardbacks sold?

Even when the statements are combined, the following cases are possible:
Case 1: hardbacks = 300, paperbacks = 540.
Case 2: hardbacks = 200, paperbacks = 640.
Since more than 280 hardbacks are sold in Case 1, but fewer than 280 hardbacks are sold in Case 2, the two statements combined are NSUFFICIENT.

The correct answer is E.

For more practice with alligation, check here:
https://www.beatthegmat.com/ratios-fract ... 15365.html

tabsang wrote:

Cedagmat wrote:146. There are less than 50 books are to be divided by students. If the books are divided by 7 students, one book will be left. How many books are there?
(1) If the books are divided by 9 students, 8 books will be left.
(2) If the books are divided by 5 students, 3 books will be left.
[spoiler] Answer: A[/spoiler]

Let x = number of books.
x=7a+1

S1: x = 9b+8
set equations to equal each other to find integer possibilites
7a+1=9b+8
7a=9b+7
a=9b/7+1

pick b =7, a = 10, plug in a values. x=7(10)+1 = 71. Insufficient since question mentions no. of books <50.

S2: x=5c+3
7a+1=5c+3
a=(5c+2)/7
a works for values c at 1 and 6. Either 8 or 43. Not sufficient.

Does anyone have suggestions here?

From the question stem we have:
No. of books (n) = 7a + 1
Possible values of n = 1,8,15,22,29,36 and 43 ....(i)

Consider Statement 1:
n=9b+8
Possible values of n = 8,17,26,35 and 44 ....(ii)

Thus, number of books = 8 (common to both i & ii)
SUFFICIENT

Consider Statement 2:
n=5c+3
Possible values of n = 3,8,13,18,23,28,33,38,43 and 48 ....(iii)

There are 2 values that are common to i & iii i.e. 8 and 43
Thus NOT SUFFICIENT.

Ans: [spoiler](A)[/spoiler]

Hope this helps.

P.S: If you find this post to be helpful, please take a moment to hit the "Thanks" button

Hi, I have a question here, when you say that 8 is a possibility in S1, doesn't it mean that b will be equal to zero? Will that be the correct way of approaching the statement? I got confused at this step, all the rest of it, I solved just the same manner.

Q:
The original cost for paperback copy and hardback copy is $8 and $9.5, respectively. The sales price for the paperback copy and the hardback copy is $10 and $13, respectively. If a total of 834 books were sold out, was the total profit greater than $2,000?

1. More paperback copied were sold.
2. At least 100 hardback copies were sold.

A:
First - This is a yes/no question, so looking to see if the answers satisfy the ALWAYS YES or ALWAYS NO requirements in determining sufficiency.

Determined the profit for each book, as the question we're trying to answer involves determining profits from the sale of two kinds of books - $2 per paperback ("p") and $3.5 per hardcover ("h").

Will let the variables p and h stand for the quantity of each book sold.

So the condition in the problem that must be fulfilled: Is 2p +3.5h > 2000? We also know that the total combined quantity of books sold is 834: p + h = 834.

#1 tells me more paperbacks are sold than hardcovers. I look at the two extremes. First, if the bare minimum of paperbacks vs. hardcovers are sold to meet this condition (i.e. 418 paperbacks + 416 hardcovers = 834 books sold), we get a profit of:

2*(417) + 3.5*(416) = ?
836 + 1456 = A number greater than 2000, so yes, this meets the condition that profit must be greater than $2000.

However, I must also look at the second extreme. What if 834 paperbacks were sold and no hardcovers? Simple math says profit will be 2*(834) + 3.5*(0) = 1668. This is less than the $2000 in profit as a part of the condition.

Conclusion: A does not give us enough information to determine if profit is always greater or always less than $2000. We can eliminate A and D.

#2 At least 100 hardcovers were sold. Again, look at the two extremes. If 834 hardcovers were sold and 0 paperbacks, profit would be over $2000 ($3.5 * 834 = some number greater than $2400). On the other end of the spectrum, if we meet the bare minimum to satisfy this condition, 100 hardcovers and 734 paperbacks are sold:

2*(734) + 3.5*(100) = 1468 + 350 = a number less than $2000.

Conclusion: B does not give us enough information to say B is always greater or always less than $2000. We can eliminate B.

Lastly, if we combine the two, we have already done the work above to determine the two extremes. If the bare minimum of hardcovers is sold (100), then the number of paperbacks sold is 734, which is greater than the number of hardcovers, satisfying both conditions. The profit would be less than $2000 as we determined earlier. On the other end of the spectrum, we want to sell the most amount of hardcovers over the minimum of 100, while still making sure more paperbacks are sold than hardcovers. We solved this in #1: 2*(418) + 3.5*(416) = greater than $2000. The two extremes tell us profits can be above or below $2000, and thus, A and B combined is not enough information to determine if the original condition is ALWAYS YES or ALWAYS NO. We can eliminate C.

So, E is correct.