## Word Problems with Ratios and Remainders

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### Word Problems with Ratios and Remainders

by Cedagmat » Fri Nov 12, 2010 11:30 am
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.

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?

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by [email protected] » Sat Nov 13, 2010 12:50 am
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. Consider first (1) alone. It means minimum number of paperback copies sold was 834/2 + 1 = 418 and the maximum number sold was 834. If number of paperback copies sold was 418, the number of hardback copies sold was 416. So the total profit is (10 - 8)*418 + (13 - 9.5)*416 = 2292 > 2000. If the total number of paperback copies sold was 834, the number of hard back copies sold was 0. So the total profit is 2*834 = 1668 < 2000. Since nothing definite can be said, (1) alone is not sufficient. Next consider (2) alone. So the minimum number of hardback copies sold were 100 and the maximum number of hardback copies sold was 834. If no. of hardback copies sold was 100, the number of paperback copies sold was 734. So the total profit is 3.5*(100) + 2 * 734 = 1818 < 2000. If hard back copies sold were 834, the number of paperback copies sold was 0. The total profit is 3.5 * 834 = 2919 > 2000. Since nothing definite can be said, (2) alone is not sufficient. Next combine both statements together and check. On combining we have that the minimum number of hard back copies sold is 100 and the maximum is 416. When it is 100, total profit is 1818 < 2000 and when it is 416, total profit is 2292 > 2000. Again nothing definite can be said from both statements combined. The correct answer is hence (E). Rahul Lakhani Quant Expert Gurome, Inc. https://www.GuroMe.com On MBA sabbatical (at ISB) for 2011-12 - will stay active as time permits 1-800-566-4043 (USA) +91-99201 32411 (India) Junior | Next Rank: 30 Posts Posts: 21 Joined: 04 May 2010 Thanked: 4 times by yuliawati » Sat Nov 13, 2010 5:27 am 146. The question can be rephrased as "if X is divided by 7, the remainder is 1" and X<50. Lets try to make a list: The only possibilities are 8, 15, 22, 29, 36, and 43. Statement 1. X is divided by 9, the remainder is 8. The only possibilities are 8, 17, 26, 35, and 44. 8 is the only number in both list. Thus, there are 8 books (Sufficient). Statement 2. X is divided by 5, the remainder is 3. Again, make a list: 8, 13, 18, 23, 28, 33, 38, and 43. The number of books can be either 8 and 43. Hence, not sufficient. Answer: A For the first problem, I prefer to simplify the question first. We know that Profit = Revenue - Cost Lets say X = paperback copy; Y = hardback copy Total Cost = 8X + 9.5Y Total Revenue = 10X + 13Y Profit = Total cost - Total profit = 10X + 13Y - (8X + 9.5Y) = 2X + 3.5Y The question can be rephrased as: "Is 2X + 3.5Y > 2000"? Statement 1. More paperback copies were sold. (X>Y). Not sufficient. X and Y can be any number. Statement 2. At least 100 hardback copies (Y) were sold. Using the formula above, If Y = 100, X = 2, Profit < 2000. But if Y = 10.000, Y = 0, sufficient. The statement gives us Yes and No, thus insufficient. Statement 1 and 2, we know X>Y, and Y at least 100. If Y = 100, X = 200, the profit is <2000. But, if Y = 1000, X = 2000, the profit > 2000. Thus, insufficient. Answer: E Junior | Next Rank: 30 Posts Posts: 16 Joined: 03 Sep 2010 by Cedagmat » Sat Nov 13, 2010 3:29 pm Going over your posts, I realized one of the biggest mistakes I always make in reading world problems, not reading the information correctly. (i.e. I calculated the wrong profit margins and threw my entire solution off). Newbie | Next Rank: 10 Posts Posts: 3 Joined: 07 Sep 2010 Thanked: 3 times GMAT Score:690 by anilmac » Fri Dec 10, 2010 1:17 pm Here's how i approached the question - we are given P + H = 834; we are asked is 2P + 3.5H > 2000. S1 - P > H meaning P could be between 833 and 417; using the two extermes in the expression (from the stem) will give us if P = 418 then H = 416 using the values in 2P + 3.5H > 2000 gives us a result > 2000 if P = 833 then H = 1 using the values in 2P + 3.5H > 2000 gives us a result < 2000 so S1 is not suff. S2 - H >= 100 meaning H could be between 834 and 100; using the two extermes in the expression (from the stem) will give us if H = 100 then P = 734 using the values in 2P + 3.5H > 2000 gives us a result < 2000 if H = 834 then P = 0 using the values in 2P + 3.5H > 2000 gives us a result > 2000 so S2 is not suff. S1 + S2 combined H could be between 416 and 100 P could be between 734 and 418 using the two extermes in the expression (from the stem) will give us values > 2000 and < 2000 so the two combined are insuff. Hope this is helpfull. MAC Senior | Next Rank: 100 Posts Posts: 43 Joined: 06 Nov 2010 by Woozler » Tue Dec 14, 2010 11:10 am "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. " Where is this problem from? It should be "fewer than 50 books" and I doubt "divided by 7 students" is a phrase that's ever going to appear on GMAT. It should be "divided equally among 7 students", or something along those lines. Given that, I solved it just by plugging-n-chugging. There are only 6 numbers below 5 that are divisible by 7 with a remainder of 1. Senior | Next Rank: 100 Posts Posts: 43 Joined: 15 Jan 2011 Thanked: 9 times by tgou008 » Mon Mar 21, 2011 7:10 am IMO E. This is how I approached the problem. After reading the question I wrote down the following: P + H = 834 ($10P - $8P) + ($13H - $9.50H) < 2000? This can be further simplified down to$2P + $3.50H < 2,000? Where P = paperback and H = hardback I actually then turned to Stmt (2) H >= 100, therefore P <= 734 (100 x 3.5) + (734 x 2) = 1,818, which is <2k However, if say H = 834 and P = 0 then (834 x 3.5) + (0 x 2) = 2,549 which is > 2k Therefore Stmt 2 is insufficient - Cross out B and D Now, Statement 1 P > H We can test this quickly using extremes e.g.1 , P = 834, H = 0. Then profit = 834 x 2 = <2K e.g.2, P = 418, H = 416. Then profit = (418 x 2) + ( 416 x 3.5) = > 2k Therefore Stmt 1 is INSUFF. Cross out A Looking at Stmt 1 and 2 together we know P > H P <= 734 H >= 100 Therefore, 418<= P <=734 and 100<= H <=416 We know from above that if P is 418 and H is 416 then the profit >$2k
AND we also know that if H >= 100, therefore P <= 734
(100 x 3.5) + (734 x 2) = 1,818, which is <2k

We can therefore cross out C. The correct answer is E

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by ramapra » Sat Nov 26, 2011 11:25 pm
This is how I approached it .

For each Paper back copy sold , the profit is $2 For each Hardback copy sold , the profit is$3.5

If all books sold were PP , then profit = 1668
if all books sold were HB , then profit = 2919 .

The data provided in A and B are each not sufficient to conclusively decide one way or another , hence E

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by amit2k9 » Sun Nov 27, 2011 9:47 pm
total profit = 2p + 3.5(834-p)
1.5 p + 3.5 * 834 > 2000 is the condition.

none of the options as what is p.

hence E.
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by immaculatesahai » Fri Dec 09, 2011 12:23 am
+1 for E.

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by ronnie1985 » Fri Mar 30, 2012 10:14 am
S1 does not give any clue at all
S2 makes sure that the profit is at least 1818.
Combining does not help
(E)

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by surmilsehgal » Sat Aug 11, 2012 1:48 am

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by varun289 » Fri Oct 26, 2012 7:27 am
can any body answer , what will be GMAT score for this type questions particularly ?

is it tough question ?

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by Kshitij Sharma » Tue Dec 25, 2012 9:04 pm
E it is!

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by tabsang » Wed Dec 26, 2012 5:21 am
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.

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.

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