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15 Nov 2023, 02:30
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
57% (02:09) correct
43%
(02:07)
wrong
based on 176
sessions
History
Date
Time
Result
Not Attempted Yet
Mr. Smith purchases books from the bargain bin. He buys only books that cost either $1, $2, or $7 dollars. How many $7 books he buy?
(1) Mr. Smith spends $24 on bargain books.
(2) Mr. Smith buys a total of 10 books.
(1) Mr. Smith spends $24 on bargain books.
(2) Mr. Smith buys a total of 10 books.
15 Nov 2023, 04:48
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Mr. Smith purchases books from the bargain bin. He buys only books that cost either $1, $2, or $7 dollars. How many $7 books he buy?
(1) Mr. Smith spends $24 on bargain books.
(2) Mr. Smith buys a total of 10 books.
(1) Only
Mr. Smith could buy 0, 1, 2, or 3 $7 books.
Insufficient
(2) Only
Mr. Smith could buy 0, 1, 2, ... or 10 $7 books.
Insufficient.
(1)(2)Together
Mr. Smith could buy 2 $7 books. Then, 6 $1 books and 2 $2 books.
OR
Mr. Smith could buy 1 $7 book. Then, 1 $1 books and 8 $2 books.
Insufficient
The answer is thus (E).
(1) Mr. Smith spends $24 on bargain books.
(2) Mr. Smith buys a total of 10 books.
(1) Only
Mr. Smith could buy 0, 1, 2, or 3 $7 books.
Insufficient
(2) Only
Mr. Smith could buy 0, 1, 2, ... or 10 $7 books.
Insufficient.
(1)(2)Together
Mr. Smith could buy 2 $7 books. Then, 6 $1 books and 2 $2 books.
OR
Mr. Smith could buy 1 $7 book. Then, 1 $1 books and 8 $2 books.
Insufficient
The answer is thus (E).
14 Feb 2026, 02:55
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This is a really well-designed DS problem that tests integer constraint reasoning. The question type is "find a specific value," so each statement (or the combination) needs to give us exactly one possible answer.
Let a = number of $1 books, b = number of $2 books, c = number of $7 books. We need the value of c.
Statement 1: a + 2b + 7c = 24
We need to find all non-negative integer solutions for c. Testing: c = 0 works (e.g., a = 24, b = 0). c = 1 works (e.g., a = 15, b = 1). c = 2 works (e.g., a = 6, b = 2). c = 3 works (a = 3, b = 0). Multiple values of c, so this is not sufficient.
Statement 2: a + b + c = 10
This just tells us he bought 10 books total. c could be anything from 0 to 10. Not sufficient.
Statements 1 + 2 together:
From the two equations, subtract: (a + 2b + 7c) - (a + b + c) = 24 - 10, giving us b + 6c = 14. Now find non-negative integer solutions:
- c = 0: b = 14, but then a = 10 - 0 - 14 = -4. Not valid.
- c = 1: b = 8, a = 10 - 1 - 8 = 1. Valid.
- c = 2: b = 2, a = 10 - 2 - 2 = 6. Valid.
- c = 3: b = -4. Not valid (and anything higher won't work either).
We still get c = 1 or c = 2 — two possible answers. Not sufficient even together.
Answer: (E)
Common trap: Students often stop testing after finding one valid combination and assume it's sufficient, especially under time pressure. On DS "find the value" questions, you MUST keep checking until you either find a second valid answer (insufficient) or prove no other answer exists. The habit of always asking "could there be another solution?" is what separates 700+ scorers on these problems.
Takeaway: When you have two linear equations with three unknowns restricted to non-negative integers, simplify to one equation in two variables first, then systematically test integer values — don't just check one and move on.
Let a = number of $1 books, b = number of $2 books, c = number of $7 books. We need the value of c.
Statement 1: a + 2b + 7c = 24
We need to find all non-negative integer solutions for c. Testing: c = 0 works (e.g., a = 24, b = 0). c = 1 works (e.g., a = 15, b = 1). c = 2 works (e.g., a = 6, b = 2). c = 3 works (a = 3, b = 0). Multiple values of c, so this is not sufficient.
Statement 2: a + b + c = 10
This just tells us he bought 10 books total. c could be anything from 0 to 10. Not sufficient.
Statements 1 + 2 together:
From the two equations, subtract: (a + 2b + 7c) - (a + b + c) = 24 - 10, giving us b + 6c = 14. Now find non-negative integer solutions:
- c = 0: b = 14, but then a = 10 - 0 - 14 = -4. Not valid.
- c = 1: b = 8, a = 10 - 1 - 8 = 1. Valid.
- c = 2: b = 2, a = 10 - 2 - 2 = 6. Valid.
- c = 3: b = -4. Not valid (and anything higher won't work either).
We still get c = 1 or c = 2 — two possible answers. Not sufficient even together.
Answer: (E)
Common trap: Students often stop testing after finding one valid combination and assume it's sufficient, especially under time pressure. On DS "find the value" questions, you MUST keep checking until you either find a second valid answer (insufficient) or prove no other answer exists. The habit of always asking "could there be another solution?" is what separates 700+ scorers on these problems.
Takeaway: When you have two linear equations with three unknowns restricted to non-negative integers, simplify to one equation in two variables first, then systematically test integer values — don't just check one and move on.
