Question:

The only gift certificates that a certain store sold yesterday were worth either 100 each or 10 each. If the store sold a total of 20 gift certificates yesterday, how many gift certificates worth \$10 each did the store sell yesterday?
(1) The gift certificates sold by the store yesterday were worth a total of between 1,650 and 1,800.
(2) Yesterday the store sold more than 15 gift certificates worth 100 each.

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When solving problems with multiple conditions, sometimes combining the statements allows you to narrow down the solution, but each statement alone might not be enough.

Updated On: Oct 1, 2025

Statement (1) ALONE is sufficient, but statement (2) alone is not sufficient.
Statement (2) ALONE is sufficient, but statement (1) alone is not sufficient.
BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
EACH statement ALONE is sufficient.
Statements (1) and (2) TOGETHER are not sufficient

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The Correct Option is C

Solution and Explanation

Step 1: Analyze statement (1).
Let the number of \$100 certificates sold be $ x $, and the number of \$10 certificates sold be $ 20 - x $. The total value of the certificates sold is: \[ 100x + 10(20 - x) = 100x + 200 - 10x = 90x + 200 \] Statement (1) tells us that the total value of the certificates sold is between \$1,650 and \$1,800. This gives the inequality: \[ 1650 \leq 90x + 200 \leq 1800 \] Subtract 200 from both sides: \[ 1450 \leq 90x \leq 1600 \] Divide by 90: \[ \frac{1450}{90} \leq x \leq \frac{1600}{90} \quad \implies \quad 16.1 \leq x \leq 17.8 \] Since $ x $ must be an integer, $ x = 17 $. Therefore, the number of \$10 certificates sold is $ 20 - 17 = 3 $.
Thus, statement (1) alone is sufficient.
Step 2: Analyze statement (2).
Statement (2) tells us that more than 15 certificates worth \$100 each were sold. Thus, $ x>15 $. Since the total number of certificates sold is 20, the number of \$10 certificates sold is $ 20 - x $. If $ x = 17 $, then $ 20 - x = 3 $.
Thus, statement (2) alone is sufficient.
\[ \boxed{C} \]

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LIST-I (Infinite Series)	LIST-II (Nature of Series)
(A) $ 12 - 7 - 3 - 2 + 12 - 7 - 3 - 2 + \dots $	(II) oscillatory
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Match List-I with List-II and choose the correct option:

LIST-I (Set)	LIST-II (Supremum/Infimum)
(A) $ S = \{2, 3, 5, 10\} $	(III) Sup S = 10, Inf S = 2
(B) $ S = (1, 2] \cup [3, 8) $	(IV) Sup S = 8, Inf S = 1
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LIST-I (Infinite Series)	LIST-II (Nature of Series)
(A) \( 12 - 7 - 3 - 2 + 12 - 7 - 3 - 2 + \dots \)	(II) oscillatory
(B) \( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \)	(IV) conditionally convergent
(C) \( \sum_{n=0}^{\infty} \left( (n^3+1)^{1/3} - n \right) \)	(I) convergent
(D) \( \sum_{n=1}^{\infty} \frac{1}{n \left( 1 + \frac{1}{n} \right)} \)	(III) divergent

LIST-I (Set)	LIST-II (Supremum/Infimum)
(A) \( S = \{2, 3, 5, 10\} \)	(III) Sup S = 10, Inf S = 2
(B) \( S = (1, 2] \cup [3, 8) \)	(IV) Sup S = 8, Inf S = 1
(C) \( S = \{2, 2^2, 2^3, \dots, 2^n, \dots\} \)	(II) Sup S = 5, Inf S = -5
(D) \( S = \{x \in \mathbb{Z} : x^2 \le 25\} \)	(I) Inf S = 2

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The Correct Option is C

Solution and Explanation

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