Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized for You
we will pick new questions that match your level based on your Timer History
Track Your Progress
every week, we’ll send you an estimated GMAT score based on your performance
Practice Pays
we will pick new questions that match your level based on your Timer History
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:
A collection of 16 coins, each with a face value of either 10 cents or [#permalink]
Show Tags
14 Jun 2016, 14:09
5
This post received KUDOS
7
This post was BOOKMARKED
00:00
A
B
C
D
E
Difficulty:
15% (low)
Question Stats:
82% (01:20) correct 18% (01:25) wrong based on 827 sessions
HideShow timer Statistics
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?
A) 3 B) 5 C) 7 D) 9 E) 11
One solution is to start TESTING answer choices.
A) 3 If there are 3 quarters (worth $0.25 each), and we have a total of 16 coins, the remaining 13 coins must be dimes (worth $0.10 each) 3 quarters = $0.75 and 13 dimes = $1.30 So the TOTAL value = 0.75 + $1.30 = $2.05 NO GOOD - we want a total of $2.35
B) 5 If there are 5 quarters (worth $0.25 each), and we have a total of 16 coins, the remaining 11 coins must be dimes (worth $0.10 each) 5 quarters = $1.25 and 11 dimes = $1.10 So the TOTAL value = 1.25 + $1.10 = $2.35 BINGO!
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?
A) 3 B) 5 C) 7 D) 9 E) 11
Another approach is to use algebra:
Let Q = # of quarters ($0.25) So, 16 - Q = # of dimes ($0.10)
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?
A) 3 B) 5 C) 7 D) 9 E) 11
We can let the number of 10-cent coins = d and the number of 25-cent coins = q. We are given that there are 16 total coins; thus, d + q = 16.
We are also given that the total face value is $2.35, thus:
0.1d + 0.25q = 2.35
10d + 25q = 235
Isolating d in our first equation, we have d = 16 - q. We can substitute 16 - q for d in our second equation and we have:
10(16 - q) + 25q = 235
160 - 10q + 25q = 235
15q = 75
q = 5
There are 5 coins with a face value of 25 cents.
Answer: B
_________________
Scott Woodbury-Stewart Founder and CEO
GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35.
Let x: no coins with 10 cents as face value Let y: no coins with 25 cents as face value
I am clear about forming x + y = 16
How can I be sure to form which of two below algebraic equations?
0.1 x + 0.25 y = 2.35 OR
0.25 x + 0.1 y = 2.35
Since I am getting y = 5 (or x= 11) in first case and (x=5 or y=11) in second case. What mistake I am making since I must get an unique solution from answer choice in PS.
_________________
It's the journey that brings us happiness not the destination.
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35.
Let x: no coins with 10 cents as face value Let y: no coins with 25 cents as face value
I am clear about forming x + y = 16
How can I be sure to form which of two below algebraic equations?
0.1 x + 0.25 y = 2.35 OR
0.25 x + 0.1 y = 2.35
Since I am getting y = 5 (or x= 11) in first case and (x=5 or y=11) in second case. What mistake I am making since I must get an unique solution from answer choice in PS.
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35.
Let x: no coins with 10 cents as face value Let y: no coins with 25 cents as face value
I am clear about forming x + y = 16
How can I be sure to form which of two below algebraic equations?
0.1 x + 0.25 y = 2.35 OR
0.25 x + 0.1 y = 2.35
Since I am getting y = 5 (or x= 11) in first case and (x=5 or y=11) in second case. What mistake I am making since I must get an unique solution from answer choice in PS.
A collection of 16 coins, each with a face value of either 10 cents or [#permalink]
Show Tags
18 Mar 2018, 07:47
AbdurRakib wrote:
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?
A) 3 B) 5 C) 7 D) 9 E) 11
We know x + y = 16 ----- I
x = 16 - y -----(II)
Now, 0.10x + 0.25y = 2.35
0.10(16 - y) + 0.25y = 2.35
1.6 - 0.10y +0.25y = 2.35
1.6 - 0.15y = 2.35
y = 5
(B)
_________________
"Success is not as glamorous as people tell you. It's a lot of hours spent in the darkness."
Re: A collection of 16 coins, each with a face value of either 10 cents or [#permalink]
Show Tags
07 Apr 2018, 07:02
GMATPrepNow wrote:
AbdurRakib wrote:
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?
A) 3 B) 5 C) 7 D) 9 E) 11
One solution is to start TESTING answer choices.
A) 3 If there are 3 quarters (worth $0.25 each), and we have a total of 16 coins, the remaining 13 coins must be dimes (worth $0.10 each) 3 quarters = $0.75 and 13 dimes = $1.30 So the TOTAL value = 0.75 + $1.30 = $2.05 NO GOOD - we want a total of $2.35
B) 5 If there are 5 quarters (worth $0.25 each), and we have a total of 16 coins, the remaining 11 coins must be dimes (worth $0.10 each) 5 quarters = $1.25 and 11 dimes = $1.10 So the TOTAL value = 1.25 + $1.10 = $2.35 BINGO!
GMATPrepNow, How and When do you we know that we can solve a word problem (or any other problem for that matter) by using Testing "The Answer Choices Method"
A collection of 16 coins, each with a face value of either 10 cents or 25 cents, has a total face value of $2.35. How many of the coins have a face value of 25 cents?
A) 3 B) 5 C) 7 D) 9 E) 11
One solution is to start TESTING answer choices.
A) 3 If there are 3 quarters (worth $0.25 each), and we have a total of 16 coins, the remaining 13 coins must be dimes (worth $0.10 each) 3 quarters = $0.75 and 13 dimes = $1.30 So the TOTAL value = 0.75 + $1.30 = $2.05 NO GOOD - we want a total of $2.35
B) 5 If there are 5 quarters (worth $0.25 each), and we have a total of 16 coins, the remaining 11 coins must be dimes (worth $0.10 each) 5 quarters = $1.25 and 11 dimes = $1.10 So the TOTAL value = 1.25 + $1.10 = $2.35 BINGO!
GMATPrepNow, How and When do you we know that we can solve a word problem (or any other problem for that matter) by using Testing "The Answer Choices Method"
For most word problems, you should at least consider the possibility of testing the answer choices. HOWEVER this doesn't necessarily mean that testing the answer choices will be the best/fastest approach.
I'd say that, in most cases, testing the answer choices is not the best approach. That said, it's a great strategy if you can't think of any other ways to solve the question.
In other situations, testing the answer choices is impossible. For example:
In how many ways can we arrange the letters in the word FAST? A) 4 B) 8 C) 12 D) 16 E) 24
As you can see, there is no way to actually test the answer choices for this question.
82% (01:20) correct 
