|
|
GMAT Club Daily Prep
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
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
26 May 2017, 04:44
Kudos
Bookmarks
We make 4 digit codes from the numbers 2, 3, 4, and 5. If none of each numbers are used more than once - for example, 2345 can be a code but 2234 cannot be a code - what is the sum of all the possible codes?
A. 92,324
B. 92,624
C. 93,124
D. 93,324
E. 93,424
A. 92,324
B. 92,624
C. 93,124
D. 93,324
E. 93,424
26 May 2017, 04:45
Kudos
Bookmarks
Official Solution:
We make 4 digit codes from the numbers 2, 3, 4, and 5. If none of each numbers are used more than once - for example, 2345 can be a code but 2234 cannot be a code - what is the sum of all the possible codes?
A. 92,324
B. 92,624
C. 93,124
D. 93,324
E. 93,424
For this question, since it means that the sum of all 4-digit integers are formed using 2, 3, 4, and 5, you can calculate the thousands digit of the code. If you set the thousands digit,
From \(2( )( )( ), 3*2*1 = 6\) outcomes are possible for each blank. In other words, there can be 6 integers that become 2,000.
From \(3( )( )( )( ), 3*2*1 = 6\) outcomes are possible for each blank. In other words, there can be 6 integers that become 3,000. Also,
\(4( )( )( )\) can have 6 integers that become 4,000 and
\(5( )( )( )\) can have 6 integers that become 5,000, so if you calculate the total sum where the thousands digit is set, you get \(2,000*6 + 3,000*6 + 4,000*6 + 5,000*6 = 14,000*6\).
You get the same result when the hundreds digits have set 2, 3, 4, and 5 and the sum is \(200*6 + 300*6 + 400*6 + 500*6 = 1,400*6\).
You get the same result when the tens digits have set 2, 3, 4, and 5 and the sum is \(20*6 + 30*6 + 40*6 + 50*6 = 140*6\).
You get the same result when the units digits have set 2, 3, 4, and 5, and the sum is \(2*6 + 3*6 + 4*6 + 5*6 = 14*6\).
If you add all the sums above, you get
\(14,000*6 + 1,400*6 + 140*6 + 14*6 = (14,000 + 1,400 + 140 + 14)6 = 93,324\). D is the answer.
Answer: D
We make 4 digit codes from the numbers 2, 3, 4, and 5. If none of each numbers are used more than once - for example, 2345 can be a code but 2234 cannot be a code - what is the sum of all the possible codes?
A. 92,324
B. 92,624
C. 93,124
D. 93,324
E. 93,424
For this question, since it means that the sum of all 4-digit integers are formed using 2, 3, 4, and 5, you can calculate the thousands digit of the code. If you set the thousands digit,
From \(2( )( )( ), 3*2*1 = 6\) outcomes are possible for each blank. In other words, there can be 6 integers that become 2,000.
From \(3( )( )( )( ), 3*2*1 = 6\) outcomes are possible for each blank. In other words, there can be 6 integers that become 3,000. Also,
\(4( )( )( )\) can have 6 integers that become 4,000 and
\(5( )( )( )\) can have 6 integers that become 5,000, so if you calculate the total sum where the thousands digit is set, you get \(2,000*6 + 3,000*6 + 4,000*6 + 5,000*6 = 14,000*6\).
You get the same result when the hundreds digits have set 2, 3, 4, and 5 and the sum is \(200*6 + 300*6 + 400*6 + 500*6 = 1,400*6\).
You get the same result when the tens digits have set 2, 3, 4, and 5 and the sum is \(20*6 + 30*6 + 40*6 + 50*6 = 140*6\).
You get the same result when the units digits have set 2, 3, 4, and 5, and the sum is \(2*6 + 3*6 + 4*6 + 5*6 = 14*6\).
If you add all the sums above, you get
\(14,000*6 + 1,400*6 + 140*6 + 14*6 = (14,000 + 1,400 + 140 + 14)6 = 93,324\). D is the answer.
Answer: D
Moderator:
