How many different three-digit numbers between 100 and 1,000 have 5 as the tens digit?
A. 9
B. 10
C. 50
D. 89
E. 90
What is the answer? and please explain, this question is really driving me crazy.
A. 9
B. 10
C. 50
D. 89
E. 90
What is the answer? and please explain, this question is really driving me crazy.
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E. 90
If you think about it, every number that has 5 as the tens digit will either be 50, 51, 52, and so on.
From 50-59, there are 10 numbers, and from 100-1000, there are 9 "sets" of these 50-59's, i.e.150-159, 250-259, etc.
Knowing this, you can just multiply 9 and 10, which gives you 90.
I think I'm right, but if I'm not, I'm terribly sorry... Good luck!
If you think about it, every number that has 5 as the tens digit will either be 50, 51, 52, and so on.
From 50-59, there are 10 numbers, and from 100-1000, there are 9 "sets" of these 50-59's, i.e.150-159, 250-259, etc.
Knowing this, you can just multiply 9 and 10, which gives you 90.
I think I'm right, but if I'm not, I'm terribly sorry... Good luck!
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This question is technically asking you how many three-digit numbers have 5 as the second digit.
So We have the following:
150 - 159 = Total of 10 numbers
Now there are 900 total three digit numbers between 100 and 1000, so just mutiply the 10 by 9 to get your answer.
10*9 = 90 numbers
E
So We have the following:
150 - 159 = Total of 10 numbers
Now there are 900 total three digit numbers between 100 and 1000, so just mutiply the 10 by 9 to get your answer.
10*9 = 90 numbers
E
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You are dealing with 9 sets of 100 (i.e. 100, 200, 300, etc.).
For each set of 100, you have a range from 50-59 (10 numbers) where there is a 5 as the tens digit.
So, 10*9=90
For each set of 100, you have a range from 50-59 (10 numbers) where there is a 5 as the tens digit.
So, 10*9=90