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16 Mar 2020, 23:43
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
75% (02:06) correct
25%
(02:16)
wrong
based on 343
sessions
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Time
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Not Attempted Yet
What is the sum of every integer between 100 and 300 that has a units digit of 3 ?
A) 2,730
B) 3,960
C) 4,450
D) 5,120
E) 7,920
A) 2,730
B) 3,960
C) 4,450
D) 5,120
E) 7,920
17 Mar 2020, 00:06
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CAMANISHPARMAR
Numbers to be added are
103 + 113 + 123 + 133 + ..... + 293
It's an Arithmetic Progression
Sum of Arithmetic Progression = (n/2) [First term + Last term]
where n - number of terms
Every 100 numbers have 10 such terms with unit digit 3
i.e. n = 20 as we are dealing with 200 consecutive numbers
hence, Sum = (20/2) [103+293] = 3960
Answer: Option B
General Discussion
09 Apr 2020, 04:43
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I solve as follow:
I if sum: 3+13+26+...93= 480,with this number I have to sum 100 in each one that is equal to 100*10+480=1480,
similar to 200= 480+2000, and 300=480+3000, the sum of 1480+2480+3480= 7740.
Please help me I'm confused and my numbers look consistently.
I if sum: 3+13+26+...93= 480,with this number I have to sum 100 in each one that is equal to 100*10+480=1480,
similar to 200= 480+2000, and 300=480+3000, the sum of 1480+2480+3480= 7740.
Please help me I'm confused and my numbers look consistently.
22 Aug 2020, 15:46
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jjcastiblancopa
jjcastiblancopa
The thing is that you are adding the 3XX numbers (303, 313, 323 ...) and the question asks until 300
meaning that the last number to add has to be 293.
because I didn't know what the formula was, I wrote all the numbers
103, 113, 123, 133, 143, 153, 163, 173, 183 193
203, 213, 223, 233, 243, 253, 263, 273, 283, 293
and add the as fast as I could. (it took me a little less than 2 minutes).
Hope this (not so efficient) way helps.
Will study the formula written above (
) 
