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01 Oct 2018, 19:53
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
61% (02:41) correct
39%
(02:17)
wrong
based on 274
sessions
History
Date
Time
Result
Not Attempted Yet
How many three digit numbers are there whose first digit equals the sum of the second and third digits?
A. 18
B. 27
C. 36
D. 45
E. 54
A. 18
B. 27
C. 36
D. 45
E. 54
01 Oct 2018, 20:19
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gracie
In such questions, it is good to write down the ways as it does not entail too much of work...
Let the number be xyz
1) x is 9, so y+z=9
Ways : (9,0); (8,1);(7,2);(6,3);(5,4) ..
Each of the above 5 ways can be taken as 2 different arrangements.. 981 or 918
So 5*2=10
2) x is 8, so y+z=8
Ways : (8,0);(7,1);(6,2);(5,3);(4,4) ..
4 of the above 5 ways can be taken as 2 different arrangements..
So 4*2+1=9
3) x is 7, so y+z=7
Ways : (7,0);(6,1);(5,2);(4,3)
Each of the above 4 ways can be taken as 2 different arrangements..
So 4*2
4) x is 6, so y+z=6
Ways : (6,0);(5,1);(4,2);(3,3)....
3 of the above 4 ways can be taken as 2 different arrangements..
So 3*2+1=7
And so on
There is a pattern...
So 10+9+8+7+6+5+4+3+2=54
Last 2 is when x is 1...
101 or 110
E
02 Oct 2018, 16:33
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gracie
Find a pattern, and start with numbers that don't have many addends.
After you write the first digit, write all the second digits in ascending order, then all the third digits in descending order.
101
110
202
211
220
303
312
321
330
404
413
422
431
440
The pattern is
100s have 2 such numbers
200s have 3 " "
300s have 4
400s have 5
900s will have 10
Total = sum of integers 2 - 10 inclusive
(9 groups, and each group has +1 more than its first digit))
SUM of consecutive integers?
(Average) * (# of terms, n)
Average = \(\frac{First+ Last}{2}\)
\(n=\)(Last-First) + 1 =
[(10 -2 ) + 1]= 9
(\(\frac{First+ Last}{2}*n)\)
SUM: \(\frac{2+10}{2}*9)=(6*9)=54\)
Answer E
