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18 Jan 2022, 05:15
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There are 6 notes, one $50, one $25, one $15, and three $10 notes. If 3 notes have to be chosen from this. The number of unique possible sums of the chosen 3 notes is?
A. 8
B. 10
C. 20
D. 24
E. 120
Are You Up For the Challenge: 700 Level Questions
A. 8
B. 10
C. 20
D. 24
E. 120
Are You Up For the Challenge: 700 Level Questions
20 Jan 2022, 00:48
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Bunuel
A quick observation would get you down to two options.
The least SUM would be 10+10+10=30 and maximum would be 50+25+15=90.
With all the notes multiple of 5, we will have SUM too as multiple of 5, so at the MAX, distinct possibilities are \(\frac{90-30}{5}+1=13\).
Only A and B are left.
Let us do it in a simple way by taking different cases:-
(1) If only 10s are chosen, the sum is 30...........30
(2) If we choose 15 and other two from 10s, then only possibility will be 15+10+10 or 35..........35
(3) If we choose 25 and the other two from 10s and 15, then maximum will be 25+15+10 or 50 and least will be 25+10+10 or 45.............45 and 50
(4) If we choose 50 and the other two from 10s, 15 and 25, then maximum will be 50+25+15 or 90 and least will be 50+10+10 or 70.............
cases we get - 70, 75, 80, 85, 90
50+10+10=70
50+10+15=75
50+10+25=85
50+25+15=90
Total 8 cases, so A.
NOTE: Some more analysis, may help somewhere else
All three EVEN numbers: None of the terms are multiple of 4, so TWO of them will add up to a multiple of 4, but THREE even numbers will always add upto a non multiple of 4.
One odd and two even: Will give odd integer.
Two odd and one even: Possibility 25+15+10 or 25+15+50.
Thus we can straight way discard multiples of 4, that is 40, 60 and 80
07 May 2024, 22:20
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