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Numerical Ability
Algebra
How many different integers can be expressed as the sum of three distinct numbers from the set
{3, 8, 13, 18, 23, 28, 33, 38, 43, 48}?
Read Solution (Total 9)
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- 22
No of terms = {(last - first )/ (difference)} + 1
={(129 - 24) / 5 } +1
=22
- 9 years agoHelpfull: Yes(8) No(4)
- From first set of integers sum of3+8+13=24
Last set sum of 38+43+48=129
So all integers expressed should be in between 24 and 129.
And all sum of integers have unit place of 9 and 4 so ...no.of itegers=(129-24)/(9-4)=21
Ans:21
- 9 years agoHelpfull: Yes(6) No(4)
- 53....The given series is alternate differance of 10..
- 9 years agoHelpfull: Yes(2) No(1)
- 10 p 3=720 is answer
- 9 years agoHelpfull: Yes(1) No(1)
- since we have to select combination of 3 distinct nos.
So , ans is 10c3=120 - 9 years agoHelpfull: Yes(1) No(1)
- sala jisne iss side ko banaya h wo bhi chutiya h , campusgate se seekho kuch , answer sahit h wo ,
- 9 years agoHelpfull: Yes(0) No(2)
- Simply 10c3
- 9 years agoHelpfull: Yes(0) No(1)
- 24,.......,129
24-29- 2 nos
30-39- 2 nos
.
.
.
last 120-129- 2
total= 24
- 9 years agoHelpfull: Yes(0) No(1)
- L-a/d+1=129-24/5+1=25+1=26...l=largest of three numbers 48,43,38...a=smallest of 3 numbers..
- 9 years agoHelpfull: Yes(0) No(0)
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