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Numerical Ability
Sequence and Series
Q. A student starts at the year 2011 and counts backwards, 7 at a time, giving the sequence of years: 2011, 2004, 1997, ...
A year which she will count is
Read Solution (Total 14)
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- It backward by 7 so 2011/7 =287 & remainder will be 2 & it is greater than 0
or when we 287*7=2009 then 2011-2009=2 so she will count upto 2 - 11 years agoHelpfull: Yes(17) No(2)
- actually wat s de question here ? wat do we need to find ?
- 11 years agoHelpfull: Yes(7) No(0)
- 1990 Because there is a 7 year gap in between.The years are given at an interval of 7 years.
- 11 years agoHelpfull: Yes(2) No(2)
- 2 year........
- 11 years agoHelpfull: Yes(1) No(0)
- a+(n-1)d=l here a = 2011,d=-7 so solving this max value of n is 287 and the last year is 3
- 11 years agoHelpfull: Yes(1) No(2)
- up to last year is 3...........
- 11 years agoHelpfull: Yes(0) No(1)
- The answer should be 116
- 11 years agoHelpfull: Yes(0) No(1)
- 1990
1997-7=1990 - 11 years agoHelpfull: Yes(0) No(0)
- 1997-7=1990
- 11 years agoHelpfull: Yes(0) No(0)
- the question is not clear
- 11 years agoHelpfull: Yes(0) No(0)
- Answer is 2-year
- 11 years agoHelpfull: Yes(0) No(0)
- Ans : up to 2 years...student can count up to 286 times back
- 11 years agoHelpfull: Yes(0) No(0)
- can ne1 explain the ques clearly..
- 10 years agoHelpfull: Yes(0) No(0)
- arey bhai kehna jya
- 10 years agoHelpfull: Yes(0) No(0)
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