Elitmus
Exam
Numerical Ability
Number System
Let Q be any number less than 200 which leaves remainder 2 when divisible by 5 or 7. What is the sum of all those numbers of Q?
Read Solution (Total 6)
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- least number which gives remainder of 2 when divided by 5 or 7 is the LCM of 5 and 7 plus 2 i.e 35+2=37.
now,second number will 72 and by this way,we can say that sum all number less than 200 is
37+72+107+142+177=535 - 8 years agoHelpfull: Yes(26) No(2)
- divisible by 5:2,7,12......197.sum of divisible 5 is 3980.
divisible by 7:2,9,16.....198.sum of divisible 7 is 2900.
divisible by 5,7:2,37,72,107,142,177. and sum is 537.
so answer is 3980+2900-537=6347.. - 8 years agoHelpfull: Yes(6) No(7)
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pbsv -54% is there any chance of getting calls - 8 years agoHelpfull: Yes(5) No(10)
- Numbers that leave the remainder 2 when divided by 5 are 7,12,17,22,27,32........,197 in AP where a=7,d=5
By calculating using the formula of nth term, we get n= 39
By using sum of n terms, s=n/2(2a+(n-1)d),s=3978
Numbers that leave the remainder when 2 divided by 7 are 9,16,23,30,........,198 in AP where a=9,d=7
By calculating using the formula of nth term, we get n= 28
By using sum of n terms, s=n/2(2a+(n-1)d),s=2898
Total sum=3978+2898=6876
- 8 years agoHelpfull: Yes(3) No(2)
- Suraj kumar dubey the question says when divisible by 5 or 7 i.e there is an or between 5 and 7. Your solution would have been correct if there was an and between 5 and 7. I hope every one understands the difference
- 8 years agoHelpfull: Yes(1) No(0)
- least number which gives remainder of 2 when divided by 5 or 7 is the LCM of 5 and 7 plus 2 i.e 35+2=37.
now,second number will 72 and by this way,we can say that sum all number less than 200 is
2+37+72+107+142+177=537 - 8 years agoHelpfull: Yes(0) No(0)
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