How many positive multiples of 10 that are less than 1000 are the sum of 4 consecutive integers
a)51 b)50 c)49 d)none

• Hi @Anika
  i ll explain the solution,after that let me know ur status
  
  in the question what they given is we have to find the sum of the 4 consecutive numbers that sum should be multiple of like 10 ie..10,20.....and should be less than 1000.
  so that we can conclude tat last 10 multiple before 1000 is 990
  then w.k.t 1+2+3+4=10
  6+7+8+9=30--->2
  11+12+13+14=50
  16+17+18+19=70
  like wise it will be repeat,
  based upon the given we need to find upto 990 how many time it occurs is
  so that use this formula to find last number in AP
  sn=a+(n-1)d
  we knw sn=990=>last number
  a=10=>initial number
  d=20==>diff b/w two consecutive numbers
  990=10+(n-1)20
  990=10+20n-20
  990=20n-10
  20n=1000
  n=50
  
  tat i hope,u ll get the concept now..
  Thanks
  JP
• 8 years agoHelpfull: Yes(25) No(1)
• 1+2+3+4=10; 6+7+8+9=30; likewise the series ll repeat..so try finding last number...using a+(n-1)d ..u ll endup getting
  990..then 990-10/20+1= 50 is the ans
• 8 years agoHelpfull: Yes(12) No(0)
• Guys 50 is the correct answer ....
  We can confirm it like upto 1000 ... so maximum = 246+247+248+249 = 990 ... so from 1 to 10 we get two multiples i.e 10 and 30 so from 1 to 100 we get 20 multiples ( 2 in each row so 10*2 =20) .... 101 to 200 we get antoher 20 multiples and from 201 to 250 we get another 10 mutiples ... 20+20+10 = 50
• 8 years agoHelpfull: Yes(10) No(1)
• 1+2+3+4=10
  6+7+8+9=30
  11+12+13+14=50
  16+17+18+19=70
  .................repeating a.p series
  till 246+247+248+249=990
  so,
  990=10+(n-1)20
  990=10+20n-20
  n=1000/20
  n=50
• 8 years agoHelpfull: Yes(7) No(0)
• mr ramlal cd is 20 not 40..answer will be 50..
• 8 years agoHelpfull: Yes(1) No(0)
• Why not 5+6+7+8
• 6 years agoHelpfull: Yes(1) No(0)
• ans d)none
  the sum of 4 consecutive integers will be multiple of 10, when the sum result's unit digit will b 0(zero).
  so the 1st result=1+2+3+4=10;
  2nd will be=11+12+13+14=50;
  3rd will be=21+22+23+24=90;
  .
  .
  last will be 990;
  so, this is an AP series.....which 1st digit a=10
  last digit l=990
  common diff d=40
  so if number of terms will be n then...
  l=a+(n-1)*d
  990=10+(n-1)*40
  n-1=980/40=49/2
  n=1+49/2=51/2=25.something=25(ans)
  
  
• 8 years agoHelpfull: Yes(0) No(13)
• guys please tell the simple solution.
  i didnt get anyone of urs.
  
  pls pls do the needful
• 8 years agoHelpfull: Yes(0) No(1)