CTS
Company
Numerical Ability
Arithmetic
The mean of 100 items was found to be 30. If at the time of calculation two items were wrongly taken as 32 and 12 instead of 23 and 11.What is the correct mean?
Read Solution (Total 11)
-
- the sum of 100 no=30*100=3000
as 23 is added as 32 & 11 is added as 12 so actually 10 is added more than the actual sum
so the sum should be 3000-10=2990
correct mean=29.90
- 11 years agoHelpfull: Yes(68) No(1)
- x/100=30
x=3000
3000-32-12+23+11=2990
2990/100=29.90
Ans:(29.90) - 11 years agoHelpfull: Yes(17) No(2)
- Ans is 29.9
- 11 years agoHelpfull: Yes(3) No(1)
- can you please explain me?
- 11 years agoHelpfull: Yes(3) No(3)
- x/100=30
x=3000
3000-9-1=2990
2990/100=29.9 - 11 years agoHelpfull: Yes(3) No(1)
- mean of 100 items (3000/100)is 30
now,difference is 10(32+12-23+11)
so,3000-10/100=2990/100=29.9 - 11 years agoHelpfull: Yes(3) No(1)
- no. of item = 100,
then mean = 30, so we get 100*30=3000 & reduce the wrong value so we obtained
3000-(32=12)=>ans=(23+11)=>2990/100=>29.9
answer=29.9 - 11 years agoHelpfull: Yes(2) No(3)
- sum of those 100 items/(100)=30
i.e, sum of those items=30*100=3000
now in this sum we have 2 incorrect values that is is 32 &12, remove them and add the correct values 23 & 11, then we get the sum of 100 items as 2990. for avg. divide it by the no. of items 100. - 11 years agoHelpfull: Yes(1) No(1)
- let the sum of 100 nos be x.
then x/100=30 be the first equation so x=3000
again (32-23)=9 and (12-11)=1 so 10 should be subtracted from sum to get the correct sum now,x-10/100=correct mean ...so correct mean =2990/100=29.90 - 10 years agoHelpfull: Yes(1) No(0)
- no. of item = 100,then mean(avg) = 30
so we get the sum of 100 items=30*100=3000
23 is added as 32 & 11 is added as 12
i.e., 3000-32-12+23-11=2990
correct mean=2990/100=29.90
Ans:(29.90) - 10 years agoHelpfull: Yes(1) No(1)
- x/100=30
x=3000
3000-32-12+23+11=2990
2990/100=29.90
Ans:(29.90) - 10 years agoHelpfull: Yes(0) No(0)
CTS Other Question