If we add 53 and 53 like wise 40 and 40then subtract 106-80 the diff. Is 26 and at last adding 27 and 27 sums to 54 then subtract 26 it's 28 and 28 comes adding 14, 14.
Payal Mahajan said:
(Sat, May 21, 2011 01:55:05 AM)
Difference between 53 - 40 is 13 & 40 - 27 is 13.
So 27 - 13 = 14 is answer.
Anil said:
(Wed, May 25, 2011 06:57:34 AM)
What is the number after this series: 6, 3 ,3, 4.5, 9, 22.5, _?
Sundar said:
(Wed, May 25, 2011 09:43:21 AM)
Given:
6, 3, 3, 4.5, 9, 22.5, ?
Write the diff b/w numbers of the given series:
-3, 0, 1.5, 4.5, 13.5,
This is clue to find answer.
V.Vinodkumar said:
(Fri, Jul 8, 2011 01:51:01 AM)
@Anil
Your question is very good. The answer is 67.5
The series is 6,3,3,4.5,9,22.5,67.5
Explanation:
second number is divided by first number and so on that is
3/6=1/2=0.5
3/3 =1
4.5/3 =1.5
9/4.5 =2
22.5/9 =2.5
The division of difference is increased by o.5
The calculation of last number:
x/22.7 =3 because after 2.5 the difference is 3 thats why
x=3*22.7
x=67.5 this is answer
Majid Yaseen said:
(Mon, Dec 5, 2011 09:18:23 PM)
6,3,3,4.5,9,22.5 . . .?
The first value i.e 6 is multiplied by 0.5 to get the next number i.e. 3, now to get the next number after this 3 is multiplied by (0.5+0.5=1) to have the third number 3, and in a similar fashion it is multiplied by (0.5+0.5+0.5=1.5) to have the next number i.e. 3*1.5=4.5 and then 4.5 is multiplied by (0.5+0.5+0.5+0.5=2) to get 9 and finally now 9 is multiplied by (0.5+0.5+0.5+0.5+0.5=2.5) to get 22.5; now if you people look at the whole process, it is clear that the number which we multiplied to get the next number is increasing by 0.5,
So we must multiply (0.5+0.5+0.5+0.5+0.5+0.5=3) with 22.5 to get the next number which is 67.5
Simanchal said:
(Thu, Dec 15, 2011 04:32:03 PM)
53-13=40
40-13=27
27-13=14
Sahil Suman said:
(Tue, Apr 10, 2012 10:07:25 AM)
It is so easy. Subtract 13 from 53, 40, 27-13=14ans.
Cherry said:
(Mon, Jun 4, 2012 12:11:19 AM)
Here 53-13=40and then taken this no.twice.
40-13=27 next also taken twice.
27-13=14 is answer check it.