Page 100 - New Medley 3 Semester 1
P. 100
We cannot subtract 1 from 0, so 0 has no predecessor.
Example 9: (a) Give the successor of 3195.
(b) Give the predecessor of 4007.
(c) Give two numbers just after 7099.
(d) Give two numbers just before 8078.
(e) Which number comes in between 5192 and 5194.
(f) Write the two numbers which comes in between 9176
and 9179.
Solution: (a) Successor of 3195 = 3195 + 1 = 3196.
(b) Predecessor of 4007 = 4007 – 1 = 4006.
(c) Successor of 7099 = 7099 + 1 = 7100.
Successor of 7100 = 7100 + 1 = 7101.
Required numbers are 7100 and 7101.
(d) Predecessor of 8078 = 8078 – 1 = 8077.
predecessor of 8077 = 8077 – 1 = 8076.
Required numbers are 8077 and 8076.
(e) 5192, 5193, 5194
Required number is 5193.
(f) 9176, 9177, 9178, 9179
Required numbers are 9177 and 9178.
Skip Counting
In skip counting we start with a given number and write more numbers
with a fixed gap between the successive numbers.
Counting by Twos
Gap of two between every two successive numbers. Suppose we start with
3583. Then counting by two’s we get: 3583, 3585, 3587 and so on.
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