plz anyone can solve then
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3 digit number is expressed as: 100a+10b+c
Here,c=0
100a+10b+0 is our number
Given, when left hand and middle digits are interchanged, the number diminishes by 180.
100a + 10 b + 0......(1)
100b + 10a + 0......(2)
90(a-b) = 180.... [given 1-2 =180]
a-b = 2 [so a>b]
(possible values: (8,6), (6,4), (4,2) )
So the number can be either of: 860,640,420.
Now, the second condition: left-hand digit is halved, 10's place and 1's place are interchanged, the number gets diminished by 454. Which means, our number has to be greater than 450(then only there is a possibility to obtain the required difference of 454)
Taking 860: 8/2=4..6 & 0 interchanged, so new number=406...860-406=454..condition satisfied..Hence the number is 860..
[Don't know whether this was the solving pattern you expected or you required.Hope it helps]
Here,c=0
100a+10b+0 is our number
Given, when left hand and middle digits are interchanged, the number diminishes by 180.
100a + 10 b + 0......(1)
100b + 10a + 0......(2)
90(a-b) = 180.... [given 1-2 =180]
a-b = 2 [so a>b]
(possible values: (8,6), (6,4), (4,2) )
So the number can be either of: 860,640,420.
Now, the second condition: left-hand digit is halved, 10's place and 1's place are interchanged, the number gets diminished by 454. Which means, our number has to be greater than 450(then only there is a possibility to obtain the required difference of 454)
Taking 860: 8/2=4..6 & 0 interchanged, so new number=406...860-406=454..condition satisfied..Hence the number is 860..
[Don't know whether this was the solving pattern you expected or you required.Hope it helps]
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