Let a and b be positive integers such that 90<a+b<99 and 0.9<a/b<0.91 Find ab/46
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Given : a and b be positive integers such that 90<a+b<99 and 0.9<a/b<0.91
To Find : ab/46
Solution:
90<a+b<99
0.9<a/b<0.91
0.9b < a
a < 0.91b
0.9b < a
a+b<99
=> a < 99-b
=> 0.9b < 99-b
=> 1.9b < 99
=> b < 52.1
a < 0.91b
90<a+b
=> 90 - b < a
=> 90 - b < 0.91b
=> 90 < 1.91b
=> 47.12 < b
47.12 < b < 52.1
=> b = 48 , 49 , 50 , 51 , 52
0.9b < a < 0.91b
b = 48
=> 43.2 < a < 43.68 no integer values of a
b = 49
=> 44.1 < a < 44.59 no integer values of a
b = 50
=> 45 < a < 45.5 no integer values of a
b = 51
=> 45.9 < a < 46.41 so a = 46
b = 52
=> 46.8 < a < 47.32 so a = 47
but 52 + 47 not less than 99.
b = 51 and a = 46 only satisfy
ab /46 = 46 * 51/46 = 51
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