The least number which on division by 25 leaves the remainder 15, on division by 35 leaves the remainder 25 and on division by 45 leaves the remainder 35 is
Explanation:
This is a direct application of LCM Model II, as the complement of the remainder is the same in all the three cases.
Complement of remainder = Difference between divisor and remainder
Case I d1 - r1 = 25 – 15 = 10 = x (say) Case II d2 - r2 = 35 – 25 = 10 Case III d3 - r3 = 45 – 35 = 10
Hence, smallest number = LCM(d1, d2, d3) - x
In this case, LCM of 25, 35 and 45 = 1575
Required number = 1575 – 10 = 1565
Alternatively, The answer should be such a number that when 10 is added to it, it is divisible by 25, 35 as well as 45, i.e. by 1575.
Add 10 to each option and check which option is divisible by 1575.
Hence, option (a).
» Your doubt will be displayed only after approval.
Help us build a Free and Comprehensive Preparation portal for various competitive exams by providing us your valuable feedback about Apti4All and how it can be improved.