ABC26GN0555 · H.C.F. and L.C.M. of Numbers
Subject: General Aptitude · Chapter: H.C.F. and L.C.M. of Numbers · Exam: 2011 · Marks: · Difficulty:
The least number, which when divided by 48, 60, 72, 108 and 140 leaves 38, 50, 62, 98 and 130 as remainders respectively is
(a)11115
(b)15110
(c)15120
(d)15210
Answer
Explanation
Here $(48-38)=10,(60-50)=10,(72-62)=10$ $$(108-98)=10 \&(140-130)=10 .$$ $\therefore \quad$ Required number $=($ L.C.M. of $48,60,72,108,140)$ $$\text { - } 10 \text { = } 15120 \text { - } 10 \text { = } 15110 .$$
Explanation as extracted from the printed page; notation may be imperfect.
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