Quantitative Aptitude
HCF AND LCM MCQs
Problems On Hcf And Lcm, H.C.F And L.C.M. Of Numbers, Lcm & Hcf, Hcf And Lcm
L.C.M. of 12, 18, 21 30 2 | 12 - 18 - 21 - 30
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= 2 x 3 x 2 x 3 x 7 x 5 = 1260. 3 | 6 - 9 - 21 - 15
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Required number = (1260 ÷ 2) | 2 - 3 - 7 - 5
= 630.
Let the numbers be 3x and 4x. Then, their H.C.F. = x. So, x = 4.
So, the numbers 12 and 16.
L.C.M. of 12 and 16 = 48.
Required number = (L.C.M. of 12,16, 18, 21, 28) + 7
= 1008 + 7
= 1015
Clearly, 252 = 2 x 2 x 3 x 3 x 7.
Since the numbers are co-prime, they contain only 1 as the common factor.
Also, the given two products have the middle number in common.
So, middle number = H.C.F. of 551 and 1073 = 29;
First number = \(\left(\frac{551}{29}\right) = 19\)
Third number = \(\left(\frac{1073}{29}\right) = 37.\)
Therefoer Required sum = (19 + 29 + 37) = 85.
36 = 22 x 32
84 = 22 x 3 x 7
Therefore H.C.F. = 22 x 3 = 12.
L.C.M. of 8, 16, 40 and 80 = 80.
\(\frac{7}{8}=\frac{70}{80}; \frac{13}{16}=\frac{65}{80}; \frac{31}{40}=\frac{62}{80}\)
since,\(\frac{70}{80}>\frac{65}{80}>\frac{63}{80} so, \frac{7}{8}>\frac{13}{16}>\frac{69}{80}>\frac{31}{40}\)
\(so,\frac{7}{8}is the largest\)
Required number = (L.C.M. of 12, 15, 20, 54) + 8
= 540 + 8
= 548.
Required number = H.C.F. of (1657 - 6) and (2037 - 5)
= H.C.F. of 1651 and 2032 = 127.
99 = 1 x 3 x 3 x 11
101 = 1 x 101
176 = 1 x 2 x 2 x 2 x 2 x 11
182 = 1 x 2 x 7 x 13
So, divisors of 99 are 1, 3, 9, 11, 33, .99
Divisors of 101 are 1 and 101
Divisors of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.