HCF and LCM Problems (Basic)

Sol : Option 2
25 = 5 × 5, 30 = 5 × 3 × 2, 35 = 5 × 7, 40 = 2 × 2 × 2 × 5
Required LCM = 2 × 2 × 2 × 5 × 5 × 3 × 7 = 4200

Sol : Option 4
H.C.F. of 639 and 1065 is 213. H.C.F. of 213 and 1491 is 213.

Sol : Option 3
H.C.F of 4/9, 10/21 and 20/63 = H.C.F of 4,10 and 20 / L.C.M of 9,21 and 63
= H.C.F of 4, 10 and 20 = 2 & L.C.M. of 9, 21 and 63 = 63. Required H.C.F. = 2/63

Sol : Option 4
The difference of requisite numbers must be 12 and each should be divisible by 12. Checking the options given, only the fourth option satisfies.

Sol : Option 4
Let the numbers be 29a and 29b. Then, 29a + 29b= 174 or 29(a + b) = 174 or, a + b = 174/29 = 6. Values of co-primes (with sum 6) is(1, 5).
So, the possible pairs of numbers is (29 x 1, 29 x 5) i.e. 29 and 145, which will become the answer.

Sol : Option 4
The side of the square slab is the H.C.F. of 544 and 374 cm i.e. 34.

Sol : Option 5
Let the numbers be 13x and 13y. So, 13X ×13 Y = 6760 X×Y=40.
Possible values of (x, y) are (1, 40); (2, 20); (4, 10); (5, 8) Only two acceptable values are (1, 40) and (5, 8).

Sol : Option 2
L.C.M. of 2, 3, 4, 6 = 12 ; 12 = 2² × 3. So, the required number =2² × 3 x 2 x 3² = 216.

Sol : Option 4
L.C.M. of 3, 4, 5, 6, 8 = 120 ; Required number =2 x 2 x 2 x 2 x 3 x 3 x 5 x 5 = 3600.

Sol : Option 5
Maximum possible length of such rod = (H.C.F. of 44, 22, 55)c m = 11cm.