HCF and LCM MCQ Questions

Solution: Required number = H.C.F of (127 – 55), (175 – 127) and (175 – 55)

= H.C.F of 72, 48 and 120 = 24

Rules: The greatest number that will divide x, y and z leaving remainders a, b and c respectively = H.C.F of (x – a), (y – b) and (z – c).

Solution:  N = H.C.F of (4665 – 1305), (6905 – 4665) and (6905 – 1305)

= H.C.F. of 3360, 2240 and 5600 = 1120

∴ Sum of digits in N = (1 + 1 + 2 + 0) = 4

Rules: The greatest number that will divide x, y and z leaving remainders a, b and c respectively = H.C.F of (x – a), (y – b) and (z – c).

Solution: Required number = H.C. F of (410 – 7), (751 – 7) and (1030 – 7)

= H. C. F of 403, 744 and 1023 = 31

Solution: L.C.M. of 12, 18, 21, 30 = 2 × 3 × 2 × 3 × 7 × 5 = 1260

∴ Required number = (1260 ÷ 2) = 630

Solution: Least number of 5 digits is 10,000.

L.C.M. of 12, 15 and 18 is 180.

On dividing 10000 by 180, the remainder is 100.

∴ Required number = 10000 + (180 – 100) = 10080.

Solution: Let the numbers are 3x, 4x and 5x then their L.C.M = 60x

So, 60x = 2400 (given)

⇒x = 40

∴ The number are (3×40), (4×40), (5×40) = 120, 160, 200

Hence required H.C.F. = 40

Solution: we know, Product of two numbers = (L.C.M× H.C.F)

⇒ 297 × other number = 2376 × 33

⇒ Other number = $\frac{2376\times 33}{297}=264$

Solution: m×n = 6×210

Now, $\frac{1}{\mathrm{m}}+\frac{1}{\mathrm{n}}\,\,=\,\,\frac{\mathrm{m}+\mathrm{n}}{\mathrm{mn}}\,\,=\frac{72}{6\times 210}=\frac{2}{35}$

Solution: LCM of 16, 18 and 20 = 720

Required number = 720k + 4 where, k is a natural number to be divisible by 7,

(720 k + 4) will be a multiple of 7. Smallest value of k = 4

∴ Required number = 720 X 4 + 4 = 2884

Solution: Here, R = (48 – 38) = 10, (64 – 54) = 10, (90 – 80) = 10 and (120 – 110) = 10.

∴ Required number = (L.C.M of 48, 64, 90 and 120) – 10 = 2870

Rules: The least number which when divided by x, y and z leaves the remainder a, b and c respectively=L.C.M of (x, y and z) – R where R = (x – a) = (y – b) = (z – c)