HCF and LCM MCQ Questions

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Practice HCF and LCM MCQ Questions with Details Solution

 

Q11. Find the greatest number that will divide 55, 127 and 175, so as to leave the same remainder in each case.

(a) 11

(b) 16

(c) 18

(d) 24

Answer: (d) 24

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).

Q12. Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits is N is:

(a) 4

(b) 5

(c) 6

(d) 8

Answer: (a) 4

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).

Q13. The greatest number which will divide 410, 751 and 1030 leaving a remainder 7 in each case is

(a) 29

(b) 31

(c) 17

(d) 37

Answer: (b) 31

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

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

Q14. What will be the least number which when doubled will be exactly divisible by 12, 18, 21 and 30?

(a) 196

(b) 630

(c) 1260

(d) 2520

Answer: (b) 630

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

∴ Required number = (1260 ÷ 2) = 630

Q15. The least number of five digits which is exactly divisible by 12, 15 and 18, is:

(a) 10010

(b) 10051

(c) 10020

(d) 10080

Answer: (a) 10010

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.

Q16. Three numbers are in the ratio of 3: 4: 5 and their L.C.M is 2400. Their H.C.F is:

(a) 40

(b) 80

(c) 120

(d) 200

Answer: (a) 40

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

Q17. The LCM of two numbers is 2376 while their HCF is 33. If one of the number is 297, then the other number is

(a) 216

(b) 264

(c) 642

(d) 792

Answer: (b) 264

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

⇒ 297 × other number = 2376 × 33

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

Q18. The HCF and LCM of two numbers m and n are respectively 6 and 210. If m + n = 72 then

$\frac{1}{\mathrm{m}}+\frac{1}{\mathrm{n}}$ is equal to

(a) $\frac{1}{35}$

(b) $\frac{2}{35}$

(c) $\frac{3}{35}$

(d) $\frac{5}{37}$

(e) None of the above

Answer: (b) $\frac{2}{35}$

Solution: m×n = 6×210

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

Q19. Find the least number which when divided by 16, 18 and 20 leaves a remainder 4 in each case, but is completely divisible by 7.

(a) 2884

(b) 2256

(c) 865

(d) 3332

(e) None of the above

Answer: (a) 2884

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

Practice HCF and LCM MCQ Questions with Details Solution

 

Q20. The least number which when divided by 48, 64, 90, 120 will leave the remainders 38, 54, 80, 110 respectively, is

(a) 2870

(b) 2860

(c) 2890

(d) 2880

Answer: (b) 2860

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)

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