Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M
1.
The LCM of three different numbers is 120. Which of the following cannot be their HCF ?
(A) 8
(B) 12
(C) 24
(D) 35
Solution:
Since HCF is always a factor of LCM, we cannot have three numbers with HCF 35 and LCM 120.
2.
The greatest number which when divides 989 and 1327 leave remainder 5 and 7 respectively = ?
(A) 8
(B) 16
(C) 24
(D) 32
Solution:
989 - 5 = 984 1327 - 7 = 1320 Subtract the remainder from the number. HCF = (984, 1320) = 24 for greatest number take HCF of the numbers.
3.
LCM of two prime numbers x and y (x > y) is 161. Then the value of 3y - x is = ?
(A) -2
(B) -1
(C) 1
(D) 2
Solution:
HCF of two prime numbers is 1 Product of numbers = (1 × 161) = 161 Let the numbers be a and b. Then, ab = 161 Now, co-primes with product 161 are (1, 161) and (7, 23) Since x and y are prime numbers and x > y, we have x = 23 and y = 7 ∴ 3y - x = (3 × 7) - 23 = - 2
4.
The HCF of two numbers is 29, and the other factors of their LCM are 15 and 13. The larger of the two numbers is:
(A) 435
(B) 377
(C) 406
(D) 464
Solution:
Let the numbers are a and b HCF = 29 LCM = HCF (a × b) LCM = 29(15 × 13) Largest number = 15 × 29 = 435
5.
Which greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively ?
(A) 15
(B) 30
(C) 45
(D) 60
Solution:
Required number = HCF if (3026 - 11) and (5053 - 13) HCF of 3015 and 5040 = 45
6.
The LCM and ratio of four numbers are 630 and 2 : 3 : 5 : 7 respectively. The difference between the greatest and least numbers is = ?
(A) 6
(B) 14
(C) 15
(D) 21
Solution:
Let the numbers be 2x, 3x, 5x and 7x respectively Then their LCM = (2 × 3 × 5 × 7)x = 210x.[∴ 2, 3, 5, 7 are prime numbers] So, 210x = 630 or x = 3 ∴ The numbers are 6, 9, 15 and 21 Required difference = 21 - 6 = 15
7.
Product of two co-prime numbers is 117. Then their LCM is = ?
(A) 117
(B) 9
(C) 13
(D) 39
Solution:
HCF of co-prime number is always 1 ∴ Let number are = x & y respectively Product of number = xy xy = 117 (given) ∴ Product of number = LCM × HCF ⇒ LCM × 1 = 117 ⇒ LCM = 117
8.
When Seeta made necklaces of either 16 beads, 20 beads or 36 beads, not a single bead was left over. What could be the least number of beads Seeta had = ?
(A) 700
(B) 720
(C) 750
(D) 780
Solution:
Required number of beads = LCM of 16, 20, 36 = 720
9.
The least number which when increased by 5 is divisible by each one of 24, 32, 36 and 54 is = ?