Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M
81.
The LCM of two prime numbers x and y (x > y) is 161. The value of (3y - x):
(A) -2
(B) -1
(C) 1
(D) 2
Solution:
LCM of prime number x and y = 161 (given) ⇒ xy = 161 ⇒ xy = 23 × 7 ⇒ x = 23, y = 7 (∵ x > y given) So, the value of (3y - x) = 3 × 7 - 23 = 21 - 23 = -2
82.
Three number 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
Solution:
Let the numbers be 3x, 4x and 5x Then, their L.C.M. = 60x So, 60x = 2400 or x = 40 ∴ The numbers are (3 x 40), (4 x 40) and (5 x 40) Hence, required H.C.F. = 40
83.
Four bells begin to toll together and toll respectively at intervals of 6, 7, 8 and 9 seconds. In 1.54 hours, how many times do they toll together and in what interval (seconds)?
(A) 14, 504
(B) 14, 480
(C) 12, 504
(D) 16, 580
Solution:
Interval after which the bells will toll together = (LCM of 6, 7, 8, 9) sec = 504 sec. In 1.54 hours, they will toll together $$\eqalign{ & {\text{ = }}\left[ {\left( {\frac{{1.54 \times 60 \times 60}}{{504}}} \right) + 1} \right] \cr & {\text{ = 12 times}} \cr} $$
84.
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.
85.
The HCF and LCM of two numbers are 21 and 84 respectively. If the ratio of the two numbers is 1 : 4 then larger of the two numbers is = ?
(A) 48
(B) 12
(C) 84
(D) 108
Solution:
We know, LCM × HCF = 1st number × 2nd number Let 1st number = K 2nd number = 4K K × 4K = 21 × 84 K = 21 Then number = 21, 84 So, larger number = 84
86.
Three numbers are in the proportion of 3 : 8 : 15 and their LCM is 8280. What is their HCF?
(A) 60
(B) 69
(C) 76
(D) 57
Solution:
Three numbers are 3n, 8n, 15n Its LCM = 360n 360n → 8280 n → $$\frac{{8280}}{{360}}$$ n = 23 Only option B is the product of 23 ∴ 69 is the answer
87.
The HCF and LCM of two numbers are 12 and 924 respectively. Then the number of such pair is = ?
(A) 0
(B) 1
(C) 2
(D) 3
Solution:
HCF = 12 Let numbers are 12x and 12y respectively LCM ⇒ 12xy = 924(given) ⇒ xy = 77 ⇒ Possible pairs are = (1 × 77) (7 × 11) ∴ Only two pairs are possible
HCF of 3240, 3600, and a third number is 36 and their LCM is 2 ÃÂÃÂÃÂÃÂÃÂÃÂ 3 ÃÂÃÂÃÂÃÂÃÂÃÂ 5 ÃÂÃÂÃÂÃÂÃÂÃÂ 7 . The third number is = ?4522
(A) 22 × 35 × 72
(B) 22 × 53 × 72
(C) 25 × 52 × 72
(D) 23 × 35 × 72
Solution:
3240 = 23 × 34 × 5 3600 = 24 × 32 × 52 HCF = 36 = 22 × 32 Since H.C.F. is the product of lowest powers of common factors, so the third number must have (22 × 32) as its factors. Since L.C.M. is the product of highest powers of common prime factors, so the third number must have 35 and 72 as its factors. ∴ Third number = 22 × 35 × 72
90.
The least multiple of 13, which on dividing by 4, 5, 6, 7 and 8 leaves remainder 2 in each case is
(A) 2520
(B) 842
(C) 2522
(D) 840
Solution:
LCM of (4, 5, 6, 7, 8) = 4 × 5 × 6 × 7 = 840 ⇒ Required number = 840k + 2, which is divisible by 13 $$\eqalign{ & {\text{For }}\frac{{840k + 2}}{{13}},\,\,\left( {{\text{remainder}} = 0} \right) \cr & {\text{Remainder}} = \frac{{8k + 2}}{{13}} \cr} $$ Put k = 3 Then, remainder = 0 For least multiple value of k is minimum ⇒ at k = 3 we get 840k + 2 = 840 × 3 + 2 = 2520 + 2 = 2522