Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M
21.
The product of two numbers is 4107. If the H.C.F. of these numbers is 37, then the greater number is:
(A) 101
(B) 107
(C) 111
(D) 185
Solution:
Let the numbers be 37a and 37b Then, 37a x 37b = 4107 ⇒ ab = 3 Now, co-primes with product 3 are (1, 3) So, the required numbers are (37 x 1, 37 x 3) i.e., (37, 111) ∴ Greater number = 111
22.
Four metal rods of lengths 78 cm, 104 cm, 117 cm and 169 cm are to be cut into parts of equal length. Each part must be as long as possible. What is the maximum number of pieces that can be cut ?
(A) 27
(B) 36
(C) 43
(D) 480
Solution:
Maximum length of each part = HCF of 78 cm, 104 cm, 117 cm, 169 cm = 13 cm ∴ Number of pieces $$\eqalign{ & {\text{ = }}\left( {\frac{{78 + 104 + 117 + 169}}{{13}}} \right) \cr & = \frac{{468}}{{13}} \cr & = 36 \cr} $$
23.
If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to:
The greatest 4-digit number exactly divisible by 10, 15, 20 is = ?
(A) 9990
(B) 9960
(C) 9980
(D) 9995
Solution:
LCM of (10, 15, 20) ⇒ 5 × 2 × 3 × 2 = 60 ⇒ Largest 4-digit number = 9999 divide 9999 by LCM of given number ⇒ We get remainder = 39 ⇒ So, to divide completely subtract it from (9999 - 39) = 9960 ∴ 9960 is the largest four digit number which is completely divided by the given numbers.
25.
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.
26.
The ratio of two numbers is 13 : 15 and their LCM is 39780 . The numbers are = ?
(A) 884, 1020
(B) 884, 1040
(C) 670, 1340
(D) 2652, 3060
Solution:
Let the numbers be 13x and 15x Then, their LCM = 195x So, 195x = 39780 x = 204 ∴ The numbers are 2652 and 3060
27.
The sum of two numbers is 528 and their HCF is 33. The number of pairs of numbers satisfying the above conditions is = ?
(A) 4
(B) 6
(C) 8
(D) 12
Solution:
Let the required numbers be 33a and 33b. Then 33a + 33b = 528 ⇒ a + b = 16 Now co - primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9) ∴ Required numbers are (33 × 1, 33 × 15), (33 × 3, 33 × 13), (33 × 5, 33 × 11), (33 × 7, 33 × 9) The number of such pairs is 4.
28.
The ratio of two numbers is 3 : 4 and their LCM is 120. The sum of numbers is = ?
(A) 70
(B) 140
(C) 35
(D) 105
Solution:
Let the numbers are 3x and 4x So HCF = x ∴ HCF × LCM = Product of numbers ⇒ x × 120 = 3x × 4x ⇒ x × 120 = 12x2 ⇒ 120 = 12x x = 10 ∴ Numbers are 30 and 40 ∴ Sum of two numbers = 30 + 40 = 70
29.
Six bells commence tolling together and toll at intervals of 2, 4, 6, 8, 10 and 12 seconds respectively. In 30 minutes , how many times do they toll together ?
(A) 4
(B) 10
(C) 15
(D) 16
Solution:
LCM of 2, 4, 6, 8, 10, 12 is 120 So the bells will toll together after every 120 seconds, i.e., 2 minutes In 30 minutes, they will toll together $$\eqalign{ & = \left[ {\left( {\frac{{30}}{2}} \right) + 1} \right] \cr & = {\text{16 times}} \cr} $$
30.
A number between 1000 and 2000 which when divided by 30, 36 and 80 gives a remainder 11 in each case is = ?
(A) 11523
(B) 1451
(C) 1641
(D) 1712
Solution:
LCM of 30, 36, 80 = 720 Number = 720 × K + 11 (K = 2) Then the number = 720 × 2 + 11 = 1440 + 11 = 1451