Practice MCQ Questions and Answer on Problems on H.C.F and L.C.M
61.
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 LCM of two numbers is 495 and their HCF is 5. If the sum of the numbers is 100, then their difference is = ?
(A) 10
(B) 46
(C) 70
(D) 90
Solution:
LCM = 495 HCF = 5 ( given ) ∴ Let numbers are = 5x and 5y ∴ LCM = 5xy ⇒ 5xy = 495 ⇒ xy = 99 ∴ Possible co-prime factors are 1, 99 9, 11 ∴ Possible numbers are 5x, 5y 45, 55 5, 495 Now given that sum of numbers = 100 So, required numbers are = (45, 55) ∴ Difference of numbers = 55 - 45 = 10
63.
The HCF of $$\frac{9}{{10}}{\text{,}}\frac{{12}}{{25}}{\text{,}}\frac{{18}}{{35}}$$ ÃÂÃÂ and $$\frac{{21}}{{40}}$$ is = ?
(A) $$\frac{3}{5}$$
(B) $$\frac{{252}}{5}$$
(C) $$\frac{3}{{1400}}$$
(D) $$\frac{{63}}{{700}}$$
Solution:
$$\eqalign{ & {\text{Required HCF }} \cr & {\text{ = }}\frac{{{\text{HCF of 9,12,18,21}}}}{{{\text{LCM of 10,25,35,40}}}} \cr & = \frac{3}{{1400}} \cr} $$
64.
A person has to completely put each of three liquids: 403 litres of petrol, 465 litres of diesel and 496 litres of Mobil Oil in bottles of equal size without mixing any of the above three types of liquids such that each bottle is completely filled. What is the least possible number of bottles required ?
(A) 34
(B) 44
(C) 46
(D) None of these
Solution:
For the least number of bottles, the capacity of each bottle must be maximum. Capacity of each bottle = HCF of 403 litres, 465 litres and 496 litres = 31 litres. Hence, required number of bottles $$\eqalign{ & {\text{ = }}\left( {\frac{{403 + 465 + 496}}{{31}}} \right) \cr & = \frac{{1364}}{{31}} \cr & = 44 \cr} $$
65.
If the product of three consecutive numbers is 210, then the sum of the smaller numbers is = ?
(A) 3
(B) 4
(C) 5
(D) 11
Solution:
210 = 21 × 10 = 7 × 3 × 2 × 5 take 2 and 3 with together then we find number is 5, 6, 7 which is consecutive number so, 1st + 2nd = 5 + 6 = 11
66.
Two numbers are in the ratio 3 : 4. The product of their HCF and LCM is 2028. The sum of the numbers is = ?
(A) 68
(B) 72
(C) 86
(D) 91
Solution:
Let the numbers are = 3x, 4x respectively ∴ HCF = x LCM = 3 × 4 × x = 12x given that = HCF × LCM => x × 12x = 2028 => 12x2 = 2028 => x2 = 169 => x = 13 ∴ sum of numbers = 3x + 4x = 7x = 7 × 13 = 91
67.
The least number, which when divided by 12, 15, 20 and 54 leaves in each case a remainder of 8 is:
(A) 504
(B) 536
(C) 544
(D) 548
Solution:
Required number = (L.C.M. of 12, 15, 20, 54) + 8 = 540 + 8 = 548
68.
The LCM of two numbers is 1820 and their HCF is 26. If one number is 130 then the other number is ?
(A) 70
(B) 1690
(C) 364
(D) 1264
Solution:
$$\eqalign{ & {\text{LCM = 1820}} \cr & {\text{HCF = 26}} \cr & {{\text{1}}^{{\text{st}}}}{\text{ number = 130}} \cr & {\text{LCM}} \times {\text{HCF }} \cr & {\text{ = Product of numbers}} \cr & \Rightarrow {\text{Let the other number is }}x{\text{ }} \cr & 130 \times x = 1820 \times 26 \cr & x = \frac{{1820 \times 26}}{{130}} = 364 \cr} $$
69.
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
70.
The maximum number of students among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets the same number of pens and same number of pencils is = ?
(A) 91
(B) 910
(C) 1001
(D) 1911
Solution:
Required number of students = HCF of 1001 and 910 = 91