model 4 addition, subtraction, multiplication and division with lcm & hcf Practice Questions Answers Test with Solutions & More Shortcuts

Answer: (b)

Let the numbers be 7x and 7y

where x and y are co-prime.

Now, LCM of 7x and 7y = 7xy

∴ 7xy = 140 ⇒ xy = $140/7$ = 20

Now, required values of x and y whose product is 50 and are coprime, will be 4 and 5.

∴ Numbers are 28 and 35 which lie between 20 and 45.

∴ Required sum = 28 + 35 = 63.

Answer: (b)

Using Rule 1,

1st number × 2nd number = L.C.M. × H.C.F

Let the larger number be x.

Smaller number = x – 2

First number × Second number = HCF × LCM

→ x (x – 2) = 24

→ $x^2$ – 2x – 24 = 0

→ $x^2$ – 6x + 4x – 24 = 0

→ x (x – 6) + 4 (x – 6) = 0

→ (x – 6) (x + 4) = 0

→ x = 6 because x ≠ – 4

Answer: (c)

HCF = 12

∴ Numbers = 12x and 12y where x and y are prime to each other.

∴ 12x + 12y = 84

⇒ 12 (x + y) = 84

⇒ x + y = $84/12$ = 7

∴ Possible pairs of numbers satisfying this condition

= (1,6), (2,5) and (3,4). Hence three pairs are of required numbers.

Answer: (d)

Let the numbers be 10x and 10y

where x and y are prime to each other.

∴ LCM = 10 xy

→ 10xy = 120

→ xy = 12

Possible pairs = (3, 4) or (1, 12)

∴ Sum of the numbers = 30 + 40 = 70

Answer: (b)

Let the numbers be 48x and 48y

where x and y are co-primes.

48x + 48y = 384

→ 48 ( x + y) = 384

→ x + y = $384/48$ = 8 ........... (i)

Possible and acceptable pairs of x and y satisfying this condition are : (1, 7) and (3, 5).

∴ Numbers are : 48 × 1 = 48 and 48 × 7 = 336

and 48 × 3 = 144 and 48 × 5 = 240

∴ Required difference = 336 – 48 = 288