model 4 addition, subtraction, multiplication and division with lcm & hcf Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (d)

Firstly, we find the LCM of 30, 36 and 80.

LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720

Required number = Multiple of 720 = 720 × 5 = 3600;

because 3000 < 3600 < 4000

Answer: (b)

Let the numbers be 3x and 3y.

∴ 3x + 3y = 36

⇒ x + y = 12 ... (i)

and 3xy = 105 ... (ii)

Dividing equation (i) by (ii), we have

$\text"x"/ \text"3xy" + \text"y"/ \text"3xy" = 12/105$

⇒ $1/\text"3y" + 1/\text"3x" = 4/35$

Answer: (a)

Using Rule 1,

Let the HCF of numbers = H

Their LCM = 12H

According to the question,

12H +H = 403

⇒ 13H = 403

⇒ H = $403/13$ =31

⇒ LCM = 12 × 31

Now, First number × second number

= HCF × LCM

= 93 × Second Number

= 31 × 31 × 12

Second number = ${31 × 31 × 12}/93$ = 124

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: (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: (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