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

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: (a)

Using Rule 1,

Suppose 1st number is x then, 2nd number = 100 – x

∴ LCM × HCF = 1st number × 2nd number

→ 495 × 5 = x × (100 – x)

→ 495 × 5 = 100x – $x^2$

→ $x^2$ – 55x – 45x – 2475 = 0

→ (x – 45) (x – 55) = 0

→ x = 45 or x = 55

Then, difference = 55 – 45 = 10

Answer: (d)

210 = 2 × 3 × 5 × 7 = 5 × 6 × 7

∴ Required answer = 5 + 6 = 11

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 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

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