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

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 21x and 21y

where x and y are prime to each other.

21x + 21y = 336

21 (x + y) = 336

x + y = $336/21$ = 16

∴ Possible pairs = (1, 15), (5, 11), (7, 9), (3, 13)

Answer: (a)

We find LCM of 30, 36 and 80.

LCM = 2 × 2 × 3 × 3 × 4 × 5 = 720

∴ Required number = 2 × 720 + 11

= 1440 + 11 = 1451

Answer: (d)

Let no. are x and y and HCF = A, LCM = B

Using Rule, we have

xy = AB

⇒ x + y = A + B (given) ...(i)

$(x–y)^2$ = $(x+y)^2$ – 4xy

or, $(x–y)^2$ = $(A+B)^2$ – 4 AB

$(x–y)^2$ = $(A–B)^2$

(x–y) = A – B ...(ii)

Using (i) and (ii), we get

x = A and y = B

∴ $A^3 + B^3 = x^3 + y^3$