Question 1

For any integers a and b with HCF (a, b) = 1, what is HCF (a + b, a - b) equal to?

Accepted Answer

Putting arbitrary values of a and b. IIIustration 1: Let a = 9 and b = 8 HCF (8 + 9, 9 - 8) HCF (17, 1) = 1 IIIustration 2: Let a = 23 and b =17 HCF (17 + 23, 23 -17) HCF ( 40, 6) = 2 Hence, HCF (a + b, a - b) can either be 1 or 2.

Question 2

6 bells commence tolling together and toll of intervals are 2, 4, 6, 8, 10 and 12 s, respectively. In 1 h, how many times, do they toll together?

Accepted Answer

LCM of 2, 4, 6, 8, 10, 12 =(2 x 2 x 3 x 2 x 5) = 120 ∴ After every 2 min, they toll together. ∴ Number of times they toll together in one hour = (60/2) + 1 times = 31 times

Question 3

The HCF and LCM of two number are 21 and 4641, respectively. If one of the numbers lies between 200 and 300, then find the two numbers.

Accepted Answer

Let the numbers be 21a and 21b, where a and b are co-primes. Then, 21a x 21b = (21 x 4641) ⇒ ab = 221 Two co-primes with product 221 are 13 and 17. ∴ Required number = (21 x 13, 21 x 17) = (273, 357)

Question 4

If (x - 6) is the HCF of x 2 - 2x - 24 and x 2- kx - 6, then what is the value of k?

Accepted Answer

Given that, (x - 6) is the HCF of x2 - 2x - 24 and x2 - kx - 6 i.e., (x - 6) is a factor of both expressions. Let f (x1) = x12 - 2x1 - 24 and f (x2) = x22 - kx2 - 6 Now f(x1) = f (x2) at (x1 = x2 = 6) ⇒ (6)2 - 2(6) - 24 = (6)2 - k(6) - 6 [by condition] ⇒ 0 = 30 - 6k ⇒ 6 k = 30 ∴ k = 5

Question 5

If a and b are positive integers, then what is the value of (a / HCF (a,b) , b/HCF (a,b) )?

Accepted Answer

HCF (a /HCF (a, b) , b/HCF (a,b) it value is always equal to 1; IIIustration 1 Let the two positive integers be a = 24 and b = 36. ∴ HCF (24/HCF (24,36) , 36/HCF (24,36) ⇒ HCF (24/12 , 36/12) ⇒ HCF (2, 3) = 1 IIIustration 2 Let the two positive integers be a = 13 and b = 17 ∴ HCF (13 / HCF (13,17) , 17/HCF (13,17) ⇒ HCF (13/1 , 17/1) = 1

Question 6

What is the HCF of 8 (N 5 - N 3 + N ) and 28 (N 6 + 1)?

Accepted Answer

Let p(N) = 8 (N5 - N3 + N) = 4 x 2 x N (N4 - N2 + 1) and q(N) = 28 (N6 + 1) = 7 x 4 [( N2)3 + (1)3] = 4 x 7 (N2 + 1) ( N4 - N2 + 1) ∴ HCF of p(N) and q (N) = 4 (N4 - N2 + 1)

Question 7

The HCF of (x 4 - y 4) and (x 6 - y 6) is

Accepted Answer

Let f (x) = (x4 - y4) = (x2 - y2) (x2 + y2) = (x - y) (x + y) (x2 + y2) and g(x) = (x6 - y6) = (x3)2 - (y3)2 = (x3 + y3) (x3 - y3) = (x + y) (x2 - xy + y2) (x - y)(x2 + xy + y2) = (x - y) (x + y) (x2 - xy + y2) (x2 + xy + y2) ∴ HCF of [f(x), g(x)] = (x - y) (x + y) = x2 - y2

Question 8

The HCF of (x 3 - y 2 - 2x) and (x 3 + x 2) is

Accepted Answer

Let f(x) = x3 - y2 - 2x = x( x2 - x- 2) = x( x2 - 2x + x - 2) = x {x (x - 2) + 1 (x -2)} = x (x +1) (x - 2) and g(x) = x3 + x2 = x2 (x +1) = x. x (x +1) ∴ LCM of [ f(x), g(x)] = x (x +1) . x . (x - 2) = x2(x + 1)(x - 2) = x2(x2 - x - 2) = x4 - x3 - 2x2

Question 9

What is the HCF of a 2 b 4 + 2 a 2 b 2 and (ab) 7 - 4a 2 b 2?

Accepted Answer

a2 b4 + 2 a2 b2 = a2 b2 ( b2 +2).......................(i) and (ab)7 - 4 a2 b9 = a7 b7 - 4 a2 b9 = a2 b2(a5 b5 - 4b7)................(ii) from Eqs. (i) and (ii) HCF = a2 b2

Question 10

Find the largest number which divides 1305, 4665 and 6905 leaving same remainder in each case. Also, find the common Remainder.

Accepted Answer

Given that x = 1305, y = 4665, z = 6905 Then , |x - y| = |1305 - 4665| = 3360 |y - z] = |4665 - 6905| = 2240 |z - x| =|6905 -1305| = 5600 ∴ Required number = HCF of 3360, 2240 and 5600 = 1120 On dividing 1305 by 1120, remainder is 185. On dividing 4665 by 1120, remainder is 185. On dividing 6905 by 1120, remainder is 185. ∴ Common remainder = 185

Question 11

LCD of 12 and 18

Accepted Answer

36

Question 12

Find the lowest common multiple of 24, 36 and 40.

Accepted Answer

2 | 24  -  36  - 40
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 2 | 12  -  18  - 20
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 2 |  6  -   9  - 10
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 3 |  3  -   9  -  5
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   |  1  -   3  -  5
   
L.C.M.  = 2 x 2 x 2 x 3 x 3 x 5 = 360.

Problems on H.C.F and L.C.M Questions