LCM and HCF Questions Answers

129 − 3 = 126
and 281 − 5 = 276
Now, the H.C.F. of 126 and 276 are required number i.e., 6

LCM of the Fractions

$= frac{text{LCM of } 1,3,4,5}{text{HCF of } 2,7,21} = frac{60}{1} = 60$

$text{G.C.M } =frac{396times576}{6336} = 36$

Subtract 7 from 2930 and 11 from 3250, we have 2923 and 3239. The greatest number is the H.C.F. of 2923 and 3239 which is 79.

$begin{aligned} text{Given HCF } &= (x + 2) text{LCM } &= x^3 + 2x^2 - x - 2 &= (x + 2)(x - 1)(x + 1) text{ Now } p(x) q(x) &= text{ HCF and LCM} therefore p(x) &= (x + 2)(x - 1) = x^2 + x - 2 therefore q(x ) &= (x + 2) (x + 1) = x^2 + 3x + 2 end{aligned}$

Required number = HCF of (400-9, 435-10, 541-14)
= HCF of (391, 425, 527) = 17

Put arbitrary values of a and b.
Illustration 1 Let a = 9 and b = 8.
 $therefore$ HCF (8 + 9, 9 − 8) $Rightarrow$ HCF (17,1) = 1

Illustration 2 Let a = 23 and b = 17.
 $therefore$ HCF (17 + 23, 23 − 17) $Rightarrow$ HCF (40,6) = 2
Hence, HCF (a + b, a − b) can either be 1 or 2.

The maximum capacity of the vessel = HCF of 1653, 2261 and 2527 = 19

Given $x = 2^3 times 3^2 times 5^4$

and $y = 2^2 times 3^2 times 5 times 7$

 $therefore$ HCF $= 2^2 times 3^2 times 5 = 4 times 9 times 5 = 180$

Here, say $a = 2^3 times 3 times 5$   and
$b = 2^4 times 5 times 7$ , then
LCM $= 2^4 times 3 times 5 times 7$

The numbers are $13x$ and $15x$.
So, $x$ is the HCF. Now,
HCF $times$ LCM = Product of numbers

$begin{aligned} & x times 39780 = 13x times 15x & Rightarrow x times 39780 = 13x times 15 times x^2 & Rightarrow x = dfrac{39780}{13 times 15} = 204 end{aligned}$

$therefore$ Numbers are $13 × 204 = 2652$ and $15 × 204 = 3060$

Here, we know that
144 = 12 × 2 × 2 × 3
and 192 = 12 × 2 × 2 × 2 × 2

By taking option (d), 48 = 12 × 2 × 2

Hence, the value of $x$ will not be 48 otherwise the HCF of given numbers becomes 48.

LCM of 6, 9 and 12 = 36
So, number is the form of 36$p$ + 4.
Since, the required numbers are between 300 and 400.
$therefore p$ = 9 and 10
 $therefore$ Required sum = 328 + 364 = 692

Here,
$4 - 3 = 5 - 4 = 8 - 7 = 9 - 8 = 1$

Now, $4 = 2 times 2, 5 = 5$
$8 = 2 times 2 times 2, 9 = 3 times 3$

$therefore$ LCM $= 5 times 2 times 2 times 2 times 3 times 3 = 360$
Required number = 360 − 1 = 359

$a^2 b^4 + 2a^2b^2 = a^2b^2 (b^2 + 2)$  ...(i)

and $(ab)^7 - 4a^2b^9 = a^7b^7 - 4a^2b^9$

$= a^2b^2 (a^5b^5 - 4b^7)$ ...(ii)

From Eqs. (i) and (ii), we get HCF $= a^2b^2$

$begin{aligned} & text{Here, } 52 - 33 = 78 - 59 &= 117 - 98 = 19 &text{Now, } 52 = 13 times 2 times 2 &Rightarrow 78 = 13 times 2 times 3 &Rightarrow 117 = 13 times 3 times 3 &therefore text{LCM } = 13 times 2 times 2 times 3 times 3 = 468 end{aligned}$

 $therefore$ Required number $= 468 - 19 = 449$

Hence, the sum of the digits is 17.

Since, $m = 2n + 1$ is an odd integer, so its factors may be 1 or 3 and $k = 9 n + 4$ its factors may be 1, 2 and 4.
Hence, HCF of ($m, k$) is 1.