Question

1) The HCF and LCM of two numbers are 12 and 360. If one of the number is 60 find the other
number?​

Answer 1

Given:-

• LCM of two numbers:- 360
• HCF of two numbers:- 12
• One number:- 60

To Find:-

• Another Number ?

Solution:-

$\bold{\underline{\underline{\boxed{\sf{\orange{HCF × LCM = First \; number(a) × Second \; number(b)}}}}}}</u></p><p></p><p><u>[tex]• </u><u>\:</u><u> </u><u> </u><u>12 × 360 = 60 × b$

$• \: \sf \dfrac{12 × 360}{60} = b$

$• \: \sf \dfrac{12 × \cancel{360}}{ \cancel{60}}$

$\dag\large\bold{\underline{\underline{\boxed{\sf{\red{72}}}}}}$

$\\ \implies\large\bold{\underline{\underline{\sf{\purple{Hence, \; Another \; number \; is \; 72.}}}}}$

★ Verification:-

• HCF × LCM = a × b
• 12 × 360 = 60 × 72
• 4320 = 4320

$\implies \sf {LHS = RHS}$

$\implies\large\bold{\underline{\underline{\sf{\pink{Hence, \; Verified!}}}}}$

Answer 2

Answer:

Given:-

LCM of two numbers:- 360

HCF of two numbers:- 12

One number:- 60

To Find:-

Another Number ?

Solution:-

\bold{\underline{\underline{\boxed{\sf{\orange{HCF × LCM = First \; number(a) × Second \; number(b)}}}}}} < /u > < /p > < p > < /p > < p > < u > [tex]• < /u > < u > \: < /u > < u > < /u > < u > < /u > < u > 12 × 360 = 60 × b

HCF×LCM=Firstnumber(a)×Secondnumber(b)

</u></p><p></p><p><u>[tex]•</u><u></u><u></u><u></u><u>12×360=60×b

• \: \sf \dfrac{12 × 360}{60} = b•

60

12×360

=b

• \: \sf \dfrac{12 × \cancel{360}}{ \cancel{60}}•

60

12×

360

\dag\large\bold{\underline{\underline{\boxed{\sf{\red{72}}}}}}†

72

\begin{gathered} \\ \implies\large\bold{\underline{\underline{\sf{\purple{Hence, \; Another \; number \; is \; 72.}}}}}\end{gathered}

⟹

Hence,Anothernumberis72.

★ Verification:-

HCF × LCM = a × b

12 × 360 = 60 × 72

4320 = 4320

\implies \sf {LHS = RHS}⟹LHS=RHS

\implies\large\bold{\underline{\underline{\sf{\pink{Hence, \; Verified!}}}}}⟹

Hence,Verified!