Aptitude HCF AND LCM Practice QA-Easy

Required number = L.C.M. of $\left(12,16, 18, 21, 28\right)$ + 7

   = 1008 + 7

   = 1015

L.C.M. of 2, 4, 6, 8, 10, 12 is 120.

So, the bells will toll together after every 120 seconds [2 minutes].

In 30 minutes, they will toll together$ \dfrac{30}{2} $+ 1 = 16 times.

Required number = H.C.F. of $\left(1657 - 6\right)$ and $\left(2037 - 5\right)$

  = H.C.F. of 1651 and 2032 = 127.

36 = 2² x 3²

84 = 2² x 3 x 7

$\therefore$ H.C.F. = 2² x 3 = 12.

Required Number

= LCM of [8, 12, 22 and 24] + 12

= 264 + 12 = 276

Clearly, the numbers are $\left(23 \times 13\right)$ and $\left(23 \times 14\right)$.

$\therefore$ Larger number = $\left(23 \times 14\right)$ = 322.

It is clear that 504 = 2 × 2 × 2 × 3 × 3 × 7

Let the numbers be 2 x and 3 x .

Then, their L.C.M. = 6 x .

So, 6 x = 48 or x = 8.

$\therefore$ The numbers are 16 and 24.

Hence, required sum = $\left(16 + 24\right)$ = 40.

HCF of fractions = $\dfrac{\text{HCF of Numerators}}{\text{LCM of Denominators}}$

HCF of $\dfrac{1}{3}$ , $\dfrac{2}{3}$ and $\dfrac{1}{4}$

$=\dfrac{1}{12}$

Product of two numbers = Product of their HCF and LCM.

Let one number $=x$

=> $25 \times x = 5 \times 150$

=> $x=\dfrac{5 \times 150}{25}=30$

Required length

= HCF of 800 cm, 420 cm, 1220 cm

= 20 cm

Required number = L.C.M. of $\left(12, 15, 20, 54\right)$ + 8

   = 540 + 8

   = 548.

Required length = H.C.F. of 700 cm, 385 cm and 1295 cm = 35 cm.

Required number
= HCF of $\left(1223 - 90\right)$ and $\left(2351 - 85\right)$
= HCF of 1133 and 2266
= 1133