Question

find the smallest whole number which is exactly divisible by 3/2, 4/3, 9/4, 7/2, 21/5​

Answer 1

Answer:

Take lcm of denominators which is 60 . Now convert you fraction in

$\frac{90}{60} \: \: \frac{80}{60} \: \: \frac{135}{60} \: \: \frac{210}{60} \: \: \frac{252}{60}$

Take LCM of Numerators which is 15120 . So your number will be

$\frac{15120}{60} = 252$

Answer 2

Given,

The fractions 3/2, 4/3, 9/4, 7/2, and 21/5 are given.

To find,

We have to find the smallest whole number which is exactly divisible by 3/2, 4/3, 9/4, 7/2, 21/5​.

Solution,

The smallest whole number that is exactly divisible by 3/2, 4/3, 9/4, 7/2, 21/5​ is 252.

We can simply find the smallest whole number which is exactly divisible by 3/2, 4/3, 9/4, 7/2, 21/5​ by taking the LCM of the given fractions,

      = 3/2 , 4/3 , 9/4 , 7/2 , 21/5

      = 90/60 , 80/60 , 135/60 , 210/60 , 252/60

Taking LCM of the numerator, we get 15120

      = 15120/60

      = 252

Hence, the smallest whole number that is exactly divisible by 3/2, 4/3, 9/4, 7/2, 21/5​ is 252.