Question

the numbers p,q,r and s satisfy the following equations:
p=2q+3r+4s=k 4p=3q=2r=s

What is the smallest value of 'k' for which p,q,r and are all positive integres?

(A-20//B-24//C-25//D-77//E-154)

Answer 1

Given

• $k=p+2q+3r+4s$

The value of $k$.

• $p,q,r\in \mathbb{N}$

Natural number condition.

• $4p=3q=2r=s$

$s$ has $2, 3, 4$ as a factor, which says must be a multiple of the LCM. Thus, it is a multiple of $12$.

Solution

We choose the minimum value of $s=12n$$(n\in \mathbb{N})$

If $s=12$, the remaining values are $\begin{cases} &p =3 \\ &q = 4\\ &r = 6\end{cases}$ and thus the minimum value of $k=25$.

Answer 2

Value of pq + rs = 84

For full explanation see the attachment

And the options are wrong