a-2-2-b-3-3-c-4-4-and-a-b-c-192-faind-a-b-c-

Question Number 170064 by mathlove last updated on 15/May/22

\frac{(a - 2)}{2} = \frac{(b - 3)}{3} = \frac{(c - 4)}{4}

a n d a \cdot b \cdot c = 192

f a i n d a + b + c = ?

Answered by Rasheed.Sindhi last updated on 15/May/22

\frac{(a - 2)}{2} = \frac{(b - 3)}{3} = \frac{(c - 4)}{4}

a n d a \cdot b \cdot c = 192

f i n d a + b + c = ?

\frac{a - 2}{2} + 1 = \frac{b - 3}{3} + 1 = \frac{c - 4}{4} + 1

\frac{a}{2} = \frac{b}{3} = \frac{c}{4} = \frac{a + b + c}{2 + 3 + 4} = k (s a y)

\frac{a}{2} = \frac{b}{3} = \frac{c}{4} = \frac{a + b + c}{9} = k (s a y)

a = 2 k, b = 3 k, c = 4 k, a + b + c = 9 k

a \cdot b \cdot c = 2 k . 3 k . 4 k = 192

24 k^{3} = 192

k^{3} = 8 \Rightarrow k = 2

a + b + c = 9 k = 9 (2) = 18

Answered by som(math1967) last updated on 15/May/22

l e t \frac{a - 2}{2} = \frac{b - 3}{3} = \frac{c - 4}{4} = k

\Rightarrow a = 2 (k + 1), b = 3 (k + 1), c = 4 (k + 1)

2 \times 3 \times 4 \times {(k + 1)}^{3} = a b c

\Rightarrow {(k + 1)}^{3} = \frac{192}{24} = 8

\Rightarrow k + 1 = 2 \Rightarrow k = 1

a + b + c = 2 (k + 1) + 3 (k + 1) + 4 (k + 1)

= 4 + 6 + 8 = 18 a n s