Question

Aflați nr a,b,c stiind ca 2a/3=3b/5=5c/4 si a+b+c=119.repede va rog!! De preferat rezolvarea cu K

Answer 1

2a/3 = 3b/5 = 5c/4 = k

2a/3=k, a=3k/2

3b/5=k b=5k/3

5c/4=k. c=4k/5

Inlocuim in a + b + c = 119⇒ 3k/2+5k/3+4k/5=119 (numitor comun 30)

Inmultim relatia cu 30 : 45k+50k+24k=119×30⇒119k=119×30

⇒k=30

a=3k/2=(3×30)/2=90/2=45

b=5k/3=(5×30)/3=50

c=4k/5 =(4×30):5=24

a = 45, b = 50 si c = 24

verificare

a+b+c=119

45+50+24=119

119=119

Answer 2

2a/3 = 3b/5 = 5c/4 = k

a = 3k/2

b = 5k/3

c = 4k/5

3k/2 + 5k/3 + 4k/5 = 119

45k + 50k + 24k = 119k = 3570

=> k = 3570 : 119 = 30

a = 3k/2 = (3 × 30) : 2 = 90/2 = 45

b = 5k/3 = (5 × 30) : 3 = 150 : 3 = 50

c = 4k/5 = (4 × 30) : 5 = 120 : 5 = 24