Question

Sa se imparta numarul 111 in parti invers proportionale cu numerele 21/5 si 14/9.

Answer 1

Răspuns:

Explicație pas cu pas:

21a/5 = 14b/9 = k

a = 5k/21

b = 9k/14

a + b = 111

5k/21 + 9k/14 = 111

10k/42 + 27k/42 = 111

37k/42 = 111

k = 42*111/37 = 126

a = 5*126/21 = 30

b = 9*126/14 = 81

Answer 2

$\it a+b=111\ \ \ \ \ (*)\\ \\ \{a,\ \ b\}\ i.\ p.\ \Big\{\dfrac{21}{5},\ \ \dfrac{14}{9}\Big\} \Rightarrow \{a,\ \ b\}\ d.p.\ \Big\{\dfrac{5}{21},\ \ \dfrac{9}{14}\Big\} \Rightarrow\\ \\ \\ \Rightarrow \dfrac{a}{ \dfrac{5}{21}}= \dfrac{b}{ \dfrac{9}{14}}= \dfrac{a+b}{ \dfrac{5}{21}+ \dfrac{9}{14}}\stackrel{(*)}{=}\ \dfrac{111}{ \dfrac{37}{42}}=111\cdot \dfrac{42}{37}=3\cdot37\cdot \dfrac{42}{37}=126$

$\it \dfrac{a}{ \dfrac{5}{21}}=126 \Rightarrow a=126\cdot \dfrac{5}{21}=6\cdot5=30 \\ \\ \\ \dfrac{b}{ \dfrac{9}{14}}=126 \Rightarrow b=126\cdot \dfrac{9}{14}=9\cdot9=81$