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Un arquitecto ha pagado en un corralón un total de $ 130 por 20 Ladrillos, 5 kg de Arena y 10 kg de Cemento. Calcular el precio de cada articulo, sabiendo que 1 kg de Cemento cuesta el triple que 1 Ladrillo y que 1 kg de Arena cuesta igual que 4 kg de Cemento más 4 Ladrillos. Siendo: número de ladrillos y: kg de arena kg de cemento

Sagot :

akiran007

Answer:

1 ladrillo cuesta 1 dólar.    1 brick costs 1 dollars.

1 kg de cemento cuesta 3 dólares: 1 kg cement costs 3 dollars

1 kg de arena cuesta $ 16: 1 kg sand costs $ 16

Step-by-step explanation:

Spanish

Deje que los ladrillos, la arena y el cemento se denoten por b, sy c. Luego, de acuerdo con la condición dada

20b + 5s + 10c = $ 130 --- ecuación 1

1c = 3b ------ ecuación 2

1s = 4c + 4b ------ ecuación 3

Poniendo la ecuación 2 en la ecuación 3

1s = 4 (3b) + 4b

1s = 16b ---------- ecuación 4

20b + 5s + 10 (3b) = $ 130

50b + 5s = 130 --- ecuación 5

Poniendo la ecuación 4 en la ecuación 5

50 b + 5 (16b) = 130

50 b + 80b = 130

130 b = 130

b = 130/130 = $ 1 ------- ecuación a

1 ladrillo cuesta 1 dólar.

Poner el valor de costo de 1 ladrillo en estas ecuaciones

1c = 3b ------ ecuación 2

1s = 4c + 4b ------ ecuación 3

1c = 3 (1) = $ 3 -------- ecuación b

1 kg de cemento cuesta 3 dólares

1s = 4c + 4b

1 s = 4 (3) + 4 (1)

1 s = 12 + 4

1s = $ 16 ------ ecuación c

1 kg de arena cuesta $ 16

Poner valores de la ecuación a, b, c en la ecuación 1

20b + 5s + 10c = $ 130

20 (1) + 5 (16) + 10 (3) = 130

20 + 80 + 30 = 130

130 = 130 ------------ por lo tanto probado

English

Let the bricks , sand and cement be denoted by b, s and c. Then according to the given condition

20b+ 5s+ 10c= $130--- equation 1

1c= 3b------ equation 2

1s= 4c+4b------ equation 3

Putting equation 2 in equation 3

1s= 4(3b) + 4b

1s= 16b---------- equation 4

20b+ 5s+ 10(3b)= $130

50b + 5s= 130--- equation 5

Putting equation 4 in equation 5

50 b + 5 (16b) = 130

50 b+ 80b= 130

130 b= 130

b= 130/130= $1------- equation a

1 brick costs 1 dollars.

Putting the cost value of 1 brick in these equations

1c= 3b------ equation 2

1s= 4c+4b------ equation 3

1c= 3(1)= $3-------- equation b

1 kg cement costs 3 dollars

1s= 4c+4b

1s= 4(3) + 4(1)

1s= 12+4

1s= $16 ------ equation c

1 kg sand costs $ 16

Putting values from equation a, b, c in equation 1

20b+ 5s+ 10c= $130

20 (1) + 5 (16) + 10 (3)= 130

20 + 80 + 30 = 130

130 = 130   ------------ hence proved

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