Question

two types of boxes A B are placed in a truck having capacity of 10 tons when 150 boxes of a type A and 100 boxes of type B are loaded in the truck it weighs 10 tons
but when 260 boxes of type A are loaded in the truck it can still accommodate 40 boxes of type b so that it is fully loaded. find the weight of each type of box...we can solve by any . method as eliminating coificint..

Answer 1

heya...!!!

let the of weight of each A types box = x

and the mass of each B types of box = y

according to question,

case--(1)

150 boxes of type A and 100 boxes of type

B weighs =10 tons 

=> 150x + 100y = 10

=> 15x + 10y = 1--------(1)

case--(1)

260 boxes of type A and 40 boxes of type B 

weighs = 10 tons

=> 260x + 40y = 10

=> 26x + 4y = 1-----(2)

from--(1) and--(2)
multiply by (2) in--(1) and by (5)in --(2)

30x + 20y = 2
130x + 20y = 5
(-)____(-)____(-)
—————————
-100x = -3

=> x = 3/100tons [ put in--(1)]

we get,

15(3/100) + 10y = 1

=> 9/20 + 10y = 1

=> (9 + 200y)/20 = 1

=> 9 + 200y = 20

=> 200y = 11

=> y = 11/200tons

weight of each A type of box = 3/100 × 1000kg

=> 30kg

AND,

the weight of each B type of box = 11/200 × 1000kg

=> 11×5 = 55kg

hope it's help you

Answer 2

let the of weight of each A types box = A

and the mass of each B types of box = B


according to question,


150 boxes of type A and 100 boxes of type B weighs =10 tons 

= 150A + 100B = 10

= 15A + 10B = 1--------( 1 )


Now,


260 boxes of type A and 40 boxes of type B 

weighs = 10 tons

= 260A + 40B = 10

= 26A + 4B = 1-----( 2 )



from--(1) and--(2)
multiply by 2 * --(1) and by 5 *--(2)


30A + 20B = 2
130A + 20B = 5
(-)____(-)____(-)
------------------------
-100A = -3

A = 3/100tons ( put in--(1) )



15(3/100) + 10B = 1

= 9/20 + 10B = 1


= (9 + 200B)/20 = 1

= 9 + 200B = 20

= 200B = 11

=> B = 11/200tons



weight of each A type of box = 3/100 × 1000kg

= 30kg


the weight of each B type of box = 11/200 × 1000kg

= 11×5 = 55kg




I HOPE ITS HELP YOU DEAR,
THANKS