a person has 4 boxes of different intergal but unknown weights. the person weighted the boxes in triplets. he obtained the wieghts in kgs as 150, 170, 180, 190. how much would the heaviest box weigh? (in kgs)
Answers
Answer:
The weight of the heaviest box is- 80 Kg
Let the weight of 4 boxes be A,B,C,D
Then, A+B+C=150
B+C+D=170
C+D+A=180
A+B+D=190
Now, add all the four equations, you get-
3(A+B+C+D)= 150+170+180+190
A+B+C+D=230
A+(B+C+D)=230
A+ 170= 230
A=60
Similarly, B= 230-(A+C+D)
B=50
So, C=40 and D=80
(All units in Kg)
So, A=60 Kg
B=40 Kg
C=50 Kg
D=80 Kg
I hope this helps. Please mark as the brainliest :)
Step-by-step explanation:
a+ b+ c = 150 .........EQ 1
b+c+d = 170 ..............EQ 2
a+b+d = 180 ..............EQ 3
a+c+d = 190 .................EQ 4
from EQ 1
a= 150 - b-c .........EQ 5
from EQ 3
a= 180-b-d ...........EQ 6
from EQ 4
a = 190 - c-d ..........EQ 7
from EQ 5 and 6
150-c = 180 -d
d= 30+ c
put in EQ 2 and 4
b + 2c = 140 .........EQ 8
a+ 2c = 160 ............EQ 9
EQ 9 - EQ 8
a-b = 20
a = 20 +b
put in EQ 1
2b+ c = 130 ..........EQ 10
solve eq 8 and 10 we get
b = 40
c = 50
d = 30+c= 80
a = 20+b = 60
so last box is heaviest