Eight class 2.2 exercise
tokaians:
???
wite a question here you think is difficult for you. It will help one see what fundamentals are to be explained to you. Do not go for solutions just like that. It is an elderly advice to a student who is eager to learn.
write the questions
Vinay bought a pizza and cut it into three pieces. When he weighted the first pizza piece he found that it was 7g lighter than the second piece and 4g. Heavier than the third piece. If the whole pizza weighted 300g. How much did each of three pieces weight ?
Answers
Answered by
11
Let weight of first, second and third piece be a, b, c.
a = b - 7 ------------(1)
a = c + 4 ------------(2)
from equations (1) and (2) ==> b - 7 = c + 4
or b = c + 11 -----------(3)
from equations (1) and (3) ==> a = b - 7 = (c + 11) - 7 = c + 4 -------------(4)
Now total weight = a + b + c = 300 ------(5)
Now substitute values of 'a' in terms of 'c' from equation (4), and value 'b' in terms of 'c' from equation (3) into equation (5), we have:=
(c + 4) + (c + 11) + c = 300
This gives c = 95.
Putting this value of 'c' in equations (4) and (3), we get 'a' = 99, 'b' = 106.
a = b - 7 ------------(1)
a = c + 4 ------------(2)
from equations (1) and (2) ==> b - 7 = c + 4
or b = c + 11 -----------(3)
from equations (1) and (3) ==> a = b - 7 = (c + 11) - 7 = c + 4 -------------(4)
Now total weight = a + b + c = 300 ------(5)
Now substitute values of 'a' in terms of 'c' from equation (4), and value 'b' in terms of 'c' from equation (3) into equation (5), we have:=
(c + 4) + (c + 11) + c = 300
This gives c = 95.
Putting this value of 'c' in equations (4) and (3), we get 'a' = 99, 'b' = 106.
Answered by
5
let the weight of 1st 2nd 3rd things be x y z
x = y - 7 ------------(1)
a = z + 4 ------------(2)
from equations (1) and (2) ==> y - 7 = z+ 4
y=4+7+z
or y = z+ 11 -----------(3)
from equations (1) and (3) ==> x= y - 7 = (z + 11) - 7 = z + 4 -------------(4)
Now total weight = x + y + z = 300 ------(5)
Now substitute values of 'x' in terms of 'z' from equation (4), and value 'y' in terms of 'z' from equation (3) into equation (5), we have:=
(z + 4) + ( z+ 11) + z = 300
3z+15=300
3z=285
z=285/3=95
This gives z = 95.
Putting this value of 'z' in equations (4) and (3), we get 'x' = 99, 'y' = 106.
x = y - 7 ------------(1)
a = z + 4 ------------(2)
from equations (1) and (2) ==> y - 7 = z+ 4
y=4+7+z
or y = z+ 11 -----------(3)
from equations (1) and (3) ==> x= y - 7 = (z + 11) - 7 = z + 4 -------------(4)
Now total weight = x + y + z = 300 ------(5)
Now substitute values of 'x' in terms of 'z' from equation (4), and value 'y' in terms of 'z' from equation (3) into equation (5), we have:=
(z + 4) + ( z+ 11) + z = 300
3z+15=300
3z=285
z=285/3=95
This gives z = 95.
Putting this value of 'z' in equations (4) and (3), we get 'x' = 99, 'y' = 106.
Similar questions