Zero Error Scales
Lucy stands on a set of faulty scales that state her mass as 48 kg. David's mass from the same scales is recorded as 56 kg and their combined mass is recorded as 107 kg.
Assuming the magnitude and nature of the errors was constant, what is the true mass of David?
Image credit: Wikipedia Sigurdas
The answer is 59.
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5 solutions
yes, and the error value is 3 kg.
what if i put Y+Z value then its mass will be 51
typo *Weight
Let us assume mass of Lucy as
l
and mass of David as
d
, while error is represented as
e
.
From 1st sentence of problem,
l
+
e
=
4
8
→
E
q
.
1
From 2nd sentence of problem,
d
+
e
=
5
6
→
E
q
.
2
Adding
1
&
2
,
⇒
l
+
d
+
2
e
=
1
0
4
→
E
q
.
3
From 3rd sentence of problem,
l
+
d
+
e
=
1
0
7
→
E
q
.
4
{Here, error is introduced only once as the number of instances of taking observation is only 1.}
Subtracting
3
from
4
, we get,
⇒
e
−
2
e
=
(
1
0
7
−
1
0
4
)
⇒
−
e
=
3
⇒
e
=
−
3
k
g
So, there is an error of
e
=
−
3
k
g
in the set of scales.
Putting value of error
e
in
Eq.2
,
d
−
3
=
5
6
⇒
d
=
5
9
k
g
Suppose that Lucy is on the scale, and then David steps on. The measured change in mass is the same no matter what the offset is, so David's mass is just the combined mass minus Lucy's mass. 1 0 7 − 4 8 = 5 9
Let the zero error = a. Then, a + Lucy's mass = 48. And, a + Lucy's mass + David's mass = 107. So David's mass = 59. Note that the second condition isn't even used.
48+56=104, 107-104=3, 56+3=59
why is it 56+3 instead of 56-3? the scales were measuring too large a value, as 48+56 = 104 not 107, so why add the error value again?
Lucy's true mass = x Kg
David's true mass = y Kg
Error = z kg [negative/positive]
=>x + z = 48........(1)
=>y + z = 56........(2)
=> x + y + z = 107...........(3)
Putting the value of x + z from (1) in (3), we get the value of y = 59 Kg.
Hence, David's age = y = 59 Kg.