Problem on Change the Subject of a Formula | Solved Questions on Change of Subject of Formula PDF

The process of rearranging formulae and the method of substitution to find the value of one variable is known as changing the subject of a formula. Are you wondering how to change the subject of an equation or formula? go through this guide ie., Problem on Change the Subject of a Formula.

This is the correct place to find Changing the Subject in an Equation or Formula Question and Answers with comprehensive explanation. Keep an eye on every single problem and fully understand the process of how to change of subject of formula or https://ccssanswers.com/equation easily?

Problems on Change the Subject of a Formula

Here we will be solving various kinds of problems on changing the subject of a formula using mathematical operations & inverse calculations. The subject of a formula is an unknown quantity whose relation with other known quantities of the context is desired.

Let us know the better version of understanding the concept of changing the subject formula by practicing with the presented examples on changing the subject of a formula.

Changing the subject of a formula Questions and Answers

Example 1:
In the formula A = \(\frac { 1 }{ 2 } \) bh, A is the subject. Make the formula with b as the subject where A = 20 cm² and h = 5 cm
Solution:
Given A = \(\frac { 1 }{ 2 } \) bh
2A = bh
\(\frac { 2A }{ h } \) = b
Now, substitute the given A and h values in the rearranged formula and find the value of subject pf a formula ie.,
b = \(\frac { 2A }{ h } \)
b = \(\frac { 2(20) }{ 5 } \)
b = \(\frac { 40 }{ 5 } \)
b = 8 cm

Example 2:
Make r the subject of the below Adding and subtracting formula:
(i) t = s + r
(ii) t = r – s
Solution:
(i) Given that t = s+ r
Subtract s on both sides
t – s = s + r – s
t – s = r
The final result is r = t – s
(ii) Given that t = r – s
Add s on both sides
t + s = r – s + s
t + s = r
The final result is r = t + s

Example 3:
Make x the subject for each of the following formulae or equations:
(i) y = def + x
(ii) y = – b + x
Solution:
(i) Given that y = def + x
subtract def on both sides
y – def = def + x – def
y – def = x
Hence, x = y – def
(ii) Given that y = – b + x
Add b on both sides
y + b = – b + x + b
y + b = x
Hence, x = y + b

Example 4:
Rearrange equation x = y + 5 with y as a subject of the equation and find the value of the subject where x=15.
Solution:
Given that x = y + 5
Now, subtract 5 from both sides of the expression and simplify,
x = y + 5
x – 5 = y + 5 – 5
x – 5 = y
Which would then be written as:
y = x – 5
Now, substitute the x value in the equation and get the result of y;
y = 15 – 5
y = 10

Example 5:
Make v the subject of the formula f = \(\frac { uv }{ u+v } \).
Solution:
Given that f = \(\frac { uv }{ u+v } \)
⟹ \(\frac { 1 }{ f } \) = \(\frac { u+v }{ uv } \)
⟹ \(\frac { 1 }{ f } \) = \(\frac { 1 }{ u } \) + \(\frac { 1 }{ v } \)
⟹ \(\frac { 1 }{ v } \) = \(\frac { 1 }{ f } \) – \(\frac { 1 }{ u } \)
⟹ \(\frac { 1 }{ v } \) = \(\frac { u-f }{ fu } \)
⟹ v = \(\frac { fu }{ u-f } \), Here v is the subject.

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