# Worksheet on Changing the Subject in an Equation or Formula | Rearranging the Subject in an Equation or Formula Worksheets

Worksheet on Changing the Subject of a Formula or Equation will help you learn how to change the subject of formula and find the value of a variable easily using the substitution method. Practice the questions in the Changing the Subject in an Equation or Formula Worksheet PDF on a regular basis and improve your subject knowledge of the concept. Download the Changing the Subject of a Formula or Equation Practice Worksheet for free and clear your doubts anytime and anywhere.

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## Changing the Subject of a Formula Worksheet with Answers PDF

I. Discount percentage=Discount/marked price × 100 find discount when discount percent=30 and marked price=1000.

Solution:

Given that,
Discount percentage=Discount/marked price × 100
Discount percentage/100=Discount/marked price
Discount percentage/100 × marked price=Discount
Also given discount percent=30 and marked price=1000.
Discount=Discount percentage/100 × marked price
Discount=30/100 × 1000
=300
Therefore, Discount=Rs 300.

II. Area of Trapezium=1/2(a+b).h find h when Area of trapezium is 60 square inches, a=10,b=5.

Solution:

Given that,
Area of Trapezium=1/2(a+b).h
A=1/2(a+b).h
2A=(a+b).h
2A/(a+b)=h
Also given A=60 sq inches,a=10,b=5.
2(60)/(10+5)=h
120/15=h
Therefore, h=8.

III. Profit percentage=profit/cp × 100 find profit when cost price is 500 and profit percentage is 20.

Solution:

Given that,
Profit percentage=profit/cp × 100
Profit percentage/100=profit/cp
Profit percentage/100 × cp=profit
Also given cost price is 500 and profit percentage is 20.
20/100 × 500=profit
1/5 × 500=profit
profit=100
Therefore, Profit=Rs 100.

Solution:

Given that,
Also given diameter=20
raius=20/2=10.

V. F=9/5 × c + 32 find c when F=500.

Solution:

Given that,
F=9/5 × c + 32
F-32=9/5 × c
c=(F-32)5/9
Also given F=500
c=(50 – 32)5/9
=18 × 5/9
=100
Therefore,c=100.

VI. S=D/T find d when s=80 kmph and time=2 hours.

Solution:

Given that,
S=D/T
D=S × T
Also given s=80 km per hour and time=2 hours.
D=80 × 2
=160
Therefore, distance D=160 km.

VII. principal=Amount – Interest. Find Amount when principal=1000 and interest=40.

Solution:

Given that,
principal=Amount – Interest
Amount=principal + Interest
Also given principal=1000 and interest=40.
Amount=1000 + 40
=1040
Therefore, the Amount is Rs 1040.

VIII. A=wl find l when A=200 and w=50

Solution:

Given that,
A=wl
A/w=l
Also given A=200 and w=50
200/50=l
l=4
Therefore, l=4.

IX. If y=3x+6 find x when y=8

Solution:

Given that,
y=3x+6
y-6=3x
x=y-6/3
Also given y=8
x=8-6/3
=2/3
Therefore, x=2/3.

x. v=u+at find a when v=8,u=4,t=2

Solution:

Given that,
v=u+at
v-u=at
v-u/t=a
Also given  v=8,u=4,t=2
8-4/2=a
a=4/2=2
Therefore, a=2.

xi. If p=2(l+b) find b when l=5 and p=40.

Solution:

Given that,  p=2(l+b)
p=2l+2b
p-2l=2b
b=p-2l/2
Also given l=5 and p=40.
b=40-2(5)/2
=40-10/2
=30/2=15
Therefore,b=15.

xii. Discount=marked price-sales price find marked price when discount=300 and sales price=800.

Solution:

Given that,
Discount=marked price- sales price
marked price=Discount + sales price
Also given discount=300 and sales price=800.
marked price=300 + 800
marked price=1100
Therefore, marked price=1100.