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.

**IV. **Diameter=2 × radius Find the radius when diameter=20.

**Solution:**

Given that,

Diameter=2 × radius

radius=Diameter/2

Also given diameter=20

raius=20/2=10.

Therefore, radius=10.

**V. **F=9/5 × c + 32 find c when F=50^{0}.

**Solution:**

Given that,

F=9/5 × c + 32

F-32=9/5 × c

c=(F-32)5/9

Also given F=50^{0}

c=(50 – 32)5/9

=18 × 5/9

=10^{0}

Therefore,c=10^{0}.

**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.