Worksheet: Changing the Subject of a Formula

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In this worksheet, we will practice rearranging simple formulas using inverse operations.

Q1:

Solve 𝑦=2π‘₯+13π‘₯+4 with an expression for π‘₯ in terms of 𝑦.

  • Aπ‘₯=4𝑦+13π‘¦βˆ’2
  • Bπ‘₯=4π‘¦βˆ’12βˆ’3𝑦
  • Cπ‘₯=4π‘¦βˆ’13π‘¦βˆ’2
  • Dπ‘₯=4𝑦+12βˆ’3𝑦
  • Eπ‘₯=4π‘¦βˆ’12βˆ’π‘¦

Q2:

Rearrange 𝑦=3π‘₯βˆ’5 to make π‘₯ the subject.

  • Aπ‘₯=𝑦+53
  • Bπ‘₯=3𝑦+15
  • Cπ‘₯=𝑦+5
  • Dπ‘₯=π‘¦βˆ’5
  • Eπ‘₯=π‘¦βˆ’53

Q3:

In 1897, Amos Dolbear derived a formula linking the number of cricket chirps and the temperature. The law states that the temperature 𝑇, in degrees Celsius, is related to the number of cricket chirps in a minute 𝑁 by the formula 𝑇=10+π‘βˆ’407.

Rearrange the formula to make 𝑁 the subject.

  • A𝑁=(7𝑇+10)βˆ’40
  • B𝑁=(7π‘‡βˆ’10)+40
  • C𝑁=7(𝑇+10)βˆ’40
  • D𝑁=7(𝑇+10)+40
  • E𝑁=7(π‘‡βˆ’10)+40

Given that the temperature on a particular day is 22∘C, estimate the number of cricket chirps that you would expect to hear in a minute.

Q4:

The final velocity of an object with uniform acceleration can be calculated using the formula 𝑣=𝑒+π‘Žπ‘‘. Make 𝑑 the subject.

  • Aπ‘£π‘Žβˆ’π‘’
  • Bπ‘£π‘Ž+𝑒
  • Cπ‘Ž(π‘£βˆ’π‘’)
  • D𝑣+π‘’π‘Ž
  • Eπ‘£βˆ’π‘’π‘Ž

Q5:

The displacement, 𝑠, of an object traveling with uniform acceleration, π‘Ž, can be calculated using the formula 𝑠=𝑒𝑑+12π‘Žπ‘‘οŠ¨, where 𝑒 is the initial velocity and 𝑑 is the time.

Make π‘Ž the subject.

  • Aπ‘Ž=π‘ βˆ’π‘’π‘‘2π‘‘οŠ¨
  • Bπ‘Ž=2π‘ βˆ’π‘’π‘‘π‘‘οŠ¨
  • Cπ‘Ž=2(𝑠+𝑒𝑑)2π‘‘οŠ¨
  • Dπ‘Ž=2(π‘ βˆ’π‘’π‘‘)π‘‘οŠ¨
  • Eπ‘Ž=2𝑠+π‘’π‘‘π‘‘οŠ¨

Calculate the acceleration of an object that has started from rest and traveled 200 m in 8 s.

Q6:

The variables π‘₯ and 𝑦 where π‘₯ is non-negative are related by the formula 𝑦=6(7βˆ’π‘₯)11. Rewrite the formula with π‘₯ as the subject.

  • Aπ‘₯=ο„ž7βˆ’611𝑦
  • Bπ‘₯=ο„ž7+11𝑦6
  • Cπ‘₯=ο„ž7βˆ’11𝑦6
  • Dπ‘₯=ο„ž7βˆ’116𝑦
  • Eπ‘₯=ο„ž7+116𝑦

Q7:

The variables π‘Ž and 𝑏 are related by the formula π‘Ž(𝑏+5)=π‘˜(π‘βˆ’19π‘Ž). Make 𝑏 the subject.

  • A𝑏=5π‘Ž+19π‘˜π‘Žπ‘˜βˆ’π‘Ž
  • B𝑏=5+19π‘Žπ‘˜βˆ’π‘Ž
  • C𝑏=5π‘Žβˆ’19π‘˜π‘Žπ‘˜βˆ’π‘Ž
  • D𝑏=5π‘Žβˆ’19π‘˜π‘Žπ‘˜+π‘Ž
  • E𝑏=5+19π‘Žπ‘˜+π‘Ž

Q8:

Albert Einstein’s famous formula 𝐸=π‘šπ‘οŠ¨, where the constant 𝑐 is the speed of light, relates the energy, 𝐸, contained in matter and its mass, π‘š. Rearrange the formula to make 𝑐 the subject.

  • A𝑐=ο„žπ‘šπΈ
  • B𝑐=ο„žπΈπ‘š
  • C𝑐=𝐸2π‘š
  • D𝑐=πΈπ‘š
  • E𝑐=π‘šπΈ

Q9:

The variables 𝑑 and π‘š are related by the formula 𝑑=π‘šπ‘š+1. Make π‘š the subject.

  • Aπ‘š=π‘‘π‘‘βˆ’1
  • Bπ‘š=𝑑+1𝑑
  • Cπ‘š=βˆ’π‘‘βˆ’1𝑑
  • Dπ‘š=βˆ’π‘‘π‘‘βˆ’1
  • Eπ‘š=𝑑𝑑+1

Q10:

The circumference 𝐢 of a circle can be estimated using the formula 𝐢=447π‘Ÿ, where π‘Ÿ is the radius. Find an estimate of the radius of a circle with 𝐢=67.1. Round your answer to the nearest tenth.

Q11:

The circumference of a circle as a function of its radius is given by 𝐢(π‘Ÿ)=2πœ‹π‘Ÿ. Express the radius of a circle as a function of its circumference, denoting it by π‘Ÿ(𝐢), and then find π‘Ÿ(36πœ‹).

  • Aπ‘Ÿ(𝐢)=πΆπœ‹, 18
  • Bπ‘Ÿ(𝐢)=𝐢2πœ‹, 72
  • Cπ‘Ÿ(𝐢)=𝐢2πœ‹, 36
  • Dπ‘Ÿ(𝐢)=𝐢2πœ‹, 18
  • Eπ‘Ÿ(𝐢)=πΆπœ‹, 36

Q12:

The volume, 𝑉, of a cylinder with radius π‘Ÿ and height β„Ž is 𝑉=πœ‹π‘Ÿβ„ŽοŠ¨. Given that a cylinder has a height of 6 meters, write an equation for the radius of the cylinder as a function of 𝑉, and then use this to find the radius of the cylinder if its volume is 300 cubic meters. Give your answer to two decimal places.

  • Aπ‘Ÿ=βˆšπ‘‰6πœ‹, 0.92 meters
  • Bπ‘Ÿ=π‘‰βˆš6πœ‹, 69.10 meters
  • Cπ‘Ÿ=ο„žπ‘‰6πœ‹, 3.99 meters
  • Dπ‘Ÿ=𝑉6πœ‹, 15.92 meters
  • Eπ‘Ÿ=1πœ‹ο„žπ‘‰6, 2.25 meters

Q13:

The volume of a right circular cone with radius π‘Ÿ and height β„Ž is 𝑉=13πœ‹π‘Ÿβ„ŽοŠ¨. First, write an equation for the radius of a cone with a height of 12 inches as a function of 𝑉. Then, use this to find the radius of the cone to the nearest whole number given that its volume is 50 cubic inches.

  • Aπ‘Ÿ=πœ‹π‘‰36, 9 inches
  • Bπ‘Ÿ=ο„žπ‘‰36πœ‹, 1 inch
  • Cπ‘Ÿ=ο„žπœ‹π‘‰36, 3 inches
  • Dπ‘Ÿ=ο„žπ‘‰4πœ‹, 2 inches
  • Eπ‘Ÿ=𝑉4πœ‹, 4 inches

Q14:

The volume, 𝑉, of a sphere with radius length π‘Ÿ is given by 𝑉=43πœ‹π‘ŸοŠ©. Find the radius length of a sphere with a volume of 4.851Γ—10 cm3. (Take πœ‹=227.)

Q15:

Use the formula 𝐴=12π‘β„Ž to determine the height, β„Ž, of a triangle given that its area, 𝐴, is 4.5 and its base, 𝑏, is 2.

  • A335
  • B412
  • C145
  • D214
  • E3

Q16:

A room’s temperature ranges from 25∘C to 30∘C. Determine its temperature range in degrees Fahrenheit, using the formula πΉβˆ’32=1.8𝐢, where 𝐹 is the temperature in degrees Fahrenheit, and 𝐢 is the temperature in degrees Celsius.

  • A62∘F to 86∘F
  • B86∘F to 347∘F
  • C62∘F to 347∘F
  • D62∘F to 77∘F
  • E77∘F to 86∘F

Q17:

The surface area, 𝐴, of a cylinder in terms of its radius, π‘Ÿ, and height, β„Ž, is given by 𝐴=2πœ‹π‘Ÿ+2πœ‹π‘Ÿβ„ŽοŠ¨. Express the radius, π‘Ÿ, of a cylinder with a height of 4 feet as a function of 𝐴. Find, to the nearest foot, the radius of such a cylinder whose surface area is 200 square feet.

  • Aπ‘Ÿ=ο„žπ΄+8πœ‹2πœ‹+2, 6 feet
  • Bπ‘Ÿ=ο„žπ΄+8πœ‹2πœ‹βˆ’2, 4 feet
  • Cπ‘Ÿ=ο„žπ΄βˆ’4πœ‹2πœ‹+2, 7 feet
  • Dπ‘Ÿ=ο„žπ΄+8πœ‹2πœ‹βˆ’2, 6 feet
  • Eπ‘Ÿ=ο„žπ΄+8πœ‹2πœ‹+2, 8 feet

Q18:

The surface area, 𝐴, of a sphere in terms of its radius, π‘Ÿ, is given by 𝐴(π‘Ÿ)=4πœ‹π‘ŸοŠ¨. Express π‘Ÿ as a function of 𝐴 and find, to the nearest tenth of an inch, the radius of a sphere whose surface area is 1,000 square inches.

  • Aπ‘Ÿ=ο„ž4πœ‹π΄, 0.1 inches
  • Bπ‘Ÿ=𝐴4πœ‹, 79.6 inches
  • Cπ‘Ÿ=4πœ‹βˆšπ΄, 0.4 inches
  • Dπ‘Ÿ=√𝐴4πœ‹, 2.5 inches
  • Eπ‘Ÿ=ο„žπ΄4πœ‹, 8.9 inches

Q19:

The volume, 𝑉, of a right circular cone with radius length π‘Ÿ is given by 𝑉=13πœ‹π‘Ÿβ„ŽοŠ¨. Find the height of a right circular cone with volume 4,312 cm3 and base diameter length 28 cm. (Take πœ‹=227.)

Q20:

The formula for the circumference of a circle is 𝐢=2πœ‹π‘Ÿ. Rearrange this formula to make πœ‹ the subject.

  • Aπœ‹=2πΆπ‘Ÿ
  • Bπœ‹=𝐢2π‘Ÿ
  • Cπœ‹=2πΆπ‘Ÿ
  • Dπœ‹=πΆπ‘Ÿ
  • Eπœ‹=2π‘ŸπΆ

Q21:

Using the formulae for the circumference and area of a circle, eliminate the variable π‘Ÿ to find a formula that allows you to calculate the circumference of a circle from its area.

  • A𝐢=4πœ‹βˆšπ΄
  • B𝐢=√𝐴4πœ‹
  • C𝐢=2πœ‹βˆšπ΄
  • D𝐢=πœ‹ο„žπ΄2πœ‹
  • E𝐢=2πœ‹ο„žπ΄πœ‹

Q22:

Using the formulae for the circumference and area of a circle, eliminate the variable π‘Ÿ to find a formula that allows you to calculate the area of a circle from its circumference.

  • A𝐴=𝐢2πœ‹ο‰οŠ¨
  • B𝐴=𝐢4πœ‹ο‰οŠ¨
  • C𝐴=𝐢4πœ‹
  • D𝐴=𝐢4πœ‹οŠ¨
  • E𝐴=𝐢2πœ‹οŠ¨

Q23:

The picture shows the design of a logo which is formed from two semicircles with a common center.

Work out the perimeter of the logo, giving your answer in terms of πœ‹.

  • A16πœ‹βˆ’8
  • B16πœ‹+8
  • C8πœ‹+4
  • D8πœ‹βˆ’16
  • E16πœ‹

Work out the area of the logo, giving your answer in terms of πœ‹.

  • A11πœ‹
  • B7πœ‹
  • C17πœ‹
  • D13πœ‹
  • E10πœ‹

Q24:

The volume 𝑉 of a right circular cone in terms of its height β„Ž and base radius π‘Ÿ is 𝑉=13πœ‹π‘Ÿβ„ŽοŠ¨. Give a formula for the radius π‘Ÿ in terms of 𝑉 and β„Ž.

  • Aπ‘Ÿ=ο„žπ‘‰β„Ž
  • Bπ‘Ÿ=ο„žπœ‹π‘‰3β„Ž
  • Cπ‘Ÿ=ο„žπ‘‰3πœ‹β„Ž
  • Dπ‘Ÿ=ο„ž3π‘‰πœ‹β„Ž
  • Eπ‘Ÿ=3π‘‰πœ‹β„Ž

Q25:

A container holds 100 mL of a solution that is 25 mL acid. If 𝑛 mL of a solution that is 60% acid is added, the function 𝐢 gives the concentration, 50%, as a function of the number of milliliters added, 𝐢=25+0.6𝑛100+𝑛. Express 𝑛 as a function of 𝐢 and determine the number of milliliters needed to have a solution that is 60% acid.

  • A𝑛=100πΆβˆ’25𝐢, 50 mL
  • B𝑛=25+100𝐢𝐢+0.6, 68 mL
  • C𝑛=100πΆβˆ’250.6βˆ’πΆ, 250 mL
  • D𝑛=100𝐢+250.6βˆ’πΆ, 750 mL
  • E𝑛=100πΆβˆ’25𝐢+0.6, 23 mL

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