# Worksheet on Framing the Formula | Framing the Formula Practice Worksheets

Worksheet on Framing the Formula will let you know about converting a math sentence into a formula. Math statements can be easily converted to formulas using different symbols and literals. On this page, we included Framing the Formula Problems along with explanations for each and every problem. Practice all problems and get a perfect grip on Framing the Formula concept.

## Steps for Framing the Formula

1. Firstly, read the given statement.
2. Note down the symbols and variables that need to use to frame a formula.
3. Finally, write the formula.

### Framing the Formula Solved Examples

1. Write formulas for the following mathematical statements.

(a) The area of a box is equal to one-third the product of base and height.
(b) The diameter of a matchbox is twice the radius of the mobile box.
(c) Prove that the area of a table is half the sum of the parallel sides.
(d) The difference between the x and y is the addition of 10 and z.
(e) The volume of a cone is equal to the product of its breadth, length, and height.

Solution:

(a) Given that the area of four walls of a room(a) Given that the area of a box is equal to one-third the product of base and height.
Let the area of the box is ‘a’, Base = b, Height = h.
If the area of a box is equal to one-third the product of base and height, then a = 1/3 × bh

The final answer is a = 1/3 . bh

(b) Given that the diameter of a matchbox is twice the radius of the matchbox.
Let the diameter of a matchbox is d, the radius of the matchbox is r
If the diameter of a matchbox is twice the radius of the matchbox, then d = 2r

The final answer is d = 2r.

(c) Given that the area of a table is half the sum of the parallel sides.
Let the Area of a table = a, sum of the parallel sides = x
If the area of a table is half the sum of the parallel sides, then a = x/2.

The final answer is a = x/2.

(d) Given that the difference between the x and y is the addition of 10 and z.
x – y = 10 + z

The final answer is x – y = 10 + z

(e) The volume of a cone is equal to the product of its breadth, length, and height.
Let the volume of a cone is v, breadth is b, the length is l, and height is h.
If the volume of a cone is equal to the product of its breadth, length, and height, then v = blh.

The final answer is v = lbh.

2. Write formulas for the following mathematical statements.

(a) The surface area of the cube is 8 times the square of its edge.
(b) A body of material is moved with a force is equal to the product of the mass of that material body and the acceleration produced by it.
(c) The speed of a car x is equal to the distance (m) traveled by it upon the time (t) taken to cover this distance.
(d) Multiplication of x, y, z is equal to half of m.
(e) The area of four walls of a room is equal to two times the product of height and perimeter of the room.

Solution:

(a) The surface area of the cube is 8 times the square of its edge.
Let the surface area of the cube is S, the square of its edge R, then S = 8R.

The final answer is S = 8R.

(b) A body of material is moved with a force is equal to the product of the mass of that material body and the acceleration produced by it.
Let the force is F, the mass of that material body is m, and the acceleration of the body is a.
If the body of material is moved with a force is equal to the product of the mass of that material body and the acceleration produced by it, then F = ma.

The final answer is F = ma.

(c) Given that the speed of a car x is equal to the distance (m) traveled by it upon the time (t) taken to cover this distance.
Speed = distance/time.
x = m/t m/sec.

The final answer is x = m/t m/sec.

(d) Given that the Multiplication of x, y, z is equal to half of m.
xyz = m/2.

The final answer is xyz = m/2.

(e) Given that the area of four walls of a room is equal to two times the product of height and perimeter of the room.
Let the area of four walls of a room is a, height is h, and perimeter of the room is p.
If the area of four walls of a room is equal to two times the product of height and perimeter of the room.
a = 2hp

The final answer is a = 2ph.

3. Using math literals and symbols, write the formula for the following mathematical statements.

(a) Six subtracted from a number gives two.
(b) Five added to a number is 10.
(c) Four-thirds of a number is 12.
(d) One-fifth of a number is 10 more than 2.
(e) The sum of four times a number and 12 is 64.
(f) The sum of three consecutive odd numbers is 72.
(g) The sum of two multiples of 5 is 65.
(h) One number is 5 less than four times the other, if their sum is increased by 3 the result is 24.
(i) If 2/3 is subtracted from a number and the difference is multiplied by 5, the result is 2.
(j) The numerator of a fraction is 6 less than the denominator. If 1 is added to both numerator and denominator, the fraction becomes 3/4.

Solution:

(a) Given that Six subtracted from a number gives two.
Let the unknown number is x.
x – 6 = 2.

The final answer is x – 6 = 2.

(b) Given that Five added to a number is 10.
Let the unknown number is m.
m + 5 = 10.

The final answer is m + 5 = 10.

(c) Given that Four-thirds of a number is 12.
Let the unknown number is n.
4/3 . n = 12
n = 12 . 3/4
n = 9

The final answer is n = 9.

(d) Given that One-fifth of a number is 10 more than 2.
Let the unknown number is p.
1/5 . p = 10 + 2
p/5 = 12
p = 12 . 5
p = 60

The final answer is p = 60.

(e) Given that the sum of four times a number and 12 is 64.
Let the unknown number is y.
4y + 12 = 64.
4y = 64 – 12
4y = 52
y = 52/4
y = 13

The final answer is y = 13.

(f) Given that the sum of three consecutive odd numbers is 72.
Let the unknown odd number is z.
z + (z + 2) + (z + 4) = 72
3z + 6 = 72
3z = 72 – 6
3z = 66
z = 66/3
z = 22.

The final answer is z = 22.

(g) Given that the sum of two multiples of 5 is 65.
Let the first number be b.
b + b+ 5 = 65
2b + 5 = 65
Move 5 to the right side and substract it with 65.
2b = 65 – 5
2b = 60
Move 2 to the right side and divide it with 60.
b = 60/2
b = 30.

The final answer is b = 30.

(h) Given that One number is 5 less than four times the other, if their sum is increased by 3 the result is 24.
Let the unknown number is x.
One number is 5 less than four times the other.
4x – 5.
Their sum is increased by 3 the result is 24.
x + 4x – 5 + 3 = 24
5x -2 = 24
Move 2 to the right side and add it to the 24.
5x = 24 + 2
5x = 26
Move 5 to the right side and divide it by 26.
x = 26/5

(i) Given that If 2/3 is subtracted from a number and the difference is multiplied by 5, the result is 2.
Let the unknown number is y.
2/3 is subtracted from y.
y – 2/3
The above difference is multiplied by 5.
5(y – 2/3)
The result is 2.
5(y – 2/3) = 2.
Move 5 to the right side and divide it by 2.
y – 2/3 = 2/5.
Move 2/3 to the right side and add it to 2/5.
y = 2/5 + 2/3 = 16/15
y = 16/15

The final answer is y = 16/15.

(j) Given that the numerator of a fraction is 6 less than the denominator. If 1 is added to both numerator and denominator, the fraction becomes 3/4.
Let the unknown number is x.
The numerator of a fraction is 6 less than the denominator.
Numerator = x – 6, denominator = x
(x – 6)/x = 3/4
1 is added to both numerator and denominator.
(x – 6 + 1)/(x + 1) = 3/4
(x – 5)/(x + 1) = 3/4
4(x – 5) = 3(x + 1)
4x – 20 = 3x + 3
4x – 3x = 3 + 20
x = 23.

The final answer is x = 23.

(i) If the present age of a girl is y years.
(a) What will her age be after 9 years?
(b) What was her age 7 years ago?
(c) Her father’s age is 2 more than 8 times her age. Find her father’s age.
(d) Mother is 3 years younger than her father. What is her mother’s age?
(ii) The length of a rectangle is 6m less than thrice its breadth. If the perimeter of the rectangle is 124m find its length and breadth. ​
(iii). A bus covers 168 km in 4 hours. How much distance will it cover in 80 minutes?
(iv) The weight of orange is 70 g and that of mango is 60 g. Find the total weight of x oranges and y mangoes?

Solution:

(i) If the present age of a girl is y years.
(a) What will her age be after 9 years?
After 9 years, her age will be (y + 9) years.
(b) What was her age 7 years ago?
7 years ago, her age will be (y – 7) years
(c) Her father’s age is 2 more than 8 times her age. Find her father’s age.
Her father’s age is (8y + 2) years
(d) Mother is 3 years younger than her father. What is her mother’s age?
(8y + 2 – 3)  = (8y – 1) years

(ii) The length of a rectangle is 6m less than thrice its breadth. If the perimeter of the rectangle is 124m find its length and breadth. ​
Let the length is l, breadth = b, and Perimeter = p.
length = 2b – 6
Perimeter p = 2(l + b) = 2(2b – 6 + b) = 2(3b – 6) = 6b – 12
6b – 12 = 124m
6b = 124 + 12 = 136
b = 136/6
length = 2b – 6 = 2(136/6) – 6 = 136/3 – 6 = (136 – 18)/3 = 118/3

The final answer is length = 118/3, b = 136/6.

(iii). A bus covers 168 km in 4 hours. How much distance will it cover in 80 hours?
Distance d = 168km, Time t = 4 hours.
Speed s = d/t = 168/4 km/hr = 42 km/hr.
If the time t = 80 hours, distance d = 80hrs × 42 km/hr = 3360 km
Distance d = 3360 km.

The final answer is d = 3360km.

(iv) The weight of orange is 70 g and that of mango is 60 g. Find the total weight of x oranges and y mangoes?
The weight of the orange is 70 g.
For x oranges, 70xg
The weight of the mango is 60g
For y mangoes, 60yg
The total weight of x oranges and y mangoes = 70x + 60y grams.

The final answer is 70x + 60y grams.

5. Change the following math statements using expression into statements in the ordinary language.

(a) A pencil costs \$x. An eraser costs \$2x.
(b) A pen costs \$y. A notebook costs \$7 + y.
(c) The number of girls in a class is ‘m’. The number of boys in the class is (1/3)m.
(d) Ram is y years old. His mother is (5y – 1) years old.
(e) In an arrangement of flowerpots, there are n rows. Each row has 8 flower-pots.

Solution:

(a) A pencil costs \$x. An eraser costs \$2x.
A pencil costs \$x. The eraser cost is twice the pencil cost.

(b) A pen costs \$y. A notebook costs \$7 + y.
A pen costs \$y. A notebook cost is \$7 more than that pen cost.

(c) The number of girls in a class is ‘m’. The number of boys in the class is (1/3)m.
The number of girls in a class is ‘m’. The number of boys in the class is one-third of the number of girls in a class.

(d) Ram is y years old. His mother is (5y – 1) years old.
Ram is y years old. His mother’s age is one less than five times of Ram’s age.

(e) In an arrangement of flowerpots, there are n rows. Each row has 8 flower-pots.
In an arrangement of flower pots, there are 8n rows of flowerpots available. If there are n rows, how many flowerpots available for each row?