Practice the Framing Formula Worksheet over here and get acquainted with the concept of how to frame a formula by translating mathematical statements using symbols and literals. Solve the various questions in the Worksheet on Framing a Formula and learn how to approach when given a mathematical statement. Answering the questions over here helps you to improve your math proficiency as well as attempt the exam with utmost confidence.
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Framing a Formula Worksheet with Solutions
I. Write formulas for the following mathematical statements.
(i) The difference between the cost price and the selling price is Loss.
(ii) Area of the rectangle is the product of length and width.
(iii) The perimeter of a triangle is the sum of the length of the sides of a triangle where a,b,c are sides of a triangle.
(iv) The distance traveled by vehicle is equal to the product of the speed of the vehicle and the time taken to cover the distance.
(v) Discount is calculated as the difference between the marked price and the selling price.
(vi) Area of the parallelogram is defined as the product of the base and height.
Solution:
(i) Given, The difference between the cost price and the selling price is Loss.
The difference between the cost price and the selling price is represented as the Cost price-selling Price.
Therefore, Loss=Cost price-selling Price
(ii) Given, the Area of the rectangle is the product of length and width.
The product of length and width is represented as Length × width.
Therefore, the Area of rectangle=Length × width.
(iii) In a triangle ABC, sides of the triangle are a,b,c.
The perimeter of a triangle is the sum of the length of the sides of a triangle is represented as a+b+c.
Hence,Perimeter of a triangle=a+b+c
(iv) product of the speed of the vehicle and the time taken to cover the distance is represented as = speed × time
Therefore, Distance=speed × time.
(v) The difference between the marked price and the selling price is MP-SP.
Therefore,Discount=MP-SP
(vi)Product of base and height is bh
Therefore, the Area of parallelogram=bh.
II. Using math literals and symbols, write the formula for the following mathematical statements.
(i) Three subtracted from a number gives Eight.
(ii) Seven added to a number is 10.
(iii) One-fifth of a number is 20.
(iv) One-third of a number is 6 more than 10.
(v) The sum of five times a number and 2 is 12.
(vi) The sum of three consecutive even numbers is 60.
(vii) The sum of two multiples of 3 is 27.
(viii) One number is 4 less than three times the other, if their sum is increased by 5 the result is 25.
(ix) The difference between m and n is the addition of 15 and p.
Solution:
(i) Given Three subtracted from a number gives Eight.
Let the number be N.
N-3=8
N=8 + 3=11
(ii) Given Seven added to a number is 10.
Let the number be N.
N+7=10
N=10-7
N=3
(iii) Given, One-fifth of a number is 20.
Let the number be N.
1/5(N)=20
N=20 × 5=100
(iv) Given, One-third of a number is 6 more than 10.
Let the number be N.
One-third of a number=1/3(N).
One-third of a number is 6 more than 10 is 1/3(N)=6+10
1/3(N)=16
N=16 × 3=48
(v) Given, The sum of five times a number and 2 is 12.
Let the number be N.
Five times a number is=5N
The sum of five times a number and 2 is 5N +2= 12
5N=12-2
5N=10
N=10/5=2
(vi) Given, The sum of three consecutive even numbers is 42.
Let the three consecutive even numbers be 2x,2x+2,2x+4.
The sum of three consecutive even numbers is 2x+2x+2+2x+4=42
6x+6=42
6x=42-6
6x=36
x=36/6=6
Therefore, the three consecutive even numbers are 6,8, and 10.
(Vii) Given, The sum of two consecutive multiples of 5 is 85.
Let x be the first multiple of 5 and x+5 be the second multiple of 5.
x+x+5=85
2x+5=85
2x=85-5
2x=80
x=80/2=40
Hence, x=40, x+5=40+5=45.
(viii) Given, One number is 4 less than three times the other, if their sum is increased by 5 the result is 25.
Let the first number be x.
Let the second number be y.
One number is 4 less than three times the other i.e. y=3x-4.->eq1
Also given that if their sum is increased by 5 the result is 25.
x+y+5=25
x+y=25-5
x+y=20 ->eq 2
subs y=3x-4 in eq 2.
x + 3x- 4=20
4x=20 + 4
4x=24
x=24/4
=6
Subs x=6 in eq2
6+y=20
y=20-6
y=14
Therefore, x=6,y=14.
(ix) Given, The difference between m and n is the addition of 15 and p.
The difference between the x and y is represented as=m – n
The addition of 15 and z is represented as= 15+ p
The difference between m and n is the addition of 15 and p is m-n=15+p.
Hence, The difference between m and n is the addition of 15 and p is m-n=15+p.
III. Change the following math statements using expression into statements in the ordinary language.
(i) Age of Ramya is x. The age of Sravya is 2x.
(ii) Sugar costs Rs x. Wheat costs Rs 20 + x.
(iii) The number of women in a town is ‘n’. The number of men in the town is (1/5)n.
(iv) Arshik is x years old. His mother is (4x – 4) years old.
Solution:
(i) Given, Age of Ramya is x. The age of Sravya is 2x.
The age of Ramya is x. The age of Sravya is double the age of Ramya.
(ii) Given, Sugar costs Rs x. Wheat costs Rs 20 + x.
Sugar costs Rs x. Wheat cost is 20 more than sugar.
(iii) Given, The number of women in a town is ‘n’. The number of men in the town is (1/5)n.
The number of women in a town is ‘n’. The number of men in the town is one-fifth of the number of women in the town.
(iv) Given, Arshik is x years old. His mother is (4x – 4) years old.
Arshik is x years old. His mother’s age is 4 less than four times Arshik’s age.
IV. Answer the following:
(i) If the present age of a Swapna is x years.
(a) What will her age be after 5 years?
(b) What was her age 3 years ago?
(c)Her father’s age is 3 more than three times her age. Find her father’s age.
(d) Mother is 5 years younger than her father. What is her mother’s age?
Solution:
(a) Swapna’s age after 5 years is (X+5).
(b) 3 years ago Swapna’s age is (X-3).
(c) Father’s age =(3x+3)
(d) Mother’s age=3x+3-5
=(3x-2) years
(ii) The weight of the apple is 30 g and that of sweet lime is 50 g. Find the total weight of m apples and n sweet limes?
Solution:
The weight of the apple is 30 g.
For m apples, 30mg
The weight of the sweet lime is 50g
For n sweet limes, 50ng
The total weight of m apples and n Sweet limes = 30m+ 50n grams.
(iii) The length of a rectangle is 3m less than thrice its breadth. If the perimeter of the rectangle is 128m find its length and breadth?
Solution:
Let the length is l, breadth = b, and Perimeter = p.
length = 3b – 3
Perimeter p =2(l+b)
=2(3b-3+b)
=2(4b-3)
=8b-6
8b-6=128
8b=128+6
8b=134
b=134/8=16.75
length=3(16.75-3)=41.25
Hence, length of the rectangle=41.25, Breadth=16.75.