REPREESENT VERBAL SITUATIONS ALGEBRAICALLY

Problem 1 :

The subtraction of 5 times of y from x is

(a) 5x – y      (b) y – 5x    (c) x – 5y     (d) 5y – x

Solution :

5 times of y = 5y

Subtract 5 times of y from x = x - 5y 

Problem 2 :

The side length of the top of square table is x. The expression for perimeter is:

(a) 4 + x     (b) 2x      (c) 4x       (d) 8x 12.

Solution :

Side length of square = x

Perimeter of square = x + x + x + x  = 4x

Problem 3 :

The number of scarfs of length half meter that can be made from y meters of cloth is :

(a) 2y     (b) y/2       (c) y + 2    (d) y + 1/2

Solution :

Length of scarf = 1/2 meter

Number of scarfs = y / (1/2) = 2y

Problem 4 :

The length of a side of square is given as 2x + 3. Which expression represents the perimeter of the square?

(a) 2x + 16 (b) 6x + 9 (c) 8x + 3 (d) 8x + 12

Solution :

Side length of the square = 2x + 3

Perimeter of the square = 4 side

= 4 (2x + 3)

Using distributive property, we get

= 8x + 12

Problem 5 :

Arjun bought a rectangular plot with length x and breadth y and then sold a triangular part of it whose base is y and height is z. Find the area of the remaining part of the plot

Solution :

Area of the rectangle = xy

Area of triangle = (1/2)  ⋅ base  ⋅ height

= (1/2) ⋅ y ⋅ z

Area of remaining part of plot = xy - (yz/2) 

Problem 6 :

Amisha has a square plot of side m and another triangular plot with base and height each equal to m. What is the total area of both plots?

Solution :

Area of square plot with side length m = m²

Area of triangle = (1/2)  ⋅ base  ⋅ height

= (1/2) ⋅ m ⋅ m

= m²/2

Write the following statements in the form of algebraic expressions and write whether it is monomial, binomial or trinomial.

Problem 7 :

x is multiplied by itself and then added to the product of x and y.

Solution :

x is multiplied by itself = x ⋅ x  ==> x²

Product of x and y = xy

= x² + xy

It has two terms, so it is binomial.

Problem 8 :

Three times of p and two times of q are multiplied and then subtracted from r.

Solution :

Three times of p = 3p

Two times of q = 2q

are multiplied = 3p(2q) ==> 6pq

Subtracted from r = r - 6pq

So, the required algebraic expression is r - 6pq

It has two terms, so it is binomial.

Problem 9 :

Product of p, twice of q and thrice of r .

Solution :

Twice of q = 2q

thrice of r = 3r

Product of p, 2q and 3r = p(2q)(3r)

= 6pqr

So, the required algebraic expression is 6pqr. It has one term, so it is monomial.

Problem 10 :

Sum of the products of a and b, b and c and c and a. 

Solution :

Products of a and b, b and c and c and a = ab, bc and ca

Sum of them = ab + bc + ca

It is trinomial.

Problem 11 :

Perimeter of an equilateral triangle of side x.

Solution :

Side length of equilateral triangle = x

perimeter of the triangle = x + x + x

= 3x

It is monomial.

Problem 12 :

Perimeter of a rectangle with length p and breadth q.

Solution :

Length of the rectangle = p, breadth of the rectangle = q

perimeter of rectangle = 2 (length + breadth)

= 2(p + q)

It is binomial.

Problem 13 :

Area of a triangle with base m and height n.

Solution :

base = m, height = n

Area of triangle  = (1/2)  ⋅ base  ⋅ height

 = (1/2) ⋅ m⋅n

= mn/2

Problem 14 :

Area of a square with side x

Solution :

Side length of square = x

Area of square = x²

It is monomial.

Problem 15 :

Cube of s subtracted from cube of t.

Solution :

Cube of s = s³

Cube of t = t³

= t³ - s³

It is binomial.

Problem 16 :

Quotient of x and 15 multiplied by x.

Solution :

Quotient of x and 15 = x/15

Multiplied by x = (x/15) x

= x²/15

It is monomial.

Problem 17 :

The sum of square of x and cube of z.

Solution :

square of x = x²

Cube of z = z³

Sum of x² and z³ = x² + z³

It is binomial.

Problem 18 :

Two times q subtracted from cube of q

Solution :

2 times q = 2q

cube of q = q³

The required algebraic expression is q³ - 2q

It is binomial.