# Free Algebraic Expressions Subjective Test 01 Practice Test - 7th grade

Simplify the expression: 6m + 10 − 10m.  [1 MARK]

#### SOLUTION

Solution :

6m + 10 − 10m

(6m - 10m) + 10

- 4m + 10 is the simplified form.

The product of monomial and a binomial expression is a ___________. Give an example.  [2 MARKS]

#### SOLUTION

Solution :

Example: 1 Mark

The product of a monomial and a binomial expression is a binomial.

E.g.  a(a2b2)=a3ab2 the expression has only two terms  a3 and  ab2.

Find the sum of the numerical coefficients of the algebraic expression:

2x2+31y5xy  [2 MARKS]

#### SOLUTION

Solution : Steps: 1 Mark

As per the question:
We have to find out the sum of the numerical coefficients of the expression 2x2+31y5xy
The numerical coefficient of 2x2 = 2
The numerical coefficient of 31y = 31
The numerical coefficient of 5xy = 5

Sum of the coefficients = 2 + 31 - 5 = 28
Hence, the sum of the numerical coefficients of the expression 2x2+31y5xy is 28.

Is the statement given below correct? Explain.

"In the expression x + 4y + 4, the coefficient of x is 1, y is 4 and of 4 is 1".  [2 MARKS]

#### SOLUTION

Solution :

Reason: 1 Mark

The statement is incorrect.

A  coefficient is defined as a number or symbol multiplied to a variable or an unknown quantity in an algebraic term.

It is only defined for the terms containing variable and not for constants.

In the expression, 4 is a constant and therefore cannot have a coefficient.

Hence, the statement is incorrect.

Write the coefficients and terms of the algebraic expression 10x25x6. Also, find the value of the expression at x=2. [2 MARKS]

#### SOLUTION

Solution :

Coefficients and terms: 0.5 Mark each
Value of the expression: 1 Mark

Coefficients of x2 and x are 10 and - 5 respectively

Terms are 10x2, 5x and 6
We have to find the value of the expression at x=2 is
10×2×25×26
= 40106
= 24
Hence, the value of the expression at x=2 is 24

Simplify: (a+b)(2a3b+c)(2a3b)(c). [3 MARKS]

#### SOLUTION

Solution :

Steps: 2 Marks

(a+b)(2a3b+c)(2a3b)c

=[(a+b)(2a3b+c)][(2a3b)c]

=[2a23ba+ac+2ab3b2+bc][2ac3bc]

=2a23ba+ac+2ab3b2+bc2ac+3bc

=2a23b2abac+4bc

The simplified expresion is
=2a23b2abac+4bc.

Find  (0.4p0.5q)2. If p = 5 and q = 4, evaluate the value of the expression. [3 MARKS]

#### SOLUTION

Solution :

Steps: 1 Mark
Final expression: 1 Mark
Value: 1 Mark

(0.4p0.5q)2

=(0.4p0.5q)×(0.4p0.5q)

=(0.4p)×(0.4p)(0.4p×0.5q+0.4p×0.5q)+(0.5q)×(0.5q)

=0.16p20.40pq+0.25q2

Given that
p = 5 and q = 4
on substituting the values we get,
=0.16×520.40×5×4+0.25×42
=48+4
=0

Add  2a+3b+c+d, a+bcd and 3a4b2c2d. Find the value of the final expression if a=2, b=3, c=4, d=5. [3 MARKS]

#### SOLUTION

Solution :

Forming the equation: 1 Mark
Sum: 1 Mark
Value: 1 Mark

We have to find the sum of the expressions
2a+3b+c+d, a+bcd
and 3a4b2c2d.

(2a+3b+c+d)+(a+bcd)+(3a4b2c2d)

2a+3b+c+da+bcd+3a4b2c2d

2aa+3a+3b+b4b+cc2c+dd2d

4a+0×b2c2d

Sum =4a2c2d
As per question ,
a = 2, b = 3, c = 4, d = 5
So, on substituting the values we get,
Sum = 4×22×42×5
=8810
= 10
The value of the expression is 10.

Subtract the sum of (a+b+c) and (2ab+2c) from 3ab+3c. Name the type of expression formed. [3 MARKS]

#### SOLUTION

Solution :

Steps: 1 Mark
Type: 1 Mark

According to question,

(3ab+3c)[(a+b+c)+(2ab+2c)]

=(3ab+3c)[(a+b+c+2ab+2c)]

=(3ab+3c)(3a+3c)

=(3ab+3c3a3c)

=b

The given expression contains only b, So it is a monomial.

Define monomials, binomials, and trinomials. Classify the following into monomials, binomials and trinomials:  [4 MARKS]

(i) 4y7x
(ii) y2
(iii) x+yxy
(iv) 100
(v) abab
(vi) (53t)
(vii) 4p2q4pq2
(viii) 7mn
(ix) z23z+8
(x) a2+b2
(xi) z2+z
(xii) 1+x+x2

#### SOLUTION

Solution : Definition: 1 Mark
Classification: 3 Marks

Monomials: An expression with only one term is called monomial.
Binomial: An expression with two, unlike terms, is called a binomial.
Trinomial: An expression with three, unlike terms, is called a trinomial.

S.NOExpressionType of Polynomial(i)4y7zBinomial(ii)y2Monomial(iii)x+yxyTrinomial(iv)100Monomial(v)ababTrinomial(vi)53tBinomial(vii)4p2q4pq2Binomial(viii)7mnMonomial(ix)z23z+8Trinomial(x)a2+b2Binomial(xi)z2+zBinomial(xii)1+x+x2Trinomial

From the sum of

4+3x and 54x+2x2,

subtract the sum of

3x25x and x2+2x+5.  [4 MARKS]

#### SOLUTION

Solution : Forming the equation: 1 Mark
Steps: 2 Marks

While adding or subtracting algebraic expressions we need to be aware that we can add/subtract only like terms.
Like terms have the same algebraic factors.
According to question,

(4+3x)+(54x+2x2)(3x25x) +(x2+2x+5)

=[4+3x+54x+2x2][3x25xx2+2x+5]

=[2x2+3x4x+5+4][3x2x2+2x5x+5]

=[2x2x+9][2x23x+5]

=2x2x+92x2+3x5

=2x22x2x+3x+95

=2x+4
So, the required expression is 2x+4.

Tatesville has x inches of rainfall in April, (x+1.3) inches of rainfall in May, and (2x+0.5) inches in June. Write the expression that shows the total amount of rainfall for Tatesville for the three-month period? If the rainfall in May was 2.5 inches, find the total rainfall over the three months.  [4 MARKS]

#### SOLUTION

Solution :

Forming equation: 1 Mark
Equating equation according to question: 1 Mark
Steps: 1 Mark

Given that
Tatesville has x inches of rainfall in April, (x+1.3) inches of rainfall in May, and (2x+0.5) inches in June.
The total rainfall during these months will be the sum of these expressions.

Total amount of rainfall
=x+(x+1.3)+(2x+0.5) inches

=x+x+1.3+2x+0.5 inches

=4x+1.8 inches

Given that, rainfall in May is 2.5 inches.
Rainfall in May
(x+1.3)=2.5 inches

x+1.3=2.5

x=2.51.3

x=1.2 inches

On substituting the value of x=1.2 inches in the ecpression for total rainfall.
Total Rainfall
=4x+1.8=4(1.2)+1.8=4.8+1.8=6.6 inches
So, the total rainfall over the three months is 6.6 inches

Aliyah had Rs. 24 to spend on seven pencils. After buying them she had Rs. 10 left. How much did each pencil cost?  [4 MARKS]

#### SOLUTION

Solution :

Forming the equation: 1 Mark
Steps: 2 Marks
Result: 1 Mark

Given that:

Total amount Aliyah had =Rs. 24

Total number of pencils Aliyah bought =7

Total amount left after buying pencil =Rs. 10

Total amount she spent
=Rs. 24Rs. 10=Rs. 14

Total money spent on pencil = Rs 14
Let the cost each pencil be x
7x=Rs. 14

Cost of each pencil ,x=147=Rs.2
The cost of each pencil is Rs 2.

The perimeter of a triangle is 6p24p+9 and the length of two of its adjacent side are p22p+1 and  3p25p+3. Find the length if the third side of the triangle. If p = 3, find the perimeter of the triangle. [4 MARKS]

#### SOLUTION

Solution :

Forming the equation: 1 Mark
Steps: 1 Marks
Equation for perimeter: 1 Mark
Perimeter: 1 Mark

Perimeter of the triangle = sum of the sides of the triangle

6p24p+9=p22p+1+3p25p+3+3rd side

3rd side=6p24p+9[(p22p+1)+(3p25p+3)]

3rd side=6p24p+9(4p27p+4)

3rd side=2p2+3p+5
The 3rd side of the triangle has length is 2p2+3p+5

Given that
p = 3
The perimeter of the triangle is 6p24p+9 .
On substituting the values we get:
=6×324×3+9
=5412+9
=51

The perimeter of the triangle is 51 units.

a) What should be subtracted from 3x24y2+5xy+20 to obtain x2y2+6xy+20?
b) Find the final expression when the algebraic expression x2+2xy+y2 is multiplied by xy. [4 MARKS]

#### SOLUTION

Solution : Each option: 1.5 Marks

a) Let q be the expression to be subtracted.

Then according to the question,

3x24y2+5xy+20q=x2y2+6xy+20

q=3x24y2+5xy+20(x2y2+6xy+20)

q=3x24y2+5xy+20+x2+y26xy20

q=3x2+x24y2+y2+5xy6xy+2020

q=4x23y2xy+0

Hence, 4x23y2xy should be subtracted from 3x24y2+5xy+20 to obtain x2y2+6xy+20.

b) As per the question:
x2+2xy+y2 is multiplied by xy

So, xy× x2+2xy+y2

=xy×x2+xy×2xy+xy×y2

=x2+1y+2x1+1y1+1+xy1+2

am×an=am+n

=x3y+2x2y2+xy3