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

### Question 1

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

#### SOLUTION

Solution :

→ 6m + 10 − 10m

→ (6m - 10m) + 10

→ - 4m + 10 is the simplified form.

### Question 2

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

#### SOLUTION

Solution :Answer: 1 Mark

Example: 1 Mark

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

E.g. a(a2−b2)=a3−ab2 the expression has only two terms a3 and ab2.

### Question 3

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

2x2+31y−5xy [2 MARKS]

#### SOLUTION

Solution :Steps: 1 Mark

Answer: 1 Mark

As per the question:

We have to find out the sum of the numerical coefficients of the expression 2x2+31y−5xy

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+31y−5xy is 28.

### Question 4

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 :Answer: 1 Mark

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.

### Question 5

Write the coefficients and terms of the algebraic expression 10x2−5x−6. 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

10×2×2−5×2−6

= 40−10−6

= 24

Hence, the value of the expression at x=2 is 24

### Question 6

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

#### SOLUTION

Solution :Steps: 2 Marks

Answer: 1 Mark

(a+b)(2a−3b+c)−(2a−3b)c

=[(a+b)(2a−3b+c)]−[(2a−3b)c]

=[2a2−3ba+ac+2ab−3b2+bc]−[2ac−3bc]

=2a2−3ba+ac+2ab−3b2+bc−2ac+3bc

=2a2−3b2−ab−ac+4bc

The simplified expresion is

=2a2−3b2−ab−ac+4bc.

### Question 7

Find (0.4p−0.5q)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

(0.4p−0.5q)2

=(0.4p−0.5q)×(0.4p−0.5q)

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

=0.16p2−0.40pq+0.25q2

Given that

p = 5 and q = 4

on substituting the values we get,

=0.16×52−0.40×5×4+0.25×42

=4−8+4

=0

### Question 8

Add 2a+3b+c+d, −a+b−c−d and 3a−4b−2c−2d. 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+b−c−d

and 3a−4b−2c−2d.

(2a+3b+c+d)+(−a+b−c−d)+(3a−4b−2c−2d)

2a+3b+c+d−a+b−c−d+3a−4b−2c−2d

2a−a+3a+3b+b−4b+c−c−2c+d−d−2d

4a+0×b−2c−2d

Sum =4a−2c−2d

As per question ,

a = 2, b = 3, c = 4, d = 5

So, on substituting the values we get,

Sum = 4×2−2×4−2×5

=8−8−10

= −10

The value of the expression is −10.

### Question 9

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

#### SOLUTION

Solution :Answer: 1 Mark

Steps: 1 Mark

Type: 1 Mark

According to question,

(3a−b+3c)−[(a+b+c)+(2a−b+2c)]

=(3a−b+3c)−[(a+b+c+2a−b+2c)]

=(3a−b+3c)−(3a+3c)

=(3a−b+3c−3a−3c)

=−b

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

### Question 10

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

(i) 4y−7x

(ii) y2

(iii) x+y−xy

(iv) 100

(v) ab−a−b

(vi) (5−3t)

(vii) 4p2q−4pq2

(viii) 7mn

(ix) z2−3z+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)4y−7zBinomial(ii)y2Monomial(iii)x+y−xyTrinomial(iv)100Monomial(v)ab−a−bTrinomial(vi)5−3tBinomial(vii)4p2q−4pq2Binomial(viii)7mnMonomial(ix)z2−3z+8Trinomial(x)a2+b2Binomial(xi)z2+zBinomial(xii)1+x+x2Trinomial

### Question 11

From the sum of

4+3x and 5−4x+2x2,

subtract the sum of

3x2−5x and −x2+2x+5. [4 MARKS]

#### SOLUTION

Solution :Forming the equation: 1 Mark

Steps: 2 Marks

Answer: 1 Mark

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)+(5−4x+2x2)⌋−⌊(3x2−5x) +(−x2+2x+5)⌋

=[4+3x+5−4x+2x2]−[3x2−5x−x2+2x+5]

=[2x2+3x−4x+5+4]−[3x2−x2+2x−5x+5]

=[2x2−x+9]−[2x2−3x+5]

=2x2−x+9−2x2+3x−5

=2x2−2x2−x+3x+9−5

=2x+4

So, the required expression is 2x+4.

### Question 12

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

Answer: 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=2.5−1.3

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

### Question 13

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. 24−Rs. 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.

### Question 14

The perimeter of a triangle is 6p2−4p+9 and the length of two of its adjacent side are p2−2p+1 and 3p2−5p+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

6p2−4p+9=p2−2p+1+3p2−5p+3+3rd side

3rd side=6p2−4p+9−[(p2−2p+1)+(3p2−5p+3)]

3rd side=6p2−4p+9−(4p2−7p+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 6p2−4p+9 .

On substituting the values we get:

=6×32−4×3+9

=54−12+9

=51

The perimeter of the triangle is 51 units.

### Question 15

a) What should be subtracted from 3x2−4y2+5xy+20 to obtain −x2−y2+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,

3x2−4y2+5xy+20−q=−x2−y2+6xy+20

⇒q=3x2−4y2+5xy+20−(−x2−y2+6xy+20)

⇒q=3x2−4y2+5xy+20+x2+y2−6xy−20

⇒q=3x2+x2−4y2+y2+5xy−6xy+20−20

⇒q=4x2−3y2−xy+0

Hence, 4x2−3y2−xy should be subtracted from 3x2−4y2+5xy+20 to obtain −x2−y2+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