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

Question 1

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

SOLUTION

Solution :

$\to$ 6m + 10 − 10m

$\to$ (6m - 10m) + 10

$\to$ - 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. $a (a^{2} - b^{2}) = a^{3} - a b^{2}$ the expression has only two terms $a^{3}$ and $a b^{2}$ .

Question 3

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

$2 x^{2} + 31 y - 5 x y$ [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 $2 x^{2} + 31 y - 5 x y$
The numerical coefficient of $2 x^{2}$ = $2$
The numerical coefficient of $31 y$ = $31$
The numerical coefficient of $- 5 x y$ = $- 5$

Sum of the coefficients = 2 + 31 - 5 = 28
Hence, the sum of the numerical coefficients of the expression $2 x^{2} + 31 y - 5 x y$ 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 $10 x^{2} - 5 x - 6$ . 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 $x^{2}$ and $x$ are 10 and - 5 respectively

Terms are $10 x^{2}$ , $- 5 x$ and $- 6$
We have to find the value of the expression at $x = 2$ is
$10 \times 2 \times 2 - 5 \times 2 - 6$
= $40 - 10 - 6$
= $24$
Hence, the value of the expression at $x = 2 i s 24$

Question 6

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

SOLUTION

Solution :
Steps: 2 Marks
Answer: 1 Mark

$(a + b) (2 a - 3 b + c) - (2 a - 3 b) c$

$= [(a + b) (2 a - 3 b + c)] - [(2 a - 3 b) c]$

$= [2 a^{2} - 3 b a + a c + 2 a b - 3 b^{2} + b c] - [2 a c - 3 b c]$

$= 2 a^{2} - 3 b a + a c + 2 a b - 3 b^{2} + b c - 2 a c + 3 b c$

$= 2 a^{2} - 3 b^{2} - a b - a c + 4 b c$

The simplified expresion is
$= 2 a^{2} - 3 b^{2} - a b - a c + 4 b c$ .

Question 7

Find $(0.4 p - 0.5 q)^{2}$ . 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

$(0.4 p - 0.5 q)^{2}$

$= (0.4 p - 0.5 q) \times (0.4 p - 0.5 q)$

$= (0.4 p) \times (0.4 p) - (0.4 p \times 0.5 q + 0.4 p \times 0.5 q) + (0.5 q) \times (0.5 q)$

$= 0.16 p^{2} - 0.40 p q + 0.25 q^{2}$

Given that
p = 5 and q = 4
on substituting the values we get,
$= 0.16 \times 5^{2} - 0.40 \times 5 \times 4 + 0.25 \times 4^{2}$
$= 4 - 8 + 4$
$= 0$

Question 8

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

$(2 a + 3 b + c + d) + (- a + b - c - d) + (3 a - 4 b - 2 c - 2 d)$

$2 a + 3 b + c + d - a + b - c - d + 3 a - 4 b - 2 c - 2 d$

$2 a - a + 3 a + 3 b + b - 4 b + c - c - 2 c + d - d - 2 d$

$4 a + 0 \times b - 2 c - 2 d$

Sum $= 4 a - 2 c - 2 d$
As per question ,
a = 2, b = 3, c = 4, d = 5
So, on substituting the values we get,
Sum = $4 \times 2 - 2 \times 4 - 2 \times 5$
= $8 - 8 - 10$
= $- 10$
The value of the expression is $- 10$ .

Question 9

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

SOLUTION

Solution :
Answer: 1 Mark
Steps: 1 Mark
Type: 1 Mark

According to question,

$(3 a - b + 3 c) - [(a + b + c) + (2 a - b + 2 c)]$

$= (3 a - b + 3 c) - [(a + b + c + 2 a - b + 2 c)]$

$= (3 a - b + 3 c) - (3 a + 3 c)$

$= (3 a - b + 3 c - 3 a - 3 c)$

$= - b$

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]

$(i) 4 y - 7 x$
$(i i) y^{2}$
$(i i i) x + y - x y$
$(i v) 100$
$(v) a b - a - b$
$(v i) (5 - 3 t)$
$(v i i) 4 p^{2} q - 4 p q^{2}$
$(v i i i) 7 m n$
$(i x) z^{2} - 3 z + 8$
$(x) a^{2} + b^{2}$
$(x i) z^{2} + z$
$(x i i) 1 + x + x^{2}$

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.

$\begin{matrix} S . N O & E x p r e s s i o n & T y p e o f P o l y n o m i a l (i) & 4 y - 7 z & B i n o m i a l (i i) & y^{2} & M o n o m i a l (i i i) & x + y - x y & T r i n o m i a l (i v) & 100 & M o n o m i a l (v) & a b - a - b & T r i n o m i a l (v i) & 5 - 3 t & B i n o m i a l (v i i) & 4 p^{2} q - 4 p q^{2} & B i n o m i a l (v i i i) & 7 m n & M o n o m i a l (i x) & z^{2} - 3 z + 8 & T r i n o m i a l (x) & a^{2} + b^{2} & B i n o m i a l (x i) & z^{2} + z & B i n o m i a l (x i i) & 1 + x + x^{2} & T r i n o m i a l \end{matrix}$

Question 11

From the sum of

$4 + 3 x a n d 5 - 4 x + 2 x^{2}$ ,

subtract the sum of

$3 x^{2} - 5 x a n d - x^{2} + 2 x + 5$ . [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,

$⌊ (4 + 3 x) + (5 - 4 x + 2 x^{2}) ⌋ - ⌊ (3 x^{2} - 5 x) + (- x^{2} + 2 x + 5) ⌋$

$= [4 + 3 x + 5 - 4 x + 2 x^{2}] - [3 x^{2} - 5 x - x^{2} + 2 x + 5]$

$= [2 x^{2} + 3 x - 4 x + 5 + 4] - [3 x^{2} - x^{2} + 2 x - 5 x + 5]$

$= [2 x^{2} - x + 9] - [2 x^{2} - 3 x + 5]$

$= 2 x^{2} - x + 9 - 2 x^{2} + 3 x - 5$

$= 2 x^{2} - 2 x^{2} - x + 3 x + 9 - 5$

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

Question 12

Tatesville has $x$ inches of rainfall in April, $(x + 1.3) i n c h e s$ of rainfall in May, and $(2 x + 0.5) i n c h e s$ 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, $(x + 1.3) i n c h e s$ of rainfall in May, and $(2 x + 0.5) i n c h e s$ in June.
The total rainfall during these months will be the sum of these expressions.

Total amount of rainfall
$= x + (x + 1.3) + (2 x + 0.5) i n c h e s$

$= x + x + 1.3 + 2 x + 0.5 i n c h e s$

$= 4 x + 1.8 i n c h e s$

Given that, rainfall in May is 2.5 inches.
Rainfall in May
$(x + 1.3) = 2.5 i n c h e s$

$x + 1.3 = 2.5$

$x = 2.5 - 1.3$

$x = 1.2 i n c h e s$

On substituting the value of $x = 1.2 i n c h e s$ in the ecpression for total rainfall.
Total Rainfall
$= 4 x + 1.8 = 4 (1.2) + 1.8 = 4.8 + 1.8 = 6.6 i n c h e s$
So, the total rainfall over the three months is $6.6 i n c h e s$

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 $= R s . 24$

Total number of pencils Aliyah bought $= 7$

Total amount left after buying pencil $= R s . 10$

Total amount she spent
$= R s . 24 - R s . 10 = R s . 14$

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

Cost of each pencil , $x = \frac{14}{7} = R s . 2$
The cost of each pencil is $R s 2$ .

Question 14

The perimeter of a triangle is $6 p^{2} - 4 p + 9$ and the length of two of its adjacent side are $p^{2} - 2 p + 1$ and $3 p^{2} - 5 p + 3$ . 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

$6 p^{2} - 4 p + 9 = p^{2} - 2 p + 1 + 3 p^{2} - 5 p + 3 + 3^{r d} s i d e$

$3^{r d} s i d e = 6 p^{2} - 4 p + 9 - [(p^{2} - 2 p + 1) + (3 p^{2} - 5 p + 3)]$

$3^{r d} s i d e = 6 p^{2} - 4 p + 9 - (4 p^{2} - 7 p + 4)$

$3^{r d} s i d e = 2 p^{2} + 3 p + 5$
The $3^{r d}$ side of the triangle has length is $2 p^{2} + 3 p + 5$

Given that
p = 3
The perimeter of the triangle is $6 p^{2} - 4 p + 9$ .
On substituting the values we get:
$= 6 \times 3^{2} - 4 \times 3 + 9$
$= 54 - 12 + 9$
$= 51$

The perimeter of the triangle is 51 units.

Question 15

a) What should be subtracted from $3 x^{2} - 4 y^{2} + 5 x y + 20$ to obtain $- x^{2} - y^{2} + 6 x y + 20$ ?
b) Find the final expression when the algebraic expression $x^{2} + 2 x y + y^{2}$ is multiplied by $x y$ . [4 MARKS]

SOLUTION

Solution : Each option: 1.5 Marks

a) Let q be the expression to be subtracted.

Then according to the question,

$3 x^{2} - 4 y^{2} + 5 x y + 20 - q = - x^{2} - y^{2} + 6 x y + 20$

$\Rightarrow q = 3 x^{2} - 4 y^{2} + 5 x y + 20 - (- x^{2} - y^{2} + 6 x y + 20)$

$\Rightarrow q = 3 x^{2} - 4 y^{2} + 5 x y + 20 + x^{2} + y^{2} - 6 x y - 20$

$\Rightarrow q = 3 x^{2} + x^{2} - 4 y^{2} + y^{2} + 5 x y - 6 x y + 20 - 20$

$\Rightarrow q = 4 x^{2} - 3 y^{2} - x y + 0$

Hence, $4 x^{2} - 3 y^{2} - x y$ should be subtracted from $3 x^{2} - 4 y^{2} + 5 x y + 20$ to obtain $- x^{2} - y^{2} + 6 x y + 20$ .

b) As per the question:
$x^{2} + 2 x y + y^{2}$ is multiplied by $x y$

So, $x y$ $\times$ $x^{2} + 2 x y + y^{2}$

$= x y \times x^{2} + x y \times 2 x y + x y \times y^{2}$

$= x^{2 + 1} y + 2 x^{1 + 1} y^{1 + 1} + x y^{1 + 2}$

$∵ a^{m} \times a^{n} = a^{m + n}$

$= x^{3} y + 2 x^{2} y^{2} + x y^{3}$

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

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

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