Free Algebraic Expressions Subjective Test 02 Practice Test - 7th grade

Question 1

Simplify: $- 2 (3 y + 6)$ [1 MARK]

SOLUTION

Solution :

$- 2 (3 y + 6) = (- 2 \times 3 y) + (- 2 \times 6)$

$= - 6 y + (- 12)$

$= - 6 y - 12$

Question 2

Which type of expression does $(3 x - 8 y + 4) - (x - y)$ give after simplification? [2 MARKS]

SOLUTION

Solution :
Simplified equation: 1 Mark
Name of expression: 1 Mark

The given equation is:
$(3 x - 8 y + 4) - (x - y)$

$= 3 x - 8 y + 4 - x + y$

$= 2 x - 7 y + 4$

where $2 x, 7 y$ and $4$ are the three terms.
So, the simplified equation is a trinomial.

Question 3

What will be the sum of the numerical coefficients in the below expression: [2 MARKS]
$4 x^{2} + 12 x y - 5 y^{2} - 11 y$

SOLUTION

Solution : Numerical Coefficients: 1 Mark
Sum: 1 Mark

The numerical coefficient of $4 x^{2}$ = 4
The numerical coefficient of $12 x y$ = 12
The numerical coefficient of $- 5 y^{2}$ = -5
The numerical coefficient of $- 11 y$ = -11

Sum of the coefficients = 4 + 12 - 5 - 11 = 0
Hence, the sum of all the numerical coefficients in the above expression is zero.

Question 4

Find the value of the polynomial $a^{3} + 3 a^{2} b + 2 a b - b^{2}$ at $a = - 2$ and $b = 3$ . [2 MARKS]

SOLUTION

Solution :
Steps: 1 Mark
Answer: 1 Mark

The value of the polynomial $a^{3} + 3 a^{2} b + 2 a b - b^{2}$ at a = -2 and b = 3

$a^{3} + 3 a^{2} b + 2 a b - b^{2}$
On substituting the values of a and b in the above expression we get:

$= (- 2)^{3} + 3 \times (- 2)^{2} \times 3 + 2 \times (- 2) \times 3 - 3^{2}$

$= - 8 + 36 - 12 - 9 = 7$

So, the value of the polynomial $a^{3} + 3 a^{2} b + 2 a b - b^{2}$ at $a = - 2$ and $b = 3$ is 7.

Question 5

The number 5 is added to three times the product of two numbers, m, and n. What is the expression formed? What is the value of the expression if m is 4 and n is 5? [2 MARKS]

SOLUTION

Solution :
Forming the expression: 1 Mark
Value of expression: 1 Mark

The product of two numbers, m, and n is
m $\times$ n.

Three times the product is 3mn.

Adding 5 to three times the product
= 5 + 3mn

The expression formed is 5 + 3mn

The value of the product at m = 4 and n = 5
$= 5 + 3 \times 4 \times 5 = 65$

Hence, the value of the expression if m = 4 and n = 5 is 65.

Question 6

(i) Will the value of $2 x^{2} + 12 x + 3$ for x = - 4 be greater than or less than 15?

(ii) If $x^{2}$ is added to $2 x^{2} + 12 x + 3$ and 15 is subtracted from it then evaluate its value at x = -4? [3 MARKS]

SOLUTION

Solution :
Each part: 1.5 Marks

(i) Value of the expression for $x = - 4$

$2 x^{2} + 12 x + 3$

$2 (- 4)^{2} + 12 (- 4) + 3$

$2 (16) - 48 + 3$

$32 - 48 + 3 = - 15$

Hence, the value of expression $2 x^{2} + 12 x + 3$ at $x = - 4$ is lesser than 15.

(ii) As per question

$x^{2}$ is added to $2 x^{2} + 12 x + 3$ and 15 is subtracted from it

So, the final expression is:

$2 x^{2} + 12 x + 3 + x^{2} - 15$

$3 x^{2} + 12 x - 12$

On putting x = -4 in the above equation we get:

$3 \times (- 4)^{2} + 12 \times (- 4) - 12$

$= 48 - 48 - 12$

$= - 12$

Question 7

From the sum of 3x - y + 11 and - y - 11, subtract 3x – y – 11. Write the algebraic expression. [3 MARKS]

SOLUTION

Solution : Forming the equation: 1 Mark
Steps: 1 Mark
Result: 1 Mark

According to question,

= (3x - y + 11) + (- y - 11) - (3x - y - 11)

= 3x - y + 11 - y - 11 - 3x + y + 11

= 3x - 3x - y - y + y + 11 - 11 + 11

= (3 - 3)x - (1 + 1 - 1)y + 11 + 11 -11

= 0x - y + 11

= -y +11
The required expression is -y +11.

Question 8

Write the steps which should be followed to check if two terms are like or unlike. Are $x^{2} y$ and $x y$ like terms? [3 MARKS]

SOLUTION

Solution :
Steps: 2 Marks
Answer: 1 Mark

Steps to be followed to decide whether the given terms are like or not are as followed:
i) Ignore the numerical coefficients.
ii) Check the variables in the terms. They must be same otherwise they are unlike terms.
iii) Next, check the powers of each variable in the terms. They should be the same and if they are not then they are not like terms.

We have to check if $x^{2} a n d x y$ are like terms.

The variables in both these expressions are $x a n d y$ .

The power of $y$ is equal in both the expressions.

But the power of $x$ in the term $x^{2} y$ is 2, while in the term $x y$ it is one.

$∴$ The terms are not like.

Question 9

Two adjacent sides of a rectangle are $5 x^{2} - 3 y^{2}$ and $x^{2} - 2 x y$ . Find its perimeter. [3 MARKS]

SOLUTION

Solution :
Formula: 1 Mark
Steps: 1 Mark
Answer: 1 Mark

Given that:
Two adjacent sides of a rectangle are $5 x^{2} - 3 y^{2}$ and $x^{2} - 2 x y$ .

The perimeter of a rectangle is given by:
Perimeter of a rectangle = 2 (Length + Breadth)

$= 2 [(5 x^{2} - 3 y^{2}) + (x^{2} - 2 x y)]$

$= [10 x^{2} - 6 y^{2}) + (2 x^{2} - 4 x y)]$

$= 12 x^{2} - 6 y^{2} - 4 x y$

So, the perimeter of the rectangle is $12 x^{2} - 6 y^{2} - 4 x y$ .

Question 10

Each symbol given below represents an algebraic expression: [4 MARKS]

Find the expression represented by the above symbols.

SOLUTION

Solution :
Visualizing the question and forming equation: 2 Marks
Steps: 1 Mark
Answer: 1 Mark

According to question:

$= [(2 x^{2} + 3 y) + (5 x^{2} + 3 x)] - (8 y^{2} + 3 x^{2} + 2 x + 3 y)$

$= [2 x^{2} + 3 y + 5 x^{2} + 3 x] - (8 y^{2} + 3 x^{2} + 2 x + 3 y)$

$= [7 x^{2} + 3 y + 3 x] - (8 y^{2} + 3 x^{2} + 2 x + 3 y)$

$= [7 x^{2} + 3 y + 3 x - 8 y^{2} - 3 x^{2} - 2 x - 3 y$

$= 4 x^{2} + x - 8 y^{2}$

Hence, the expression represented by the above symbols is $4 x^{2} + x - 8 y^{2}$ .

Question 11

Sonu and Raju have to collect different kinds of leaves for a science project. They went to a park where Sonu collected 12 leaves and Raju collected $x$ leaves. After sometime Sonu lost 3 leaves and Raju collects $2 x$ leaves. Write an algebraic expression to find the total number of leaves collected by both of them. If they collected an equal number of leaves then find the value of $x$ . [4 MARKS]

SOLUTION

Solution :
Final number of leaves by both: 2 Mark
Steps: 1 Mark
Answer: 1 mark

The first case, Sonu collected 12 leaves and Raju collected x leaves.

Sum of leaves collected by Sonu = 12

Sum of leaves collected by Raju = $x$

After some time, Sonu lost 3 leaves and Raju collected $2 x$ leaves.

Sum of leaves collected by Sonu = 12 - 3 = 9

Sum of leaves collected by Raju $= 2 x + x = 3 x$

Algebraic expression for the total number leaves collected by both $= 9 + 3 x$

It is given that, both of them collected equal number of leaves,
So, $3 x = 9$
Or $x = 9 \div 3 = 3$
Hence, the value of $x$ is $3$ .

Question 12

Add
$t - t^{2} - 14$ , $15 t^{3} + 13 + 9 t - 8 t^{2}$ , $12 t^{2} - 19 - 24 t$ and $4 t - 9 t^{2} + 19 t^{3}$ .

If $t = - 1$ find the value of the expression. [4 MARKS]

SOLUTION

Solution :
Forming equation: 1 Mark
Steps: 1 Mark
Answer: 1 Mark
Value: 1 Mark

Here while adding the algebraic expressions we need to know that we can only add the like terms.
Like terms are those terms which have the same algebraic factors.

Sum

$= (t - t^{2} - 14) + (15 t^{3} + 13 + 9 t - 8 t^{2}) + (12 t^{2} - 19 - 24 t) + (4 t - 9 t^{2} + 19 t^{3})$
$= t - t^{2} - 14 + 15 t^{3} + 13 + 9 t - 8 t^{2} + 12 t^{2} - 19 - 24 t + 4 t - 9 t^{2} + 19 t^{3}$
$= (t + 9 t - 24 t + 4 t) + (- t^{2} - 8 t^{2} + 12 t^{2} - 9 t^{2}) + (- 14 + 13 - 19) + (15 t^{3} + 19 t^{3})$
$= - 10 t - 6 t^{2} - 20 + 34 t^{3}$

Hence, the required expression is $- 10 t - 6 t^{2} - 20 + 34 t^{3}$ .

Given that:
$t = - 1$
On substituting the values we get:
$- 10 t - 6 t^{2} - 20 + 34 t^{3}$
$= - 10 \times (- 1) - 6 \times (- 1)^{2} - 20 + 34 \times (- 1)^{3}$
$= 10 - 6 - 20 - 34$
$= - 50$

Question 13

Remit's mother gave him $R s . 3 x y^{2}$ and his father gave him $R s 5 (x y^{2} + 2)$ .Out of this total money, he spent $R s . (10 - 3 x y^{2})$ on his birthday party. How much money is left with him? If $x = 2 a n d y = 3$ find the total amount left with him. [4 MARKS]

SOLUTION

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

As per the question,

The amount of money given by Remits' mother $R s 3 x y^{2}$ .

The amount of money given by Remits' father $R s 5 (x y^{2} + 2)$ .

Total amount $= 3 x y^{2} + 5 x y^{2} + 10 = 3 x y^{2} + 5 x y^{2} + 10 = 8 x y^{2} + 10$

Total amount Remit spent $= 10 - 3 x y^{2}$

Amount of money left $= (8 x y^{2} + 10) - (10 - 3 x y^{2}) = 8 x y^{2} + 10 - 10 + 3 x y^{2} = R s . 11 x y^{2}$

Now if $x = 2 a n d y = 3$ , then on substituting the values we get,
Amount of money left with
= $11 \times 2 \times \times 3 \times 3$ = $R s 198$

Hence, the amount of money left with Remit is $R s 198$ .

Question 14

A number is 12 more than the other if their sum is 48. What are the numbers? Also, find the product of these numbers. [4 MARKS]

SOLUTION

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

Given that
A number is 12 more than the other number.
Let the number be $x$ , then the other number will be $x + 12$ .

As per question
$x + (x + 12) = 48$

$\Rightarrow 2 x + 12 = 48$

$\Rightarrow 2 x = 48 - 12$

$\Rightarrow x = \frac{36}{2} = 18$

Then, the other number is $x + 12 = 18 + 12 = 30$ .

Therefore, the numbers are 18 and 30.
The product of these numbers $= 18 \times 30 = 540$ .

Question 15

The sum of two consecutive numbers with a difference of 5 is 55. Find the numbers. [4 MARKS]

SOLUTION

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

Let the numbers be $x$ and $x + 5$

Therefore, $x + x + 5 = 55$

$\Rightarrow 2 x + 5 = 55$

$\Rightarrow 2 x = 55 - 5$

$\Rightarrow 2 x = 50$

$\Rightarrow x = \frac{50}{2}$

$\Rightarrow x = 25$

Therefore, the requirednumbers are $x = 25, a n d x + 5 = 30$

Therefore, the two numbers with a difference of 5 whose sum is 55 are 25 and 30.

Free Algebraic Expressions Subjective Test 02 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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