Algebraic Expressions(7th Grade > Mathematics ) Questions and answers for exam Preparation

7th Grade > Mathematics

ALGEBRAIC EXPRESSIONS MCQs

Total Questions : 116 | Page 8 of 12 pages

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]

Discuss Question

Answer: Option A. ->
:

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 72.

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

Discuss Question

Answer: Option A. ->
:

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 73.

Find $(0.4 p - 0.5 q)^{2}$ . If p = 5 and q = 4, evaluate the value of the expression. [3 MARKS]

Discuss Question

Answer: Option A. ->
:

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 74.

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]

Discuss Question

Answer: Option A. ->
:

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 75.

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]

Discuss Question

Answer: Option A. ->
:

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 76.

Define monomials, binomials, and trinomials. Classify the following into monomials, binomials and trinomials: [4 MARKS]
$(i) 4 y - 7 x$
$(i i) y^{2}$
$(i i i) x + y - x y$
$(i v) 100$
$(v) a b - a - b$
$(v i) (5 - 3 t)$
$(v i i) 4 p^{2} q - 4 p q^{2}$
$(v i i i) 7 m n$
$(i x) z^{2} - 3 z + 8$
$(x) a^{2} + b^{2}$
$(x i) z^{2} + z$
$(x i i) 1 + x + x^{2}$

Discuss Question

Answer: Option A. ->
:
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 77.

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]

Discuss Question

Answer: Option A. ->
:
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 78.

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]

Discuss Question

Answer: Option A. ->
:

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 79.

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]

Discuss Question

Answer: Option A. ->
:

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 80.

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]

Discuss Question

Answer: Option A. ->
:

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.