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Mathematics Paper 2 (Algebra - I) - CTET & State TET MCQ


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10 Questions MCQ Test - Mathematics Paper 2 (Algebra - I)

Mathematics Paper 2 (Algebra - I) for CTET & State TET 2024 is part of CTET & State TET preparation. The Mathematics Paper 2 (Algebra - I) questions and answers have been prepared according to the CTET & State TET exam syllabus.The Mathematics Paper 2 (Algebra - I) MCQs are made for CTET & State TET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Mathematics Paper 2 (Algebra - I) below.
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Mathematics Paper 2 (Algebra - I) - Question 1

What is the value of (27x3 - 58x2y + 31xy2 - 8y3), when x = -5 and y = -7?

Detailed Solution for Mathematics Paper 2 (Algebra - I) - Question 1

Calculation:

(3x - 2y)3 = (3x)3 - (2y)3 - 3 × (3x)2 × (2y) + 3 × 3x × (2y)2

⇒ 27x3 - 8y3 - 54x2y - 4x2y + 36 xy2 - 5xy2 = (3x - 2y)3 - 4x2y - 5xy2

⇒ 27x3 - 58x2y + 31xy2 - 8y3 = (3x - 2y)3 - 4x2y - 5xy2

Now we put the value of x = - 5 and y = - 7

(3x - 2y)3 - 4x2y - 5xy2 

⇒ {3 × (- 5) - 2 × (- 7)}3 - 4 × (- 5)2 × (- 7) - 5 × (- 5) × (- 7)2

 {- 15 + 14}3 + 4 × 25 × 7 + 25 × 49

⇒ (-1)3 + 700 + 1225

⇒ 1925 - 1 = 1924

∴ The correct answer is 1924. 

Mathematics Paper 2 (Algebra - I) - Question 2

Simplify,

x42x2+1x22x+1

Detailed Solution for Mathematics Paper 2 (Algebra - I) - Question 2

Formula used:

(a - b)2 = a2 + b2 - 2ab

(a2 - b2) = (a + b) × (a - b)

(a + b)2 = a2 + b2 + 2ab

Calculation:

⇒ {(x2)2 + 1 - 2 × (x2) × 1}/(x - 1)2

⇒ (x2 - 1)2/(x - 1)2

⇒ {(x + 1)2(x - 1)2}/(x - 1)2

⇒ (x + 1)2 = x2 + 2x + 1

∴ The correct option is 3.

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Mathematics Paper 2 (Algebra - I) - Question 3

Simplify the given expression. 

Detailed Solution for Mathematics Paper 2 (Algebra - I) - Question 3

Given:

Concept used:

(a + b)3 = a3 + b3 + 3ab (a + b)

(a - b)3 = a3 - b3 - 3ab (a - b)

Calculation:

According to the question,

The expression is =  

We know,

( x + 3)2 = x3 + 33 + 3 × (x) × 3 (x + 3)

⇒ ( x + 3)2 = x3 + 27 + 9x(x + 3)

⇒( x + 3)2 = x3 + 27 + 9x2 + 27x

⇒( x + 3)2 = x3 + 9x2 + 27x + 27                    ---------(1)

Similarly,

( x - 3)2 = x3 - 33 - 3 × (x) × 3 (x - 3)

⇒ (x - 3)2 = x3 - 27 - 9x(x - 3)

⇒( x - 3)2 = x3 - 27 - 9x2 + 27x

⇒( x - 3)2 = x3 - 9x2 + 27x - 27                        ---------(2)

Put the value of eq. (1) and eq. (2) in given expression

Mathematics Paper 2 (Algebra - I) - Question 4

Simplify the expression: 

(c + d)2 - (c - d)2

Detailed Solution for Mathematics Paper 2 (Algebra - I) - Question 4

Formula Used :

x- y2 = (x + y) (x - y)

Calculation:

⇒ (c + d)2 - (c - d)2

⇒ (c + d + c - d) (c + d -(c - d))

⇒ 2c × (c + d - c + d)

⇒ 2c × 2d = 4cd

∴ The simplified value is 4cd.

Alternate Method

Formula Used :

(x + y)2 = x2 + y2 + 2xy

(x - y)2 = x2 + y2 - 2xy

Calculation:

⇒ (c + d)2 - (c - d)2

⇒ c+ d+ 2cd - (c2 + d2 - 2cd)

⇒ c2 + d2 + 2cd - c2 - d2 + 2cd

⇒ 4cd

∴ The simplified value is 4cd.

Mathematics Paper 2 (Algebra - I) - Question 5
If (p - q) = 8, then what is the value of q3 - p3 + 24pq?
Detailed Solution for Mathematics Paper 2 (Algebra - I) - Question 5

Given:

(p - q) = 8,

Formula used:

(a - b)3 = a3 - b3 - 3ab(a - b)

Calculation:

According to the question,

⇒ (p - q) = 8

Take cube on both sides, 

⇒ (p - q)3 = 83

⇒ p3 - q3 - 3pq(p - q)= 83

⇒ p3 - q3 - 3pq(8) = 512

⇒ p3 - q3 - 24pq = 512

It can be written as:

⇒  q3 - p3 + 24pq = -512

Therefore, "-512" is the required answer.

Mathematics Paper 2 (Algebra - I) - Question 6
The square of the difference between two given natural numbers is 324, while the product of these two given numbers is 144. Find the positive difference between the squares of these two given numbers.
Detailed Solution for Mathematics Paper 2 (Algebra - I) - Question 6

Given:

The square of the difference between two given natural numbers is 324, while the product of these two given numbers is 144. 

Calculation:

Let the numbers are x and y

(x - y)2 = 324

So, x - y = 18, xy = 144

(x + y)2 = (18)2 + 4× 144

⇒ 900

⇒ x + y = 30

Then, x is (30 + 18) / 2 = 24 and y = 6

So , x2 - y2 = 242 - 62

⇒ 576 - 36 = 540

∴ The correct  option is 2

Mathematics Paper 2 (Algebra - I) - Question 7

Find the value of x:

Detailed Solution for Mathematics Paper 2 (Algebra - I) - Question 7

Given:

Concept used:

BODMAS rule

Calculation:

We know from BODMAS the rule that,

Addition is solved first

Then subtraction is solved.

According to the BODMAS rule  -

 = 479

 = 479

  = 479

On solving for x, we get,

x = 4790

Therefore, the correct option is 3 that is 4790.

Mathematics Paper 2 (Algebra - I) - Question 8
If  then what is (x + y) equal to?
Detailed Solution for Mathematics Paper 2 (Algebra - I) - Question 8

Calculation:

⇒ 

⇒ 

⇒ 6x - 9y + 3 = 2x + 8y + 16

⇒ 4x - 17y = 13     ------(1)

⇒ 

⇒ 10x - 15y + 5 = 8x - 14y + 4

⇒ (2x - y = -1) × 2

⇒ 4x - 2y = -2        -------(2)

Subtracting equation (1) & (2), we get

⇒ 15y = -15

⇒ y = -1

By putting the value of y in equation (2), we get

⇒ x = -1

⇒ x + y = -1 + (-1) = -2

∴ The correct answer is -2.

Mathematics Paper 2 (Algebra - I) - Question 9

Simplify the expression  provided (s + t + u) ≠ 0.

Detailed Solution for Mathematics Paper 2 (Algebra - I) - Question 9

Calculation:

∴ The correct option is 1.

Mathematics Paper 2 (Algebra - I) - Question 10
If (d + e + f) = 14, (d2 + e2 + f2) = 96, then find the value of (de + ef + fd).
Detailed Solution for Mathematics Paper 2 (Algebra - I) - Question 10

Given:

(d + e + f) = 14

(d2 + e2 + f2) = 96

Formula Used:

(a + b + c)2 = 2 × (ab + bc + ca) + a2 + b2 + c2

Calculation:

⇒ (d + e + f)2 = 2 × (de + ef + fd) + d2 + e2 + f2

⇒ ( 14)2 = 2× (de + ef + fd) + 96

⇒ 2 × (de + ef + fd) = 196 - 96 = 100

⇒ (de + ef + fd) = 100/2 = 50

∴ The value of (de + ef + fd) is 50.

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