New Type of Equation Comparison Practice Questions for Mains Exam

RBI Assistant Mains is an upcoming important exam which is just around the corner. The exam is falling on 31st December thereby not leaving much time for practice. We are here trying to help the aspirants who are preparing for any mains examination especially RBI Assistant mains 2023. Here we are providing a few new type of Equation Comparison practice questions for mains exam which will be helpful for RBI Assistant mains. 

New Type of Equation Comparison Practice Questions for Mains Exam

Directions (1 – 3): Answer the questions based on the information given below.

I. 2x² + 7x + k = 0

II. (ay + b)² = 0

Smaller root of equation I is -2.

Larger root of equation I is a root of equation II.

Question 1: If p = -2 × √(k + 3), then find the value of ‘p’.

A) 3

B) 6

C) -6

D) 9

E) None of these

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Question 2: Find the value of ‘b’.

A) 3

B) -3

C) 0

D) -1

E) Can’t be determined

Question 3: Find the other root of equation (I).

A) 1.25

B) -1.25

C) 1.5

D) -1.5

E) 0.25

Question 4: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.

I. x³ – 14² + (32² – 24²) = 154 + 21²

II. y²– 12y + 35 = 0

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

Question 5: In the questions, three equations I, II and III are given. You have to solve the equations to establish the correct relation. Which of the following pair of symbols will define the relation between x and y and between y and z?

I. 3x + 2y + 4z = 61

II. x + 5y + z = 53

III. 5x + y + z = 49

A. =

B. >

C. <

D. >=

E. <=

F. # (relation cannot be established)

A) B, C

B) D, C

C) F, D

D) C, B

E) F, C

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Question 6: In the given table there are two columns I and II. Column I contains three equations and column II contains the roots of the equations given in column I, not necessarily in the same order. Study the given table carefully and answer the questions accordingly.

Which of the following relation is correct?

A) I – a, II – b, III – c

B) I – a, II – c, III – b

C) I – c, II – b, III – a

D) I – c, II – a, III – b

E) I – b, II – a, III – c

Question 7: Which of the following pair of symbols will define the relation between x and y and between y and z respectively?

A. =

B. >

C. <

D. ≥

E. ≤

F. # (relation cannot be established)

I. 4x² – 20x + 21 = 0

II. 2y² = y + 3

III. 5z² – 24z + 28 = 0

A) C, D

B) F, C

C) C, B

D) D, C

E) F, B

Question 8: Which of the following pair of symbols will define the relation between x and y and between y and z respectively?

A. =

B. >

C. <

D. ≥

E. ≤

F. # (relation cannot be established)

I. x + 7y – 2z = 28

II. 9x – 4y + 5z = 76

III. 5x – 6y + 4z = 32

A) B, C

B) A, C

C) B, A

D) C, B

E) C, A

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Question 9: In the question, two equations I and II are given. You have to solve both the equations to establish the correct relation between x and y and choose the correct option.

I. x² + (7 + 6√2)x + 42√2 = 0

II. y² + 13√2y + 84 = 0

A) x > y

B) x < y

C) x = y or the relationship cannot be established

D) x ≥ y

E) x ≤ y

निर्देश (1 – 3): नीचे दी गई जानकारी के आधार पर प्रश्नों के उत्तर दें।

I. 2x² + 7x + k = 0

II. (ay + b)² = 0

समीकरण I का छोटा सूत्र -2 है।

समीकरण I का बड़ा सूत्र समीकरण II का सूत्र है।

प्रश्न 1: यदि p = -2 × √(k + 3) है, तो ‘p’ का मान ज्ञात करें।

A) 3

B) 6

C) -6

D) 9

E) इनमें से कोई नहीं

प्रश्न 2: ‘b’ का मान ज्ञात करें।

A) 3

B) -3

C) 0

D) -1

E) निर्धारित नहीं किया जा सकता है

प्रश्न 3: समीकरण (I) की दूसरा सूत्र खोजें।

A) 1.25

B) -1.25

C) 1.5

D) -1.5

E) 0.25

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प्रश्न 4: प्रश्न में, दो समीकरण I और II दिए गए हैं। आपको x और y के बीच सही संबंध स्थापित करने और सही विकल्प चुनने के लिए दोनों समीकरणों को हल करना होगा।

I.x³ – 14² + (32² – 24²) = 154 + 21²

II. y²– 12y + 35 = 0

A) x > y

B) x < y

C) x = y या संबंध स्थापित नहीं किया जा सकता है 

D) x ≥ y

E) x ≤ y

प्रश्न 5: प्रश्नों में, तीन समीकरण I, II और III दिए गए हैं। आपको सही संबंध स्थापित करने के लिए समीकरणों को हल करना होगा। निम्नलिखित में से कौन सा प्रतीक x और y और y और z के बीच के संबंध को परिभाषित करेगा?

I. 3x + 2y + 4z = 61

II. x + 5y + z = 53

III. 5x + y + z = 49

A. =

B. >

C. <

D. >=

E. <=

F. # (संबंध स्थापित नहीं किया जा सकता)

A) B, C

B) D, C

C) F, D

D) C, B

E) F, C

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प्रश्न 6: दी गई तालिका में दो कॉलम I और II हैं। कॉलम I में तीन समीकरण हैं और कॉलम II में कॉलम I में दिए गए समीकरणों के सूत्र हैं, ज़रूरी नहीं कि इसी क्रम में हों। दी गई तालिका का ध्यानपूर्वक अध्ययन करें और उसके अनुसार प्रश्नों के उत्तर दें।

निम्नलिखित में से कौन सा संबंध सही है?

A) I – a, II – b, III – c

B) I – a, II – c, III – b

C) I – c, II – b, III – a

D) I – c, II – a, III – b

E) I – b, II – a, III – c

प्रश्न 7: निम्नलिखित में से कौन सा प्रतीक क्रमशः x और y और y और z के बीच के संबंध को परिभाषित करेगा?

A. =

B. >

C. <

D. ≥

E. ≤

F. # (संबंध स्थापित नहीं किया जा सकता)

I. 4x² – 20x + 21 = 0

II. 2y² = y + 3

III. 5z² – 24z + 28 = 0

A) C, D

B) F, C

C) C, B

D) D, C

E) F, B

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प्रश्न 8: निम्नलिखित में से कौन सा प्रतीक क्रमश : x और y और y और z के बीच के संबंध को परिभाषित करेगा?

A. =

B. >

C. <

D. ≥

E. ≤

F. # (संबंध स्थापित नहीं किया जा सकता)

I. x + 7y – 2z = 28

II. 9x – 4y + 5z = 76

III. 5x – 6y + 4z = 32

A) B, C

B) A, C

C) B, A

D) C, B

E) C, A

प्रश्न 9: प्रश्न में, दो समीकरण I और II दिए गए हैं। आपको x और y के बीच सही संबंध स्थापित करने और सही विकल्प चुनने के लिए दोनों समीकरणों को हल करना होगा।

I. x² + (7 + 6√2)x + 42√2 = 0

II. y² + 13√2y + 84 = 0

A) x > y

B) x < y

C) x = y या संबंध स्थापित नहीं किया जा सकता है

D) x ≥ y

E) x ≤ y

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ANSWER KEYS and SOLUTIONS:

Solution 1: C)

Since, (-2) is a root of equation I,

2 × (-2)² 7 × (-2) k = 0

Or, 2 × 4 (-14) k = 0

Or, 8 – 14 k = 0

Or, k = 6

So, 2x² 7x 6 = 0

Or, 2x² 4x – 3x 6 = 0

Or, 2x(x 2) 3(x 2) = 0

Or, (x 2) (2x 3) = 0

So, x = -2 or x = -(3/2)

Since, k = 6, required value = -2 × √(6 + 3) = -2 × 3 = -6

Hence, option c.

Solution 2: E)

Given = (ay + b)² = 0

Or, ay + b = 0

So, y = -(b/a)

Or, -(b/a) = -(3/2)

Or, (b/a) = (3/2)

Therefore, actual value of ‘b’ cannot be determined.

Hence, option e.

Solution 3: D)

So the other root of equation (I) is -3/2 = -1.5.

Hence, option d.

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Solution 4: D)

From I:

x³ – 14² + (32² – 24²) = 154 + 21²

x³ – 196 + (32 + 24)(32 – 24) = 154 + 441

x³ – 196 + 448 = 595

x³ = 343

x = 7

From II:

y² – 12y + 35 = 0

y² – 7y – 5y +35 = 0

y(y – 7) – 5(y – 7) = 0

(y – 7)(y – 5) = 0

y = 7, 5

So, x ≥ y.

Hence, option d.

Solution 5: D)

From I and II,

3x + 2y + 4z = 61

z = (61 – 3x – 2y)/4 ———-(a)

And, x + 5y + z = 53

z = 53 – x – 5y ———(b)

Solving (a) and (b), we get

(61 – 3x – 2y)/4 = 53 – x – 5y

x + 18y = 151 ——-(i)

From II and III,

x + 5y + z = 53

z = 53 – x – 5y ——-(c)

And, 5x + y + z = 49

z = 49 – 5x – y ———(d)

Solving (c) and (d), we get

53 – x – 5y = 49 – 5x – y

y – x = 1 ———(ii)

Solving (i) and (ii), we get

y = 8 and x = 7,

So, z = 6

Hence, option d.

Solution 6: D)

From I:

x² – 3x = 28

Or, x² – 3x – 28 = 0

Or, x² – 7x + 4x – 28 = 0

Or, x(x – 7) + 4(x – 7) = 0

Or, (x – 7)(x + 4) = 0

So, x = 7 or x = -4

From II:

y² + 104 = 21y

Or, y² – 21y + 104 = 0

Or, y² – 13y – 8y + 104 = 0

Or, y(y – 13) – 8(y – 13) = 0

Or, (y – 13)(y – 8) = 0

So, y = 13 or y = 8

From III:

z² = 4z + 32

Or, z² – 4z – 32 = 0

Or, z² – 8z + 4z – 32 = 0

Or, z(z – 8) + 4(z – 8) = 0

Or, (z – 8)(z + 4) = 0

So, z = 8 or z = -4

Therefore, correct relation is I – c, II – a, III – b.

Hence, option d.

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Solution 7: D)

From I:

4x² – 20x + 21 = 0

4x² – 14x – 6x + 21 = 0

2x(2x – 7) – 3(2x – 7) = 0

(2x – 7)(2x – 3) = 0

x = 7/2, 3/2

From II:

2y² = y + 3

2y² – y – 3 = 0

2y²– 3y + 2y – 3 = 0

y(2y – 3) + 1(2y – 3) = 0

(y + 1)(2y – 3) = 0

y = -1, 3/2

From III:

5z² – 24z + 28 = 0

5z² – 14z – 10z + 28 = 0

z(5z – 14) – 2(5z – 14) = 0

(5z – 14)(z – 2) = 0

z = 2, 14/5

So, x ≥ y and y < z.

Hence, option d.

Solution 8: C)

Using equations (I) and (II):

9 × (28 – 7y + 2z) – 4y + 5z = 76

252 – 63y + 18z – 4y + 5z = 76

67y – 23z = 176

y = (176 + 23z)/67 ———-(i)

Using equations (I) and (III):

5 × (28 – 7y + 2z) – 6y + 4z = 32

140 – 35y + 10z – 6y + 4z = 32

41y – 14z = 108

Using (I), 41 × (176 + 23z)/67 – 14z = 108

7216 + 943z – 938z = 7236

5z = 20

z = 4

y = (176 + 23 × 4)/67 = 268/67 = 4

x = 28 – 7 × 4 + 2 × 4 = 28 – 28 + 8 = 8

So, x = 8, y = 4 and z = 4

So, x >y and y = z.

Hence, option c.

Solution 9: D)

From I:

x² + (7 + 6√2)x + 42√2 = 0

x² + 7x + 6√2x + 42√2 = 0

x(x + 7) + 6√2(x + 7) = 0

(x + 7)(x + 6√2) = 0

x = -7,-6√2

From II:

y² + 13√2y + 84 = 0

y² + 7√2y + 6√2y + 84 = 0

y(y + 7√2) + 6√2(y + 7√2) = 0

(y + 7√2)(y + 6√2) = 0

y = -7√2, -6√2

So, x ≥ y.

Hence, option d.

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