Target Exam

CUET

Subject

General Aptitude Test

Chapter

Quantitative Reasoning

Topic

Algebra

Question:

Match List - I with List - II.

List – I

List – II

(A) $14-\frac{x-1}{10}=\frac{x+5}{6}-3$

(I) 4

(B) $(x-5)^2-(x+3)^2=48$

(II) $-\frac{23}{2}$

(C) $6(x-4)=4(x-3)-(3x-8)$

(III) 61

(D) $(2x-1) (2x+3)=(2x-7) (2x+7)$

(IV) -2

Choose the most appropriate answer from the options given below:

Options:

(A)-(I), (B)-(II), (C)-(III), (D)-(IV)

(A)-(II), (B)-(I), (C)-(III), (D)-(IV)

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

(A)-(III), (B)-(IV), (C)-(II), (D)-(I)

Correct Answer:

(A)-(III), (B)-(IV), (C)-(I), (D)-(II)

Explanation:

14− (x1)/10=(x+5)/−  3

⇒ (x+5)/6 + (x1)/10 = 17

⇒ 10x + 50 + 6x - 6 = 17*60 ⇒ 16x = 1020 - 44 ⇒ x = 61

 

(x5)2(x+3)2=48

⇒ x2+25-10x-x2-9-6x = 48

⇒ 16-16x = 48 ⇒ 16x = -32 ⇒ x = -2

 

6(x4)=4(x3)(3x8)

⇒ 6x-24 = 4x-12-3x+8

⇒ 6x-4x+3x = 24-12+8

⇒ 5x = 20 ⇒ x = 4

 

(2x1)(2x+3)=(2x7)(2x+7)

⇒ 4x2+6x-2x-3 = 4x2-14x+14x-49

⇒ 4x-3 = -49 ⇒ x = -46/4 = -23/2

The correct answer is Option (3) → (A)-(III), (B)-(IV), (C)-(I), (D)-(II)