Quantitative Reasoning

Miscellaneous

Hard

If $\frac{x}{(b-c)(b+c-2a)} = \frac{y}{(c-a)(c+a-2b)} = \frac{z}{(a-b)(a+b-2c)}$ then value of $x + y + z$ is:

$a + b + c$

$a^2 + b^2 + c^2$

✅ Correct Option: 3

Let each ratio equal $k$ . Then $x = k(b-c)(b+c-2a)$ , $y = k(c-a)(c+a-2b)$ , $z = k(a-b)(a+b-2c)$ . Expanding: $x = k(b^2 - c^2 - 2ab + 2ac)$ , $y = k(c^2 - a^2 - 2bc + 2ab)$ , $z = k(a^2 - b^2 - 2ac + 2bc)$ . Adding, all terms cancel: $x + y + z = 0$ .

2022: 24 Aug Shift 1

Difference in the number of applications received for admission in B.Com. and B.Sc. is :

2022: 24 Aug Shift 1

How many students failed in both subjects English and Hindi ?

2026: 29 May Shift 1

The algebraic identity in List-I and List-II are given below. Match List-I with List-II for correct relation between them.

List-I	List-II
(A) $(a+b)^2$	(I) $a^2 + b^2 - 2ab$
(B) $(a-b)^2$	(II) $a^2 + b^2 + 2ab$
(C) $(a+b)^3$	(III) $a^3 - b^3 - 3ab(a-b)$
(D) $(a-b)^3$	(IV) $a^3 + b^3 + 3ab(a+b)$

2026: 18 May Shift 1

Which of the following list of numbers is/are in Geometric Progressions ?

(A) 6, 12, 24, 48...

(B) 2, 6, 18, 20...

(D) 4, 12, 36...

Choose the correct answer from the options given below:

2022: 8 Aug Shift 1

The number of people owing books more than 20 but less than 60 is

2023: 30 May Shift 3

The 10th term of the A.P. 1, 5, 9, 13, ..., is :

If x(b−c)(b+c−2a)=y(c−a)(c+a−2b)=z(a−b)(a+b−2c)\frac{x}{(b-c)(b+c-2a)} = \frac{y}{(c-a)(c+a-2b)} = \frac{z}{(a-b)(a+b-2c)}(b−c)(b+c−2a)x​=(c−a)(c+a−2b)y​=(a−b)(a+b−2c)z​ then value of x+y+zx + y + zx+y+z is:

Solution

Related questions:

Keyboard Shortcuts

CUET General Test Past Year Questions (Topic-Wise)

If $\frac{x}{(b-c)(b+c-2a)} = \frac{y}{(c-a)(c+a-2b)} = \frac{z}{(a-b)(a+b-2c)}$ then value of $x + y + z$ is: