CUET · MATHS · PYQ PAPER 2023

Given \(\sum p_0 q_0=800, \sum p_0 q_1=1500, \sum p_1 q_1=1300\) and \(\sum p_1 q_0=900\), where subscripts 0 and 1 are used for base year and current year respectively. The Paasche's

A 88.9
B 115.4
C 144.4
D 86.7

Verified Solution

Answer & Solution

Correct Answer

(D) 86.7

Step-by-step Solution

Detailed explanation

\(P_P = \frac{\sum p_1 q_1}{\sum p_0 q_1} \times 100\) \(P_P = \frac{1300}{1500} \times 100\) \(P_P = 86.666...\) \(P_P \approx 86.7\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From CUET

Match List-I with List-II

List-I (Differential Equation)	List-II (Order and Degree)
(A) \(\left(\frac{d^2 y}{d x^2}\right)^2-e^x\left(\frac{d y}{d x}\right)^4+1=0\)	(I) order \(=1\) and degree \(=2\)
(B) \(\left(\frac{d y}{d x}\right)^2+x y=0\)	(II) order \(=2\) and degree \(=1\)
(C) \(\left(1+\frac{d y}{d x}\right)^{3 / 2}=4\left(\frac{d^2 y}{d x^2}\right)^2\)	(III) order \(=2\) and degree \(=2\)
(D) \(\sqrt{\frac{d^2 y}{d x^2}}+1=\frac{d y}{d x}\)	(IV) order \(=2\) and degree \(=4\)

Choose the correct answer from the options given below :

CUET 2025 Easy

Explore more questions on app