Q.6 (A)(ii) Given that Σ p₀q₀ = 220, Σ p₀q₁ = 380, Σ p₁q₁ = 350 and Marshall-Edgeworth's Price Index Number is 150, find Laspeyre's Price Index Number.

Answer: Laspeyre's Price Index Number = 250 Marshall-Edgeworth's index: P₀₁(M-E) = (Σ p₁q₀ + Σ p₁q₁)/(Σ p₀q₀ + Σ p₀q₁) × 100 150 = (Σ p₁q₀ + 350)/(220 + 380) × 100 ⇒ Σ p₁q₀ + 350 = (150 × 600)/100 = 900 ⇒ Σ p₁q₀ = 550 Laspeyre's index: P₀₁(L) = (Σ p₁q₀)/(Σ p₀q₀) × 100 = 550/220 × 100 = 250

Chapter: Index Numbers· Mathematics & Statistics (Commerce) Part 2

Given that \(\Sigma p_0q_0 = 220\), \(\Sigma p_0q_1 = 380\), \(\Sigma p_1q_1 = 350\) and Marshall-Edgeworth's Price Index Number is 150, find Laspeyre's Price Index Number.

Q.6 (A)(ii) Given that \(\Sigma p_0q_0 = 220\), \(\Sigma p_0q_1 = 380\), \(\Sigma p_1q_1 = 350\) and Marshall-Edgeworth's Price Index Number is 150, find Laspeyre's Price Index Number.

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Answer: Laspeyre's Price Index Number \(= 250\)

Marshall-Edgeworth's index: \(P_{01}(M\text{-}E) = \dfrac{\Sigma p_1q_0 + \Sigma p_1q_1}{\Sigma p_0q_0 + \Sigma p_0q_1} \times 100\)

\(150 = \dfrac{\Sigma p_1q_0 + 350}{220 + 380} \times 100 \Rightarrow \Sigma p_1q_0 + 350 = \dfrac{150 \times 600}{100} = 900 \Rightarrow \Sigma p_1q_0 = 550\)

Laspeyre's index: \(P_{01}(L) = \dfrac{\Sigma p_1q_0}{\Sigma p_0q_0} \times 100 = \dfrac{550}{220} \times 100 = 250\)

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Given that \(\Sigma p_0q_0 = 220\), \(\Sigma p_0q_1 = 380\), \(\Sigma p_1q_1 = 350\) and Marshall-Edgeworth's Price Index Number is 150, find Laspeyre's Price Index Number.

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