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Exercises · Q6

Q.Verify that Fisher's Ideal Index Number satisfies the Time Reversal Test, using the data of Commodities A, B, C, D from the worked examples above (p0,p1,q0,q1p_0, p_1, q_0, q_1 as given there).

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Step 1 — Recall P01FP_{01}^{F} from the worked example above.

P01F≈115.46(ratio 1.1546)P_{01}^{F}\approx115.46 \quad(\text{ratio } 1.1546)

Step 2 — Compute P10FP_{10}^{F} by swapping every p0↔p1p_0\leftrightarrow p_1 and q0↔q1q_0\leftrightarrow q_1 together.

Laspeyres10=Σp0q1Σp1q1×100=400461×100=86.77\text{Laspeyres}_{10}=\dfrac{\Sigma p_0q_1}{\Sigma p_1q_1}\times100=\dfrac{400}{461}\times100=86.77

Paasche10=Σp0q0Σp1q0×100=300347×100=86.46\text{Paasche}_{10}=\dfrac{\Sigma p_0q_0}{\Sigma p_1q_0}\times100=\dfrac{300}{347}\times100=86.46

P10F=86.77×86.46≈86.61(ratio 0.8661)P_{10}^{F}=\sqrt{86.77\times86.46}\approx86.61 \quad(\text{ratio } 0.8661)

Step 3 — Multiply the two ratios (dual-check against the algebraic identity).

P01F×P10F=1.1546×0.8661≈1.0000P_{01}^{F}\times P_{10}^{F}=1.1546\times0.8661\approx1.0000

This also follows algebraically: P01F×P10F=Σp1q0Σp0q0⋅Σp1q1Σp0q1⋅Σp0q1Σp1q1⋅Σp0q0Σp1q0=1=1P_{01}^F\times P_{10}^F=\sqrt{\dfrac{\Sigma p_1q_0}{\Sigma p_0q_0}\cdot\dfrac{\Sigma p_1q_1}{\Sigma p_0q_1}\cdot\dfrac{\Sigma p_0q_1}{\Sigma p_1q_1}\cdot\dfrac{\Sigma p_0q_0}{\Sigma p_1q_0}}=\sqrt{1}=1 exactly — the small 0.0000-level gap above is only rounding.

✓Final answer

P01F×P10F≈1.0000P_{01}^{F}\times P_{10}^{F}\approx1.0000, so Fisher's Ideal Index satisfies the Time Reversal Test.

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