Given P01(M-E) = 120, ∑p1q1 = 300, ∑p0q0 = 120, ∑p0q1 = 320, Find P01(L) - Mathematics and Statistics

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Sum

Given P01(M-E) = 120, `sum"p"_1"q"_1` = 300, `sum"p"_0"q"_0` = 120, `sum"p"_0"q"_1` = 320, Find P01(L)

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Solution

Given, P01(M-E) = 120, `sum"p"_1"q"_1` = 300, `sum"p"_0"q"_0` = 120, `sum"p"_0"q"_1` = 320

P01(M-E) = `(sum"p"_1"q"_0 + sum"p"_1"q"_1)/(sum"p"_0"q"_0 + sum"p"_0"q"_1) xx 100`

∴ 120 = `(sum"p"_1"q"_0 + 300)/(120 + 130) xx 100`

∴ 120 = `(sum"p"_1"q"_0 + 300)/440 xx 100`

∴ `sum"p"_1"q"_0 + 300 = (120 xx 440)/100`

∴ `sum"p"_1"q"_0 + 300` = 528

∴ `sum"p"_1"q"_0` = 528 – 300

∴ `sum"p"_1"q"_0` = 228

Laspeyre’s Price Index Number:

P01(L) = `(sum"p"_1"q"_0)/(sum"p"_0"q"_0) xx 100`

= `228/120 xx 100`

= 190

Concept: Construction of Index Numbers - Weighted Aggregate Method
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Chapter 2.5: Index Numbers - Q.4

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