Given the following data:

Find X if the ratio between Laspeyre's and Paasche's index number is 28:27.

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Solution

$\begin{matrix} Items & P_{0} & Q_{0} & P_{1} & Q_{1} & P_{0} Q_{0} & P_{1} Q_{0} & P_{0} Q_{1} & P_{1} Q_{1} A & 1 & 10 & 2 & 5 & 10 & 20 & 5 & 10 B & 1 & 5 & X & 2 & 5 & 5 X & 2 & 2 X \sum P_{0} Q_{0} = 15 & \sum P_{1} Q_{0} = 20 + 5 X & \sum P_{0} Q_{1} = 7 & \sum P_{1} Q_{1} = 10 + 2 X \end{matrix}$

Laspeyre's Price Index:

$L = \frac{\sum P_{1} Q_{0}}{\sum P_{0} Q_{0}} \times 100 = \frac{20 + 5 X}{15} \times 100 . . . (i)$

Paasche's Price Index:

$P = \frac{\sum P_{1} Q_{1}}{\sum P_{0} Q_{1}} \times 100 = \frac{10 + 2 X}{7} \times 100 . . . (i i)$

Dividing (i) by (ii),we get,

$\frac{L}{P} = \frac{20 + 5 X}{15} \div \frac{10 + 2 X}{7} = \frac{20 + 5 X}{10 + 2 X} \times \frac{7}{15}$

Given: $\frac{L}{P} = \frac{28}{27}$

$\Rightarrow \frac{28}{27} = \frac{20 + 5 X}{10 + 2 X} \times \frac{7}{15}$

$\frac{28 \times 15}{27 \times 7} = \frac{20 + 5 X}{10 + 2 X}$

or $\frac{28 \times 15}{27 \times 7} = \frac{20 + 5 X}{10 + 2 X}$

$\frac{20}{9} = \frac{20 + 5 X}{10 + 2 X}$

$\Rightarrow 9 (20 + 5 X) = 20 (10 + 2 X)$

$\Rightarrow 5 X = 20$

$\Rightarrow X = 4$

Thus,missing figure=4

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Given the following data: Find X if the ratio between Laspeyre's and Paasche's index number is 28:27.

Given the following data:

Find X if the ratio between Laspeyre's and Paasche's index number is 28:27.