Question

The prices and the quantities of three commodities are given are :

Commodity	Price (₹)		Quantities
	in Year 2006	in Year 2009	in Year 2006	in Year 2009
P	100	90	12	10
Q	80	\(x\)	8	7
R	60	50	4	6

The Laspeyre's price index number for year 2009 with year 2006 as base is 200. The value of \(x\) is:

• A. 320
• B. 360
• C. 140
• D. 260

Answer 1

The Correct Option is B

The Laspeyre's price index is calculated using the formula:

\( I_L = \frac{\sum(P_1 \cdot Q_0)}{\sum(P_0 \cdot Q_0)} \times 100 \)

where:

• \( P_0 \) = Price in the base year (2006 here)
• \( P_1 \) = Price in the current year (2009 here)
• \( Q_0 \) = Quantity in the base year (2006 here)

Given that \( I_L = 200 \), we use the values for commodities P, Q, and R.

The data is:

• Commodity P: \( P_0 = 100, P_1 = 90, Q_0 = 12 \)
• Commodity Q: \( P_0 = 80, P_1 = x, Q_0 = 8 \)
• Commodity R: \( P_0 = 60, P_1 = 50, Q_0 = 4 \)

Applying the formula:

\( \frac{(90 \times 12) + (x \times 8) + (50 \times 4)}{(100 \times 12) + (80 \times 8) + (60 \times 4)} \times 100 = 200 \)

Simplify each term:

• \( 90 \times 12 = 1080 \)
• \( 50 \times 4 = 200 \)
• \( 100 \times 12 = 1200 \)
• \( 80 \times 8 = 640 \)
• \( 60 \times 4 = 240 \)

Substituting these into the equation gives:

\( \frac{1080 + 8x + 200}{1200 + 640 + 240} \times 100 = 200 \)

Calculating the denominators and numerators:

• Denominator: \( 1200 + 640 + 240 = 2080 \)
• Numerator: \( 1280 + 8x \)

Revisit the price index equation:

\( \frac{1280 + 8x}{2080} \times 100 = 200 \)

Solving for \( x \):

\( \frac{1280 + 8x}{2080} = 2 \)

\( 1280 + 8x = 2 \times 2080 = 4160 \)

Solve for \( x \):

\( 8x = 2880 \)

\( x = \frac{2880}{8} = 360 \)

Thus, the value of \( x \) is 360.